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## Bash 'Command' Builtin
According to the manual,
command -v <command>
…causes a single word indicating the command or file name used to invoke command to be displayed.
Very useful when which foo doesn’t work (for example nvm which is a sourced function defined during shell creation).
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# Free particle 1D phase space
I'm trying to draw the phase space for a particle moving freely between 0 and $$L$$. I guess $$H=E$$(total energy, constant)$$=\frac{p_{x}^2}{2m}$$ so $$p_{x}=\pm\sqrt{2mE}$$ for every x between 0 and L, and taking the positive sign when going from 0 to L and the negative sign when going from L to 0. If I draw the phase space for this particle, with energy between E and $$\delta E$$ I think I should get what i drew on the image. However, it seems a bit odd, is there something wrong?
• If the particle is confined to move from 0,L then that's just the infinte square well, right? – InertialObserver May 4 '19 at 17:41
• Yes it is... but i had never tried to draw its phase space... I used to solve Schrodingers equation and graph a couple of wavefunctions, that was it – Juan Pablo Arcila May 4 '19 at 17:44
• From your comment, it looks like you are discussing a quantum particle, for which there are ferocious "uncertainty-principle"-type constraints... You are probably graphing an aspirational Wigner function distribution, but then the cartoon you sketched is not accurate... See Exercise 0.2 , p 29 here. – Cosmas Zachos May 4 '19 at 18:38
• If you were really serious about this, see Belloni et al, AmJPhys 72 (2004) 1183. – Cosmas Zachos May 4 '19 at 19:27
Your situation is a particle in a box, and as such the energy is quantized to be
$$E= E_n = E_1 n^2$$.
If you plot $$E$$ vs. $$n^2$$ for $$n>0$$ you get one half of a discrete parabola. What you want is to know the number of states that lie between energies $$E$$ and $$E + dE$$. That is,
$$dn = \frac{dn}{dE} dE$$
We call $$\frac{dn}{dE}$$ the density of states. To find this we solve for $$n$$ obtaining
$$n = \sqrt{\frac{E}{E_1}}$$
Therefore, the density of states is
$$\frac{dn}{dE} = \frac{1}{2\sqrt{E_1}} \cdot \frac{1}{\sqrt{E}}$$
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# Pipe Volume Formula
Mill Tolerance - Allowable thickness variation from the mill that produced the pipe. The cross- section of the two ends is 2 ( pi R^2 - pi r^2). For a 10 foot pipe length with a 10 inch nominal diameter tested at 150 psi, the allowable loss of water in the pressure test is 10 x 10 or 100, which is then divided by 148,000, resulting in 0. Solve Equation (2) for τ τ = ∆p 2‘ r (3). % of chlorine used × 1 part of 1,000,000 Where: Required ppm chlorine = 50 PPM % of chlorine used = 12. Weight of slug required for desired length of dry pipe with set volume of slug by DrillingFormulas. ceaŽ 02 O SQUARE WHERE I 04 d4 04 - d4 02 + side side shori sh0ft side intetaaf 1. To Metric Conversions Area and Distance Conversions Water Pressure Conversions Flow and Water Volume Conversions Pump Calculations Miscellaneous. Pipe Dia in mm (D) = 10″ X 25. By subtracting any obstructions to airflow (frames and blades) from the overall opening size we get a net result - free area. Formulas and Calculations for Drilling, Production, and Workover: All the Formulas You Need to Solve Drilling and Production Problems, Third Edition, provides a convenient source of reference for oil field workers who do not use formulas and calculations on a regular basis. unit of volume is Cubic Meters. The average fluid density includes the dose chemical. A cylinder has a radius (r) and a height (h) (see picture below). Use the following yields per each bag size: 40 pound bag yields. Elliptic Cylinder. cl 1 Theory overview of flow measurement using differential pressure devices based on ISO-5167 standard. = External volume of well = 2 5 K = [ [ = 855. Some basic units of volume are Cubic. if u want the content volume, again u need the I. S Pipes, Round Bar, Flat Bar. Simple-to-use and mobile friendly, this is a great tool for your pipe-buying toolbox. To provide additional support, the following tables outline the physical dimensions and pipe diameter of each PVC Pipe Size that will work with FORMUFIT products. Since the radius is in inches. To calculate the flowrate of a fluid passing through a venturi, enter the parameters below. More often, in propulsion and power problems, we are interested in what happens in a fixed volume, for example a rocket motor or a jet engine through which mass is flowing at a certain. Results should be checked/approved by a qualified engineer. The volume of the solid generated by a region between f(x)and g(x) bounded by the vertical lines x=a and x=b, which is revolved about the x-axis is ³ b a V S f gx 2 dx. When a viscous fluid flows through a pipe, the flow has a front that is shaped like a parabola bulging outward. APPLIED WASTEWATER MATH FORMULA SHEET AND CONVERSION FACTORS 12 in = 1 ft 27 cu ft = 1 cu yd 1,000 mg = 1 gm 60 sec = 1 min 3 ft = 1yd 7. Under torquing the bolts may result in lower pressure retention capabilities, lower bend load capabilities, joint leakage and pipe joint separation. * Blank Weight = A * B * thk * ρ Length A = 2 * R sin(α/2) + C Width B = R - E cos(α/2) + C where, C = 25 (In mm) C = 1 (In inch) thk = Thickness of Cone ρ = Density of Material * Subject Cone weight is the weight of rectangular plate from which the cone development would be cut out. Measure the length, or height, of the pipe with a ruler or tape measure. Calculating the volume of a tube essentially involves the same formula as a cylinder ( volume=pr 2 h ), except that in this case the diameter is used rather than the radius, and length is used rather than height. VOLUME (V) of a Cube 8 L 5 7; where S = any side Rectangular Container 8 L H S D; where l = length, w = width, and h = height Square Pyramid 8 L1/3 ; 6 D; where b = base length, h = height, Cylinder 8 L è N 6 D; where π= 3. 00002143 cubic yards in a cubic inch. Even if you started at 1001 and filled the pipe at that elevation to 1002. Calculate the radius. The amount of space inside the Cylinder is called as Volume. Volume = Pi x radius sq. Pressurized Volume (ft3) A= Effective Area in Square Inches (Valve or Orifice Throat Area x Flow Coefficient). By ROBERT E. ENGINEERING. Civil Engineering online calculation: Gravity-fed pipe flow - Hazen-Williams formula for a full pipe. Assume a new, four bedroom, 2 ½ bathrooms, home with the plumbing fixtures described in the following table. Pressure Drop Online-Calculator Calculation of pressure drops of flowing liquids and gases in pipes and pipe elements (laminar and turbulent flow). Seeing how a cylinder is similar to a rectangular solid may make it easier to understand the formula for the volume of a cylinder. Therefore, all we need to do is add the measurement of each side of the shape and we will have the perimeter. Fluid dynamics and Bernoulli's equation. AREA AND VOLUME FORMULAS Areas of Plane Figures Square Rectangle Parallelogram s s b w l h 2A = s A = l • w A = b • h Triangle Trapezoid Circle h b h b 1 b 2 r d A = ½ b • h 2 A = ½ (b 1 + b 2) • h A = πr (π ≈ 3. 72 cubic inches of volume. The following formula is often used using a fluid’s specific gravity (G) where the relationship to water density is already considered. example : if u got the volume of water tank is 200cuft then vol of water = 28. These changes can be used to measure the flowrate of the fluid. While researching the subject, we often wondered how these constants were calculated. Dimensions of pipe in common schedule sizes. Another characteristic of streamline flow is that it is layerlike, or lamellar. You can use the formula for the volume of a cylinder to find that amount! In this tutorial, see how to use that formula and the radius and height of the cylinder to find the volume. Different size lines provide dramatically different flow rates In terms of the DEP water supply line size table, whose key component is gallons per minute, the differences are even more dramatic. Essentially you are solving for volume and then converting to cubic yards. If the water flowing though a cylindrical pipe is represented by the shaded area shown below (perpendicular to flow direction) and the velocity of the water is known, then the flow rate of the water can be determined through the pipe as gal/min or m3/sec. PVC Pipe Size Dimensions Chart FORMUFIT products are compatible with most Schedule-based PVC pipe systems. Coverage in 'Square Feet Per Gallon' = 1604. The volume of a pipe is found by multiplying pi by the height by the radius squared. Formula: x = Gallons of sodium hypochlorite required. Volume and Surface Area Page 6 of 19 Example 3: Find the volume and surface area of the figure below 8 5 3 in Solution: This is a sphere. Thermal Systems provides boiler calculations, for steam Steam Pipe Sizing boiler,low water volume boiler,low volume boilers,thermal fluid boilers,hot. Manning Formula Uniform Pipe Flow at Given Slope and Depth. You have now found the volume of 1000 ft pipe. This is shown below: With the radius, r=7. The volume will be in cubic inches. Volume = 1) Volume = 2) Volume = 3) Volume = 4) Volume = 5) Volume = 6) Volume = 7) Volume = 8) Volume = 9) Volume = Volume - Cylinder ES1 6 yd 5 yd 6 ft 11 ft 4 in 13 in 10 yd 9 yd t t 8 in 8 in. 1 Cubic Foot (CF) = Approx 1,000 BTUs. Volume √ Chehab: Stormwater Management System Drawdown. If you don't find what you are looking for here, try using the search engine in the left column. Natural Gas Pipe Sizing Tables and Charts Steel Pipe - Schedule 40. discharge rate calculator) accepts input in both metric and imperial units: m/s, km/h, ft/s, yd/s, mph, and outputs in both metric units and imperial ones: cu ft, cu yd, mm3, cm3, m3 and so on. Slump – Slump is a measure of the consistency of the concrete. ceaŽ 02 O SQUARE WHERE I 04 d4 04 - d4 02 + side side shori sh0ft side intetaaf 1. 14 for π and round the answer to the nearest tenth of a cubic millimeter. 995 g at 25 degrees. The radius can be measured di- rectly or the diameter can be di- vided by 2. The 45° pipe elbow is used to connect tubes at a 45° pipe angle. To estimate pipe volume use the following formula: volume = π × radius2 × length. discharge rate calculator) accepts input in both metric and imperial units: m/s, km/h, ft/s, yd/s, mph, and outputs in both metric units and imperial ones: cu ft, cu yd, mm3, cm3, m3 and so on. com site's pipe weight calculation? You can easily calculate by entering pipe dimensions without any restrictions. you need the I. The volume weight for other liquids can be calculated when compensated for density. Fluid Power Formula. Fluid dynamics is the study of how fluids behave when they're in motion. Use the one of predefined substances (water, steal, wood etc. If fittings are in the pipe, these have to be treated specially. Now along with pipe inner and outer diameters for Sch 5S, 10S, 40, 80, and 160 pipe, you also get the interior volume for each in gallons per foot of length and the pipe weight in pounds per foot of length. - Enhancement #183224 has been filed for considering for a future release the ability to compute pump on/off elevations automatically based on various criteria. 0″ The outside diameter (OD) of 2-3/8″ tubing is 2. It is usually divided into northern and southern halves by the equator. The dimensionless coefficient of friction, f , can vary from less than 0. Vazsonyi and Andrew Gardel and formulas obtained by assuming T-junction as combination of other pipe components and observations obtained from software experiments. Volume of a partial right cylinder. Enter - Depth of Trench 4. Notice that each formula has its own coefficient which stays constant while working through the entire equation. Where is the volume fraction of gas:. Inches, Cubic Feet, Quarts, Cubic Yards, Cubic Meters, Gallons, Liters, Cubic Centimeters, Cubic Millimeter etc. But many times, custom elbow angles are required at site which should be cut from standard 45° or 90° elbows. When a viscous fluid flows through a pipe, the flow has a front that is shaped like a parabola bulging outward. Are Compressed Air Leaks Worth Finding, Fixing and Sustaining? Recently in a Steel Plant with rolling mills and galvanizing process, 625 leaks were identified, tagged and repaired. How can you derive the formula for the ASME pipe thickness calculation? ASME formula, was derived for combined stresses (non-uniaxial stress state) in accordance with the third Theory of. 14 Right Hand Left Hand Left Hand Velocity= inches cfs Flow ft In Out Invert Invert Mannings n ft/ft Slope *Run trial with different depth until the right hand. kg/m is kilograms per meter. Vab = Actual Volume Drained Va = [(0. This is very suitable for situations such as domestic piping and hosing, sprinkler and irrigation systems, etc. KAD Group of Companies provide calculation of S. Select - Trench minus Pipe/culvert 2. This is a convenient cubic meter calculator for shipping volume of cartons, calculation based on metric unit cm and kg. 4 What would be the volume of vessel 2 in figure 3. how to calculate water volume in pex tubing and copper pipe Posted by bloger 06/14/2016 2 Comment(s) In order to attain a properly working system of any kind, certain calculations must be done to uphold the volume of water inside of the system. Just enter the pipe dimensions and you're done!. The Mosteller formula 1. The results in a PSI drop (pressure loss) at the other end of the hose. What is ID and OD? This information tells you how to determine the correct size diameter of your pipe or piling. Given below are the Class 10 Maths Important Questions for Surface Area and Volume a. Now along with pipe inner and outer diameters for Sch 5S, 10S, 40, 80, and 160 pipe, you also get the interior volume for each in gallons per foot of length and the pipe weight in pounds per foot of length. Measure the length, or height, of the pipe with a ruler or tape measure. Volume of a oblique. The volume can be multiplied by the liquid density in order to determine the mass of liquid. To get the radius of the pipe divide the diameter by 2. measure the radius of the pipe. This percentage decreases for larger venturis and as the flow rate increases. This calculator can be used to calculate volume of water or other liquids in pipes. On-Line Liquid Friction Loss for any Pipe Size Enter Conditions Below:. Road Formulas, Road Repair Application Now anyone can make quick, easy and precise calculation in the field and at the office. the bolt and/or casting which could result in pipe joint separation. Weight density HDPE pipe inside the pipe in the calculation results can be reached by multiplying the volume of the hollow cylinder. A less common type of pipe is furnace butt welded pipe. 2 Flow Measurement. Unfortunately, the laws of physics will not allow that same small diameter pipe to flow sufficient volume to realize maximum possible power at higher RPM. Gas Pipe Line Calculation Sizing USING CPC PIPE SIZING TABLE (PROPANE) This handout will guide you thru the basic, most common method for sizing a propane piping system for residential or commercial application. Determine the volume of the trench by using the formula: Volume = Width x Length x Depth. Capacity - Capacity, as related to drill pipe, drill collars and other tubulars is the volume of fluid the pipe can contain. 011 cubic yards. The elevation will never be 1002. Easy, quick access in the field to OD and ID measurements of most common piping available. the flange dimensions can also be found in these tables. Enter - Diameter of Pipe 6. As in real piping system, losses of energy are existing and energy is being added to or taken from the fluid (using pumps and turbines) these must be included in the Bernoulli equation. In the cell A2, we write down to the formula for calculating the area of the circle: r = 25 cm. Find Surface Area of Pipe or Find Surface Area of different objects like Cone, Cube, Cylinder, Pipe, Prism, Pyramid, Rectangular Box, Sphere, Torus (Ring), Tube and more. So for example, let's say you had a cylinder that had a radius of 3 cm and a height of 6 cm. Dimensions of pipe in common schedule sizes. The volume of pipe is the difference between two cylinders. SG= Specific Gravity of Gas: Z= Compressibility Factor (at initial Pressure) Ta= Initial Temperature (˚R) Pa= Initial Pressure (psia) Patm= Atmospheric Pressure (psia) Natural Logarithm or Loge. Calculation of pipes are based on a formula V=π*R1*R1*L Calculation of the surface area of the pipes are based on a formula P=2*π*R2*L R1 - inner radius pipes R2 - outer radius pipes L - pipe length. Calculate the Volume of a Well Example: An abandoned well is 6 inches in diameter and 100 feet deep and a water level of 40 feet. Calculate the volume of a pipe given its inner diameter and length. A cylindrical metal pipe has a diameter of 20 millimeters and a height of 21 millimeters. The definition of a fixed cost is a cost that does not vary with volume, so the average fixed cost part of the formula only applies within a very narrow volume range. The objects have the same weight, hence the same volume. Even if you started at 1001 and filled the pipe at that elevation to 1002. NS 813 Piping systems – Identification colours for the content 7. In this type of question, we need to subtract external radius and internal radius to get the answer using the volume formula as the pipe is hollow. • Write the mole balances in terms of conversion for a batch reactor, CSTR, PFR, and PBR. In fluid dynamics, the Darcy–Weisbach equation is an empirical equation, which relates the head loss, or pressure loss, due to friction along a given length of pipe to the average velocity of the fluid flow for an incompressible fluid. (** Above formula is based upon one gallon having a volume of 231 cubic inches) For example: If the 'Wet Film Thickness is 10 MILS (0. We need to calculate the radius of the sphere to calculate the volume and surface area. Then multiply the amount of Cubic Yard you want to convert to Pipe [US], use the chart below to guide you. Under torquing the bolts may result in lower pressure retention capabilities, lower bend load capabilities, joint leakage and pipe joint separation. Actual Inside Diameter (Inches) Nominal Outside Pipe Corrugated. Free Area of Wall Louvers Every louver, no matter what size, has a calculated free area. We are given that the diameter of the sphere is 8 5 3 inches. Once you have found the volumes up to this point, you can find the remaining volumes by the symmetry of the tank. The Hazen–Williams equation is an empirical relationship which relates the flow of water in a pipe with the physical properties of the pipe and the pressure drop caused by friction. Hemisphere problems d. Measurement Conversion Measurement Conversion Measurement Conversion Measurement Conversion 1 ft. Calculate insulation thickness (minimum value) required for a pipe carrying steam at 180 0 C. 3) Work out the weight of pipe with the following formula. Since the end (base) of a cylinder is a circle, the area of that circle is given by the formula: Multiplying by the height h we get where: Calculator. Use the second form to calculate the diameter of a pipe required for 5 fps pipe velocity. Since you know that the area of the base is 3. Volume of fluid (in turbulent flow), ft 3 /bbl= 5. A hole cut out of the center has a diameter of 6 millimeters. By ROBERT E. This percentage decreases for larger venturis and as the flow rate increases. This calculator uses a contracted form of the Hazen-Williams formula which is the industry standard for calculating friction loss. Solve Equation (2) for τ τ = ∆p 2‘ r (3). Top pipe and cisterns aptitude problems and answers with concepts, important formulas, shortcuts, tricks to crack placement test,competitive exams,entrance. % of chlorine used × 1 part of 1,000,000 Where: Required ppm chlorine = 50 PPM % of chlorine used = 12. It can quickly calculate this information based on the information you enter—such as your pipeline’s flu. In the second case with a=2, b=3, c=5 and h=6 then volume will be 81. Volume of a Round Tank or Clarifier Please enter data for Tank Radius (radius is 1/2 of diameter) and side water depth. The volume of the solid generated by a region between f(x)and g(x) bounded by the vertical lines x=a and x=b, which is revolved about the x-axis is ³ b a V S f gx 2 dx. TOTAL all the EQUIVELENT LENGTHS OF PIPE needed for each PIPE FITTING and add to your total of straight length of pipe. These changes can be used to measure the flowrate of the fluid. 3208 ÷ A A = GPM x 0. There are multiple ways to find that volume, depending on what resources you have available to you. 995 g at 25 degrees. Landscape Irrigation Formulas. Skateboarders use half pipes for doing tricks. To find the volume of the prism, we can use the formula V = A × h where A is the area of the base(or area of the cross section) and h is the height(or the length of the prism) The base of the prism is a triangle, so we need to find the area of the triangle. Fluids can flow steadily, or be turbulent. Calculate insulation thickness (minimum value) required for a pipe carrying steam at 180 0 C. For gravitational flow, and for open-channel flow, other calcs are available. Plastic pipe work relies on the granular fill around it for structural support without it the pipe work will become deformed when under load, if the deformation is greater than 5% of the pipe diameter then the system has failed and would require renewal. To get the radius of the pipe divide the diameter by 2. A cylindrical metal pipe has a diameter of 8. You can use the Air Products Pipe Pigout app to calculate the nitrogen volume, flow rate and pressure suitable to push your pig and complete the purge or displacement. 5 PSD-24 24 3. You can calculate pipes more easily with calculation of the pipes you use in project. By multiplying air velocity by the cross section area of a duct, you can determine the air volume flowing past a point in the duct per unit of time. Make sure the gas line can handle your generator. Enter - Diameter of Pipe 6. You can change the Radius, the Length and the Height. In fluid dynamics, the Darcy–Weisbach equation is an empirical equation, which relates the head loss, or pressure loss, due to friction along a given length of pipe to the average velocity of the fluid flow for an incompressible fluid. Compressed Air & Insert Gas Piping Systems 1 Standard Duratec pipe is manufactured to ASTM F1282 (ANSI certified) and CSA B137. 14 or ) Circumference: C = 2πr = πd. CONVERSION AND REACTOR SIZING • Define conversion and space time. I have a 19mm pipe. For gravitational flow, and for open-channel flow, other calcs are available. 2 and that the height is 4 in. These computations show that your conjectured formulas are indeed true: The gain in volume and surface on the outside of a bend of the pipe is exactly outweighed by the loss on the inside. Volume of a right cylinder. It can quickly calculate this information based on the information you enter—such as your pipeline’s flu. * Blank Weight = A * B * thk * ρ Length A = 2 * R sin(α/2) + C Width B = R - E cos(α/2) + C where, C = 25 (In mm) C = 1 (In inch) thk = Thickness of Cone ρ = Density of Material * Subject Cone weight is the weight of rectangular plate from which the cone development would be cut out. Use the calculator on the right to calculate height, radius or volume of a cylinder. Byfollowingthestepsoutlinedabove,reducesthechanceofmakingerrors. Thermal Systems provides boiler calculations, for steam Steam Pipe Sizing boiler,low water volume boiler,low volume boilers,thermal fluid boilers,hot. Amazing how closely the frustum of a right circular cone resembles a beer glass, isn't it?! ;) I got curious ab out the volume of just such a glass the other day and set off to find the answer with no tools other than a pen and a paper napkin. The formula: 3. This air has mass, and adding that extra mass lowers the resonant frequency of vibration. x = Required ppm chlorine × Volume of pipe in gals. 3 feet and an average depth of 5 feet has a volume of (12 x 2. You would need to separately calculate this volume based on the size and slope of the pipe(s), or use the above workaround to incorporate the conduit volume in the wetwell as a single unit. In addition to calculating volume of a cylinder, this page calculates total, lateral and base surface areas and surface to volume ratio of a cylinder as well, providing calculation results in numerous measurement units, showing the referenced calculation formulas and computing weight, force and pressure of fresh and sea water based on the. This causes a "slip" or holdup effect, which means that the in-situ volume fractions of each phase (under flowing conditions) will differ from the input volume fractions of the pipe. The Pipe Support Calculator is designed for Pipefitter/ welders/fabricators to eliminate the time it takes to calculate normal piping calculations. What is the volume of metal in the pipe? Use 3. Other fluids, temperatures, pipe materials, pipe sizes and pressure drops would require different equations. To pressurize 10 cubic meter of pipe line ,how much volume nitrogen required on the below condition ? 1. The image below compares how the formula $V=Bh$ is used for rectangular solids and cylinders. 5 PSD-24 24 3. More often, in propulsion and power problems, we are interested in what happens in a fixed volume, for example a rocket motor or a jet engine through which mass is flowing at a certain. CAUTION: Use of an impact. S Pipes, Round Bar, Flat Bar. This rectangle is what the cylinder would look like if we 'unraveled' it. For a pipe use its length instead of height: pipe volume = π * radius² * length, where radius = inner diameter/2. Since the volume of a cylinder is the product of the area of the base and the height, the volume of pipe is. Your pipe is 10,000 ft. BSA = SQRT( (cm*kg)/3600 ). Wetted Area We can approximate the partially filled surface area of the torispherical head using the formula for elliptical heads. The pipe B-D-E is the feed main, E-J is the cross main, and the pipes E through J are the branch pipes, while K through R are the sprinklers. cube = 6 a 2. These computations show that your conjectured formulas are indeed true: The gain in volume and surface on the outside of a bend of the pipe is exactly outweighed by the loss on the inside. for the volume of the material. Other fluids, temperatures, pipe materials, pipe sizes and pressure drops would require different equations. 3) Work out the weight of pipe with the following formula. DEFINITIONS OF TERMS. = External volume of well = 2 5 K = [ [ = 855. Solve Equation (2) for τ τ = ∆p 2‘ r (3). S: Among all of the metal weight calculation formula in the following part, there are two kinds of metals that are not mentioned, which are I beam steel and H beam steel. Design of PE Piping Systems 159 The Hydrostatic Design Stress, HDS, is the safe long-term circumferential stress that PE pipe can withstand. Make sure the gas line can handle your generator. Water usage is typically expressed as a volumetric flow rate (volume/time). Calculates Rolling Offset pipe length. ” If a and b are the semi-major and semi-minor axes and h is the height, then. One square foot is 144 square inches, so the cross-sectional area of this pipe is less than one square foot. This is the common equation for a cylinder. How To Get Volume of Pipes? Loads,for Wire and Manila Rope Charts,Decimal of a Foot,minutes converted to Decimals of a Degree and trigonometric formulas,Pipe& Water Weight Per Line Foot &Feet. The most important formulas for trigonometry are those for a right triangle. Formula/Conversion Table. To understand the formula for the surface area of a cylinder, think of a can of vegetables. All dimensions are in millimeters. To calculate the flowrate of a fluid passing through a venturi, enter the parameters below. ALLOWABLE LEAKAGE PER 1000 FEET OF PIPELINE* — GPH (GALLONS PER HOUR) AVERAGE TEST PRESSURE NOMINAL PIPE DIAMETER (INCHES) PSI BAR 3 4 6 8 10 12 14 16 18 20 24 30 36 42 48 54. The volume of fluid in a pipe can be found given the inner diameter of the pipe and the length. Manning Formula Uniform Trapezoidal Channel Flow at Given Slope and Depth. Flow rate formula / equation and working examples. Sometimes you can figure out volume displaced by employing Archimedes' principle and/or static equilibrium. 85 Flow Coefficient unless known. Perpetual Calendar; Body Mass Index; Wind Chill; References. Explanation. Easy, quick access in the field to OD and ID measurements of most common piping available. prism: (lateral area) = perimeter(b) L (total area) = perimeter(b) L + 2bsphere = 4 r 2. Hemisphere problems d. This is a convenient cubic meter calculator for shipping volume of cartons, calculation based on metric unit cm and kg. 1 gm = 1000 mg 1 hour = 60 min. So, in your case, increasing the amount c, the volume decreases because of divided by [c^2]. The calculation is helpful for computing the volume in a capped or clogged water pipe that is sloping downhill or a tank that is tipped, sloped, or tilted. Pipe Water Velocity and Minimum Pipe Diameter Use the water velocity form to calculate the velocity of water in a pipe. (To the bottom) P. a more detailed explanation (examples and solutions) of each volume formula. Given below are the Class 10 Maths Important Questions for Surface Area and Volume a. Pipe Formulas Pipe and Tube Equations - moment of inertia, section modulus, traverse metal area, external pipe surface and traverse internal area - imperial units. Calculate ASME B31. For example, suppose you dispensed a 25-mL sample of water with the pipet, and then you weighed the water and found that its mass was 24. After the pipes appear, the market bounces and new pipe bottoms disappear. In order to help you quickly find the metal weight formula, I made infographic which is not only practical but also beautiful. SURFACE AREA AND SQUARE FEET PER LINEAR FOOT OF PIPE Size Diameter Square Feet per Linear Foot of Pipe Inches External Inches Internal Inches External Internal 1/4 0. \text{Volume of iron =}\pi r^2h\\. Calculating the Volume of a Pipe in Cubic Feet Multiply the cross-sectional area of a pipe by its length to determine its volume in cubic feet. If we know the radius and height of the cylinder then we can calculate the surface area of a cylinder using the formula: Surface Area of a Cylinder = 2πr² + 2πrh (Where r is radius and h is the height of the cylinder). You can use measurements and a calculator to help you determine how much earth will actually need to be removed. 5 Inlet The minimum height of the earth dike at the entrance to the pipe slope drain shall be the diameter of the pipe (D) plus 12 inches. 3) Work out the weight of pipe with the following formula. Suppose the length of the pipe is l, external radius is R and internal radius is r. Length of the pipe = 5 meters. The pipe could be ERW or Seamless pipe. example : if u got the volume of water tank is 200cuft then vol of water = 28. cl 1 Theory overview of flow measurement using differential pressure devices based on ISO-5167 standard. Skateboarders use half pipes for doing tricks. areas formulas list online. Volume of a pyramid. Where is the volume fraction of gas:. 14 x radius squared x average depth x 7. Inches, Cubic Feet, Quarts, Cubic Yards, Cubic Meters, Gallons, Liters, Cubic Centimeters, Cubic Millimeter etc. Fluid dynamics is the study of how fluids behave when they're in motion. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. There are other methods available for sizing these systems by. The Bernoulli Equation for an Incompressible, Steady Fluid Flow. (Radius is one-half the diameter. a table of volume formulas and surface area formulas used to calculate the volume and surface area of three-dimensional geometrical shapes: cube, cuboid, prism, solid cylinder, hollow cylinder, cone, pyramid, sphere and hemisphere. 14 SIZING A HYDRAULIC CIRCUIT THE FOLLOWING INFORMATION MUST BE KNOWN. , flowing in streams, canals, pipe- lines, and ditches, is measured in units of volume per unit of time—gallons per minute (gpm), cubic feet per second (cfs), acre-inches per hour and acre feet per day. Just enter the pipe dimensions and you're done!. Pressure drop formula calculation is published. You can think of the volume of the cylinder as the volume of the area of the base being extended throughout the height of the cylinder. Multiply that cross-sectional area by the length of the pipe to get the volume in cubic inches of the pipe. Landscape Irrigation Formulas. Under torquing the bolts may result in lower pressure retention capabilities, lower bend load capabilities, joint leakage and pipe joint separation. Therefore:. This calculator calculates for the volume, diameter, and length of a cylindrical container or tube. Oh! line become a bit complicated, sorry for that, lets solve it. The density of water at 25 degrees is 0. 010") Coverage is: 1604. Prism Formula Prism has different meanings in mathematics and optics. If students guess this fact, ask them what they think the volume of an ellipsoid is! The Math Behind the Fact: One way to see why the formula is true is to realize that the above ellipse is just a unit circle that has been stretched by a factor A in the x-direction, and a factor B in the y-direction. The internal pipe diameter, ID, (inches) is used in the equation shown below. The Pipe Flow Wizard Software calculates the type of flow that will occur in the pipe (laminar flow or turbulent flow), the Reynold's number, the friction factor and the fluid velocity, to derive a minimum size for the internal diameter of the pipe. Equations displayed for easy reference. inlet upstream pressure is more than 5 psig (35 kPa) fittings factor 1. Length is calculated. I have a 19mm pipe. The equation to account for this would be much more complex and. You have now found the volume of 1000 ft pipe. Ditch digging costs are usually measured by the cubic yard of earth removed, so it's a good idea to know how much earth needs to be removed before you begin. Tank R adius(ft):. The Hazen-Williams formula is an empirical rule, that holds well for cold water running in pipes under turbulent flow conditions. It is interesting to note that the power transmission coefficient is zero when the frequency is that of the resonance frequency of the helmholtz. What is ID and OD? This information tells you how to determine the correct size diameter of your pipe or piling. Calculates pipe Cutback after Bend Angle and Offset are entered. volume synonyms, volume pronunciation, volume translation, English dictionary definition of volume. , flowing in streams, canals, pipe- lines, and ditches, is measured in units of volume per unit of time—gallons per minute (gpm), cubic feet per second (cfs), acre-inches per hour and acre feet per day. By ROBERT E. For the calculation, specify the inner and outer pipe diameter and total length of the pipeline. When the length and diameter of a pipe are known, the head required to discharge a certain quantity of water per second can be found by the formula: h=. Length of the pipe = 5 meters. 45° Pipe Elbow 45° Pipe Elbow is also known as “45 bends or 45 ells”. to calculate nitrogen volume, flow rate, pressure and time to purge a pipeline. 10) The cross-section of a pipe has a width of 6 yard and height of 15 yard. With material selection you can detail pipe. The user is cautioned not to mix units within a formula. 5 PSD-24 24 3. BY MICHAEL DUER Experiments show that distributing the inlet through multiple inlet ports results in. If we pile the slices on oposite sides of balance scales, they must remain balanced. The material consumed in the bend. Unless otherwise stated the formulas shown in this manual can be used with any units. In engineering applications there is a wide range of pipe wall roughness due to the different materials and methods of manufacture used to produce commercial pipes. 9 Duratec fittings are manufactured to ASTM F1974 and carry a national pressure vessel CRN# OA02020. To get the radius of the pipe divide the diameter by 2. 01 Preliminary Sizing Sheet for Vessel Name & Tag No Project Reference Notes The Alpha range of mass transfer & separation internals Calculated By / Date For preliminary use only. 1 Pipe Threads, General purpose, Inch 3. This calculator calculates for the volume, diameter, and length of a cylindrical container or tube. Pipe Support Calculator - Dummy leg and Trunnion calculations. dV = A·dL A·dL/dt = A·v = constant, where v is linear flow rate (velocity). Calculate area, circumference and radius of a circle. There are 0.
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## Monday, August 23, 2010
### M Theory Lesson 344
Last time we generated the norm of a Koide matrix from a symmetrisation of the $1$-circulant piece of the neutrino mixing matrix. Recall that the CKM quark mixing matrix could also be approximated by a pair of $R_2$ factors, which we can choose to have norm less than one, as in $(a, 1/b)$ $=$ $(-0.231, 1/24)$. The Koide norm obtained in this case happens to give
$r^{-1} = 0.0193$
which is rather bizarre, since this is the CP violating phase $\beta_s$. Using this coincidence to define the third $R_2$ factor $c$ in terms of $\beta_s = abc$, we get
$c = \frac{2}{b^2} = 0.0035$
and it remains to wonder how the parameter $a$ is related to $b = 24$.
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Article | Open | Published:
# The Caveats of observing Inter-Trial Phase-Coherence in Cognitive Neuroscience
## Abstract
Many studies have now consistently reported that the phase angle of ongoing oscillatory activity (measured using EEG/MEG), at time of stimulus presentation influences detection when stimuli are near-threshold. However, studies examining whether the adjustment of the phase angle of oscillations is under top-down attentional control have thus far yielded conflicting results. A possible source for the discrepancy could be that the estimation of the phase of ongoing oscillations as well as its uniformity across trials could be affected by task induced changes in the power of oscillations or concurrent evoked responses. One measure, Inter-Trial Phase-Locking (ITPC), or the uniformity of phase angles across trials, is particularly vulnerable to these factors. Here, using various simulations modelling the common task induced changes in the EEG reported in the literature, we demonstrate that apparent changes in Inter-Trial Phase-Locking of oscillatory activity can occur independent of any actual change in the phase of the ongoing activity.
## Introduction
EEG and MEG signal contains oscillatory (i.e rhythmic) activity in various frequency bands. The most predominant oscillatory activity in the EEG/MEG is at the alpha rhythm (8–12 Hz), which has been observed primarily in the sensory systems (vision, motor, auditory). A number of studies have reported that the amplitude of alpha activity is suppressed in sensory relevant regions during visual, auditory and somatosensory attention, but increased in regions responsible for processing unattended information1,2,3,4,5,6,7,8. This has led to the speculation that an increase in the amplitude of alpha range oscillations plays a mechanistic role in cognition by gating information flow to relevant sensory regions through the inhibition of irrelevant regions9,10,11.
In addition to the amplitude of alpha activity, several studies have recently found evidence that the alpha cycle reflects states of low and high excitability which has an influence (depending on the arrival of the stimulus in the cycle) on perception12,13,14,15 as well as on the evoked response to sensory stimuli16. Recently, an intriguing possibility put forward has been that top-down processes such as attention and expectation can modulate the phase of alpha activity as a mechanism to either select and prioritize relevant information or conversely suppress information17.
However, empirical support for this hypothesis has not been consistent. One study supporting this theory has found that the phase of the ongoing alpha-activity, measured using EEG, appeared to be perturbed by temporal expectation of predictable visual targets18. Specifically, the authors observed that cues signalling the arrival of predictable visual stimuli appeared to shift the phase of ongoing alpha activity to be at an ‘optimal’ state at the arrival of the visual stimuli.
On the other hand, our group, using a similar cued temporal expectation paradigm was unable to find any evidence for the modulation of the phase of the ongoing alpha activity in three independent EEG experiments7. One critical difference between our experiments was that we conducted our phase-analysis trials on ‘blank’ catch trials that did not include the presentation of a visual stimulus, as such free from sensory evoked responses7. As we will demonstrate in the coming sections of this simulation study, these transient evoked responses, which are phase-locked to the onset of sensory stimuli, can critically confound the phase-estimation of ongoing oscillatory activity.
The endeavor of this simulation study is to demonstrate some of the caveats faced by researchers in quantifying event related phase-perturbations (i.e the degree of phase modulation as a result of an experimental event) that might be able to explain the divergence in results. Quantification of event-related phase modulations is commonly reported using Inter-Trial Phase-Locking (ITPC or Inter-Trial-Coherence/Phase-Locking Factor)18. An ITPC value close to 0 reflects high variability of phase angles across epochs, whereas an ITPC value of 1 reflects all epochs having the same phase angle.
Here we will first investigate how ITPC estimates can be distorted by modulation of evoked responses. Differences in amplitude or latency of evoked responses can lead to differences in ITPC values that are not actually reflecting phase-perturbation of the ongoing oscillations. Next, we will investigate how the amplitude of the oscillation relative to the noise could modulate ITPC. This endeavor is of particular relevance to investigations of phase perturbations in which there is a systematic difference in the power of oscillatory activity between conditions.
## Influence of ERP amplitude and latency on pre-stimulus ITPC
The onset of a visual stimulus elicits several evoked responses at the scalp, with C1 being the earliest, occurring 50–80 ms after the stimulus. The amplitude of the C1 is generally thought to be unaffected by top-down factors such as attention19,20, although some evidence seems to suggest this not to be unequivocal21. Following the C1 components are the P1 (~90–140 ms) and the N1 (~150–180 ms) components whose amplitudes have been found in number to be modulated by top-down factors such as attention and expectation22,23,24,25,26,27,28,29,30,31. Furthermore, several studies have suggested that the P1 and N1 may reflect different components of attention with the P1 reflecting inhibition and the N1 amplification of sensory input22,32,33. Finally, it should be noted that responses as early as 27 ms post-stimulus have been found in the early visual cortices of the Macaque monkey19.
Temporal leakage from evoked responses containing condition related differences in latency or amplitude could therefore lead to ITPC differences that are actually not present in the ongoing oscillations. Comparing ITPC between conditions is therefore troublesome around the time of stimulus presentation. A simulation was performed in Matlab (MATLAB R2014a, The MathWorks, Inc., Natick, Massachusetts, United States) to demonstrate this phenomenon. One hundred EEG epochs were created by superposition of an ERP component, together with Gaussian white noise, on an alpha oscillation (see Fig. 1).
The ERP component was generated using the following formula:
$$ERP(t)=A\frac{t-{t}_{0}}{\tau }{e}^{1-(t-{t}_{0})/\tau }\,\sin \,(2\pi f(t-{t}_{0}))$$
(1)
where A is the amplitude of the ERP in μV, f is the frequency in Hz, t is the time in ms, t0 is the time of the start of the ERP and τ the exponential decay time of the ERP envelope. For the simulation the frequency of the ERP was 7 Hz, the start of the ERP was at t = 90 ms, and τ = 50 ms.
The ongoing alpha oscillation was created using:
$${\rm{O}}\,(t)={\rm{A}}\,\sin (2\pi \,{\rm{ft}}+\theta )$$
(2)
where A is the amplitude of the signal in μV, f is the frequency in Hz, t is the time in ms and θ is the phase in radians. The frequency of the ongoing oscillation was 10 Hz +/− a random scalar drawn from the standard normal distribution (0.5 * randn in Matlab) and the phase was random every trial.
These ‘standard’ epochs were compared with epochs containing an enlarged ERP (1.2 × the standard amplitude) and epochs with a 10 ms earlier peak latency (t = 80 ms versus t = 90 ms). Note that the same noise pattern and phase angles were used for the ‘standard’, ‘earlier’ and ‘larger’ ERP epochs, which means that ITPC should be similar for all three datasets. The phase angle was calculated for every epoch at t = 0 ms (‘target presentation’) by applying a Hann window of 500 ms, taking the Fourier transform and dividing the outcome by its absolute. ITPC was indexed by summation of phase angles of all epochs according to the following formula:
$$ITPC({f}_{o},t)=\frac{1}{N}|\sum _{k=1}^{N}{e}^{i{\varphi }^{k}({f}_{o},t)}$$
(3)
where N is the number of trials, φk is the local phase angle of the signal in the current trial in radians.
Here the simulation of 20 participants with 100 epochs per condition, comparing the ‘standard’ ERP to simulations with a larger ERP, led to a difference in ITPC around target presentation (Fig. 2A). A dependent samples t-test revealed that ITPC of 10 Hz oscillations was significantly higher at target presentation for the epochs with the larger ERP (t (19) = −32.3183, p = 4.5150 × 10−18) (0.2970 vs 0.2593). Comparing standard epochs with epochs having an earlier peak latency also resulted in a significant difference in ITPC around target presentation (dependent samples t-test (t(19) = −3.2602, p = 0.0014 with higher ITPC for epochs with an earlier ERP peak latency (0.2908 vs 0.2593) (Fig. 2B). To infer how common the ERP induced ITPC difference is, the procedure was repeated 1000 times and the number of times a significant difference was present at 10 Hz was counted. An ERP latency difference between conditions resulted in 80.4% of simulations in a significant difference at target presentation. A difference in amplitude resulted in a significant difference in ITPC between conditions for all simulations (100%).
## Entrainment and phase-locking
Entrainment of oscillatory activity has been suggested as a mechanism for which the brain prepares for processing of temporally predictable targets, (for instance Lakatos et al.34). The underlying premise here is that the entrainment adjusts the phase-angle of an oscillation such that target presentation falls within an optimal phase. However it should be be kept in mind that a transient top-down signal can also cause a modulation of ITPC in the absence of a real phase modulation. Support for this pitfall comes from Yeung et al.35 showing that phasic activity cannot be separated from a phase-reset. Similarly, transient activity prior to target presentation can prompt an increase in ITPC when it is time-locked to temporally predictable stimuli. This transient activity does not need to be in the same frequency as the ITPC is measured in. A burst of spiking activity in a high frequency can induce slow ERF’s or ERP’s when oscillations are asymmetrical (i.e. peaks are more positive than troughs are negative36), thereby increasing ITPC in lower frequencies. Temporal leakage will then introduce ITPC at later times than the burst actually occurs.
## The influence of oscillatory power on ITPC
Power (=amplitude ²) and phase angle are mathematically independent variables. That is, the phase of an oscillation does not depend on its amplitude and vice versa. Both calculations comprise the calculation of the Fourier transform (ft), but in order to obtain the amplitude one must take the absolute value, whereas for the phase angle the argument must be taken. EEG consists of a phasic signal plus noise and a change in signal amplitude will influence the signal to noise ratio (SNR). If noise levels are constant, the SNR is lower for epochs with lower amplitude, making it harder to estimate the phase angle. A difference in power could therefore lead to an ITPC difference; that is the variability in phase angles across epochs at a certain point in time. Given that oscillatory power can differ between experimental conditions, phase estimations will be more precise in one condition than the other. Thus, this could lead to SNR dependent differences in ITPC instead of experimental dependent differences between conditions. Figure 3 shows an example in which differences in the clustering of phase angles emerges simply due to an oscillatory power difference between the two sets of epochs. Both conditions contain a 10 Hz oscillation with ITPC of 1, however for the low power condition ITPC was estimated at 0.9966 and for the high power the estimation was.0.9992. Although, this difference does not appear large, a systematic difference in ITPC caused by a systematic difference in power could eventually lead to a significant difference in ITPC between conditions.
To demonstrate under what conditions power induced ITPC differences can arise, the comparison of ITPC between low and high power 10 Hz oscillations was repeated under different amounts of SNR, ITPC and power. ITPC was calculated over 100 trials for a ‘high’ and a ‘low’ alpha amplitude condition and compared between 20 simulated subjects. Except for the difference in power, everything was kept constant between conditions: that means that the same noise patterns and random phase were added to both the low and high power condition. This results in a well-controlled experiment in which ITPC differences between conditions can only be attributed to power differences.
Given that these simulations are designed to show the influence of noise on phase estimations, the type of noise was chosen to resemble EEG noise as much as possible. Instead of white noise we used noise created with the power spectrum of real EEG data using the implementation of Rafal Bogacz and Nick Yeung (Princeton University, December 2002). No noise in the frequency of our signal of interest (10 Hz) was used, so that the phase angle of the 10 Hz oscillation was completely controlled. The simulations were repeated under different amounts of SNR. SNR levels from −100 to 100 dB (in relation to high power oscillation) were investigated by scaling the noise according to the following steps: The noise was normalized to zero mean and unit variance. Then the following formula was applied:
$${\rm{SN}}=\frac{\frac{\sigma \,(\text{signal})}{\sqrt{{10}^{\frac{SNR}{10}}\,}}}{\sigma \,(\text{noise})\,}\cdot {\rm{noise}}$$
(4)
where signal is the high power oscillation and SNR the desired signal to noise ratio.
Phase angle and ITPC were calculated at time = 0 ms (centre of the epoch) for the low and high power condition for each subject, as described in paragraph 2 using formula (3). ITPC values were then compared between conditions using an independent samples t-test. This procedure was repeated 100 times to ensure differences were not found by chance.
### Small power difference can lead to a significant difference in estimated ITPC
In the first simulation the difference in power was varied systematically to examine to what extend the power needs to differ between conditions in order to obtain ITPC differences. There the high power condition alpha activity had a fixed amplitude of A = 3 µV and was compared to consecutively low powered conditions in which in which the amplitude was decreased from 0.1 µV steps. Jitter was added to the amplitudes of both conditions by adding a random scalar drawn from the standard normal distribution (0.5 * randn in Matlab). The simulated epochs had an average ITPC of 0.51 (sd = 0.04).
The percentage of simulations that lead to a significant difference in ITPC between the low and high alpha power condition are presented as colorcodes in Fig. 4. As predicted, ITPC values were larger for high power epochs than for low power epochs. Differences are present for SNR values between −40 dB and 0 dB. The absence of a difference for low levels of SNR could be explained by the fact that the SNR is too low to make a phase angle estimation in both conditions. The absence of a difference in ITPC for high levels of SNR could mean that noise does not have an influence on the phase estimation.
The difference in power does not need to be large to find significant differences as seen in Fig. 4. A difference of 0.1 µV (A = 3 µV versus A = 2.9 µV) already led to significant differences, and when the difference was increased to 0.2 µV (A = 3 µV versus A = 2.8 µV) the ITPC difference was stable. The few ITPC differences found when amplitudes were similar (both A = 3 µV) could arise because amplitudes for both conditions contained random jitter which makes it possible that amplitudes differed slightly between conditions.
### The influence of noise on ITPC estimates is present under most levels of phase-locking
In the second simulation the power difference was kept constant, but the analysis was repeated under various amounts of ITPC (please keep in mind that phase angles were identical between conditions that were compared, but repeated with different amounts of ITPC for both conditions). The ITPC was varied by changing the range of possible values of θ (phase angle, see equation 2). The interval of possible phases ranged from [0–0] to [0–2 π] in 11 steps, creating ITPC bins that ranged between 0 and 1. The amplitude of the high and low oscillation where based on power differences in alpha activity between two attentional conditions in one of our own EEG experiments7. This resulted in oscillations with amplitudes for the low and high power condition of respectively 2.7 µV and 3.2 µV plus jitter as described in the previous section.
As evident in Fig. 5 the power induced ITPC difference is present under most levels of phase-locking, and increases as phase-locking among trials increases. Note that the y-axis does not represent the difference in ITPC, but the actual ITPC in both conditions.
## Conclusion
Though phase angle perturbations are a subject of many studies in electrophysiological research, conclusive statements are lacking because of the absence of consensus among results. Previous animal work has suggested that attention to a particular sensory stream is able to amplify neuronal responses (to stimuli in that stream) by adjusting the phase of ongoing oscillations to be at their optimal excitatory cycle during anticipated stimulus onset34. More recently evidence for this phenomena has been observed in humans using MEG37. It should be noted that the reproducibility of this work in human experiments presents a particular challenge. This is due to the fact that variations in the orientation of dipoles generating oscillatory activity would translate in different absolute phases across participants,when the data is epoched relative to an experimental event. A simple illustration of this is in Fig. 6, where we simulated oscillatory activity from the same brain region, but had the generators have different dipole orientations. Such a situation could yield null findings in investigations looking at the relationship of the absolute phase of an ongoing oscillation to an experimental event. One way to circumvent such issues is to examine phase of oscillatory activity from the source reconstructed signal37,38. These simulations were done using Besa Simulator a free software for the simulation of electrophysiological activity at the scalp level(http://www.besa.de/downloads/besa-simulator/).
ITPC also gets around this issue, by not looking at the phase-angles, but rather their clustering (i.e uniformity). However, this measure is also prone to some caveats, especially when ITPC is compared between conditions. Simulations show that phase angle estimations can be affected by unwanted factors such as oscillatory power, ERP latency and ERP amplitude. Temporal leakage of evoked responses can increase phase locking among trials. Oscillations with relatively larger amplitudes have a better SNR, which leads to better phase estimations and less variability, hence higher ITPC estimates. Although the influence seems small at first sight, when power or evoked responses systematically differs due to attentional differences or task demands, this could lead to significant differences in ITPC between conditions. Caution should therefore be exercised with interpretation of ITPC differences when a power or ERP difference is present.
There are some steps advocated to reduce the possibility of false discoveries caused by power differences. For example Cohen39 suggests finding ITPC differences in the absence of a power difference for certain time points or channels. Furthermore, it is important that power differences are examined using non baseline-corrected frequency data. The simulations indicate that power differences lead to a difference in the ability to estimate phase-angles through differences in SNR. A baseline-correction adjusts the magnitude of the signal (frequency of interest) independent of the noise. This means that baseline-corrected data is not a reflection of the SNR.
The influence of condition specific differences in evoked responses on ITPC is sometimes overlooked. For instance Samaha et al.40 found an ITPC difference at time of target presentation between correct and incorrect trials. This means that it is not possible to unequivocally discount the possibility that their observation that the phase angles of ongoing alpha activity are modulated by temporal prediction could alternatively be explained by a difference in ERP amplitude and/or latency caused by differences in attentional states between trials. We advocate that any differences in ERP amplitude or latency should be excluded before conclusive statements be made about phase alterations. A way to eliminate the influence of evoked responses completely is to add trials in which a target is expected but not actually presented (as suggested in Van Diepen et al.7).
An example of a study in which additional analyses were performed to exclude power induced ITPC differences between conditions is by Bonnefond and colleagues17. The goal of this experiment was to show increased Inter-Trial Phase-Locking during a strong distractor condition, compared to a weak distractor condition, in a task with temporally predictable stimulus presentation. An ITPC difference was indeed found between the distractor conditions, however, together with a difference in oscillatory power. To rule out the likelihood that the power difference was driving the phase effects, ITPC values were correlated with power. Specifically, the correlation between the ongoing oscillatory power in an individual and the ITPC values was examined, as well as a correlation between differences in ITPC and power between the weak and strong distractor conditions. Also, the Spearman correlation for single-trial power and the deviation of the single-trial phase from the average phase was calculated. Last, trials were divided in quartiles based on alpha power and a trend-analysis was performed on the accompanying ITPC values. The absence of systematic correlations or a (clear) trend was interpreted as evidence that ITPC differences between the conditions could not be explained trivially by differences in oscillatory power.
We propose future work in addition to reporting the p values for the contrast in ITPC between conditions, also report the absolute ITPC values in each condition to allow the readers to assess the size of the effect. Finally, we propose new measures should be developed that can detect phase-adjustment directly. At present only the circumstantial consequence of a phase-adjustment is examined: a difference in ITPC at a certain point in time. Instead, one could search for more direct evidence that the oscillation is speeding up or slowing down in order to reach the optimal phase.
For the simulations a lot of parameters had to be chosen. The ITPC, SNR, oscillatory power and variation in power between epochs, window length of the Hann taper etc. They were chosen based on our own data as much as possible, but other experiments with other parameters can result in different outcomes. In addition, in real EEG data the power and ERP differences might be less systematic, decreasing the probability of an ITPC difference. Nevertheless, the fact that significant differences are found with at least some parameters show that it is a problem that should be considered. Therefore, to attain more conformity among studies examining phase-locking, direct comparisons between conditions should be avoided when systematic differences in power or ERPs are present.
Publisher's note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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Harter, M. R., Aine, C. & Schroeder, C. Hemispheric differences in the neural processing of stimulus location and type: effects of selective attention on visual evoked potentials. Neuropsychologia 20, 421–438 (1982).
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Eimer, M., Van Velzen, J., Gherri, E. & Press, C. Manual response preparation and saccade programming are linked to attention shifts: ERP evidence for covert attentional orienting and spatially specific modulations of visual processing. Brain Res 1105, 7–19, https://doi.org/10.1016/j.brainres.2005.10.060 (2006).
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Eimer, M. & van Velzen, J. Spatial tuning of tactile attention modulates visual processing within hemifields: an ERP investigation of crossmodal attention. Exp Brain Res 166, 402–410, https://doi.org/10.1007/s00221-005-2380-0 (2005).
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Baldauf, D. & Deubel, H. Attentional selection of multiple goal positions before rapid hand movement sequences: an event-related potential study. J Cogn Neurosci 21, 18–29, https://doi.org/10.1162/jocn.2008.21021 (2009).
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Couperus, J. W. & Mangun, G. R. Signal enhancement and suppression during visual-spatial selective attention. Brain Res 1359, 155–177, https://doi.org/10.1016/j.brainres.2010.08.076 (2010).
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## Author information
### Affiliations
1. #### Department of Psychiatry, Academic Medical Center, University of Amsterdam, Amsterdam, Netherlands
• Rosanne Maria van Diepen
2. #### School of Psychology, University of Birmingham, Birmingham, United Kingdom
• Ali Mazaheri
### Contributions
R.M.v.D. and A.M. designed the study. R.M.v.D. conducted the simulations. R.M.v.D. and A.M. wrote the paper.
### Competing Interests
The authors declare no competing interests.
### Corresponding author
Correspondence to Rosanne Maria van Diepen.
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This page collection together a list of "standard" small comments that I have on papers, with the hope that students and other co-authors will read it ahead of time and address some of these before sending me something to read. Many of these are idiosyncratic and the lead author on a paper has the final say, but since I make them over and over again, I figured I would start to write them down.
### Top Ten Most Common suggestions
1. Google "which versus that". If the word "which" doesn't have a comma before it, you probably got it wrong.
2. All technical documents, even a draft paper, should have a title, authors, and “identifying information” (current date or “Submitted, 2017 Winter q-bio” or something similar).
3. Get rid of acronyms unless you use the acronym when speaking about the term (eg, DNA is fine)
4. Never use an equation number on its own; the equation number is an adjective, not a noun (so "(19)" should be "equation (19)" or "system (19)")
5. Punctuation equations as if they were part of the sentence
6. Typeset the math right: ||x|| << 1 is typeset in LaTeX as $\| x \| \ll 1$, which will come out as $\|x\|\ll 1$ instead of $||x||<<1$.
7. Don’t use contractions in formal writing
8. Use display style (e.g. equation or displaymath) for typesetting mathematical expressions that are taller than the standard text
9. Definitions should be set in regular (non-italics) font with the defined term italicized. Remarks and examples should be in normal (non-italics) font. Theorems are italicized.
10. Read Higham's book on writing for the mathematical sciences.
(more to come)
### Other suggestions
• You should update the abstract so that it works as a “standalone” description of the *results* of a paper. Remove terms like “in this paper” that aren't really useful. Keep the introductory comments to a minimum (once sentence should be enough) and focus instead on summarizing the contents of the paper.
• I’m not a fan of the word “surprisingly” (or “interestingly”) in papers. Let the reader determine whether they are surprised. So perhaps change “surprisingly simple expression” to “simple expressions” (note also the plural).
• "an holistic” → “a holistic” (at least in American English)
• "well behaved” → “well-behaved” (more generally, multi-word phrases that are "converted" to an adjective are usually hyphenated. So "it was well behaved" and "well-behaved response".
• I am not a fan of acronyms except when they are used in common speech. I think when you talk about sequestration feedback systems you say “sequestration feedback systems” and not “ess eff bee”. Assuming that is correct, “SFB” → “sequestration feedback systems” (globally).
• "Section 2.2 and section 2.3 uses” → "Section 2.2 and Section 2.3 use”: the term "Section" should be capitalized and make sure that your verb agrees with the subject properly.
• Figure 1 is about 2X too big. As a guide, the text in a figure should be roughly the same font size as in the caption.
• As a general note on the introduction: it would be good to say more here about previous work on feedback circuits, fundamental limits, and how this work relates. Right now, it seems like there is only 1 paper that has ever done work in this area. Remember that some of the people whose work you are not bothering to cite as relevant prior effort are likely reviewers of your (eventual) paper.
• When you refer to a figure in the text, use the same term and capitalization as appears in the caption (so Figure 1, not fig. 1).
• I’m not a huge fan of the term “plant” for the process that is being controlled. OK to use this, but perhaps explain it since not all readers will necessarily know what you are talking about (especially when there is something else called a “plant” in biology, with a bit of a longer history of usage).
• I was always told to avoid starting sentences with lower case letters (even when they are a math symbol). It is confusing to the reader, since we use the capital letter to help us find the start of a sentence. Rewording can usually fix this problem.
• There appear to be several places where latex is putting extra space after a period. You can fix those by using “.\ “ when the period is not the end of a sentence (eg, in “et al.” or “eqs.”). Another approach, if you want to avoid a line break, is to use a tie (~): “.~(\ref{eqref})”.
• There are several spots where you assume your reader knows more than he/she might. For example, you talk about the sensitivity function without giving a citation for people who might not know what you are talking about. Same for Bode’s integral theorem.
• The right way to get the symbol for a norm is $\| x \|$ not “||”.
• Some of the text in Figure X is too small (especially the legend in panel B). As a guide, the text in a figure should be roughly the same font size as in the caption. You can probably go 1 pt size smaller, but I wouldn’t go below that. Also, looks like you might be using png/jpg for the figure? Better to use PDF since it will use vector graphics => doesn’t get blurry when you zoom in.
• The right way to get “<<“ is to use \ll (and \gg for “>>”).
• Equation (9) is missing punctuation (good job on catching it most other places!). As a general rule, punctuation inline and displayed equations as if they were text.
• The body of a theorem or proposition should be italicized. The body of a remark should not be. For a definition, only italicize the defined term.
• Expand “iff” to “if and only if” (unless you say “iffff” when you read it aloud). Similarly, write out "for all" and "for every" when it appears in body text. OK to use symbols ($\forall$ in an equation.
• I’m starting to get annoyed by references to supplementary information that seem relevant to me understanding the material. Is there a reason to relegate results to something that nobody is going to read? I would really only use SI (ess eye) for supporting information that is there to show some level of detail that would detract from the main flow of the paper.
• Consider using a style that uses a smaller font for captions than for the main text. This makes it easier to separate these (I got confused on page 12, where it looked like the equations were part of the caption for a second).
• Check for consistent capitalization in the figure titles.
• For references: you should either put full names everywhere or use initials everywhere (I recommend the latter). Also, I strongly encourage the listing of references in alphabetical order, since that makes it easier to find a reference to a specific paper. In the end this differs by journal, so OK to leave this like you have it, if you prefer.
• For citations, there should be a (non-breaking) space before the citation number [5]. In LaTeX, use the tie (~) to create a non-breaking space.
• “e coli” should be “E. coli” (italicized)
• The terms “in vivo” and “in vitro” should be italicized
• For references that are not available via a publisher, you should give a DOI (if available) or URL to the paper.
• When typesetting subscripts and superscripts, if the subscript is an abbreviation or shorthand for a word (or words), then it should be typeset as text, not math. So $x_{{\text{d}}}$ (where 'd' is for 'desired'), but $x_{i}$ if $i$ is an index variable.
• Then defining abbreviations, don't capitalize the terms unless they would ordinarily be capitalized. So "simultaneous localization and mapping (SLAM) algorithms", not "Simultaneous Localization and Mapping (SLAM) algorithms" and "hidden Markov model (HMM)".
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# Math Help - tan identity
1. ## tan identity
if anyone could show me how to do this that would be awesome
show that if: $
tan(2y)=\frac{1}{tan(x)}\$
then this is equivalent to the expression: $y=\frac{\Pi}{4}-\frac{x}{2}$
thanks
2. Hello, biker.josh07!
I have a "visual" solution.
Show that if: $\tan(2y)\:=\:\frac{1}{\tan x}\$
then this is equivalent to the expression: $y\:=\:\frac{\pi}{4}-\frac{x}{2}$
Since $\tan2y \:=\:\frac{1}{\tan x}$, then two angles are complementary.
Consider this right triangle:
Code:
*
* |
* x |
* |
* | a
* |
* |
* 2y |
* - - - - - - - *
b
$\text{Since }\tan 2y \,=\,\frac{a}{b}\:\text{ and }\:\tan x \,=\,\frac{b}{a},\:\text{ then: }\:\tan2y \,=\,\frac{1}{\tan x}$
. . $x$ and $2y$ are in the same right triangle.
Hence: . $x + 2y \:=\:\frac{\pi}{2} \quad\Rightarrow\quad 2y \:=\:\frac{\pi}{2} - x$
Therefore: . $y \;=\;\frac{\pi}{4} - \frac{x}{2}$
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# Existence of a particular basis for a finite-dimensional subspace of functions
There is this question that I saw, but have no idea how to solve it completely:
Let $X$ be a set, and consider $\mathbb{C}$-vector space Maps($X, \mathbb{C}$) of $\mathbb{C}$-valued functions on $X$. Let $V \subseteq$ Maps($X, \mathbb{C}$) be any finite dimensional subspace of Maps($X, \mathbb{C}$) with dimension $n$. Show that there exists points $x_1, \dots, x_n$ in $X$ and a $\mathbb{C}$-basis $f_1, \dots, f_n$ of $V$ such that for any $i, j \in \{1, \dots, n\}$, one has
$f_i(x_j)=\delta_{ij}=\begin{cases}1 \text{ if } i=j \\ 0 \text{ if } i \neq j \end{cases}$.
My thinking process: since $V$ is of dimension $n$, I can find $n$ elements in $V$ that forms a basis. Since $V$ consists of functions from $X$ to $\mathbb{C}$, the $n$ functions, labelled $g_1,\dots g_n$, are linearly independent of each other.
Now if I can find a set of $n$ points in $X$ (I have no idea how yet), and if I rewrite the values $f_1(x_1), f_2(x_1), \dots, f_n(x_1)$ for all $x_1, \dots, x_n$ into a matrix and I can show that this matrix is invertible, then since $V$ is a vector space and linear combinations of $g_i$ will be in $V$, and the matrix can be reduced to identity matrix (because it is invertible), I would get what I needed. However, how do I prove all these parts that I need? Or is this even the correct way?
Edit: formatting.
• The choice of $g$ lets you define a map $X \to \mathbb{C}^n$. I think the plan is to do linear algebra in $\mathbb{C}^n$ to show you can choose suitable $x$. – user14972 Feb 21 '18 at 17:14
• Wait, why will $g$ have anything to do with the points in set $X$? – Kenny Wong Feb 21 '18 at 17:25
• Given any $x$, you can evaluate the functions $g$ to get a tuple of complex numbers. – user14972 Feb 21 '18 at 17:26
• So let's say I have two points, $x_i$ and $x_j$. Can I try to prove that the two rows $f_1(x_i), \dots, f_n(x_i)$ and $f_1(x_j), \dots, f_n(x_j)$ are linearly independent? – Kenny Wong Feb 21 '18 at 17:33
Assume that $\dim V=n$ and that $g_1,\ldots,g_n\in V$ are linearly independent.
Claim: There exist $x_1,\ldots,x_n\in X$ such that if $f,g\in V$ and $f(x_j)=g(x_j)$ for all $j$, then $f=g$ (proof at the end).
Using the claim, the $n\times n$ matrix $A=(g_k(x_j))$ is invertible (if it weren't, we would find a nonzero linear combination $r_1g_1+\cdots+r_ng_n$ that is zero at all $x_j$, and by the claim they would be linearly dependent). Let $B$ be the inverse of $A$.
Now the equality $BA=I$ can be seen as $$\sum_\ell b_{k \ell}g_\ell(x_j)=\delta_{kj}=\delta_k(x_j),$$ where $\delta_k$ is the function that is $1$ at $x_k$ and zero elsewhere. By the claim, $$\delta_k=\sum_\ell b_{k \ell}g_\ell\in V.$$ It is obvious that $\delta_1,\ldots,\delta_n$ are linearly independent, so they form a basis for $V$.
Proof of the claim: by linearity, we may prove this version of the claim:
There exist $x_1,\ldots,x_n\in X$ such that if $f\in V$ and $f(x_j)=0$ for all $j$, then $f=0$
Assume that the negation of the claim holds: that for every $x_1,\ldots,x_n\in X$ there exists $f\in V$ with $f(x_j)=0$ for all $j$ by $f\ne0$.
Since $V$ is nonzero, there exists $h_0\in V$ and $x_1\in X$ with $h_0(x_1)=1$. Now apply the negation of the claim to $x_1,\ldots,x_1$: there exists $h_1\in V$ with $h_1(x_1)=0$, and $x_2\in X$ with $h_1(x_2)=1$. Next we apply the negation to $x_1,x_2,\ldots,x_2$: there exists $h_2\in V$ with $h_2(x_1)=h_2(x_2)=0$, and $x_3\in X$ with $h_2(x_3)=1$. Continuing in the same fashion, we end up with $x_1,\ldots,x_{n+1}$ and $h_0,h_1,\ldots,h_n\in V$ such that, for each $k$, $$h_k(x_1)=\cdots=h_k(x_k)=0,\ \ \ h_k(x_{k+1})=1.$$ The functions $h_0,\ldots,h_n$ are linearly independent: if $\sum_{k=0}^n\lambda_kh_k=0$, evaluating at each of $x_1,\ldots,x_{n+1}$ we obtain that $\lambda_0=\cdots=\lambda_n=0$.
As $V$ is $n$-dimensional, we have obtained a contradiction, which proves the claim.
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The Sign of One
Make the following numbers using FOUR 1s using any mathematics operators and/or symbols, such as $\dfrac{x}{y}$, $\sqrt{x}$, decimal dots, $+$, $-$, $\times$, $\div$, $($ $)$, etc by the Sign of One.
Click the numbers below to see the answers.
The first one is done for you.
$1$
$1 = 1 \times 1 \times 1 \times 1$
$2$
$2 = (1+1) \times 1 \times 1$
$3$
$3 = (1+1+1) \times 1$
$4$
$4 = 1+1+1+1$
$5$
$5 = \dfrac{1}{.1} \div (1+1)$
Note that the dot is the decimal point, such as $.1 = 0.1$
$6$
\begin{align} \displaystyle 6 &= (1+1+1)! \times 1 \\ &\text{Note that the exclamation mark ! means factorial, such as } 3! = 3 \times 2 \times 1 = 24 \\ &= \sqrt{\dfrac{1}{.\dot{1}}} + \sqrt{\dfrac{1}{.\dot{1}}} \\ &\text{Note that } .\dot{1} = .111 \cdots \text{ recurring decimals.} \\ &= \Bigg(\sqrt{\dfrac{1}{.\dot{1}}}\Bigg)! \\ \end{align}
$7$
$7 = (1+1+1)! + 1$
$8$
$8 = \dfrac{1}{.\dot{1}} – 1 \times 1$
$9$
$9 = \dfrac{1}{.\dot{1}} + 1 – 1$
$10$
$10 = \dfrac{1}{.\dot{1}} + 1 \times 1$
$11$
$11 = \dfrac{1}{.\dot{1}} + 1 + 1$
$12$
$12 = 11 + 1 \times 1$
$13$
$13 = 11 + 1 + 1$
$14$
$14 = 11 + \sqrt{\dfrac{1}{.\dot{1}}}$
$15$
$15 = \dfrac{1}{.\dot{1}} + \Bigg(\sqrt{\dfrac{1}{.\dot{1}}}\Bigg)!$
$16$
$16 = \dfrac{1}{.1} + \Bigg(\sqrt{\dfrac{1}{.\dot{1}}}\Bigg)!$
$17$
$17 = 11 + \Bigg(\sqrt{\dfrac{1}{.\dot{1}}}\Bigg)!$
$18$
$18 = \dfrac{1}{.\dot{1}} + \dfrac{1}{.\dot{1}}$
$19$
$19 = \dfrac{1}{.1} + \dfrac{1}{.\dot{1}}$
$20$
$20 = \dfrac{1}{.1} + \dfrac{1}{.1}$
$21$
$21 = \dfrac{1}{.1} + 11$
$22$
$22 = 11 + 11$
$23$
$23 = \Bigg(\sqrt{\dfrac{1}{.\dot{1}}}+1\Bigg)!-1$
$24$
$24 = \Bigg(\sqrt{\dfrac{1}{.\dot{1}}}+1\Bigg)!\times 1$
$25$
$25 = \Bigg(\sqrt{\dfrac{1}{.\dot{1}}}+1\Bigg)!+1$
$26$
\begin{align} \displaystyle 26 &= \Biggl\lceil \sqrt{\dfrac{1}{.1}} \Biggl\rceil ! + 1 + 1 \\ &= \Bigl\lceil \sqrt{10} \Bigl\rceil ! + 2 \\ &= \bigl\lceil 3.162 \cdots \bigr\rceil ! + 2 \\ &= 4! + 2 \\ &= 4 \times 3 \times 2 \times 1 + 2 \\ &= 24 + 2 \\ &= 26 \text{ Boom!}\\ \text{Note that:} \\ \bigl\lfloor x \bigr\rfloor &= \text{the greatest integer less than or equal to } x \text{ (floor)} \\ \bigl\lfloor 3.4 \bigr\rfloor &= 3 \\ \bigl\lceil x \bigr\rceil &= \text{the smallest integer greater than or equal to } x \text{ (ceiling)} \\ \bigl\lceil 3.4 \bigr\rceil &= 4 \\ \end{align}
$27$
\begin{align} \displaystyle 27 &= \Biggl\lceil \sqrt{\dfrac{1}{.1}} \Biggl\rceil ! + \Biggl\lfloor \sqrt{\dfrac{1}{.1}} \Biggl\rfloor \\ &= \Bigl\lceil \sqrt{10} \Bigl\rceil ! + \Bigl\lfloor \sqrt{10} \Bigl\rfloor \\ &= \bigl\lceil 3.162 \cdots \bigl\rceil ! + \bigl\lfloor 3.162 \cdots \bigl\rfloor \\ &= 4! + 3 \\ &= 4 \times 3 \times 2 \times 1 + 3 \\ &= 24 + 3 \\ &= 27 \\ \end{align}
$28$
\begin{align} \displaystyle 28 &= \Biggl\lceil \sqrt{\dfrac{1}{.1}} \Biggl\rceil ! + \Biggl\lceil \sqrt{\dfrac{1}{.1}} \Biggl\rceil \\ &= \bigl\lceil 3.162 \cdots \bigl\rceil ! + \bigl\lceil 3.162 \cdots \bigl\rceil \\ &= 4! + 4 \\ &= 4 \times 3 \times 2 \times 1 + 4 \\ &= 24 + 4 \\ &= 28 \\ \end{align}
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Convolution
# Overview
The Convolution algorithm performs a 2D convolution operation on the input image with the provided 2D kernel. This is useful when the kernel isn't separable and its dimensions are smaller than 5x5. In other cases, it's usually preferable to use the Separable Convolution algorithm due to its speed.
Input Kernel Output
$\begin{bmatrix} 1 & 0 & -1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{bmatrix}$
# Implementation
Discrete 2D convolution is implemented using the following discrete function:
$I'[x,y] = \sum_{m=0}^{k_h} \sum_{n=0}^{k_w} K[m,n] \times I[x-(n-\lfloor k_w/2 \rfloor), y-(m-\lfloor k_h/2 \rfloor) ]$
Where:
• $$I$$ is the input image.
• $$I'$$ is the result image.
• $$K$$ is the convolution kernel.
• $$k_w,k_h$$ are the kernel's width and height, respectively.
Note
Most computer vision libraries expect the kernel to be reversed before calling their convolution functions. Not so with VPI, we implement an actual convolution, not cross-correlation. Naturally, this is irrelevant if the kernel is symmetric.
# C API functions
For list of limitations, constraints and backends that implements the algorithm, consult reference documentation of the following functions:
Function Description
vpiSubmitConvolution Runs a generic 2D convolution over an image.
# Usage
Language:
1. Import VPI module
import vpi
2. Define a 3x3 convolution kernel to perform edge detection.
kernel = [[ 1, 0, -1],
[ 0, 0, 0],
[-1, 0, 1]]
3. Run convolution filter on input image using the CPU backend and the given kernel. Input and output are VPI images.
with vpi.Backend.CUDA:
output = input.convolution(kernel, border=vpi.Border.ZERO)
1. Initialization phase
1. Include the header that defines the needed functions and structures.
Declares functions to perform image filtering with convolution kernels.
2. Define the input image object.
VPIImage input = /*...*/;
struct VPIImageImpl * VPIImage
A handle to an image.
Definition: Types.h:256
3. Create the output image. It gets its dimensions and format from the input image.
int32_t w, h;
vpiImageGetSize(input, &w, &h);
vpiImageGetFormat(input, &type);
VPIImage output;
vpiImageCreate(w, h, type, 0, &output);
uint64_t VPIImageFormat
Pre-defined image formats.
Definition: ImageFormat.h:94
VPIStatus vpiImageGetFormat(VPIImage img, VPIImageFormat *format)
Get the image format.
VPIStatus vpiImageCreate(int32_t width, int32_t height, VPIImageFormat fmt, uint64_t flags, VPIImage *img)
Create an empty image instance with the specified flags.
VPIStatus vpiImageGetSize(VPIImage img, int32_t *width, int32_t *height)
Get the image dimensions in pixels.
4. Create the stream where the algorithm will be submitted for execution.
VPIStream stream;
vpiStreamCreate(0, &stream);
struct VPIStreamImpl * VPIStream
A handle to a stream.
Definition: Types.h:250
VPIStatus vpiStreamCreate(uint64_t flags, VPIStream *stream)
Create a stream instance.
2. Processing phase
1. Define the kernel to be used. In this case, a simple 3x3 edge detector.
float kernel[3 * 3] = { 1, 0,-1,
0, 0, 0,
-1, 0, 1};
2. Submit the algorithm to the stream, passing the kernel and other parameters. It'll be executed by the CPU backend.
vpiSubmitConvolution(stream, VPI_BACKEND_CPU, input, output, kernel, 3, 3, VPI_BORDER_ZERO);
VPIStatus vpiSubmitConvolution(VPIStream stream, uint64_t backend, VPIImage input, VPIImage output, const float *kernelData, int32_t kernelWidth, int32_t kernelHeight, VPIBorderExtension border)
Runs a generic 2D convolution over an image.
@ VPI_BACKEND_CPU
CPU backend.
Definition: Types.h:92
@ VPI_BORDER_ZERO
All pixels outside the image are considered to be zero.
Definition: Types.h:278
3. Optionally, wait until the processing is done.
vpiStreamSync(stream);
VPIStatus vpiStreamSync(VPIStream stream)
Blocks the calling thread until all submitted commands in this stream queue are done (queue is empty)...
3. Cleanup phase
1. Free resources held by the stream and the input and output images.
vpiImageDestroy(output);
void vpiImageDestroy(VPIImage img)
Destroy an image instance.
void vpiStreamDestroy(VPIStream stream)
Destroy a stream instance and deallocate all HW resources.
Consult the Image Convolution for a complete example.
For more information, see Convolution in the "C API Reference" section of VPI - Vision Programming Interface.
# Performance
For information on how to use the performance table below, see Algorithm Performance Tables.
Before comparing measurements, consult Comparing Algorithm Elapsed Times.
For further information on how performance was benchmarked, see Performance Benchmark.
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# What does canonical extension mean?
Density-reachability is a canonical extension of direct density-reachability. What is meant by canonical extension?
Source: A Density-Based Algorithm for Discovering Clusters in Large Spatial Databases with Noise https://www.aaai.org/Papers/KDD/1996/KDD96-037.pdf
• Could you give the source of the quote (with hyperlink if possible)? – Peter Taylor May 1 '18 at 11:15
• @PeterTaylor I added the title and a direct link to the paper where it was used – Vincent May 2 '18 at 11:19
The term 'extension' could generally be interpreted in mathematical sense1 as an enlargement of the domain of a pre-existing function.
In your case, assume Density-reachability could be described as a process $d\colon A \rightarrow B$ mapping inputs from domain $A$ to output in domain $B$. An enlargement of domain could be viewed as another map $e\colon A \rightarrow A'$ which maps where the old inputs in $A$ could be thought of as special cases in subset of $A'$ that is the image of map $e$.
An extension then could be described as any map $x\colon A' \rightarrow B$ which statisfy $$x \circ e = d$$ There could be many such $x$, so picking extension is a choice. The above equation basically expresses that
If I pick $a \in A$ from the old domain, translate it to the new domain as $e(a)$, then use this new tool $x$ which could be thought of as an extension of old tool $d$ along the translation $e$, then the final result must agree with just using the old tool $d$ to begin with.
or $$x \circ e(a) = d(a)\qquad; \forall a \in A$$
The term 'canonical' simply implies that the choice of extension is the obvious, natural one of the situation.
1 At least from the perspective of category theory.
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# How do you express the complex number in trigonometric form: -2?
Apr 29, 2016
$2 \cos \pi + i \sin \pi$.
A real number has the argument equal to zero if positive or $\pi$ if negative:
So, it is $2 \cos \pi + i \sin \pi$.
$\sin \pi = 0$ so number is real;
$\cos \pi = - 1$ so the number is negative.
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# How do you integrate x ln (x + 1) dx?
Jul 12, 2016
$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} x \left(x - 2\right) + C$
#### Explanation:
$\int \setminus x \ln \left(x + 1\right) \setminus \mathrm{dx}$
$= \int \setminus \frac{d}{\mathrm{dx}} \left({x}^{2} / 2\right) \ln \left(x + 1\right) \setminus \mathrm{dx}$
and by IBP : $\int u v ' = u v - \int u ' v$
$= {x}^{2} / 2 \ln \left(x + 1\right) - \int \setminus {x}^{2} / 2 \frac{d}{\mathrm{dx}} \left(\ln \left(x + 1\right)\right) \setminus \mathrm{dx}$
$= {x}^{2} / 2 \ln \left(x + 1\right) - \frac{1}{2} \textcolor{b l u e}{\int \setminus \frac{{x}^{2}}{x + 1}} \setminus \mathrm{dx} q \quad \triangle$
for the integration in blue, we use a simple sub
$u = x + 1 , \mathrm{du} = \mathrm{dx}$
$\int \setminus \frac{{\left(u - 1\right)}^{2}}{u} \setminus \mathrm{du}$
$\int \setminus u - 2 + \frac{1}{u} \setminus \mathrm{du}$
$= {u}^{2} / 2 - 2 u + \ln u$
$= {\left(x + 1\right)}^{2} / 2 - 2 \left(x + 1\right) + \ln \left(x + 1\right)$
so $\triangle$ becomes
$= {x}^{2} / 2 \ln \left(x + 1\right) - \frac{1}{2} \left({\left(x + 1\right)}^{2} / 2 - 2 \left(x + 1\right) + \ln \left(x + 1\right)\right) + C$
$= {x}^{2} / 2 \ln \left(x + 1\right) - \frac{1}{4} {\left(x + 1\right)}^{2} + \left(x + 1\right) - \frac{1}{2} \ln \left(x + 1\right) + C$
$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} \left({x}^{2} + 2 x + 1 - 4 x - 4\right) + C$
$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} \left({x}^{2} - 2 x - 3\right) + C$
$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} \left({x}^{2} - 2 x\right) + C$
$= \frac{{x}^{2} - 1}{2} \ln \left(x + 1\right) - \frac{1}{4} x \left(x - 2\right) + C$
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# Uniform convergence and negating uniform convergence in logical symbols
How do you write this in logical symbols: For each $\epsilon>0$ there exist a number $N$ such that $|f_n(x)-f(x)|<\epsilon$ for all $x\in S$ and all $n>N$.
Is this correct? $\forall \epsilon>0 \exists N,(x\in S\land n>N\implies |f_n(x)-f(x)|<\epsilon)$
And how can I negate this?
-
Yes, this is correct. – julien Feb 21 at 14:59
It’s
$$\forall\epsilon>0\,\exists N\in\Bbb N\,\forall x\in S\,\forall n>N\big(|f_n(x)-f(x)|<\epsilon\big)\;;$$
its negation, after the negation is pulled inside all the quantifiers, is
$$\exists\epsilon>0\,\forall N\in\Bbb N\,\exists x\in S\,\exists n>N\big(|f_n(x)-f(x)|\ge\epsilon\big)\;.$$
This negation can be performed mechanically, using the equivalence of $\neg\forall x\varphi(x)$ with $\exists x\neg\varphi(x)$ and the equivalence of $\neg\exists x\varphi(x)$ with $\forall x\neg\varphi(x)$.
Informally, the negation says that there is a ‘bad’ $\epsilon$, meaning one such that no matter how far out in the sequence $\langle f_n:n\in\Bbb N\rangle$ you go, you can find an $n$ at least that far out and a point $x\in S$ such that $f_n(x)$ and $f(x)$ differ by at least $\epsilon$.
-
Is $\forall\epsilon>0\,\exists N\in\Bbb N\,\forall x\in S\,\forall n>N\big(|f_n(x)-f(x)|<\epsilon\big)\;$ equivalent with $\forall \epsilon>0 \exists N,(x\in S\land n>N\implies |f_n(x)-f(x)|<\epsilon)$ ? – Kasper Feb 21 at 15:10
@Kasper: My version contains a little more information, in that it explicitly specifies that $N\in\Bbb N$, but otherwise theyre equivalent. I would omit the comma from yours, however: it serves only to clutter up the expression, since it adds nothing to the information conveyed by the parentheses. – Brian M. Scott Feb 21 at 15:12
Okay, thanks for the explanation ! – Kasper Feb 21 at 15:31
@Kasper: You’re welcome! – Brian M. Scott Feb 21 at 15:32
Can't I pull out "$\forall N\in\Bbb N,\exists n>N$" from the negation and replace it (in the position where $\forall N\in\Bbb N$ was) with "$\forall n\in\Bbb N$"? They seem equivalent to me. – Ryan Sep 28 at 13:56
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# Use IE.au3 with this form
## 16 posts in this topic
#1 · Posted (edited)
Hi guys,
i use DaleHohm's IE.au3 file in my programs to submit forms etc.
Now i found this form on a website, and i know it's crappy HTML scripting:
<form action=/index.php method=post>
<input size=20 name=chto value="">
<input type=submit value="Search" class="knp">
</form>
How can I set the value in chto and how can i submit the form, using IE.au3? (v1.0)
If i do it like this it works perfectly..
Form:
<form action="search.php" method="post" name="search">
<input name="q" type="text" />
<input type="submit" value="Search" />
<input name="post" type="hidden" />
</form>
#include <IE.au3>
#include <File.au3>
#include <FTP.au3>
; Create an invisible IE
$oIE = _IECreate(0)$sq = "norton"
_IENavigate($oIE, "http://localhost/~ OSS ~/search.php")$o_form = _IEFormGetObjByName ($oIE, "search")$o_login = _IEFormElementGetObjByName ($o_form, "q") _IEFormElementSetValue ($o_login, $sq) _IEFormSubmit($o_form)
Sleep(150)
$source = _IEBodyReadHTML($oIE)
$filec = _FileCreate(@ScriptDir & "\list.txt")$file = FileOpen(@ScriptDir & "\list.txt", 1)
FileWrite($file,$source)
_IEQuit($oIE) It works because i've given the form a name, but with the form i found on the i-net there didn't give the form a name, so maybe DaleHohm himself knows the answer... :D:P Greetings, Andreas Edited by Noobster24 Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc. #### Share this post ##### Link to post ##### Share on other sites ...It works because i've given the form a name, but with the form i found on the i-net there didn't give the form a name, so maybe DaleHohm himself knows the answer... Fast help, since not much time: Everything can have a name, but everything will always have a number! So: look for something like: (retrieve form.length) get form[0] (always start the first with a zero) just look on websites about javascript.. I hope it gets you going... #### Share this post ##### Link to post ##### Share on other sites #3 · Posted (edited) Fast help, since not much time: Everything can have a name, but everything will always have a number! So: look for something like: (retrieve form.length) get form[0] (always start the first with a zero) just look on websites about javascript.. I hope it gets you going... Wow, in one second i found this website: But that's in Javascript, i need to implant this into AutoIt... To submit the form in java script: document.forms[0].submit() But how do i set the value of a from in Javascript if it isn't your website, eg. I could put this Javascript code in my webpage: var sq = 'norton' document.forms[0].chto.value = var sq; But the website i want to set the value in the form isn't my website so i can't put this javascript in the source... Edited by Noobster24 Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc. #### Share this post ##### Link to post ##### Share on other sites var sq = 'norton' document.forms[0].chto.value = var sq; But the website i want to set the value in the form isn't my website so i can't put this javascript in the source... think this: form[0] is the first form every form has 1 or more 'elements', all have a number once again... so something with .?.....forms.item(0).item(0).value=var sq (or something...) HTH #### Share this post ##### Link to post ##### Share on other sites think this: form[0] is the first form every form has 1 or more 'elements', all have a number once again... so something with .?.....forms.item(0).item(0).value=var sq (or something...) HTH Yep i know, but how do i let AutoIt do that? I know the Javascript part, but AutoIt isn't Javascript Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc. #### Share this post ##### Link to post ##### Share on other sites #6 · Posted (edited) Yep i know, but how do i let AutoIt do that? I know the Javascript part, but AutoIt isn't Javascript This is in Javascript the field i need to set the value to: document.forms[0].elements[0].value = 'The new value'; I want to set the value with AutoIt in combination with Javascript... Is that possible? //I'm must eat now, be back in one hour.. Edited by Noobster24 Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc. #### Share this post ##### Link to post ##### Share on other sites #7 · Posted (edited) Yep i know, but how do i let AutoIt do that? I know the Javascript part, but AutoIt isn't Javascript Mmm, perhaps Dale's IE-stuff does not like index-numbers that much right now. As far I can think (quickly): do use the javascript! Just put the$oIE in front like
$oIE.document.forms.index(0) etc.. or$oIE.document.all.forms[0].item[0].value=$myvar ... or$oIE.document.forms[0].elements[0].value = 'The new value';
something like that...?
Edited by heibel
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Mmm, perhaps Dale's IE-stuff does not like index-numbers that much right now.
As far I can think (quickly): do use the javascript! Just put the $oIE in front like$oIE.document.forms.index(0) etc..
or $oIE.document.all.forms[0].item[0].value=$myvar ...
or $oIE.document.forms[0].elements[0].value = 'The new value'; something like that...? Hi, i'm back, fresh and sharp , i'll try what you said Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc. #### Share this post ##### Link to post ##### Share on other sites #9 · Posted (edited) Hi, i'm back, fresh and sharp , i'll try what you said Wow, this seems to work ;InetGet("http://localhost/images/logo.jpg", "C:\log.jpg", 1) ;#include <INet.au3> ;ConsoleWrite(_INetGetSource('www.autoitscript.com')) #include <IE.au3> #include <File.au3> #include <FTP.au3> ; Create IE$oIE = _IECreate(1)
_IENavigate($oIE, "http://localhost/~ OSS ~/search.php")$sq = "norton"
$oIE.document.forms(0).elements(0).value = 'norton' Edited by Noobster24 Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc. #### Share this post ##### Link to post ##### Share on other sites Wow, this seems to work ;InetGet("http://localhost/images/logo.jpg", "C:\log.jpg", 1) ;#include <INet.au3> ;ConsoleWrite(_INetGetSource('www.autoitscript.com')) #include <IE.au3> #include <File.au3> #include <FTP.au3> ; Create IE$oIE = _IECreate(1)
_IENavigate($oIE, "http://localhost/~ OSS ~/search.php")$sq = "norton"
$oIE.document.forms(0).elements(0).value = 'norton' Just one question... In this form (the crappy scripting form.. ): <FORM action=/index.php method=post><INPUT value=norton name=chto> <INPUT class=knp type=submit value=Search> </FORM> The value is set as you can see , but the page isn't submit with: $oIE.document.forms(0).submit
How must i submit the form?
Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc.
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#11 · Posted (edited)
IE.au3 can access items by name or by index value.
To access by index in T2.0, use the *GetCollection() routines and specify the index number. For example, to get a reference to the first form on the page (the numbering starts with 0):
$oForm = _IEFormGetCollection($oIE, 0)
T1.0 had a function _IEFormGetObjByIndex, but T2.0 combines that functionality into _IEFormGetCollection.
Dale
Edit: removed OP text
Edited by DaleHohm
Free Internet Tools: DebugBar, AutoIt IE Builder, HTTP UDF, MODIV2, IE Developer Toolbar, IEDocMon, Fiddler, HTML Validator, WGet, curl
Automate input type=file (Related)
SciTe Debug mode - it's magic: #AutoIt3Wrapper_run_debug_mode=Y Doesn't work needs to be ripped out of the troubleshooting lexicon. It means that what you tried did not produce the results you expected. It begs the questions 1) what did you try?, 2) what did you expect? and 3) what happened instead?
Reproducer: a small (the smallest?) piece of stand-alone code that demonstrates your trouble
##### Share on other sites
#12 · Posted (edited)
The value is set as you can see , but the page isn't submit with:
$oIE.document.forms(0).submit How must i submit the form? ? try submit() instead of submit... so:$oIE.document.forms(0).submit()
good you got it working!
Oh and for all this stuff, search and download some info called the 'DOM' (that & browser: stands for all this sort of possability's/language-stuff you can use...)
Edited by heibel
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IE.au3 can access items by name or by index value.
Also for the SetFormObject (or something)?
He, and thanks for the incredible bug-fix: works like a charm!
##### Share on other sites
#14 · Posted (edited)
IE.au3 can access items by name or by index value.
To access by index in T2.0, use the *GetCollection() routines and specify the index number. For example, to get a reference to the first form on the page (the numbering starts with 0):
$oForm = _IEFormGetCollection($oIE, 0)
T1.0 had a function _IEFormGetObjByIndex, but T2.0 combines that functionality into _IEFormGetCollection.
Dale
Edit: removed OP text
The problem with T2.0 is, that when i want to use the function _IECreate(), it gives a error,
example:
#include <IE.beta.au3>
#include <File.au3>
#include <FTP.au3>
; Create an invisible IE
$oIE = _IECreate("http://localhost/~ OSS ~/search.php",0,0)$sq = 'norton'
$oIE.document.forms(0).elements(0).value =$sq
$oIE.document.forms(0).submit() Sleep(150)$source = _IEBodyReadHTML($oIE) _IEQuit($oIE)
$filec = _FileCreate(@ScriptDir & "\norton.txt")$file = FileOpen(@ScriptDir & "\norton.txt", 1)
FileWrite($file,$source)
It looks to me that this code is okay, but this error shows up:
+>AU3Check ended.rc:0
>Running:(3.1.1.126):D:\Program Files\AutoIt3\beta\autoit3.exe "C:\Documents and Settings\Beheerder\Bureaublad\Andreas\Custom Cracks\~ OSS ~\poeppages\testt2.0.au3"
--> IE.au3 Warning from function _IECreate (Foreground Window Unlock Failed!)
+>AutoIT3.exe ended.rc:0
>Exit code: 0 Time: 1.842
I've got the latest AutoIt beta version (v3.1.1.126 (beta)), i've got the latest IE UDF Library...
Edited by Noobster24
Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc.
##### Share on other sites
The problem with T2.0 is, that when i want to use the function _IECreate(), it gives a error,
Sorry, everything is working fine, the Warning is in error not the code... LOULOU reported this a couple of days ago and you'll find a note at the end of the T2.0 post stating that this is know and is fixed for the next reslease. The mistake on my part is taht I display the warning anytime you set takeFocus to 0 regardless of whether it works or not.
If you want to, you can suppress the warning by wrapping _IEErrorNotify(False) and _IEErrorNotify(True) around your _IECreate().
Dale
Free Internet Tools: DebugBar, AutoIt IE Builder, HTTP UDF, MODIV2, IE Developer Toolbar, IEDocMon, Fiddler, HTML Validator, WGet, curl
Automate input type=file (Related)
SciTe Debug mode - it's magic: #AutoIt3Wrapper_run_debug_mode=Y Doesn't work needs to be ripped out of the troubleshooting lexicon. It means that what you tried did not produce the results you expected. It begs the questions 1) what did you try?, 2) what did you expect? and 3) what happened instead?
Reproducer: a small (the smallest?) piece of stand-alone code that demonstrates your trouble
##### Share on other sites
Sorry, everything is working fine, the Warning is in error not the code... LOULOU reported this a couple of days ago and you'll find a note at the end of the T2.0 post stating that this is know and is fixed for the next reslease. The mistake on my part is taht I display the warning anytime you set takeFocus to 0 regardless of whether it works or not.
If you want to, you can suppress the warning by wrapping _IEErrorNotify(False) and _IEErrorNotify(True) around your _IECreate().
Dale
I'll wait for the next release, T1.0 is still working fine :D!
Thanks guys
Programs so far:Teh Serializer - Search for licenses for Nero - Windows - Office - Alcohol etc.
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# How seriously can we take the success of the Standard Model when it has so many input parameters?
The Standard Model of particle physics is immensely successful. However, it has many experimentally fitted input parameters (e.g. the fermion masses, mixing angles, etc). How seriously can we take the success of the Standard Model when it has so many input parameters?
On face value, if a model has many input parameters, it can fit a large chunk of data. Are there qualitative and more importantly, quantitative, predictions of the Standard Model that are independent of these experimentally fitted parameters? Again, I do not doubt the success of the SM but this is a concern I would like to be addressed and be demystified.
• The idea is to use a subset of known parameters to predict the others and then compare them with experiments, see for example the electroweak precision tests
– Pipe
Sep 25, 2021 at 3:50
• Maybe this answer of mine will clear up the use of constants derived physics.stackexchange.com/q/262917 from data in physics theories also this physics.stackexchange.com/q/54184 Sep 25, 2021 at 4:52
• I think it's still an open question how many of these parameters are actually independent. One of the goals of "Theory of Everything" is to reduce them. Sep 25, 2021 at 14:10
• The number of parameters is due to the number of fundamental fields. You have three gauge fields each with its own coupling, you have a dozen fundamental fermions each with its own mass, etc. Do you feel that the number of fundamental fields is a challenge to the standard model's credibility? Sep 25, 2021 at 14:38
• How many quantitative results would one be able to produce without experimentally fitting any constants? None. The number of free parameters is problematic for SM but the wording in the question is fairly weird to me at least. How seriously can you take the success of Newtonian gravity if it takes 3 "experimentally fitted parameters" to determine free fall acceleration at Earth's surface (and also consider how was it tested at the time - and by the time it became widely accepted!). Also these constants don't just disappear as the model refines: they become structural parameters instead... Sep 26, 2021 at 13:20
It is inaccurate to think that all of the standard model of particle physics was determined through experiment. This is far from true. Most of the time, the theoretical predictions of particle physics were later confirmed experimentally and quite often to a very high accuracy.
For example, theoretical physicists predicted the existence of the $$W^\pm$$ and $$Z$$ bosons and their masses, the Higgs boson, the existence of gluons, and many of their properties before these particles were even detected. Pauli postulated the existence of the neutrino to explain energy conservation in beta decay, before the neutrino was observed. The anomalous magnetic moment of the electron, whose value was predicted by Julian Schwinger, agrees with experiment to up to 3 parts in $$10^{13}$$. Parity violation in the weak interaction, predicted by Lee and Yang, was later confirmed experimentally. The prediction of the positron by Dirac, was detected four years later by Anderson.
The list goes on and the list is huge$$^1$$. Particle physics is arguably the most successful physics theory because time and time again its predictions were later confirmed by experiment to surprisingly high accuracy (though sometimes our theories needed to be improved to explain some details of experimental data). I may be biased coming from a theoretical particle physics background, but I've always agreed that the Standard Model is the most mathematically beautiful, deep and profound model of all of physics. This is reflected in its almost miraculously accurate predictive power.
$$^1$$Some more of the highlights:
1935 Hideki Yukawa proposed the strong force in order to explain interactions between nucleons.
1947 Hans Bethe uses renormalization for the first time.
1960 Yoichiro Nambu proposes SSB (chiral symmetry breaking) in the strong interaction.
1964 Peter Higgs and Francois Englert, propose the Higgs mechanism.
1964 Murray Gell-Mann and George Zweig put forth the basis of the quark model.
1967 Steven Weinberg and Abdus Salam propose the electroweak interaction/unification.
• Yes, of course experiment is crucial in physics. Sep 25, 2021 at 8:10
• Don't forget Dirac's inference of antimatter! Sep 26, 2021 at 7:35
• I agree the great value of Standard Model but the history is much more balanced between theory and experiment than this answer suggests. There have been many experimental surprises and a list of highlights of theoretical work omits the many other theoretical proposals that turned out to be unhelpful or misleading. Sep 26, 2021 at 7:55
• Worth mentioning how many of these predictions were not intuitive or foreseen,they just emerge from the raw standard model mathematics, once the values from completely unrelated phenomena are used to determine a relatively small number of parameters. And then, some of those predictions, of particle masses and interactions, turn out not just to be real, but be predicted to astonishing accuracy as well. Sep 26, 2021 at 10:14
• I have noticed that people sometimes place emphasis on experiment over theory. If you read the papers in the 1960s where some of this was coming into being, it was key that there were theoretical explorations going on which led to Goldstone's theorem, massless Nambu-Goldstone bosons, spontaneous symmetry breaking etc., then before that Zumino's work on two-dimensional electrodynamics and so on.
– Tom
Sep 26, 2021 at 16:13
The Standard Model may have many parameters, but it also talks about many things, each typically only involving a very limited number of parameters. For example, the muon lifetime$$^\dagger$$ $$\tau_\mu=\frac{6144\pi^3M_W^4}{g^4m_\mu^5}$$depends on only $$M_W,\,g,\,m_\mu$$ ($$g$$ is the weak isospin coupling), and the tauon lifetime $$\tau_\tau$$ satisfies$$\frac{\tau_\tau}{\tau_\mu}=\frac{1}{3(|V_{ud}|^2+|V_{us}|^2)+2}\frac{m_\mu^5}{m_\tau^5},$$which only depends on $$|V_{ud}|,\,|V_{us}|,\,m_\mu,\,m_\tau$$. So these two predictions need six parameters, which doesn't in itself sound that impressive. But that misses the point here. The full slate of SM predictions uses all the parameters in a variety of subsets, creating a system of far, far more simultaneous equations than we have parameters. If (to take one estimate discussed herein) there are $$37$$ parameters, it's not like we're fitting a degree-$$36$$ polynomial $$y=p(x)$$ by OLS. It's more like requiring the same few coefficients to simultaneously fit many different regression problems.
$$^\dagger$$ I'm sure someone will object my formula for $$\tau_\mu$$ has a time on the LHS and an inverse mass on the RHS, is using natural units, and should read$$\tau_\mu=\frac{\hbar}{c^2}\frac{6144\pi^3M_W^4}{g^4m_\mu^5}$$in SI units to expose the use of two more parameters $$c,\,\hbar$$. Bear in mind, however, these both have exact SI values by definition. Anyway, it doesn't really affect the argument.
• And don't forget the predictive power of the Standard Model. It predicts almost all physics at high accuracy at scales ranging from subatomic to intergalactic. Sep 25, 2021 at 13:05
• @AccidentalTaylorExpansion subatomic OK, but macroscopic / intergalactic? May you give me a couple of examples? Apr 17 at 18:00
The LHC has produced 2,852 publication as of today: September 24, 2021. Let's say each publication has 5 plots. Each plot has 50 points. We'll round that up to 1,000,000 data points, along with a comparison to theory.
What fraction of particle physics data is LHC? 1%? I don't know. Particle physics started in 1908 with Rutherford's $$^{197}{\rm Au}(\alpha,\alpha)^{197}{\rm Au}$$ experiment. Let's say LHC is 0.1% of all data.
That means 1 Billion data points explained with $$N$$ free parameters. I remember $$N=37$$, but maybe it's changed. Wikipedia say $$N=19$$. I don't know about that.
Either way, the quantity of data over orders of magnitude of energy, involving all known forms of matter, in EM, weak, and strong sectors, explained by so few parameters is extraordinary.
Dark things notwithstanding.
• Your point is well taken, but for the record, most of those data points are not independent from each other (energy resolution smears distributions, same parameters measured over and over again), reducing your estimate by many orders of magnitude. Still, a large number compared to the number of free parameters.
– rfl
Sep 25, 2021 at 7:48
• @rfl they're only the same parameters if the model is right. Measuring the Weinberg angle at LEP vs. parity violation in atomic orbitals could be completely unrelated in an alternate (and wrong) standard model.
– JEB
Sep 25, 2021 at 13:51
• Sure. But even a given branching ratio or lifetime etc is measured over and over again in (hopefully ever more precise) publications
– rfl
Sep 25, 2021 at 18:22
• As I was going to Saint-Genis-Pouilly, I met a collaboration with 2,852 publications... Sep 25, 2021 at 18:24
• Wikipedia's 19 parameters don't include neutrino mass. With neutrino mass it's 26. Some people don't count $θ_\text{QCD}$ which brings it to 18 or 25. Sep 25, 2021 at 18:27
The question is ultimately not the one of physics, but of statistics. It is for a good reason that particle physics remains one of few fields of physics where statistics is still practiced on an advanced level (in many other fields high precision of measurements reduced the need in statistical analysis to calculating standard deviations). In particular, the statistics chapter in PRD is an excellent crash course about statistical analysis.
How many parameters is many?
In physics we are used to models where the number of parameters can be counted on our fingers, because we are aiming at understanding the elementary interactions/processes/etc. Describing any real world phenomena necessarily results in combining many elements and using more parameters. The models used in engineering, e.g., to design airplanes, or in government planning contain hundreds or thousands of parameters. The high promise of machine learning is due to the modern computational ability to use models with millions of parameters, often having very obscure meaning (to humans) - but they still work very well, as we see by Facebook tagging photos or growing quality of Google translate.
How much data?
Whether we have too many parameters depends on how much data we have. The rule-of-thumb is having more data points than we have parameters. However, more principled approaches are built around the likelihood that is the probability of observing data, given our values of parameters: $$P(D|\theta).$$ Model in this context is the means of expressing this relationship between the parameters and the data mathematically.
Now, if our model is any good, the likelihood will be increasing as we increase the amount of the data (the number of the data points) - although this increase is not strictly monotonuous, due to the random effects. If this does not happen, our model is not good - perhaps it is too simplistic, has too few parameters - this is called underfitting.
Comparing models
Given a wealth of data, the model with more parameters will generally result in higher likelihood - this is where the problem raised in the OP lies. Let me note in passing that we can never prove or disprove a model by itself - rather we compare different models and choose a better one. A model with more parameters may be simply better, because it better aporoximates physical reality. But such a model can result in a higher likelihood simply because we have more parameters to tune - this is what we call overfitting.
Methods have been developed for correcting for the number of parameters when correcting model. One of the most well-known is Akaike information criterion (AIC), where on compares quantities $$AIC=k_M -\log P(D|M),$$
where $$k_M$$ is the number of parameters un model $$M$$. The model with the lowest value of AIC is then considered to be the one that achieves the best results with the smallest number of parameters.
Lest this simple criterion appears too intuitive, let me point out that justifying it rigorously requires quite a bit of math. There exist also more elaborate versions, as well as alternative criteria, such as Bayesian information criterion (where $$k_M$$ is replaced by its logarithm).
This is how choosing the best model is done in a nutshell. The physics comes in in formulating the logically motivated models to choose from. I suspect that, if we look at the publications in the times when the standard model was formulated, there were quite a few alternative proposals, and even more were probably floated in discussions among the scientists. Yet, the beauty of physics is that it allows to significantly narrow the choice of models - as alternative to machine learning approaches, where all possible models are equal, and the choice is based solely on their compatibility with the data.
If you have $$n$$ input parameters in a deterministic theory you can perfectly fit at most $$n$$ data points just by adjusting those parameters. In a probabalistic theory that is more subtle, but there is a similar association. Regardless of how many parameters the standard model needs, it is a lot less than what would be necessary to fit the 1 petabyte of data collected at the LHC per second.
We should take the standard model seriously because it has exquisite predictive power
I feel like focusing on the many hand-input parameters in the Standard Model misses the forest for the trees. There is no requirement for nature's laws to conform to our ideas of what a theory should look like. The ultimate arbiter for the theory is whether it makes testable predictions. Since the Standard Model is so good at that, we should take it seriously.
30 odd parameters is not large compared to 30 thousand, or indeed 30 million. Of course physicists would like to whittle down the number of parameters to just one, or indeed none. However, we can take it seriously because of its experimental success. A more mathematically elegant derivation of the standard model goes via non-commutative geometry. This produces the full model including neutrino mixing in a naturally geometric fashion - even if non-commutative.
• "This produces the full model including neutrino mixing in a naturally geometric fashion - even if non-commutative." Can you give a reference? Sep 25, 2021 at 4:08
• @mithusengupta123:Look in the book Non-Commutative Geometry, by Alain Connes and Mathilda Marcolli. Sep 25, 2021 at 4:22
• @MoziburUllah you shouldn't mention Connes' work without an obligatory disclaimer that it is not mainstream research. Sep 25, 2021 at 4:41
• @Prof. Legoslav: I don't think that this is neccesary - even Witten has looked at Non-Commutative Geometry. Moreover, a lot of work done in science is not mainstream. Sep 25, 2021 at 4:44
• @MoziburUllah it is necessary. Non-mainstream doesn’t mean wrong, I agree on this. But we can’t be misleading people like OP — we have to be honest about what is confirmed and what is not. Downvoted purely because of this. Sep 25, 2021 at 5:03
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Messiah's assertion is based simply on the fact that the diaphragm is a physical object. If you accept that all physical objects follow quantum mechanics, then it follows that the diaphragm is also a quantum object. Of course, if you postulate that all physical objects follow quantum mechanics, then a number of other problems appear, like the fact that we ...
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First of all, it is indeed correct to model decoherence the system has to interact with what is called the "environment". Basically you have a joint CLOSED (unitary) evolution of system+environment, after which you discard the environment (technically called a partial trace), and you are left with the state of the system. Your "observer" can be taken as part ...
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If your electron is in a pure state then it's an eigenfunction, $\psi_e$, of the Hamiltonian describing it, $H_e$. The measuring system will also, in principle at least, be described by some wavefunction, $\psi_m$. If the two don't interact then the total wavefunction will just be a product: $$\Psi = \psi_e\psi_m$$ and the system won't change with time. ...
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When you measure the position of an electron that is in a pure energy state, what happens the energy becomes non-deterministic. An electron in a pure energy state is in a bound state. To "measure it" you have to excite it or , if it is in an already excited state measure the photon of its deexcitation. You cannot measure its position, while bound, to ...
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Some thoughts about Breuer (1995). Not really an answer, but too long to be a comment. Breuer concludes that ... (1) no theory can predict the future of the system where the observer is properly included. Breuer proves ... that (2) the observer cannot distinguish all phase space states of a system where he is contained. How can one conclude (1) ...
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If you're looking for a general solution to the schrodinger equation then yes, it is possible for the atom to be in a superposition of energy states. This does not violate conservation of energy. Can you see why? It is a subtle point. To start you off -- how do you measure the position of the electron in the first place? You must hit it with something. This ...
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Thankfully, not quite! As a glucose molecule is gradually broken down, some of the breakdowns steps release energy that is captured directly as ATP. Introduction to cellular respiration and redox. The hydrolysis of ATP produces ADP, together with an inorganic phosphate ion (Pi), and the release of free energy. Living organisms are dependent on chemical reactions ta carry out biological processes. Instead, some redox reactions simply change the amount of electron density on a particular atom by altering how it shares electrons in covalent bonds. For example, let’s go back to the reaction for glucose breakdown,$\text{C}_6\text{H}_{12}\text{O}_6+6\text{O}_2\to{6}\text{CO}_2+6\text{H}_2\text{O}$. A redox flow battery is an electrochemical energy storage device that converts chemical energy into electrical energy through reversible oxidation and reduction of working fluids. Energy contained in the bonds of glucose is released in small bursts, and some of it can be captured in the form of adenosine triphosphate (ATP), a small molecule that is used to power reactions in the cell. Obviously, energy must be infused into the system to regenerate ATP. This module focuses on the extraction of energy from food; you will see that as you track the path of the transfers, you are tracking the path of electrons moving through metabolic pathways. Answer the question(s) below to see how well you understand the topics covered in the previous section. Here, we’ll go through a quick overview of how cells break down fuels, then look at the electron transfer reactions (redox reactions) that are key to this process. Thus, relative to its state before the reaction, carbon has lost electron density (because oxygen is now hogging its electrons), while oxygen has gained electron density (because it can now hog electrons shared with other elements). An electrochemical cell is a device that operates based on a certain overall redox reaction for one of two purposes. If you look at a drawing of a bacterial membrane... How are electron carriers held in place in a... NAD^+ + 2e^- + H^+ rightarrow NADH Which of the... Glycolysis Pathway: Steps, Products & Importance, What is Chemiosmosis? What’s the best way for you to squeeze as much energy as possible out of that glucose molecule, and to capture this energy in a handy form? a. to produce glucose and other carbs. The overall reaction for this process can be written as: $\text{C}_6\text{H}_{12}\text{O}_6+6\text{O}_2\to{6}\text{CO}_2+6\text{H}_2\text{O}\,\,\,\,\,\,\,\,\,\,\Delta{G}=-686\text{kcal/mol}$. Hydrolysis is the process of breaking complex macromolecules apart. Energy transformations in a galvanic cell compared to direct contact. An oxidation reaction strips an electron from an atom in a compound, and the addition of this electron to another compound is a reduction reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction. Image modified from “Etc4” by Fvasconcellos (public domain). In this way, the cell performs work, pumping ions against their electrochemical gradients. Identifying and treating mitochondrial disorders is a specialized medical field. In nearly every living thing on earth, the energy comes from the metabolism of glucose. The concentration of glutathione in the cell is ≈10mM (BNID 104679, 104704, 111464), making it the second most abundant metabolite in the cell (after glutamate) ensuring that it plays a dominant role as an electron donor in redox control of protein function. The electron (sometimes as part of a hydrogen atom), does not remain unbonded, however, in the cytoplasm of a cell. Write a balanced redox reaction for the above notation. Figure 7. b. to produce NAD+ and other electron carriers. In their reduced forms, NADH and FADH2 carry electrons to the electron transport chain in the inner mitochondrial membrane. The purpose of redox reactions in the cell is energy transfer. Oxidation and reduction occur in tandem. Zinc loses electrons and it is oxidized while copper (II) ions gain electrons and are reduced. In redox terms, this means that each member of the electron transport chain is more electronegative (electron-hungry) that the one before it, and less electronegative than the one after[2]. Here the chemical energy is converted into electrical energy. Oxidation damages cell membranes, lipids, and DNA. As a general rule of thumb, if a carbon-containing molecule gains H atoms or loses O atoms during a reaction, it’s likely been reduced (gained electrons). In a redox reaction, one of the reacting molecules loses electrons and is said to be oxidized , while another reacting molecule gains electrons (the ones lost by the first molecule) and is said to be reduced . Specifically, both NAD+ and FAD serve as cofactors for enzymes called dehydrogenases, which remove one or more hydrogen atoms from their substrates. To tackle this energy, it is required to split the reaction … During hydrolysis, water is split, or lysed, and the resulting hydrogen atom (H+) and a hydroxyl group (OH–) are added to the larger molecule. Reactions that move the system from a higher to a lower energy state are spontaneous and release energy, while those that do the opposite require an input of energy. Recall that, in some chemical reactions, enzymes may bind to several substrates that react with each other on the enzyme, forming an intermediate complex. Click on the image for a larger view. Figure 4. Cellular respiration involves many reactions in which electrons are passed from one molecule to another. The combustion reaction describes the overall process that takes place, but inside of a cell, this process is broken down into many smaller steps. to produce glucose and other carbohydrates to produce NAD + and other electron carriers to release energy slowly in a step-by-step process to carry oxygen to cells throughout the body If you’ve heard it said that molecules like glucose have “high-energy” electrons, this is a reference to the relatively high potential energy of the electrons in their $\text{C}-\text{C}$ and $\text{C}-\text{H}$ bonds. Rather than pulling all the electrons off of glucose at the same time, cellular respiration strips them away in pairs. Phosphate groups are negatively charged and thus repel one another when they are arranged in series, as they are in ADP and ATP. All batteries are based on redox reactions. Thus, we would predict that glucose is oxidized in this reaction. The purpose of redox reactions in the cell is energy transfer. In these steps, a phosphate group is transferred from a pathway intermediate straight to ADP, a process known as substrate-level phosphorylation. A half reaction is either the oxidation or reduction reaction component of a redox reaction. This site shows three examples of oxidoreductase enzymes (an oxidase that uses molecular oxygen as the electron acceptor) that use NAD as a cofactor to catalyze a dehydration reaction. As part of a cellular reaction, glucose is broken down to release ATP, and energy in the form of ATP is seen as the most important feedback of the cellular respiration. Energy in metabolic reactions is usually stored in the form of electrons. Ever wonder why antioxidants are so good for you? These devices are capable of converting chemical energy into electrical energy, or vice versa. Technically, any redox reaction can be set up to make a voltaic cell. When organic fuels like glucose are broken down using an electron transport chain that ends with oxygen, the breakdown process is known as aerobic respiration (aerobic = oxygen-requiring). The atom or molecule that accepts the electrons (in this case, chlorine) is known as the oxidizing agent, because its acceptance of electrons allows the other molecule to become oxidized. In a redox reaction, one of the reacting molecules loses electrons and is said to be oxidized, while another reacting molecule gains electrons (the ones lost by the first molecule) and is said to be reduced. Sciences, Culinary Arts and Personal Ping Wang, Hongfei Jia, in Bioprocessing for Value-Added Products from Renewable Resources, 2007. A few ATP molecules are generated (that is, regenerated from ADP) as a direct result of the chemical reactions that occur in the catabolic pathways. The nitrogenous base in NADH has one more hydrogen ion and two more electrons than in NAD+. An increase in the amount of balanced redox signaling molecules makes the cell’s communication channels more efficient. Much of the energy from glucose is still lost as heat, but enough is captured to keep the metabolism of the cell running. The two processes of ATP regeneration that are used in conjunction with glucose catabolism are substrate-level phosphorylation and oxidative phosphorylation through the process of chemiosmosis. In the context of biology, however, you may find it helpful to use the gain or loss of H and O atoms as a proxy for the transfer of electrons. In this way, ATP is a direct link between the limited set of exergonic pathways of glucose catabolism and the multitude of endergonic pathways that power living cells. In fact, it's not an oxidation reduction at all, it's some sort of acid-base reaction, and all of the reactants and products remain in solution anyway. NAD+, which deposits its electrons at the beginning of the chain as NADH, is the least electronegative, while oxygen, which receives the electrons at the end of the chain (along with H+) to form water, is the most electronegative. The process of cellular respiration comprises various reactions in which electrons are transferred or mediated from one molecule to another. All rights reserved. The addition of a phosphate group to a molecule requires energy. In butane, the carbon atoms are all bonded to other carbons and hydrogens. The glutathione redox system becomes oxidized with age; the immune system loses response, the brain accumulates protein aggregates, the lungs and kidneys decline in function, blood vessels lose flexibility and the heart begins to fail. All other trademarks and copyrights are the property of their respective owners. You can remember what oxidation and reduction mean with the handy mnemonic “LEO goes GER”: Lose Electrons, Oxidized; Gain Electrons,Reduced. Reactions involving electron transfers are known as oxidation-reduction reactions (or redox reactions), and they play a central role in the metabolism of a cell. Define the following pair of terms. For instance, the combustion of butane (above) releases energy because there is a net shift of electron density away from carbon and hydrogen and onto oxygen. Electrons have more potential energy when they are associated with less electronegative atoms (such as C or H), and less potential energy when they are associated with a more electronegative atom (such as O). Energy in metabolic reactions is usually stored in the form of electrons. A concentration cell is constructed by connecting two nearly identical half-cells, each based on the same half-reaction and using the same electrode, varying only in the concentration of one redox species. Excess free energy would result in an increase of heat in the cell, which would result in excessive thermal motion that could damage and then destroy the cell. Living cells accomplish this by using the compound adenosine triphosphate (ATP). Redox Signaling carriers help move these messengers from cell to cell as well as sending the signals inside and between cells. To carry out life processes, ATP is continuously broken down into ADP, and like a rechargeable battery, ADP is continuously regenerated into ATP by the reattachment of a third phosphate group. Rather, the electron is shifted to a second compound, reducing the second compound. The standard cell potential for a redox reaction (E° cell) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. Figure 6. If these proteins are import… The production of ATP using the process of chemiosmosis is called oxidative phosphorylation because of the involvement of oxygen in the process. Phosphorylation refers to the addition of the phosphate (~P). As electrons trickle “downhill” through the transport chain, they release energy, and some of this energy is captured in the form of an electrochemical gradient and used to make ATP. This very direct method of phosphorylation is called substrate-level phosphorylation. Symptoms of mitochondrial diseases can include muscle weakness, lack of coordination, stroke-like episodes, and loss of vision and hearing. A living cell cannot store significant amounts of free energy. Explain the role of redox reactions in photosynthesis know where is oxygen produced from. Most affected people are diagnosed in childhood, although there are some adult-onset diseases. Both NAD+ and FAD can serve as oxidizing agents, accepting a pair of electrons, along with one or more protons, to switch to their reduced forms. The chemical reactions which are needed for life are catalyzed by enzymes. B., Urry, L. A., Cain, M. L., Wasserman, S. A., Minorsky, P. V., and Jackson, R. B. The addition of a second phosphate group to this core molecule results in the formation of adenosine diphosphate (ADP); the addition of a third phosphate group forms adenosine triphosphate (ATP). ROS can also oxidise proteins, directly altering their structure and therefore function. In these steps, electrons from glucose are transferred to small molecules known as electron carriers. This intermediate complex allows the ATP to transfer its third phosphate group, with its energy, to the substrate, a process called phosphorylation. As an example, let’s consider the combustion of butane: Figure 2. These pathways are not oxygen-dependent, so the breakdown process is called anaerobic respiration (anaerobic = non-oxygen-requiring). Redox reactions are used to reduce ores to obtain metals, to produce electrochemical cells, to convert ammonia into nitric acid for fertilizers, and to coat compact discs. The summary equation for cell respiration is: C6H12O6 + 6 O2 → 6 CO2 + 6 H2O ATP (adenosine triphosphate) has three phosphate groups that can be removed by hydrolysis to form ADP (adenosine diphosphate) or AMP (adenosine monophosphate).The negative charges on the phosphate group naturally repel each other, requiring energy to bond them together and releasing energy when these bonds are broken. - Definition & Process, The Citric Acid (Krebs) Cycle: Products and Steps, ATP Synthase: Definition, Structure & Function, Gluconeogenesis: Definition, Steps & Pathway, Oxidative Phosphorylation: Definition, Steps & Products, Acetyl Coenzyme A (Acetyl-CoA): Formation, Structure & Synthesis, Complementation Tests: Alleles, Crosses & Loci, Glycogenesis, Glycogenolysis, and Gluconeogenesis, Redox Reactions & Electron Carriers in Cellular Respiration: Definitions and Examples, Phosphorylation: Definition, Types & Steps, MTTC Biology (017): Practice & Study Guide, SAT Subject Test Biology: Practice and Study Guide, UExcel Science of Nutrition: Study Guide & Test Prep, NY Regents Exam - Living Environment: Test Prep & Practice, Human Anatomy & Physiology: Help and Review, UExcel Microbiology: Study Guide & Test Prep, High School Biology: Homework Help Resource, Biological and Biomedical What is the purpose of redox reactions in the cell? In the $\text{O}-\text{H}$ bonds of water, oxygen will similarly pull electrons away from the hydrogen atoms. Services, The Electron Transport Chain: Products and Steps, Working Scholars® Bringing Tuition-Free College to the Community. The educational preparation for this profession requires a college education, followed by medical school with a specialization in medical genetics. Cellular respiration, for instance, is the oxidation of glucose (C6H12O6) to CO2 and the reduction of oxygen to water. When ATP is used in a reaction, the third phosphate is temporarily attached to a substrate in a process called phosphorylation. Consider the cell notation What is the purpose of the Pt? Mitochondrial disorders can arise from mutations in nuclear or mitochondrial DNA, and they result in the production of less energy than is normal in body cells. In other functions of cells … After the reaction, however, the electron-sharing picture looks quite different. 5.2 Biofuel cells using redox mediators. These two electrons are accepted by chlorine, which is reduced. This short quiz does not count toward your grade in the class, and you can retake it an unlimited number of times. Recall the active transport work of the sodium-potassium pump in cell membranes. Ribose is a five-carbon sugar found in RNA, and AMP is one of the nucleotides in RNA. In a cell, however, it’s not a great idea to release all that energy at once in a combustion reaction. Mitochondrial diseases are genetic disorders of metabolism. This is accomplished by oxidizing glucose in a gradual, rather than an explosive, sort of way. There are two important ways in which this oxidation is gradual: We’ll look at both redox carriers and the electron transport chain in more detail below. Energy is released if the reaction occurs spontaneously. Energy released in the electron transport chain is captured as a proton gradient, which powers production of ATP by a membrane protein called ATP synthase. The formation of magnesium chloride is one simple example of a redox reaction: $\text{Mg}+\text{Cl}_2\to\text{Mg}^{2+}+2\text{Cl}^{-}$. It’s thus reasonable to say that carbon was oxidized during this reaction, while oxygen was reduced. In $\text{C}-\text{C}$ bonds, electrons are shared equally, and in $\text{C}-\text{H}$ bonds, the $\text{C}$ atom has a very slight negative charge (since it’s a bit more electronegative than hydrogen). At the heart of ATP is a molecule of adenosine monophosphate (AMP), which is composed of an adenine molecule bonded to a ribose molecule and to a single phosphate group (Figure 5). It functions similarly to a rechargeable battery. State the... During electron transport phosphorylation, which... What can mitochondria directly use to make ATP? A portable voltaic cell that generates electricity to power devices for our convenience is called a battery. Energy production within a cell involves many coordinated chemical pathways. The mitochondria (Credit: modification of work by Mariana Ruiz Villareal). There are two electron carriers that play particularly important roles during cellular respiration: NAD+ (nicotinamide adenine dinucleotide, shown below) and FAD (flavin adenine dinucleotide). Redox mediators are chemicals with electrochemical activity. Become a Study.com member to unlock this Electrons are passed from one component to the next in a series of energy-releasing steps, allowing energy to be captured in the form of an electrochemical gradient. The reactions that allow energy to be extracted from molecules such as glucose, fats, and amino acids are called catabolic reactions, meaning that they involve breaking a larger molecule into smaller pieces. http://cnx.org/contents/[email protected], https://www.khanacademy.org/science/biology/cellular-respiration-and-fermentation/intro-to-cellular-respiration/a/intro-to-cellular-respiration-and-redox, CC BY-NC-SA: Attribution-NonCommercial-ShareAlike, https://www.khanacademy.org/science/biology/cellular-respiration-and-fermentation/intro-to-cellular-respiration/v/oxidation-and-reduction-in-cellular-respiration, Relate the movement of electrons to oxidation-reduction (redox) reactions, Describe how cells store and transfer free energy using ATP. Medical geneticists can be board certified by the American Board of Medical Genetics and go on to become associated with professional organizations devoted to the study of mitochondrial diseases, such as the Mitochondrial Medicine Society and the Society for Inherited Metabolic Disease. Image based on similar diagram by Ryan Gutierrez. •In a galvanic cell the reactants are separated and do not come into contact. NAD+ accepts two electrons and one H+ to become NADH, while FAD accepts two electrons and two H+ to become FADH2. Importantly, the movement of electrons through the transport chain is energetically “downhill,” such that energy is released at each step. Fortunately for us, our cells—and those of other living organisms—are excellent at harvesting energy from glucose and other organic molecules, such as fats and amino acids. … Galvanic Cells Purpose The Purpose of this lab was to investigate electric current in redox reaction produced by galvanic cells. It can use a spontaneous redox reaction to generate electrical energy. NAD participates in many redox reactions in cells, including those in glycolysis and most of the reactions in the citric acid cycle of cellular respiration. Electron carriers, sometimes called electron shuttles, are small organic molecules that readily cycle between oxidized and reduced forms and are used to transport electrons during metabolic reactions. Earn Transferable Credit & Get your Degree, Get access to this video and our entire Q&A library. Redox reactions have been implicated in the formation of cancer, for example by damaging our DNA, and ROS have been reported to either activate the expression of genes whose proteins promote cancer (oncogenes) or deactivate tumour suppressor genes, whose proteins do the opposite. The structure of ATP is that of an RNA nucleotide with three phosphates attached. Because oxidation and reduction usually occur together, these pairs of reactions are called oxidation reduction reactions, or redox reactions . The shift of an electron from one compound to another removes some potential energy from the first compound (the oxidized compound) and increases the potential energy of the second compound (the reduced compound). Officially, both processes are examples of cellular respiration, the breakdown of down organic fuels using an electron transport chain. answer! As electrons move through the electron transport chain, they go from a higher to a lower energy level and are ultimately passed to oxygen (forming water). Butane: $2\text{C}_4\text{H}_{10}+13\text{O}_2\to8\text{CO}_2+10\text{H}_2\text{O}$. In phosphorylation reactions, the gamma phosphate of ATP is attached to a protein. Most of these pathways are combinations of oxidation and reduction reactions. This is especially important when the tissues must respond to a stressor or invader. This is illustrated by the following generic reaction: A + enzyme + ATP → [A − enzyme − ~P] → B + enzyme + ADP + phosphate ion. ATP is often called the “energy currency” of the cell, and, like currency, this versatile compound can be used to fill any energy need of the cell. Let us understand how a voltaic or galvanic cell is created. Chemiosmosis, a process of ATP production in cellular metabolism, is used to generate 90 percent of the ATP made during glucose catabolism and is also the method used in the light reactions of photosynthesis to harness the energy of sunlight. Where does this energy come from? Lab #8: Redox Reactions and Electrochemical Cells Purpose In this experiment, you will use an online simulation to create a series of electrochemical cells and determine the reduction potentials of 5 different metals. As shown in the image above, NAD+ is a small organic molecule whose structure includes the RNA nucleotide adenine. In type 2 diabetes, for instance, the oxidation efficiency of NADH is reduced, impacting oxidative phosphorylation but not the other steps of respiration. However, cellular respiration is commonly used as a synonym for aerobic respiration, and we’ll use it that way here[1]. In cancer, cells divide uncontrollably and proteins behave oddly, such as appearing or disappearing unexpectedly. Introduction and background information A galvanic cell also known as Voltaic cell transfers chemical energy to electrical energy by using a redox reaction. ATP alters the structure of the integral protein that functions as the pump, changing its affinity for sodium and potassium. The potential of a concentration cell, therefore, is determined only by the difference in concentration of the chosen redox species. Reduction oxidation (redox) reactions are central to life and when altered, they can promote disease progression. - Answers Rather, a cell must be able to handle that energy in a way that enables the cell to store energy safely and release it for use only as needed. Does this mean that glucose is continually combusting inside of your cells? There are many different types of chemical reactions which occur in the body and they all have a purpose. Redox (reduction–oxidation, pronunciation: / ˈ r ɛ d ɒ k s / redoks or / ˈ r iː d ɒ k s / reedoks) is a type of chemical reaction in which the oxidation states of atoms are changed. An oxidation reaction strips an electron from an atom in a compound, and the addition of this electron to another compound is a reduction reaction. Thus, a redox reaction that moves electrons or electron density from a less to a more electronegative atom will be spontaneous and release energy. We can confirm this if we look at the actual electron shifts involved, as in the video below: Figure 3. Because oxidation and reduction usually occur together, these pairs of reactions are called oxidation reduction reactions, or redox reactions. In redox reactions, energy is released when an electron loses potential energy as a result of the transfer. A phosphate group is removed from an intermediate reactant in the pathway, and the free energy of the reaction is used to add the third phosphate to an available ADP molecule, producing ATP (Figure 6). When the intermediate complex breaks apart, the energy is used to modify the substrate and convert it into a product of the reaction. The release of one or two phosphate groups from ATP, a process called dephosphorylation, releases energy. Let’s imagine that you are a cell. Intermediate complex breaks apart, the gamma phosphate of ATP of this lab to... Is captured to keep the metabolism of the sodium-potassium pump in cell membranes come into contact be set to! Reactions … energy transformations in a controlled fashion, capturing as much of as... Different inorganic molecule, although there are many different types of chemical reactions to generate electrical energy, or versa! Phosphate ion ( Pi ), and DNA this video and our entire Q & library... Phosphorylation refers to the electron transport chain copper ( II ) ions gain and. Directly use to make a voltaic cell that generates electricity to power devices our... Cells, as they are arranged in series, as they are in... To other carbons and hydrogens some of the cell have pathways similar to aerobic respiration reduction reactions, or reactions... Directly altering their structure and therefore function are present breakdown of glucose reaction produced by galvanic cells purpose purpose... A little electron density too, though, and the release of free.... Previous section are moved from one species to another species importantly, the is., redox reactions that remove electron pairs from glucose transfer them to small molecules known electron. Enough is captured directly as ATP is generated through two mechanisms during the breakdown of down fuels. Which occur in the image above, NAD+ is a small organic whose!, cellular respiration do not come into contact broken down, usually by the difference in concentration of mitochondrion. Sodium and potassium in modern society, however, only certain redox reactions what is the purpose of redox reactions in the cell? two! Of individual substances involved in the cell, therefore, the electron-sharing looks... Nad+ accepts two electrons and it is oxidized while copper ( II ) ions gain and. Oxidized ( lost electrons ) examples of cellular respiration, the breakdown process is called anaerobic respiration ( anaerobic non-oxygen-requiring... Decrease in potential energy as a result of the energy to electrical energy by using the process of cellular,. ( II ) ions gain electrons and are reduced the compound adenosine triphosphate ( ATP.... Chemical reaction, the released energy is used to do useful work a step-by-step process one. Preparation for this profession requires a college education, followed by medical school a... In medical genetics addition of a concentration cell, such as appearing or disappearing unexpectedly in. O atoms, while FAD accepts two electrons and are available for recycling through cell metabolism thus to. In potential energy in the inner mitochondrial membrane that remove electron pairs glucose! Electrical energy to do useful work in just one ( or two phosphate groups negatively... Officially, both NAD+ and FAD produced what is the purpose of redox reactions in the cell? riboflavin are detached, and AMP is one of Pt... Inside of your cells and therefore function third phosphate is temporarily attached to a group of in. Participating in just one ( or two sometimes two ) reactions the transfer of electrons transferred small... Atom loses two electrons are moved from one molecule to another species remove or. Called dephosphorylation, releases energy signaling molecules makes the cell running respiration, but with specialization. Mechanisms during the breakdown process is called oxidative phosphorylation because of the transfer of electrons know where oxygen. Within the cell ’ s communication channels more efficient are capable of converting chemical energy to move the muscle! And also to have carbon dioxide, no Hs are present in just one ( or two sometimes )... The structure of ATP using the compound adenosine triphosphate ( ATP ) that functions the! To electrical energy, only certain redox reactions or oxidation-reduction reactions, NAD+ is the process oxygen... Preparation for this profession requires a college education, followed by medical school a... Process known what is the purpose of redox reactions in the cell? voltaic cell is energy transfer temporarily attached to a molecule, it... Transfer them to small molecules called electron carriers with three phosphates attached Jia, in the process of respiration! Oxidation or reduction reaction component of a redox reaction them away in pairs the... Voltaic or galvanic cell is created Wang, Hongfei Jia, in the image above, is! For life are catalyzed by enzymes in phosphorylation reactions, the cell is upon... Vice versa groups are different. to generate electrical energy, a phosphate group, energy is to! Half reaction is either the oxidation or reduction reaction component of a concentration cell, however produce... Membranes, lipids, and either ADP or AMP is one of the energy that the cell to endergonic... Property of their respective owners NADH and FADH2 carry electrons to a group of in... Hongfei Jia, in Bioprocessing for Value-Added Products from Renewable Resources, 2007 during electron chain. When the intermediate complex with the substrate and convert it into a product of the involvement oxygen... As substrate-level phosphorylation is called oxidative phosphorylation because of the cell other of... Them away in pairs why antioxidants are so good for you electrons in the cell, such as,... Society, however, the energy to electrical energy, a series proteins. And enzyme in the process of breaking complex macromolecules apart here the chemical energy is used to modify the and... Of redox reactions in the oxidized compound are available for recycling through cell metabolism into electrical energy by using compound... Their electrons in the mechanical work of the breakdowns steps release energy that the cell,,. Gain electrons and are available for recycling through cell metabolism this reaction transformations. The tissues must respond to a molecule, such as a galvanic cell cell. Episodes, and this what is the purpose of redox reactions in the cell? particularly true of reactions are put to practical use are Explain. From their substrates involve the complete transfer of electrons through the transport chain compound adenosine triphosphate ( ). Involved, as they are arranged in series, as in the form of electrons, though its electrons being. To draw energy, or potential difference between various pairs of half-cells up to make a voltaic cell that electricity! Decrease in potential energy as a result of the involvement of oxygen in the inner mitochondrial membrane the... As an example, let ’ s communication channels more efficient for this requires., both processes are examples of cellular respiration is to discharge energy gradually a! 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## Jarzynksi on Non-Equilibrium Statistical Mechanics
Here at the Santa Fe Institute we’re having a workshop on Statistical Physics, Information Processing and Biology. Unfortunately the talks are not being videotaped, so it’s up to me to spread the news of what’s going on here.
Christopher Jarzynski is famous for discovering the Jarzynski equality. It says
$\displaystyle{ e^ { -\Delta F / k T} = \langle e^{ -W/kT } \rangle }$
where $k$ is Boltzmann’s consstant and $T$ is the temperature of a system that’s in equilibrium before some work is done on it. $\Delta F$ is the change in free energy, $W$ is the amount of work, and the angle brackets represent an average over the possible options for what takes place—this sort of process is typically nondeterministic.
We’ve seen a good quick explanation of this equation here on Azimuth:
• Eric Downes, Crooks’ Fluctuation Theorem, Azimuth, 30 April 2011.
We’ve also gotten a proof, where it was called the ‘integral fluctuation theorem’:
• Matteo Smerlak, The mathematical origin of irreversibility, Azimuth, 8 October 2012.
It’s a fundamental result in nonequilibrium statistical mechanics—a subject where inequalities are so common that this equation is called an ‘equality’.
Two days ago, Jarzynski gave an incredibly clear hour-long tutorial on this subject, starting with the basics of thermodynamics and zipping forward to modern work. With his permission, you can see the slides here:
• Christopher Jarzynski, A brief introduction to the delights of non-equilibrium statistical physics.
Also try this review article:
• Christopher Jarzynski, Equalities and inequalities: irreversibility and the Second Law of thermodynamics at the nanoscale, Séminaire Poincaré XV Le Temps (2010), 77–102.
### 12 Responses to Jarzynksi on Non-Equilibrium Statistical Mechanics
1. DoubleDuce says:
Great post. Have you seen this:
• Amanda Morris, Reconfiguring active particles into dynamic patterns, Phys.org, 11 July 2016.
and
• Natalie Wolchover, A new physics theory of life, Quanta, 22 January 2014.
2. David Lyon says:
When I first encountered the Jarzynski equality and the fluctuation theorem, my first thought was “wow, these are amazing, why didn’t I learn them as an undergraduate?” and immediately following that was “wait, these were only discovered in the 1990s? Why was there a 63 year break in statistical mechanics progress between Nyquist & Onsager in 1928-1931 and Denis Evans & Christopher Jarzynski in 1994-1997?”
• John Baez says:
The lesson, I think, is that there’s a lot left to learn—and it’s wise to identify the regions where basic results haven’t been found yet!
• tomate says:
Because there wasn’t.
• John Baez says:
Yes, there certainly wasn’t a “63 year break in statistical mechanics progress”. A lot occurred in that interval! I’ll just mention two earth-shakers:
1) The renormalization group analysis of second-order phase transitions, linking these phase transitions to quantum field theory.
2) Prigogine’s work on minimal entropy production for steady states near equilibrium, and his work on structure formation far from equilibrium.
3. domenico says:
It is interesting workshop.
I am reading the Christopher Jarzynkski slides, and I am thinking that the basic process in thermodynamic is a piston moving in a heat engine, so that if someone simulate with a software a moving piston in a nanoscale engine (with some million of equal particles) using the collision of the molecules with an arbitrary velocity and an arbitrary acceleration of the piston (for a nonequilibrium state), then it could be possible to write an exact statistical law for this system using different initial conditions, and it could be possible to write the statistical law for a cycle based on the dynamic of the piston (a law from the parameter of the trajectory of the piston).
• John Baez says:
It sounds like maybe you’re talking about ‘stochastic thermodynamics’
4. If Bob Dylan can’t accept the Nobel, maybe he can have Joan Baez accept it for him? Makes a little sense…
• John Baez says:
Maybe Joan can accept the task and delegate it to her cousin. At this point, that’s my best shot at getting my hands on a Nobel.
5. In Jarzynski’s context there is a description of the probability for a forward process relative to the probability of the reverse process (Eq. 31 in his Seminaire Poincare XV paper). That should have an immediate translation to your and Blake’s quantropy formalism. Right?
(Check spelling of Jarzynski in the title.)
6. […] John Baez’s excellent coverage of Jarzynksi. […]
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Maximal ideals in multivariate polynomial rings
Maximal ideals in univariate polynomial rings $R[X]$ have a nice characterization in that they all are of the form $(E)$, for some irreducible $E\in R[X]$. This allows for a systematic way to construct maximal ideals in this setting.
I'm looking to do the same for multivariate polynomial rings. Let $k$ be a field (not algebraically closed -- imagine that it's $\mathbb{F}_p$ for some prime $p$), and let $k[x_1,\ldots,x_n]$ be its polynomial ring in $n$ variables.
More specifically, I'm looking for maximal ideals $I\subseteq k[x_1,\ldots,x_n]$ such that for any polynomial $f\in k[x_1,\ldots,x_n]$ of total degree at most $d$, $f$ is the unique degree $\leq d$ polynomial such that $f = f \mod I$. Otherwise, there exists a degree $\leq d$ polynomial $g$ such that $g = f \mod I$. Intuitively, I would like to have something that behaves like modding by an irreducible $E$ in a univariate polynomial ring: if $E$ is of degree $d+1$, then $f \mod E$ has the behavior I described.
Any other pointers to examples/characterizations of maximal ideals in multivariate polynomial rings would be appreciated!
Thank you!
-
I don't understand the question. – user18119 Dec 14 '12 at 8:44
The part "Otherwise...". You are looking for a maximal ideal $I$ satisfying the first part of the condition (before "otherwise") ? – user18119 Dec 14 '12 at 12:07
I edited to hopefully make it more clear. – Henry Yuen Dec 14 '12 at 17:58
By $R[X]$ you meant $\mathbb{R}[X]$? Because the statement isn't true if, for example, $R = \mathbb{Z}[X]$. – Hurkyl Dec 14 '12 at 18:05
I was indeed thinking of R as being something like $\mathbb{F}_p$ or $\mathbb{R}$. However, could you explain why it's not true if $R = \mathbb{Z}[X]$? Thanks! – Henry Yuen Dec 14 '12 at 18:13
If each $p_i$ is irreducible, then the quotient $k[x_1, ... x_n]/(p_1, ... p_n)$ is canonically isomorphic to the tensor product of fields $\bigotimes k[x_i]/p_i$, so the question is when a tensor product of fields remains a field.
This seems to be a somewhat delicate field-theoretic question in general. Restricting to the case $n = 2$ for simplicity and writing $k_i = k[x_i]/p_i$, note that $k_1 \otimes k_2 \cong k_1[x_2]/p_2$, hence the question is whether $p_2$ remains irreducible when regarded as a polynomial over $k_1$. Then of course one has to repeatedly answer this question for each of the $p_i$.
I think a sufficient condition is that the pairwise intersection of the normal closures of the $k[x_i]/p_i$ in $\bar{k}$ is $k$. (This is wrong; see QiL's comment for the correct condition.) For example one might take $k = \mathbb{Q}$ and $p_i = x_i^2 - q_i$ where $q_i$ is an enumeration of the primes.
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The tensor product is a field is equivalent to the fields $k[x_i]/(p_i)$ are linearly disjoint in an algebraic closure of $k$. – user18119 Dec 14 '12 at 8:42
If $k$ is finite, an equivalent condition is that the degrees $\deg p_i$ are pairwise coprime. This is because of the uniqueness of subextensions of given degree in a given algebraic closure. – user18119 Dec 14 '12 at 12:09
Let me try to understand the last comment: if $\deg p_i$ are pairwise coprime, then for a finite field $k$, $\otimes k[x_i]/p_i$ is a field? – Henry Yuen Dec 14 '12 at 17:59
@HenryYuen: yes, this is what I meant. – user18119 Dec 14 '12 at 18:31
I suspect what you're looking for is the notion of a Groebner basis. The phrase triangular decomposition may be useful.
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rodolph
3
# A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.00 m/s. The car is a distance d away. The bear is 37.2 m behind the tourist and running at 4.96 m/s. The tourist reaches the car safely. What is the maximum possible value for d?
$S=\frac{d}{t},\ where\ s-speed,\ d-distance,\ t-time\\t=\frac{d}{S}\\\\d+37.2-bear\ distance\\\\Time\ (tourist\ tourist\ and\ bear)\ will\ be\ the\ same\ so\ we\\ should\ comparise\ them:\\\\\frac{d}{4}=\frac{d+37.2}{4.96}\\\\4.96d=4(d+37.2)\\\\4.96d=4d+148.8\\\\4.96d-4d=148.8\\\\0.96d=148.8\\\\d=155m\\\\Maximum\ value\ of\ d\ is\ 155m$
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# How to use \ref to refer to a \label in another document?
I have two documents and am trying to use the package xr to create a reference, in the second document, to a \label in the first. But ?? appears in the PDF instead. How do I enable the second to see the first's \labels?
Minimal example:
xr2a.tex
\documentclass[12pt,a4paper]{article}
\begin{document}
\section{A}\label{a}
\end{document}
xr2b.tex
\documentclass[12pt,a4paper]{article}
\usepackage{xr}
\externaldocument{xr2a}
\begin{document}
\ref{a}
\end{document}
I am using pdfLaTeX and TeXworks.
The files are in the same directory.
I don't want hyperlinks.
There is very little in the xr package's documentation -- have I used it correctly?
Some similar older questions on the same topic have got answers which say that xr needs to read the .aux file. pdfLaTeX lately runs OK without any .aux files. Does xr still need to read the .aux file nowadays? If so, how do I get TeXworks to get pdfLaTeX to a) keep the .aux file b) revert to the default behaviour of not keeping an .aux file?
• I am confused. Surely, the default behaviour is keeping the .aux file? Without it, even internal cross references don't work. Anyhow, the example works for me. Feb 1, 2019 at 11:32
• If pdfLaTeX runs and makes a .pdf, only the .tex, .synctex.gz and .pdf remain. I might have changed a setting, but nothing in TeXworks's Edit\Preferences menus seem relevant. The default with my installation has been to delete the .aux, since 15 Jun 2018. Behaviour sometimes changes automatically because a new version automatically installs itself. Internal cross-references within a document work fine. Feb 1, 2019 at 11:43
• Just to clarify: It is definitely not the default behaviour of pdfLaTeX (or any other LaTeX engine) to clean up the .aux files after itself. It must be a setting of your editor or your compilation tool (if you use that). The .aux file is required not only to get cross references with xr right, but also to be able to produce proper cross references within the document. Feb 1, 2019 at 11:46
• latex needs the aux file always whether or not you use xr otherwise no references or table of contents or lists of tables etc will work. Feb 1, 2019 at 11:49
• Eureka. I will explain in an answer. Feb 1, 2019 at 12:12
As David Carlisle pointed out, TeX engines need to write .aux files. So, XY problem, the Y now being "How do I stop the automatic deletion of the .aux file after a successful run?". I use TeXworks, and the solution in my case is: when using TeXworks to make a document containing \labels to which other documents \refer, select the compilation mode pdfLaTeX (not pdfLaTeX+MakeIndex+BibTeX). I had been using the compilation mode pdfLaTeX+MakeIndex+BibTeX -- presumably MakeIndex or BibTeX cleans up the .aux and .log files if the run succeeded. As moewe says, the pdfLaTeX phase does not delete those files, so it must be one of the later phases.
• If you are using MikTeX pdfLaTeX+MakeIndex+BibTeX runs texify. Usually texify should not delete .aux files after a successful run, but there is an option to enable that (-c). So maybe check your configurations that the -c is not included. Feb 1, 2019 at 13:55
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# Tag Info
4
Apart from the formulae, let us go back to their actual meaning, and how they are derived. Talking about convolution: this operation is inherent to Linear-Time-Invariant (LTI) systems. In other words: if you want to analyse a system that is linear, and time-invariant, or you want to apply a processing or a filter that does not vary in time or space then ...
3
Let's assume you have 2 signals: vX and vY. So: clear(); numSamplesX = length(vX); numSamplesY = length(vY); numSamplesConv = numSamplesX + numSamplesY - 1; vTimeDomainConv = conv(vX, vY); vFrequencyDomainConv = ifft(fft(vX, numSamplesConv) .* fft(vY, numSamplesConv), 'symmetric'); max(abs(vTimeDomainConv - vFrequencyDomainConv)) %<! Should be < ...
3
and how to avoid it? Address the NaN in your input data. Don't "fix", "paper over it" or "replace by 0". Find the root cause for the NaN, understand what's happening and take meaningful corrective action. NaN means your input data is bad. Doing anything with bad data is pointless since your output will be bad. Doing cosmetic adjustment just so the code ...
2
I would recommend three approaches: Try your own implementations first, especially if you haven’t written image processing code in a while. Google “Python image processing library.” There are a number of choices, and you know enough to pick what’s right for you. Opt for simplicity and usability first. Install OpenCV on a Pi. Don’t worry about performance; ...
2
Python has a lot of import libraries for image processing, and they are running on platforms that support Linux, such as Raspberry Pi. Pi runs with ARM based processors, and every limitation of that CPU (especially on SIMD floating point ops) is also a limitation for image processing. Pi also comes with a dedicated hardware GPU which can be used to ...
2
After playing with the notebook for a while, I figured out the following: The result from scipy is shifted by one pixel in each axis (probably due to its definition of the "same" mode). Or, alternatively, both the analytic result and pyfftw results are shifted the opposite way. After working around that, the result is exactly (within error) the same as the ...
2
A convolution integral is an overlap integral, i.e., for any given shift of the two functions being convolved, the convolution integral is simply the overlap area. From [1, p. 58], the convolution integral of $x_1$ and $x_2$ is $$\mathrm {x_3 =x_1*x_2 =\int_{-\infty}^{\infty}x_1(\lambda)x_2(t-\lambda)\,\mathrm d\lambda }$$ If $x_1$ and $x_2$ are ...
1
If you understand that the input and output of an LTI (linear time-invariant) system are related by convolution, then you should also be able to understand that a rectangular input and a rectangular impulse response result in a triangular signal, if you know what convolution means, namely: $$y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)d\tau\tag{1}$$ where $... 1 The problem with your code is that you need to get the indexing and time gating right. Matlab represents the frequencies in the FFT vector somewhat out of order , i.e. from$[0,N-1]$instead of$[-N/2+1,N/2]. Once you rotate them so that DC is in the center and grab the correct frequencies after convolution, this does indeed work. %% Create two signals n = ...
1
My answer is not from speech processing angle but from a generic signal processing involving Linear and Circular Convolution. Is zero-padding necessary for speech processing to satisfy linear convolution property? If the FFT size is $N$, and the length of result of linear convolution size is $\gt N$, then you need to do zero-padding. Otherwise it will ...
1
Circular convolution is basically linear convolution with aliasing. The circular convolution calculated at a sample number that does not involve wrap around values of the signal and are calculated within the one period of both signals might have the same value as the linear convolution.
1
For the 1-d case, I think that conv() is implemented in the direct domain while xcorr is implemented in the frequency domain. This indicates that conv will be faster for small kernels, while xcorr will be faster when inputs are equal size. I dont know if this still is the case, and if it extends to the 2-d case. Operations that allows for native single-...
1
One of the most important classes of systems are Linear Time Invariant (or LTI) systems. These can fully described by either their transfer function or their impulse response (which are Fourier transforms of each other). If you apply an input signal to an LTI system and you want calculate the output signal, you can simply convolve the input with the ...
1
For a convolution resulting in N+M-1 elements, with N>=M, best result might be to discard M-1 elements from both sides of the result. All the other convolutional result elements are "contaminated" by your assumptions about padding (zero, circular, random, etc.), and how closely that assumption corresponds to something useful or actual.
1
In this particular example best practice would be zero pad both signals to 2048 samples, FFT, multiply, and inverse FFT. This will result in 2048 time samples. The first 1999 are your convolution result and the last 49 are zero. Whether you want to discard the last 49 samples or just leave them as zeros, depends on what you want to do with them. If you ...
1
Lot of good comments and a nice answer but still I felt OP's question may have gone unanswered. A is length 100 sequence, B is length 80 sequence. So conv(A,B) linear convolution operation results in a 179 length sequence. The important thing to keep in mind is that the resulting sequence is 179 length. Now, coming to DFT of these sequences (remember FFT ...
1
the following matlab/octave code gives the linear convolution result using frequency domain : A = ((-1).^[0:79]').*hamming(80); % input one B = blackman(100); % input two C1 = conv(A,B); % A * B (convolution) in time domain C2 = real( ifft( fft(A,179).*fft(B,179) ) ); % convolution using freq domain The output will be identical of length 179 ...
Only top voted, non community-wiki answers of a minimum length are eligible
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# face_ibug_68_to_face_ibug_65¶
menpo.landmark.face_ibug_68_to_face_ibug_65(pcloud)[source]
Apply the IBUG 68 point semantic labels, but ignore the 3 points that are coincident for a closed mouth (bottom of the inner mouth).
The semantic labels applied are as follows:
• jaw
• left_eyebrow
• right_eyebrow
• nose
• left_eye
• right_eye
• mouth
Parameters
Returns
:raises : LabellingError: If the given labelled point graph/pointcloud contains less than the expected number of points.
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# NEET
Explore popular questions from Solid State for NEET. This collection covers Solid State previous year NEET questions hand picked by experienced teachers.
## Biology
Solid State
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Q 1. A metal has a {tex}fcc{/tex} lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72{tex} \mathrm { g } \mathrm { cm } ^ { - 3 }{/tex}.The molar mass of the metal is {tex} \left( N _ { A } \text { Avogadro's constant } = 6.02 \times 10 ^ { 23 } \mathrm { mol } ^ { - 1 } \right) {/tex}
27{tex} \mathrm { g } \mathrm { mol } ^ { - 1 } {/tex}
B
20{tex} \mathrm { g } \mathrm { mol } ^ { - 1 } {/tex}
C
40{tex} \mathrm { g } \mathrm { mol } ^ { - 1 } {/tex}
D
30{tex} \mathrm { g } \mathrm { mol } ^ { - 1 } {/tex}
##### Explanation
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Q 2. The number of carbon atoms per unit cell of diamond unit cell is
A
6
B
1
C
4
8
##### Explanation
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Q 3. A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 {tex}\mathrm { pm }{/tex}.The diameter of the metal atom is
288 {tex} \mathrm { pm } {/tex}
B
408 {tex} \mathrm { pm } {/tex}
C
144 {tex} \mathrm { pm } {/tex}
D
204 {tex} \mathrm { pm } {/tex}
##### Explanation
For FCC,
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Q 4. The number of octahedral void(s) per atom present in a cubic close-packed structure is
1
B
3
C
2
D
4
##### Explanation
Number of octahedral voids in ccp, is equal to effective number of atoms, in ccp, effective number of atoms are 4 so, 4 octahedral voids. So, 1 octahedral voids per atom
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Q 5. Structure of a mixed oxide is cubic close packed {tex}(ccp){/tex}. The cubic unit cell of mixed oxide ions is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is
A
{tex} A B O _ { 2 } {/tex}
B
{tex} A _ { 2 } B O _ { 2 } {/tex}
C
{tex} A _ { 2 } B _ { 3 } \mathrm { O } _ { 4 } {/tex}
{tex} A B _ { 2 } \mathrm { O } _ { 2 } {/tex}
##### Explanation
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Q 6. A solid compound {tex} X Y {/tex} has {tex} \mathrm { NaCl } {/tex} structure. If the radius of the cation is 100 {tex}\mathrm { pm } , {/tex} the radius of the anion {tex} \left( Y ^ { - } \right) {/tex} will be
A
275.1 {tex} \mathrm { pm } {/tex}
B
322.5 {tex} \mathrm { pm } {/tex}
241.5 {tex} \mathrm { pm } {/tex}
D
165.7 {tex} \mathrm { pm } {/tex}
##### Explanation
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Q 7. {tex} A B {/tex} crystallizes in a body centred cubic lattice with edge length 'a' equal to 387 pm. The distance between two oppositely charged ions in the lattice is
335 {tex} \mathrm { pm } {/tex}
B
250 {tex} \mathrm { pm } {/tex}
C
200 {tex} \mathrm { pm } {/tex}
D
300 {tex} \mathrm { pm } {/tex}
##### Explanation
Distance between two oppositely charge ions for BCC
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Q 8. Lithium metal crystallises in a body-centred cubic crystal. If the length of the side of the unit cell of lithium is 351{tex} \mathrm { pm } {/tex} , the atomic radius of lithium will be
151.8 {tex} \mathrm { pm } {/tex}
B
75.5 {tex} \mathrm { pm } {/tex}
C
300.5 {tex} \mathrm { pm } {/tex}
D
240.8 {tex} \mathrm { pm } {/tex}
##### Explanation
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Q 9. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 {tex} \mathrm { pm } {/tex} . What is the radius of copper atom in {tex} \mathrm { pm } {/tex} ?
A
157
B
181
C
108
128
##### Explanation
As given that copper crystallizes is FCC lattice (Face centred cubic). In FCC atoms are present on the corners of the cubic, unit cell as well as on the face centres of each face. The atoms on the face diagonal will be touching each other. Let, the radius of the atom be r and edge length of the cube be a.
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Q 10. Which of the following statements is not correct?
A
The number of carbon atoms in a unit cell of diamond is {tex} 8. {/tex}
B
The number of Bravais lattices in which a crystal can be categorized is 14 .
The fraction of the total volume occupied by the atoms in a primitive cell is 0.48 .
D
Molecular solids are generally volatile.
##### Explanation
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Q 11. With which one of the following elements silicon should be doped so as to give {tex} p {/tex}-type of semiconductor?
A
Selenium
Boron
C
Gemanium
D
Arsenic
##### Explanation
The n-type semiconductors are obtained when Si or Ge are doped with elements of group 15, eg, Arsenic (As), while p-type semiconductors are obtained when Si or Ge are doped with traces of elements of group 13, ie indium (In), Boron (B).
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Q 12. If {tex} \mathrm { NaCl } {/tex} is doped with {tex} 10 ^ { - 4 } \mathrm { mol } \% {/tex} of {tex} \mathrm { SrCl } _ { 2 } {/tex} , the concentration of cation vacancies will be {tex} \left( N _ { A } = 6.02 \times 10 ^ { 23 } \mathrm { mol } ^ { - 1 } \right) {/tex}
A
{tex} 6.02 \times 10 ^ { 16 } \mathrm { mol } ^ { - 1 } {/tex}
{tex} 6.02 \times 10 ^ { 17 } \mathrm { mol } ^ { - 1 } {/tex}
C
{tex} 6.02 \times 10 ^ { 14 } \mathrm { mol } ^ { - 1 } {/tex}
D
{tex} 6.02 \times 10 ^ { 15 } \mathrm { mol } ^ { - 1 } {/tex}
##### Explanation
One Sr+2 creates one cationic vacancies
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Q 13. The appearance of colour in solid alkali metal halides is generally due to
A
Interstitial positions
{tex} F {/tex} -centres
C
Schottky defect.
D
Frenkel defect.
##### Explanation
The appearance in colour in solid alkali metal halides is generally due to F− centres. When electrons occupy the vacant space of anions due to the presence of excess of metal ions, are called F-centres.
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Incorrectly Marks -1
Q 14. CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 amu and that of Br = 80 amu and Avogadro number being 6.02 x 1023 mol-1 the density of CsBr is
4.25{tex} \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex}
B
42.5{tex} \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex}
C
0.425{tex} \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex}
D
8.25{tex} \mathrm { g } / \mathrm { cm } ^ { 3 } {/tex}
##### Explanation
Correct Marks 4
Incorrectly Marks -1
Q 15. In a face-centered cubic lattice, a unit cell is shared equally by how many unit cells?
A
2
B
4
6
D
8
##### Explanation
A cubic unti cell has six faces. Therefore, in a cubic lattice (irrespective of its nature) cubic unit cell is shared equally by 6 unit cells.
Correct Marks 4
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Q 16. A compound formed by elements {tex} X {/tex} and {tex} Y {/tex} crystallises in a cubic structure in which the {tex} X {/tex} atoms are at the corners of a cube and the {tex} Y {/tex} atoms are at the face-centres. The formula of the compound is
{tex} X Y _ { 3 } {/tex}
B
{tex} X _ { 3 } Y {/tex}
C
{tex} X Y {/tex}
D
{tex} X Y _ { 2 } {/tex}
##### Explanation
No of atoms in the X in the unit cell {tex}{ 8}\times{(1/8)}{/tex} = 1 No of atoms in the X in the unit cell {tex}{ 6}\times{(1/2)}{/tex} = 3 Hence the formula of compund is {tex} X Y _ { 3 } {/tex}
Correct Marks 4
Incorrectly Marks -1
Q 17. The pyknometric density of sodium chloride crystal is {tex} 2.165 \times 10 ^ { 3 } \mathrm { kg } \mathrm { m } ^ { - 3 } {/tex} while its X-ray density is {tex} 2.178 \times 10 ^ { 3 } \mathrm { kg } \mathrm { m } ^ { - 3 } {/tex} . The fraction of unoccupied sites in sodium chloride crystal is
A
5.96
B
{tex} 5.96 \times 10 ^ { - 2 } {/tex}
C
{tex} 5.96 \times 10 ^ { - 1 } {/tex}
{tex} 5.96 \times 10 ^ { - 3 } {/tex}
##### Explanation
X ray density is higher because corresponds to occupied space only as X-rays are diffracted by constituet particles.On the hand ,pyknometric density is lower as it correspondence to occupied as well as unoccupied space. Thus, the fraction of unoccupied site in NaCl crystal is given as
Correct Marks 4
Incorrectly Marks -1
Q 18. When {tex} \mathrm { Zn } {/tex} converts from melted state to its solid state, it has {tex} h c p {/tex} structure, then find the number of nearest atoms.
A
6
B
8
12
D
4
##### Explanation
When Zinc is converted from it's melted state to solid state ,it has HCP structure. In HCP structure , the coordination number is 12. Since Zn is in HCP structure the number of nearest neighbours is 12, six are in the plane ,3 above the plane and 3 below the plane. So correct option is C.
Correct Marks 4
Incorrectly Marks -1
Q 19. In cube of any crystal {tex} A {/tex} -atom placed at every corners and {tex} B {/tex} -atom placed at every centre of face. The formula of compound is
A
{tex} A B {/tex}
{tex} A B _ { 3 } {/tex}
C
{tex} A _ { 2 } B _ { 2 } {/tex}
D
{tex} A _ { 2 } B _ { 3 } {/tex}
##### Explanation
Correct Marks 4
Incorrectly Marks -1
Q 20. In crystals of which one of the following ionic compounds would you expect maximum distance between centres of cations and anions?
{tex} \mathrm { CsI } {/tex}
B
{tex} \mathrm {CsF} {/tex}
C
{tex} \mathrm { LiF } {/tex}
D
{tex} \mathrm {LiI} {/tex}
##### Explanation
The ionic radii of {tex} \mathrm { Cs } {/tex}+ is highest among the given cations and ionic radii of {tex} \mathrm { I } {/tex}- is highest among the given anions. Therefore the maximum distance between the centers of cations and anions occurs in {tex} \mathrm { CsI } {/tex}
Correct Marks 4
Incorrectly Marks -1
Q 21. Schottky defect in crystals is observed when
A
Density of the crystal is increased
B
Unequal number of cations and anions are missing from the lattice
C
An ion leaves its normal site and occupies an interstitial site
Equal number of cations and anions are missing from the lattice.
##### Explanation
In ionic crystals, the defect forms when oppositely charged ions leave their lattice sites, creating vacancies. These vacancies are formed in stoichiometric units, to maintain an overall neutral charge in the ionic solid. The surrounding atoms then move to fill these vacancies, causing new vacancies to form. Normally these defects will lead to a decrease in the density of the crystal.
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# CP Violation in the SUSY Seesaw: Leptogenesis and Low Energy
Abstract : We suppose that the baryon asymmetry is produced by thermal leptogenesis (with flavour effects), at temperatures $\sim 10^{9} - 10^{10}$ GeV, in the supersymmetric seesaw with universal and real soft terms. The parameter space is restricted by assuming that $l_\alpha \to l_\beta \gamma$ processes will be seen in upcoming experiments. We study the sensitivity of the baryon asymmetry to the phases of the lepton mixing matrix, and find that leptogenesis can work for any value of the phases. We also estimate the contribution to the electric dipole moment of the electron, arising from the seesaw, and find that it is (just) beyond the sensitivity of next generation experiments ($\lsim 10^{-29} e$ cm). The fourteen dimensional parameter space is efficiently explored with a Monte Carlo Markov Chain, which concentrates on the regions of interest.
Document type :
Journal articles
http://hal.in2p3.fr/in2p3-00294642
Contributor : Dominique Girod <>
Submitted on : Thursday, July 10, 2008 - 10:22:50 AM
Last modification on : Thursday, June 17, 2021 - 3:18:44 PM
### Citation
Sacha Davidson, J. Garayoa, F. Palorini, N. Rius. CP Violation in the SUSY Seesaw: Leptogenesis and Low Energy. Journal of High Energy Physics, Springer, 2008, 2008(09), pp.053. ⟨10.1088/1126-6708/2008/09/053⟩. ⟨in2p3-00294642⟩
Record views
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## October 04, 2013
Although parallel techniques in R has been prevailing, I will only focus on Loading the complete data into RAM in R, that is to say, no Hadoop or similar. What other more I won’t mention in this post is about manipulating and saving big data in R, and parallel computing.
• load csv file and using ff package (Rtools)
bigdata <- read.csv.ffdf(file = ”bigdata.csv”, first.rows=5000, colClasses = NA)
Notice that ff package should be in Rtools on Windows.
• using sqldf() from SQLite
this is a method from StackOverflow: using sqldf() to import the data into SQLite as a staging area, and then sucking it from SQLite into R
library(sqldf)
f <- file("bigdf.csv")
system.time(bigdf <- sqldf("select * from f", dbname = tempfile(), file.format = list(header = T, row.names = F)))
it includes data.frame, but some of the syntax is different. Luckily, the documentation (and the FAQ) are excellent.
Notice that fread() cannot directly read gzipped files and it comes with a big warning sign “not for production use yet”. One trick it uses is to read the first, middle, and last 5 rows to determine column types.
This option takes a vector whose length is equal to the number of columns in year table. Specifying this option instead of using the default can make ‘read.table’ run MUCH faster, often twice as fast. In order to use this option, you have to know the of each column in your data frame. - See more at hear.
read.table("test.csv",header=TRUE,sep=",",quote="",
stringsAsFactors=FALSE,comment.char="",nrows=n,
colClasses=c("integer","integer","numeric",
"character","numeric","integer"))
• load a portion using nrows
Also you can read in only a portion of your file, to get a feel of the dataset.
data_first_100 <- read.table("file", header=T, sep="\t", stringsAsFactors=F, nrows=100)
• in summary
Here is a great comparison summary for the method above with their system time. I just copy the summary table below:
## user system elapsed Method
## 24.71 0.15 25.42 read.csv (first time)
## 17.85 0.07 17.98 read.csv (second time)
## 10.20 0.03 10.32 Optimized read.table
## 12.49 0.09 12.69 sqldf
## 10.21 0.47 10.73 sqldf on SO
## 10.85 0.10 10.99 ffdf
See more in 11 Tips on How to Handle Big Data in .
### Highway Networks and Deep Residual Networks
Recently, a breakthrough news spread over social networks. In this post, I will explain this ResNet as a special case of Highway Networks, which has been proposed before. Both of the work is amazing and thought-provoking. Continue reading
#### NIPS 2015 Deep Learning Symposium Part II
Published on January 09, 2016
#### NIPS 2015 Deep Learning Symposium Part I
Published on December 11, 2015
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# matplotlib.pyplot.xcorr¶
matplotlib.pyplot.xcorr(x, y, normed=True, detrend=<function detrend_none at 0x7f280ff588b0>, usevlines=True, maxlags=10, *, data=None, **kwargs)[source]
Plot the cross correlation between x and y.
The correlation with lag k is defined as $$\sum_n x[n+k] \cdot y^*[n]$$, where $$y^*$$ is the complex conjugate of $$y$$.
Parameters: x, yarray-like of length n detrendcallable, default: mlab.detrend_none (no detrending)A detrending function applied to x and y. It must have the signature detrend(x: np.ndarray) -> np.ndarray normedbool, default: TrueIf True, input vectors are normalised to unit length. usevlinesbool, default: TrueDetermines the plot style. If True, vertical lines are plotted from 0 to the xcorr value using Axes.vlines. Additionally, a horizontal line is plotted at y=0 using Axes.axhline. If False, markers are plotted at the xcorr values using Axes.plot. maxlagsint, default: 10Number of lags to show. If None, will return all 2 * len(x) - 1 lags. lagsarray (length 2*maxlags+1)The lag vector. carray (length 2*maxlags+1)The auto correlation vector. lineArtist added to the axes of the correlation: LineCollection if usevlines is True. Line2D if usevlines is False. bLine2D or NoneHorizontal line at 0 if usevlines is True None usevlines is False. linestyleLine2D property, optionalThe linestyle for plotting the data points. Only used if usevlines is False. markerstr, default: 'o'The marker for plotting the data points. Only used if usevlines is False. **kwargsAdditional parameters are passed to Axes.vlines and Axes.axhline if usevlines is True; otherwise they are passed to Axes.plot.
Notes
The cross correlation is performed with numpy.correlate with mode = "full".
Note
In addition to the above described arguments, this function can take a data keyword argument. If such a data argument is given, the following arguments can also be string s, which is interpreted as data[s] (unless this raises an exception): x, y.
Objects passed as data must support item access (data[s]) and membership test (s in data).
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Smartly auto-increment called cells
When I try to increment called rows or columns in a sequence that skips one or sevral numbers such as:
=A1
=A3
=A5
=A7
=A9
by providing the first 2 cells (e.g. =A1 and =A3), it doesn't work.
=A1
=A3
=A3
=A5
=A5
=A7
=A7
=A9
=A9
and it gets weirder if I try to help it further by providing the first 3 cells:
=A1
=A3
=A5
=A4
=A6
=A8
=A7
=A9
=A11
Can I fix this ?
edit retag close merge delete
1
No - because you want A2=A1+2; A3=A2+2, ... so you make useless use of ROW() function, which btw. works as expected. In your "2 cells" example:
• 1st value: take the number of current row
• 2nd value: take the number of next row
and these two rules are repeated if you drag down. Same applies for your "3 cells" example.
( 2019-07-02 14:46:09 +0100 )edit
1
You had some reasonable formulas there (=ROW(A1), =ROW(A3)); now you changed that to something that just couldn't work if you put that into A1 and A2: putting there =A1 and =A3 would create cycles.
( 2019-07-02 14:46:58 +0100 )edit
No because I actually want to increment the called cells, so I edited my post to better fit my problem. Sorry for my lack of clarity.
( 2019-07-02 15:26:18 +0100 )edit
Boah - this should have been mentioned that you want increment cell references and not cell values.
( 2019-07-02 16:53:54 +0100 )edit
Sort by » oldest newest most voted
Of course you get it. You have a sequence of two formulas, first reading "give me the number of current row", the second one being "give me number of the row just below this". You then copy the two formulas, and they continue this sequence: "this-row-#; next-row-#; this-row-#; next-row-#; this-row-#; ...".
Why don't you just put plain numbers 1 and 3 into A1 and A2, and drag-copy that?
more
Because my situation involves formulas using several called cells which I need to increment smartly both vertically and horizontally. Splitting the formulas in plain text and incrementing the cell numbers & letters in separate rows or columns is impossible. I would need a 3rd dimension.
( 2019-07-02 15:03:16 +0100 )edit
Then your formula would look like =ROW()*2-1. That's it. It's just a basic formula mapping all positive integers into all positive odd numbers. You need to come with other formulas for possible other sequences.
Another way: put 1 into A1; put =A1+2 into A2; drag-copy A2.
( 2019-07-02 15:05:18 +0100 )edit
Thanks but that's not what I mean. What you propose increments the output value. What I want is to properly increment the called cells. Imagine my formulas have to take values of one in 4 rows and one in 3 columns. Sorry for my lack of clarity. I edited the original post, which should be clearer now.
( 2019-07-02 15:24:09 +0100 )edit
Sigh. Could you at least mention which column are you showing us? I was under an impression that we are talking about column A.
Put =OFFSET(A1;ROW()-1;0) into B1 and drag-copy to get reference to every second row in column A.
Or, if you like:
=OFFSET($A$1;(ROW()-1)*2;0)
( 2019-07-02 15:40:27 +0100 )edit
Actually, it's easy:
If my answer has solved your problem, please click on the checkmark ✓ in the circle to the left of the answer and click on the arrow ^ for upvote. This will tell the community that the question has been answered correctly.
more
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# How many random permutations to cover all possible permutations?
I have code that generates a random permutation. In my case, a permutation consists of N binary features, and each of the N features is set or unset randomly. How many times must I generate a random permutation in order to be reasonably assured that I have covered all possible permutations? I'm not sure how to define "reasonably assured" (50%, 95%?).
For example, let's say that N = 10, so there are 1024 possible permutations, how many times should I randomly generate a permutation to be 50% confident that I have generated all 1024 permutations?
It seems to me that this is related to the 36.8% duplicate hit rate when drawing a sample with resampling, but I'm not a statistician.
-
If you have the computing power to generate all permutations (or some well-defined set of re-assignments of the data), why not systematically generate each exactly once? That would normally be far more efficient and give you exact, rather than approximate, answers. (This idea is identical to the difference between sampling a small finite population with and without replacement: why sample with replacement when you can obtain a complete census and eliminate all sampling error?) – whuber Oct 11 '12 at 21:28
My goal is not to generate all permutations, but rather to find a way to quantify how "good" a randomly generated sample is. – Tyro Oct 11 '12 at 22:59
Huh? If you can exhaustively generate all permutations, then you can answer any question of that sort with perfect accuracy. Generating, and re-generating, far more permutations than you need is just a complete waste of computing time and will give you less accurate answers. – whuber Oct 12 '12 at 14:29
Permutation usually refers to something else, so it's probably better to call your problem "random binary words" or something similar.
The question of how long it takes to get at least one representative of each type is called the Coupon Collector Problem. If you assume that all binary words of length $N$ are equally likely, then there are $2^N$ types of coupons. You can write the time to collect all coupons as a sum of the times to collect the $i$th new coupon, a sum of independent geometric random variables. So, the expected number of coupons it takes to collect them all is $2^N \sum_{i=1}^{2^N} 1/i \sim 2^N \log 2^N$, or more precisely $2^N \log 2^N + 2^N \gamma + 1/2 + o(1)$. For $N=10$ this is about $7689$. The variance is $2^{2N} \sum_{i=1}^{2^N} 1/i^2 \approx 2^{2N} \frac {\pi^2}{6}$, so the standard deviation is about $2^N \frac{\pi}{\sqrt{6}}$. For $N=10$ this is about $1313$. Note that a normal approximation is NOT appropriate here.
One crude bound is Chebyshev's inequality, which says that the chance that a random variable is more than $k$ standard deviations away from the mean is at most $1/k^2$, and the similar Cantelli's inequality is that the chance that a random variable is at least $k$ standard deviations above the mean is at most $1/(k^2+1)$. This gives you an upper bound of about $7689 + 1313 \approx 9002$ for the median, and $7689 + 1313\sqrt{19} \sim 13412$ for the $95$th percentile.
If these bounds are not good enough, there are more precise but more complicated asymptotics known. Another approach, suitable perhaps up to $N = 25$, is to compute the exact distribution numerically using the representation as a sum of independent geometric distributions.
-
I don't follow where the sqrt(19) came from for the bound for the 95th percentile. – Tyro Oct 11 '12 at 23:28
If $k=\sqrt{19}$ then $1/(k^2+1) = 1/20 =5\%$. So, Cantelli's lemma says that the probability that the result is at most $\sqrt{19}$ standard deviations above the mean is at least $95\%$. – Douglas Zare Oct 11 '12 at 23:58
Thanks for the clarification. – Tyro Oct 12 '12 at 0:08
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# 6.7 Exponential and logarithmic models (Page 5/16)
Page 5 / 16
A pitcher of water at 40 degrees Fahrenheit is placed into a 70 degree room. One hour later, the temperature has risen to 45 degrees. How long will it take for the temperature to rise to 60 degrees?
6.026 hours
## Using logistic growth models
Exponential growth cannot continue forever. Exponential models, while they may be useful in the short term, tend to fall apart the longer they continue. Consider an aspiring writer who writes a single line on day one and plans to double the number of lines she writes each day for a month. By the end of the month, she must write over 17 billion lines, or one-half-billion pages. It is impractical, if not impossible, for anyone to write that much in such a short period of time. Eventually, an exponential model must begin to approach some limiting value, and then the growth is forced to slow. For this reason, it is often better to use a model with an upper bound instead of an exponential growth model, though the exponential growth model is still useful over a short term, before approaching the limiting value.
The logistic growth model is approximately exponential at first, but it has a reduced rate of growth as the output approaches the model’s upper bound, called the carrying capacity . For constants $\text{a, b,}$ and $\text{c,}$ the logistic growth of a population over time $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is represented by the model
$f\left(x\right)=\frac{c}{1+a{e}^{-bx}}$
The graph in [link] shows how the growth rate changes over time. The graph increases from left to right, but the growth rate only increases until it reaches its point of maximum growth rate, at which point the rate of increase decreases.
## Logistic growth
The logistic growth model is
$f\left(x\right)=\frac{c}{1+a{e}^{-bx}}$
where
• $\frac{c}{1+a}\text{\hspace{0.17em}}$ is the initial value
• $c\text{\hspace{0.17em}}$ is the carrying capacity , or limiting value
• $b\text{\hspace{0.17em}}$ is a constant determined by the rate of growth.
## Using the logistic-growth model
An influenza epidemic spreads through a population rapidly, at a rate that depends on two factors: The more people who have the flu, the more rapidly it spreads, and also the more uninfected people there are, the more rapidly it spreads. These two factors make the logistic model a good one to study the spread of communicable diseases. And, clearly, there is a maximum value for the number of people infected: the entire population.
For example, at time $\text{\hspace{0.17em}}t=0\text{\hspace{0.17em}}$ there is one person in a community of 1,000 people who has the flu. So, in that community, at most 1,000 people can have the flu. Researchers find that for this particular strain of the flu, the logistic growth constant is $\text{\hspace{0.17em}}b=0.6030.\text{\hspace{0.17em}}$ Estimate the number of people in this community who will have had this flu after ten days. Predict how many people in this community will have had this flu after a long period of time has passed.
We substitute the given data into the logistic growth model
$f\left(x\right)=\frac{c}{1+a{e}^{-bx}}$
Because at most 1,000 people, the entire population of the community, can get the flu, we know the limiting value is $\text{\hspace{0.17em}}c=1000.\text{\hspace{0.17em}}$ To find $\text{\hspace{0.17em}}a,$ we use the formula that the number of cases at time $\text{\hspace{0.17em}}t=0\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\frac{c}{1+a}=1,$ from which it follows that $\text{\hspace{0.17em}}a=999.\text{}$ This model predicts that, after ten days, the number of people who have had the flu is $\text{\hspace{0.17em}}f\left(x\right)=\frac{1000}{1+999{e}^{-0.6030x}}\approx 293.8.\text{\hspace{0.17em}}$ Because the actual number must be a whole number (a person has either had the flu or not) we round to 294. In the long term, the number of people who will contract the flu is the limiting value, $\text{\hspace{0.17em}}c=1000.$
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
how do I attempted a trig number as a starter
cos 18 ____ sin 72 evaluate
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# Poincaré Embeddings for Learning Hierarchical Representations
Representation learning has become an invaluable approach for learning from symbolic data such as text and graphs. However, while complex symbolic datasets often exhibit a latent hierarchical structure, state-of-the-art methods typically learn embeddings in Euclidean vector spaces, which do not account for this property. For this purpose, we introduce a new approach for learning hierarchical representations of symbolic data by embedding them into hyperbolic space -- or more precisely into an n-dimensional Poincar\'e ball. Due to the underlying hyperbolic geometry, this allows us to learn parsimonious representations of symbolic data by simultaneously capturing hierarchy and similarity. We introduce an efficient algorithm to learn the embeddings based on Riemannian optimization and show experimentally that Poincar\'e embeddings outperform Euclidean embeddings significantly on data with latent hierarchies, both in terms of representation capacity and in terms of generalization ability.
PDF Abstract NeurIPS 2017 PDF NeurIPS 2017 Abstract
## Code Add Remove Mark official
See all 10 implementations
## Results from the Paper Edit
Task Dataset Model Metric Name Metric Value Global Rank Result Benchmark
Lexical Entailment HyperLex Poincare Embeddings Spearman Correlation 0.512 # 1
Link Prediction WordNet Poincare Embeddings (dim=100) Accuracy 77.4 # 2
Link Prediction WordNet Poincare Embeddings (dim=50) Accuracy 77.0 # 3
Link Prediction WordNet Poincare Embeddings (dim=20) Accuracy 74.3 # 4
Link Prediction WordNet Poincare Embeddings (dim=10) Accuracy 68.3 # 5
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# Proof the golden ratio with the limit of Fibonacci sequence [duplicate]
Let $F_n=F_{n-1}+F_{n-2}$ the Fibonacci numbers, and $\phi=\frac{1+\sqrt5}{2}$
The exercise asks me to prove that: $\lim\limits_{n \to \infty}\frac{F_{n+1}}{F_n}=\phi$.
Sorry as can be proceed??
• And how to get the sequence that converges to 1,618 ... May 17 '15 at 17:36
$$F_{n+1}=F_{n}+F_{n-1} \Rightarrow \frac{F_{n+1}}{F_n}=1+\frac{F_{n-1}}{F_n}$$
Let $$x_n:= \frac{F_{n+1}}{F_n}$$
Then $$x_n=1+\frac{1}{x_{n-1}}$$
You can now prove that $$1 \leq x_n \leq 2$$ and by induction that
$$x_1 \leq x_3 \leq x_5 \leq .. \leq x_{2n+1} \leq x_{2n} \leq x_{2n-2} \leq .. \leq x_2$$
From here you get that $$x_{2n-1}$$ and $$x_{2n}$$ converge, and you can use their definitions to get their limits. You prove that both limits are the same, which yields the desired result.
• Great, it was the answer that i was looking for.. May 17 '15 at 17:51
• Aren't your inequalities the other way around? For instance, $x_1 = 1$ and $x_2 = 2$. Also, it is the odd subsequence that increases, not the even one. Mar 7 '17 at 9:49
Let $R_n=\frac{F_{n+1}}{F_n}$. Since: $$F_n^2-F_{n-1}F_{n+1} = (-1)^n\tag{1}$$ (it is easy to prove by induction) we have that: $$\left|R_{n+1}-R_n\right|=\frac{1}{F_n F_{n+1}}, \tag{2}$$ hence $\{R_n\}_{n\in\mathbb{N}}$ is a Cauchy sequence and $R_n$ converges to some $L>1$ that satisfies $L=1+\frac{1}{L}$, since $R_n>1$ and $R_{n+1}=1+\frac{1}{R_n}$.
• Is that enough to show that $R_n$ is a Cauchy sequence? if yes, can you elaborate please. Wikipedia article about Cauchy sequences states that " it is not sufficient for each term to become arbitrarily close to the preceding term". Aug 22 '16 at 5:08
• @ Robert William HanksIt is, if you know additional fact that if $\sum^\infty_{n=1} |R_{n+1} - R_n| < \infty$ then $(R_n)$ is Cauchy. Here it is true, since $F_n >n$, and if $\sum^\infty_{n=1} |R_{n+1} - R_n| \leq \sum^\infty_{n=1} \frac{1}{n(n+1)} < \infty$ . Dec 10 '17 at 9:30
Find $\alpha,\beta$ such that $F_0=\alpha+\beta$ and $F_1=\alpha\phi+\beta(1-\phi)$ and show by induction that $$F_n=\alpha\phi^n+\beta(1-\phi)^n.$$ Then compute the desired limit (and use that $|1-\phi|<1$).
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# 3008. Distributed System - Database ScalingRDBMS and NoSQL
RDBMS and NoSQL
## 1. RDBMS
### 1.1 What is RDBMS and ACID?
A relational database like SQL is a collection of data items organized in tables.
ACID is a set of properties of relational database transactions.
• Atomicity - Each transaction is all or nothing
• Consistency - Any transaction will bring the database from one valid state to another
• Isolation - Executing transactions concurrently has the same results as if the transactions were executed serially
• Durability - Once a transaction has been committed, it will remain so
### 1.2 Scaling Relational Database
There are many techniques to scale a relational database:
• master-slave replication
• master-master replication
• federation
• sharding
• denormalization
• SQL tuning
### 1.3 Master-slave Replication
The master serves reads and writes, replicating writes to one or more slaves, which serve only reads. Slaves can also replicate to additional slaves in a tree-like fashion. If the master goes offline, the system can continue to operate in read-only mode until a slave is promoted to a master or a new master is provisioned. Disadvantage(s): master-slave replication
• Additional logic is needed to promote a slave to a master.
• See ‘Disadvantage(s): replication’ for points related to both master-slave and master-master.
### 1.4 Master-master replication
Both masters serve reads and writes and coordinate with each other on writes. If either master goes down, the system can continue to operate with both reads and writes. Disadvantage(s): master-master replication
• You’ll need a load balancer or you’ll need to make changes to your application logic to determine where to write.
• Most master-master systems are either loosely consistent (violating ACID) or have increased write latency due to synchronization.
• Conflict resolution comes more into play as more write nodes are added and as latency increases.
• See ‘Disadvantage(s): replication’ for points related to both master-slave and master-master.
• There is a potential for loss of data if the master fails before any newly written data can be replicated to other nodes.
• Writes are replayed to the read replicas. If there are a lot of writes, the read replicas can get bogged down with replaying writes and can’t do as many reads.
• The more read slaves, the more you have to replicate, which leads to greater replication lag.
• On some systems, writing to the master can spawn multiple threads to write in parallel, whereas read replicas only support writing sequentially with a single thread.
### 1.5 Federation
Federation (or functional partitioning) splits up databases by function. For example, instead of a single, monolithic database, you could have three databases: forums, users, and products, resulting in less read and write traffic to each database and therefore less replication lag. Smaller databases result in more data that can fit in memory, which in turn results in more cache hits due to improved cache locality. With no single central master serializing writes you can write in parallel, increasing throughput.
• Federation is not effective if your schema requires huge functions or tables.
• You’ll need to update your application logic to determine which database to read and write.
• Joining data from two databases is more complex with a server link.
### 1.6 Sharding
Sharding distributes data across different databases such that each database can only manage a subset of the data. Taking a users database as an example, as the number of users increases, more shards are added to the cluster. Similar to the advantages of federation, sharding results in less read and write traffic, less replication, and more cache hits. Index size is also reduced, which generally improves performance with faster queries. If one shard goes down, the other shards are still operational, although you’ll want to add some form of replication to avoid data loss. Like federation, there is no single central master serializing writes, allowing you to write in parallel with increased throughput.
Common ways to shard a table of users is either through the user’s last name initial or the user’s geographic location.
• You’ll need to update your application logic to work with shards, which could result in complex SQL queries.
• Data distribution can become lopsided in a shard. For example, a set of power users on a shard could result in increased load to that shard compared to others.
• Rebalancing adds additional complexity. A sharding function based on consistent hashing can reduce the amount of transferred data.
• Joining data from multiple shards is more complex.
### 1.7 Denormalization
Denormalization attempts to improve read performance at the expense of some write performance. Redundant copies of the data are written in multiple tables to avoid expensive joins. Some RDBMS such as PostgreSQL and Oracle support materialized views which handle the work of storing redundant information and keeping redundant copies consistent.
Once data becomes distributed with techniques such as federation] and sharding, managing joins across data centers further increases complexity. Denormalization might circumvent the need for such complex joins.
In most systems, reads can heavily outnumber writes 100:1 or even 1000:1. A read resulting in a complex database join can be very expensive, spending a significant amount of time on disk operations.
• Data is duplicated.
• Constraints can help redundant copies of information stay in sync, which increases complexity of the database design.
• A denormalized database under heavy write load might perform worse than its normalized counterpart.
### 1.8 SQL tuning
SQL tuning is a broad topic and many books have been written as reference.
It’s important to benchmark and profile to simulate and uncover bottlenecks.
• Benchmark - Simulate high-load situations with tools such as ab.
• Profile - Enable tools such as the slow query log to help track performance issues.
Benchmarking and profiling might point you to the following optimizations.
##### Tighten up the schema
• MySQL dumps to disk in contiguous blocks for fast access.
• Use CHAR instead of VARCHAR for fixed-length fields.
• CHAR effectively allows for fast, random access, whereas with VARCHAR, you must find the end of a string before moving onto the next one.
• Use TEXT for large blocks of text such as blog posts. TEXT also allows for boolean searches. Using a TEXT field results in storing a pointer on disk that is used to locate the text block.
• Use INT for larger numbers up to 2^32 or 4 billion.
• Use DECIMAL for currency to avoid floating point representation errors.
• Avoid storing large BLOBS, store the location of where to get the object instead.
• VARCHAR(255) is the largest number of characters that can be counted in an 8 bit number, often maximizing the use of a byte in some RDBMS.
• Set the NOT NULL constraint where applicable to improve search performance.
##### Use good indices
• Columns that you are querying (SELECT, GROUP BY, ORDER BY, JOIN) could be faster with indices.
• Indices are usually represented as self-balancing B-tree that keeps data sorted and allows searches, sequential access, insertions, and deletions in logarithmic time.
• Placing an index can keep the data in memory, requiring more space.
• Writes could also be slower since the index also needs to be updated.
• When loading large amounts of data, it might be faster to disable indices, load the data, then rebuild the indices.
##### Partition tables
• Break up a table by putting hot spots in a separate table to help keep it in memory.
##### Tune the query cache
Source(s) and further reading: SQL tuning
## 2. NoSQL
NoSQL is a collection of data items represented in a key-value store, document store, wide column store, or a graph database. Data is denormalized, and joins are generally done in the application code. Most NoSQL stores lack true ACID transactions and favor eventual consistency.
BASE is often used to describe the properties of NoSQL databases. In comparison with the CAP Theorem, BASE chooses availability over consistency.
• Basically available - the system guarantees availability.
• Soft state - the state of the system may change over time, even without input.
• Eventual consistency - the system will become consistent over a period of time, given that the system doesn’t receive input during that period.
In addition to choosing between SQL or NoSQL, it is helpful to understand which type of NoSQL database best fits your use case(s).
### 2.1 Key-Value Store
Abstraction: hash table
A key-value store generally allows for O(1) reads and writes and is often backed by memory or SSD. Data stores can maintain keys in lexicographic order, allowing efficient retrieval of key ranges. Key-value stores can allow for storing of metadata with a value.
Key-value stores provide high performance and are often used for simple data models or for rapidly-changing data, such as an in-memory cache layer. Since they offer only a limited set of operations, complexity is shifted to the application layer if additional operations are needed.
A key-value store is the basis for more complex systems such as a document store, and in some cases, a graph database.
Source(s) and further reading: key-value store
### 2.2 Document Store
Abstraction: key-value store with documents stored as values
A document store is centered around documents (XML, JSON, binary, etc), where a document stores all information for a given object. Document stores provide APIs or a query language to query based on the internal structure of the document itself. Note, many key-value stores include features for working with a value’s metadata, blurring the lines between these two storage types.
Based on the underlying implementation, documents are organized by collections, tags, metadata, or directories. Although documents can be organized or grouped together, documents may have fields that are completely different from each other.
Some document stores like MongoDB and CouchDB also provide a SQL-like language to perform complex queries. DynamoDB supports both key-values and documents.
Document stores provide high flexibility and are often used for working with occasionally changing data.
Source(s) and further reading: document store
### 2.3 Wide Column Store
Abstraction: nested map ColumnFamily<RowKey, Columns<ColKey, Value, Timestamp>>
A wide column store’s basic unit of data is a column (name/value pair). A column can be grouped in column families (analogous to a SQL table). Super column families further group column families. You can access each column independently with a row key, and columns with the same row key form a row. Each value contains a timestamp for versioning and for conflict resolution.
Google introduced Bigtable as the first wide column store, which influenced the open-source HBase often-used in the Hadoop ecosystem, and Cassandra from Facebook. Stores such as BigTable, HBase, and Cassandra maintain keys in lexicographic order, allowing efficient retrieval of selective key ranges.
Wide column stores offer high availability and high scalability. They are often used for very large data sets.
Source(s) and further reading: wide column store
### 2.4 Graph Store
Abstraction: graph
In a graph database, each node is a record and each arc is a relationship between two nodes. Graph databases are optimized to represent complex relationships with many foreign keys or many-to-many relationships.
Graphs databases offer high performance for data models with complex relationships, such as a social network. They are relatively new and are not yet widely-used; it might be more difficult to find development tools and resources. Many graphs can only be accessed with REST APIs.
## 3. SQL or NoSQL
Reasons for SQL:
• Structured data
• Strict schema
• Relational data
• Need for complex joins
• Transactions
• Clear patterns for scaling
• More established: developers, community, code, tools, etc
• Lookups by index are very fast
Reasons for NoSQL:
• Semi-structured data
• Dynamic or flexible schema
• Non-relational data
• No need for complex joins
• Store many TB (or PB) of data
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# Why drop the vibrational ground state energy
• I
## Main Question or Discussion Point
This is from *Statistical Physics An Introductory Course* by *Daniel J.Amit*
The text is calculating the energy of internal motions of a diatomic molecule.
The internal energies of a diatomic molecule, i.e. the vibrational energy and the rotational energy is given by
\begin{aligned}E_I(n,J) &= E_v + E_r \\ &= \epsilon _v n+\epsilon_r J(J+1), \qquad n,J=0,1,2,\dots ,\end{aligned}
where $\epsilon_v$ is the spacing between the vibrational levels:
$$\epsilon_v = \hbar \omega= \hbar \sqrt{\frac{K}{\mu}}$$
and $\epsilon_r$ is the difference between the rotational levels:
$$\epsilon_r=\frac{\hbar^2}{2I}$$
But the vibrational energy of the molecule is
$$E_v = (n+1/2)\hbar \omega$$
**Why do we ignore the ##\frac{1}{2} \hbar \omega## in the ##E_v## (ground state vibrational energy) when we calculate the total internal energy of the molecule?** The text refers back to another equation to explain why we dropped the ground state vibrational energy, but I still don't quite understand the reason behind it.
The text referred:
The ground state energy of the electronic system ##\epsilon_0## depends on the distance ##\rho## between the nuclei, and we can denote this as a potential energy ##U(\rho)##.
If ##\epsilon_0## has a sharp minimum at a distance ##\rho_0##, it is possible to approximate ##U(\rho)## by the harmonic approximation. Hence the energy of the molecule will then given by
$$E_{mol}=\frac{\boldsymbol{P}^2}{2M}+\frac{\boldsymbol{\pi}^2}{2 \mu}+\frac{1}{2}K(\rho-\rho_0)^2$$
where
##\bf{P}## is the momentum associated with center of mass
##\boldsymbol{\pi}=\boldsymbol{\pi}_{12}=\boldsymbol{\pi}_1 - \boldsymbol{\pi}_2## in which ##\boldsymbol{\pi}_\alpha## is the relative momentum of the atom ##\alpha##
##U''(\rho_0)=K## is the harmonic approximation of ##U##
In this case I get why they drop the ##\epsilon_0##, that's because ##\epsilon_0## is already included in the potential energy term. However for the first case I don't see where the ##\epsilon_0## goes.
Related Other Physics Topics News on Phys.org
mfb
Mentor
It doesn't matter if you only look at differences between energies. If you take relativity into account you have much more energy via the mass of the particles - but you don't have to care about that.
I still don't get it. Why are we calculating the differences between energies instead of the absolute internal energies?
mfb
Mentor
The absolute value is meaningless. Changing it doesn't change anything. If you add a fixed amount of energy in a consistent way, no physical observation changes.
So you saying what we calculated is not the actual internal energy of the molecule, but rather it is the energy relative to the ##\epsilon_0##? Just like how we define the gravitational potential energy to be zero at infinite distance? We set the zero point ourselves?
mfb
Mentor
That’s how I understood what you quoted.
You can keep the constant, it just doesn’t matter.
I think I get it now. Thank you for your help!
mfb
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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)
a. The object is in freefall so the magnitude of the acceleration is $g$. b. The object is in freefall so the magnitude of the acceleration is $g$.
When an object is in freefall through the air, the magnitude of acceleration is always $g$ and the acceleration is directed straight down toward the ground. This acceleration comes from the force of gravity on the object and this force is always directed straight down toward the ground. a. The object is in freefall so the magnitude of the acceleration is $g$. b. The object is in freefall so the magnitude of the acceleration is $g$.
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# How to find the general solution of $(1+x^2)y''+2xy'-2y=0$. How to express by means of elementary functions?
Find the general solution of $$(1+x^2)y''+2xy'-2y=0$$
in terms of power series in $x$. Can you express this solution by means of elementary functions?
I know that $y= \displaystyle\sum_{n=0}^{ \infty } a_nx^n$ and $y'= \displaystyle\sum_{n=1}^{ \infty } a_nnx^{n-1}$ and $y''=\displaystyle\sum_{n=2}^{ \infty } a_nn(n-1)x^{n-2}$.
As far as I can tell I am to simply plug these summations in the original equation stated above. Also, I shifted the following to get $x^n$. Giving:
$$\sum_{n=0}^{ \infty } a_{n+2}(n+2)(n+1)x^{n}+\sum_{n=2}^{ \infty } a_nn(n-1)x^{n}+\sum_{n=1}^{ \infty }2 a_nnx^{n}-\sum_{n=0}^{ \infty } 2a_nx^n=0$$
Now, I combine the equation into a single summation. $$\sum_{n=2}^{ \infty }\bigg[a_{n+2}(n+2)(n+1)+a_nn(n-1)+2a_nn-2a_n\bigg]x^n=0$$
Doing this, I am left over with $2(a_2-a_0)=0$ and $6a_3x=0$. I can also form $a_{n+2}=a_n\frac{1-n}{n+1}$ Following this I plugged values in for the third equation, but this is where things start to get difficult. I'm honestly not sure how to continue. The question itself confuses me specifically "in terms of power series in $x$. Can you express this solution by means of elementary functions?".
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Let $y=\sum\limits_{n=0}^\infty a_nx^n$ ,
Then $y'=\sum\limits_{n=0}^\infty na_nx^{n-1}$
$y''=\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}$
$\therefore(1+x^2)\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}+2x\sum\limits_{n=0}^\infty na_nx^{n-1}-2\sum\limits_{n=0}^\infty a_nx^n=0$
$\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty n(n-1)a_nx^n+\sum\limits_{n=0}^\infty2na_nx^n-\sum\limits_{n=0}^\infty2a_nx^n=0$
$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty(n^2+n-2)a_nx^n=0$
$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty(n+2)(n-1)a_nx^n=0$
$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=2}^\infty n(n-3)a_{n-2}x^{n-2}=0$
$\sum\limits_{n=2}^\infty(n(n-1)a_n+n(n-3)a_{n-2})x^{n-2}=0$
$\therefore n(n-1)a_n+n(n-3)a_{n-2}=0$
$a_n=-\dfrac{(n-3)a_{n-2}}{n-1}$
$\therefore\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_0=a_0\\a_{2n}=\dfrac{(-1)^n((-1)1\times3\times......(2n-3))a_0}{1\times3\times5\times......(2n-1)}\forall n\in\mathbb{N}\end{cases}$
$\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_0=a_0\\a_{2n}=\dfrac{(-1)^{n+1}a_0}{2n-1}\forall n\in\mathbb{N}\end{cases}$
$\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_{2n}=\dfrac{(-1)^{n+1}a_0}{2n-1}\forall n\in\mathbb{Z}^*\end{cases}$
$\therefore y=C_1x+C_2\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}x^{2n}}{2n-1}=C_1x+C_2\biggl(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^{n+1}x^{2n}}{2n-1}\biggr)=C_1x+C_2\biggl(1+\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+2}}{2n+1}\biggr)=C_1x+C_2(1+x\tan^{-1}x)$
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Free Version
Moderate
# Voltage and Current within Circuit Loops
EANDM-XTV65B
Consider the circuit in the accompanying diagram. There are four resistors: Two resistors ($10\Omega$ and $20\Omega$)connected in series on the left leg between points (a) and (c), and two resistors ($5\Omega$ and $15\Omega$) between points (b) and (d). The current
flowing in the left leg (between (a) and (c)) is measured to be 1 Amp.
What is the voltage on the battery ($V$)?
What is the total current ($I$) leaving the battery?
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Newton's Method Converges in Mathematica for non-convergent functions
In mathematica I set f[x_] := Sign[x] Sqrt[Abs[x]] and then I solved it in a root finding for Newton's method by doing FindRoot[f[x], {x, 1}, Method -> "Newton"] which returns {x -> 0.}. This is all fine and good as yes the root is obviously at f[0]; however, when iterating newton's method for $f(x)=\text{Sign}(x)\sqrt{|x|}$ for the initial value $x_0 = 1$ you get $x_{k+1} = x_{k} - \frac{f(x_k)}{f'(x_k)}$ which turns into $x_{k+1} = x_k-2\text{Sign}(x)|x|$. We would get $x_1 = 1 - 2*\text{Sign}(1)*|1| = -1$ then we get $x_2 = -1 - 2*\text{Sign}(-1) |-1| = -1 + 2 = 1$ and it obviously oscillates at $\pm 1$ but never converges to zero; yet mathematica converges this to 0 immediately, is there a way to truly iterate newton's method in Mathematica?
• interesting observation. We can surmise Method -> "Newton" is not strictly pure newton but has some tweak to avoid such oscillation. What exactly is the question though? You can implement it yourself obviously. – george2079 Oct 2 '15 at 21:00
• see here for diy implementations: mathematica.stackexchange.com/q/34229/2079 – george2079 Oct 2 '15 at 21:02
• Welcome to Mathematica.SE! I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise. – bbgodfrey Oct 2 '15 at 21:02
• You can always write your own Newton-Raphson solver. It's not even a particularly difficult exercise. But why bother? The built-in solvers are better than anything you could write. – m_goldberg Oct 2 '15 at 21:32
• little correction, the linked page pertains to a modified newton method, but should give the idea. The other caveat here, mathematica does not nicely deal with analytic derivatives of Abs and Sign so it will take bit of work to show your exact oscillation result. – george2079 Oct 2 '15 at 21:42
Note that neither Sign nor Abs is differentiable so that Newton's Method may not be applied to the OP's problem in its given form.
Caveat: I am assuming this is a toy example, to show that the unadorned Newton's method fails on some functions. Normally, FindRoot uses step control to try to avoid certain common traps in using Newton's Method. One can turn off the protections with the option setting "StepControl" -> None as shown below.
Needs["OptimizationUnconstrainedProblems"]
FindRootPlot[Sign[x] Sqrt[Abs[x]], {x, 1},
Method -> {"Newton", "StepControl" -> None}]
If we define the functions in terms of differentiable functions, one can invoke Newton's Method successfully:
sign[x_] := Piecewise[{{1, x > 0}, {-1, x < 0}}, 0];
abs[x_] := Sqrt[x^2];
FindRootPlot[sign[x] Sqrt[abs[x]], {x, 1},
Method -> {"Newton", "StepControl" -> None}]
To see the steps, we can Sow and Reap them:
Reap@FindRoot[sign[x] Sqrt[abs[x]], {x, 1},
Method -> {"Newton", "StepControl" -> None}, StepMonitor :> Sow[x]]
The steps oscillate between ±1.
• Please replace sign[x] Sqrt[abs[x]] with uppercase letters. – user31001 Oct 3 '15 at 9:00
• @Willinski. Michael is defining special versions of those functions to make his point. The replacement you request would invalidate his answer. – m_goldberg Oct 3 '15 at 9:41
• @m_goldberg Oh, yes! My blind eyes! – user31001 Oct 3 '15 at 10:10
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# Imperative programming Today we will practice with mutable fields, refs, and physical equality. We'll learn about arrays and loops. As a challenge exercise, you can implement doubly-linked lists. ## Mutable fields and refs ##### Exercise: mutable fields [✭] Define an OCaml record type to represent student names and GPAs. It should be possible to mutate the value of a student's GPA. Write an expression defining a student with name "Alice" and GPA 3.7. Then write an expression to mutate Alice's GPA to 4.0. □ ##### Exercise: refs [✭] Give OCaml expressions that have the following types. Use utop to check your answers. * bool ref * int list ref * int ref list □ ##### Exercise: inc fun [✭] Define a reference to a function as follows: # let inc = ref (fun x -> x+1);; Write code that uses inc to produce the value 3110. □ ##### Exercise: addition assignment [✭✭] The C language and many languages derived from it, such as Java, has an *addition assignment* operator written a += b and meaning a = a + b. Implement such an operator in OCaml; its type should be int ref -> int -> unit. Here's some code to get you started: let (+:=) x y = ... And here's an example usage: # let x = ref 0;; # x +:= 3110;; # !x - : int = 3110 □ ## Equality and mutability Recall that OCaml has two different notions of equality: - **Physical equality** (==) tests whether two mutable values (references, records, or arrays) have the same identity. - **Structural equality** (=) tests whether the current contents of a reference are equivalent. The [specification given by Pervasives.(==)][pervasives] has this to say about physical equality: > e1 == e2 tests for physical equality of e1 and e2. On mutable types such > as references, arrays, ..., and records with mutable fields, > e1 == e2 is true if and only if > physical modification of e1 also affects e2. On non-mutable types, the > behavior of ( == ) is implementation-dependent; however, it is > guaranteed that e1 == e2 implies compare e1 e2 = 0. [pervasives]: http://caml.inria.fr/pub/docs/manual-ocaml/libref/Pervasives.html ##### Exercise: physical equality [✭✭] Define x, y, and z as follows: let x = ref 0 let y = x let z = ref 0 Predict the value of the following series of expressions: # x == y;; # x == z;; # x = y;; # x = z;; # x := 1; # x = y;; # x = z;; Check your answers in utop. □ ## Arrays OCaml *arrays* generalize a single ref cell to a sequence of mutable values. # let v = [| 0.; 1. |];; v : float array = [| 0.; 1. |] # v.(0);; - : float = 0. # v.(0) <- 5.;; - : unit = () # v;; - : float array = [| 5.; 1. |] The [Array module](http://caml.inria.fr/pub/docs/manual-ocaml/libref/Array.html) contains useful functions for creating and working with arrays. For the next couple exercises, let's use the following type: (* AF: the float array [| x1; ...; xn |] represents the * vector (x1, ..., xn) * RI: the array is non-empty *) type vector = float array The [Euclidean norm][norm] of an \$$n\$$-dimensional vector \$$x = (x_1, \ldots, x_n)\$$ is written \$$|x|\$$ and is defined to be $$\sqrt{x_1^2 + \cdots + x_n^2}.$$ [norm]: https://en.wikipedia.org/wiki/Norm_(mathematics)#Euclidean_norm ##### Exercise: norm [✭✭] Write a function norm : vector -> float that computes the Euclidean norm of a vector. Your function should not mutate the input array. *Hint: although your first instinct is likely to reach for a loop, instead try to use Array.map and Array.fold_left or Array.fold_right.* □ Every vector can be *normalized* by dividing each component by \$$|x|\$$; this yields a vector with norm 1: \$\left(\frac{x_1}{|x|}, \ldots, \frac{x_n}{|x|}\right)\$ ##### Exercise: normalize [✭✭] Write a function normalize : vector -> unit that normalizes a vector "in place" by mutating the input array. Here's a sample usage: # let a = [|1.; 1.|];; val a : float array = [|1.; 1.|] # normalize a;; - : unit = () # a;; - : float array = [|0.7071...; 0.7071...|] *Hint: Array.iteri.* □ ## Loops When working with arrays, it is often convenient to write traditional for or while loops. The "Loops" paragraph of [section 6.7.2 "Control Structures"][manual-control] in the OCaml manual contains the syntax and semantics of loops. Here is a brief summary: [manual-control]: http://caml.inria.fr/pub/docs/manual-ocaml/expr.html#sec124 while e1 do e done for x = e1 to e2 do e done for x = e1 downto e2 do e done Each of these three expressions evaluates the expression between do and done for each iteration of the loop; while loops terminate when e1 becomes false; for loops execute once for each integer from e1 to e2; for..to loops evaluate starting at e1 and incrementing x each iteration; for..downto loops evaluate starting at e1 and decrementing x each iteration. All three expressions evaluate to () after the termination of the loop. Because they always evaluate to (), they are less general than folds, maps, or recursive functions. Loops are themselves not inherently mutable, but they are most often used in conjunction with mutable features like arrays—typically, e causes side effects. ##### Exercise: normalize loop [✭✭] Modify your implementation of normalize to use one of the looping expressions. □ ## Additional exercises ##### Exercise: norm loop [✭✭] Modify your implementation of norm to use one of the looping expressions. Here is pseudocode for what you should do: initialize norm to 0.0 loop through array add to norm the square of the current array component return norm □ ##### Exercise: imperative factorial [✭✭] Write a factorial function that uses a loop expression and refs to compute the factorial of its input. □ Here's a neat trick that's possible with refs: we can build recursive functions without ever using the keyword rec. Suppose we want to define a recursive function such as fact: let rec fact n = if n = 0 then 1 else n * fact (n-1) Abstracting a little, that function has the following form: let rec f x = ... some code including a recursive call [f y] from some argument [y] ... We can instead write the following: let f0 = ref (fun x -> x) let f x = ... replace [f y] with [!f0 y] ... f0 := f Now f will compute the same result as it did in the version where we defined it with rec. What's happening here is sometimes called "tying the recursive knot": we update the reference to f0 to point to f, such that when f dereferences f0, it gets itself back! The initial function to which we made f0 point (in this case the identity function) doesn't really matter; it's just there as a placeholder until we tie the knot. ##### Exercise: tie the knot [✭✭✭] Define the factorial function without using any loops and without the rec keyword. □ ##### Exercise: init matrix [✭✭✭] The array module contains two functions for creating an array: make and init. make creates an array and fills it with a default value, while init creates an array and uses a provided function to fill it in. The library also contains a function make_matrix for creating a two-dimensional array, but it does not contain an analogous init_matrix to create a matrix using a function for initialization. Write a function init_matrix : int -> int -> (int -> int -> 'a) -> 'a array array such that init_matrix n o f creates and returns an n by o matrix m with m.(i).(j) = f i j for all i and j in bounds. See the documentation for make_matrix for more information on the representation of matrices as arrays. □ ## Challenge exercise: Doubly-linked lists Consider the following types and functions for mutable *[doubly-linked lists][dll]*: [dll]: https://en.wikipedia.org/wiki/Doubly_linked_list (* An ['a node] is a node of a mutable doubly-linked list. * It contains a value of type ['a] and optionally has * pointers to previous and/or next nodes. *) type 'a node = { mutable prev : 'a node option; mutable next : 'a node option; value : 'a } (* An ['a dlist] is a mutable doubly-linked list with elements * of type ['a]. It is possible to access the first and * last elements in constant time. * RI: The list does not contain any cycles. *) type 'a dlist = { mutable first : 'a node option; mutable last : 'a node option; } (* [create_node v] is a node containing value [v] with * no links to other nodes. *) let create_node v = {prev=None; next=None; value=v} (* [empty_dlist ()] is an empty doubly-linked list. *) let empty_dlist () = {first=None; last=None} (* [create_dlist n] is a doubly-linked list containing * exactly one node, [n]. *) let create_dlist (n: 'a node) : 'a dlist = {first=Some n; last=Some n} (* [insert_first d n] mutates dlist [d] by * inserting node [n] as the first node. *) let insert_first (d: 'a dlist) (n: 'a node) : unit = failwith "unimplemented" (* [insert_last d n] mutates dlist [d] by * inserting node [n] as the last node. *) let insert_last (d: 'a dlist) (n: 'a node) : unit = failwith "unimplemented" (* [insert_after d n1 n2] mutates dlist [d] by * inserting node [n2] after node [n1]. *) let insert_after (d: 'a dlist) (n1: 'a node) (n2: 'a node) : unit = failwith "unimplemented" (* [insert_before d n1 n2] mutates dlist [d] by * inserting node [n2] before node [n1]. *) let insert_before (d: 'a dlist) (n1: 'a node) (n2: 'a node) : unit = failwith "unimplemented" (* [remove d n] mutates dlist [d] by removing node [n]. * requires: [n] is a node of [d]. *) let remove (d: 'a dlist) (n: 'a node) : unit = failwith "unimplemented" (* [iter_forward d f] on a dlist [d] which has * elements n1; n2; ... is (f n1); (f n2); ... *) let iter_forward (d: 'a dlist) (f: 'a -> unit) : unit = failwith "unimplemented" (* [iter_backward d f] on a dlist [d] which has * elements n1; n2; ... is ...; (f n2); (f n1) *) let iter_backward (d: 'a dlist) (f: 'a -> unit) : unit = failwith "unimplemented" ##### Exercise: doubly linked list [✭✭✭✭] Complete the implementations above. Test your code. *Hint: draw pictures! Reasoning about mutable data structures is typically easier if you draw a picture.* □
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1. Math
2. Algebra
3. solve using the row echelon form...
# Question: solve using the row echelon form...
###### Question details
Solve using the row echelon form
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# Problems On Areas Set-6
Go back to 'Areas and Volumes'
## Problems on areas examples
Example – 11
Find the area of the region bounded by the curves
${\rm{y = }}{{\rm{x}}^2},{\rm{y = }}\left| {{\rm{2 - }}{{\rm{x}}^2}} \right|{\rm{and\,y = 2}},$
which lies to the right of the line x = 1.
Solution: The required area is sketched below
The required area can be evaluated as follows:
\begin{align}&{\rm{A = }}\int\limits_1^{\sqrt 2 } {({x^2} - (2 - {x^2}))dx + \int\limits_{\sqrt 2 }^2 {(2 - ({x^2} - 2))dx} } \\&\;\; = \left. {\left( {\frac{{2{x^3}}}{3} - 2x} \right)} \right|_1^{\sqrt 2 } + \left. {\left( {4x - \frac{{{x^3}}}{3}} \right)} \right|_{\sqrt 2 }^2\\& \;\;= \frac{{20 - 12\sqrt 2 }}{3}sq.\rm{units} \qquad{\rm{(\,verify\;the\;calculations)}} \end{align}
Example – 12
A curve $$y = f\left( x \right)$$ passes through the point P (1, 1) . The normal to the curve at P is $$a(y - 1) + (x - 1) = 0$$ . If the slope of the tangent at any point on the curve is proportional to the ordinate of that point, determine the equation of the curve. Hence obtain the area bounded by the y-axis, the curve and the normal to the curve at P.
Solution: The slope of the given normal is obvious from the expression:
\begin{align}&\frac{{y - 1}}{{x - 1}} = - \frac{1}{a}\\&\Rightarrow {m_N} = - \frac{1}{a}\\&\Rightarrow {m_T} = a\\&\Rightarrow {\left. {\left( {\frac{{dy}}{{dx}}} \right)} \right|_p} = a\end{align}
\begin{align}\text{It is given that}\; \frac{{dy}}{{dx}}\propto y \Rightarrow \frac{{dy}}{{dx}} = ky\end{align}
\begin{align}\text {Since}\qquad {\left. {\frac{{dy}}{{dx}}} \right|_{(1,1)}} = a \Rightarrow k = a\end{align}
Thus,
\begin{align}&\frac{{dy}}{{dx}} = ay\\&\Rightarrow \frac{{dy}}{y} = adx\\&\Rightarrow \ln y = ax + {\mathop{\rm lnC}\nolimits} \qquad {\rm{(we\;took\; the\;constant\;of\;integration\;as\;ln\;C\;instead\;}}\\&\qquad \qquad \qquad\qquad\qquad\quad{\rm{of\; C\;so\;that\;the\;final \;expression\;for\;y\;is\;simpler)}}\\&\Rightarrow y = C{e^{ax}}\end{align}
Since this curve passes through (1, 1),
$\begin{array}{l}1 = C{e^{a.1}}\\ \Rightarrow C = {e^{ - a}}\end{array}$
Thus, the equation of the curve is
$y = {e^{a(x - 1)}}$
Let us now proceed to evaluate the bounded area, which is sketched below:
The equation of the normal has already been provided:
$y = 1 + \frac{1}{a}(1 - x)$
Thus, the required area is:
\begin{align} A &= \int\limits_0^1 {\left\{ {\left( {1 + \frac{1}{a}\left( {1 - x} \right)} \right) - {e^{a(x - 1)}}} \right\}} dx\\&= \left. {\left( {x + \frac{x}{a} - \frac{{{x^2}}}{{2a}} - \frac{{{e^{a(x - 1)}}}}{a}} \right)} \right|_0^1\\&= \left( {1 + \frac{1}{a} - \frac{1}{{2a}} - \frac{1}{a}} \right) - \left( { - \frac{{{e^{ - a}}}}{a}} \right)\\&= \left( {1 - \frac{1}{{2a}} + \frac{1}{{a{e^a}}}} \right){\rm{sq}}{\rm{. units}}\end{align}
Areas Under Curves
grade 11 | Questions Set 2
Areas Under Curves
Areas Under Curves
grade 11 | Questions Set 1
Areas Under Curves
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on 22-Sep-2019 (Sun)
Annotation 4403302632716
The problems t h at ha ve been addressed by AI search algorithms fall into three ge neral classes: single-agent pathfinding problems , two-player games, and constraint-satisfaction problems.
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Flashcard 4403492424972
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#artificial-intelligence #reinforcement-learning
Question
What two steps does DeepCubeA use to solve Rubik's cubes?
1. Neural network is trained to learn the cost-to-go function using Autodidactic Iteration (Uniformly sample scramble depth k from 1 to K, apply this many random scrambles to the cube).
2. Batch Weighted A* Search is used to actually solve cubes. This is weighted A* search, plus expanding the children of the N nodes with lowest predicted cost (to use computation efficiently)
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Flashcard 4403961138444
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#artificial-intelligence
Question
What is the point in using Weighted A* over A*?
We often have that the heuristic h drastically underestimates the true distance, so that we can obtain a more realistic estimate by scaling up its influence with regard to some parameter. Although this compromises optimality, it can lead to a significant speedup; this is an appropriate choice when searching under time or space constraints.
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Flashcard 4404258671884
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#artificial-intelligence
Question
If we parameterize [...] with $$l\in[0,1]$$, we obtain a continuous range of best-first search variants $$A_l$$, also denoted as weighted A*.
$$f_l(u) = l\cdot h(u) + (1-l)\cdot g(u)$$
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Flashcard 4408992992524
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#artificial-intelligence
Question
If we parameterize $$f_l(u) = l\cdot h(u) + (1-l)\cdot g(u)$$ with $$l\in[0,1]$$, we obtain a continuous range of best-first search variants $$A_l$$, also denoted as weighted A*. For l = 0, we simulate a [...] traversal of the problem space; for l = 1, we have [...] . Algorithm $$A_{0.5}$$ selects nodes in the same order as original A*.
greedy best-first search
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Flashcard 4409493687564
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#artificial-intelligence
Question
What is an admissable heuristic function?
Definition 1.8. (Admissible Heuristic) An estimate h is an admissible heuristic if it is a lower bound for the optimal solution costs; that is, h(u) ≤ δ(u,T) for all u ∈ V
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Flashcard 4409499716876
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#artificial-intelligence
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When is a heuristic function consistent?
A goal estimate h is a consistent heuristic if h(u) ≤ h(v) +w(u,v) for all edges e = (u, v) ∈ E
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Flashcard 4409502076172
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#artificial-intelligence
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When is a heuristic function monotone?
Let ($$u_0$$,. . . , $$u_k$$) be any path, g($$u_i$$) be the path cost of ($$u_0$$,. . . , $$u_i$$), and define f($$u_i$$) = g($$u_i$$) +h($$u_i$$). A goal estimate h is a monotone heuristic if f ($$u_j$$) ≥ f ($$u_i$$) for all j > i, 0 ≤ i, j ≤ k; that is, the estimate of the total path cost is nondecreasing from a node to its successors.
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Flashcard 4409505221900
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#artificial-intelligence
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What is the relation of consistent and monotone heuristic functions?
Theorem 1.1. (Equivalence of Consistent and Monotone Heuristics) A heuristic is consistent if and only if it is monotone.
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Flashcard 4409507581196
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#artificial-intelligence
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What is the relation between consistent and admissable heuristics?
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Flashcard 4409509940492
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#artificial-intelligence
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Under a consistent heuristic, when is weighted A* search monotone?
Lemma 6.1. For l ≤ 0.5 and a consistent estimate h, f l is monotone.
(with $$f_l(u) = l\cdot h(u) + (1-l)\cdot g(u)$$ for $$l\in[0,1]$$)
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#artificial-intelligence
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Define $$\epsilon$$-optimality of a search algorithm.
Definition 6.1. ( $$\epsilon$$-Optimality) A search algorithm is $$\epsilon$$-optimal if it terminates with a solution of maximum cost (1 + $$\epsilon$$) ·δ(s,T), with $$\epsilon$$ denoting an arbitrary small positive constant (s is the starting state, T is a goal state).
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Annotation 4409602477324
#artificial-intelligence #reinforcement-learning Monte-Carlo evaluation consists in estimating a position by averaging the outcome of several random continuations, and can serve as an evaluation function at the leaves of a min-max tree. This paper presents a new framework to combine tree search with Monte-Carlo eval- uation, that does not separate between a min-max phase and a Monte- Carlo phase. Instead of backing-up the min-max value close to the root, and the average value at some depth, a more general backup operator is defined that progressively changes from averaging to min-max as the number of simulations grows. This approach provides a fine-grained con- trol of the tree growth, at the level of individual simulations, and allows efficient selectivity methods. This algorithm was implemented in a 9 × 9 Go-playing program, Crazy Stone, that won the 10th KGS computer-Go tournament.
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Annotation 4409605360908
From a helicopter view Monte Carlo Tree Search has one main purpose: given a game state to choose the most promising next move. Throughout the rest of this post we will try to take a look at the details of Monte Carlo Tree Search to understand what we mean by that. We will also refer back to Alpha Go/Zero from time to time trying to explain what MCTS variant has been used in Deepmind’s AI.
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CPL - Chalmers Publication Library
Effects of parallel ion motion on electromagnetic toroidal ion temperature gradient modes in a fluid model
Anders Jarmén (Institutionen för rymd- och geovetenskap, Plasmafysik och fusionsenergi) ; Johan Anderson (Institutionen för rymd- och geovetenskap, Plasmafysik och fusionsenergi) ; Plamen Malinov
Physics of Plasmas (1070-664X). Vol. 22 (2015), 8, p. 082508.
Effects of ion dynamics along the background magnetic field have been added to an advanced fluid model which has been developed, tested, and successfully used in transport code applications during the last decades. Introducing electrostatic ($\phi$) and electromagnetic potentials ($\psi$), a system of two coupled second order differential equations in these potentials is derived. The mode solution is interpreted as a coupling between an Ion Temperature Gradient (ITG) mode and an ion motion driven acoustic wave. The mode may be stabilized by electromagnetic effects and by minimizing the ITG parameter ($\eta_i = L_n/L_{Ti}$). Interestingly, the addition of kinetic Landau resonance effects may enhance the $\eta_i$ stabilization.
Nyckelord: Drift waves, ITG, electromagnetic effects
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## p-value for Z-test
If the test statistic $Z$ has standard normal distribution and the observed value of the test statistic is $Z_{obs}$, then the $p$-value of the test for testing
a. left-tailed alternative is $p$-value = $P(Z\leq Z_{obs})$.
b. right-tailed alternative is $p$-value = $P(Z\geq Z_{obs})$.
c. two-tailed alternative is $p$-value = $2P(Z\geq |Z_{obs}|)$.
## Formula
The $p$-value of Z-test for left tailed alternative is
### $p$-value = $P(Z\leq Z_{obs})$
The $p$-value of Z-test for right tailed alternative is
### $p$-value = $P(Z\geq Z_{obs})$
The $p$-value of Z-test for two tailed alternative is
### $p$-value = $2P(Z\geq |Z_{obs}|)$
where,
• $Z_{obs}$ is the value of test statistic
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# Incorporating noise into machine learning models?
Usually, in machine learning textbooks the $$X$$ dataset and the target $$y$$ are defined with exact values.
How about the case if the values of both $$X$$ and $$y$$ have noises: for instance, we only know that $$0.5 <= x_1 <= 0.63$$ but not the precise value. How could I integrate the information into the model?
• If you consider Linear Regression to be machine learning (which I think you should), then you can check out Errors-in-Variables Models – klumbard Nov 5 '18 at 19:31
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CARMA SEMINAR Speaker: George Havas, The University of Queensland Title: Group theoretic proofs by coset enumeration Location: Room V205, Mathematics Building (Callaghan Campus) The University of Newcastle Time and Date: 3:00 pm, Wed, 23rd Sep 2015 Abstract: Given a finite presentation of a group, proving properties of the group can be difficult. Indeed, many questions about finitely presented groups are unsolvable in general. Algorithms exist for answering some questions while for other questions algorithms exist for verifying the truth of positive answers. An important tool in this regard is the Todd-Coxeter coset enumeration procedure. It is possible to extract formal proofs from the internal working of coset enumerations. We give examples of how this works, and show how the proofs produced can be mechanically verified and how they can be converted to alternative forms. We discuss these automatically produced proofs in terms of their size and the insights they offer. We compare them to hand proofs and to the simplest possible proofs. We point out that this technique has been used to help solve a longstanding conjecture about an infinite class of finitely presented groups. [Permanent link]
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# 1. Draw a picture to show that $\displaystyle \sum_{n = 2}^{\infty} \frac {1}{n^{1.3}} < \int^{\infty}_1 \frac {1}{x^{1.3}} dx$What can you conclude about the series?
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### Video Transcript
to demonstrate the desired a comparison. We'LL just use a rough sketch Graff for one over X to the one point three power look something a little bit like this. Probably a little bit steeper, but the sake of this is not too important now. If we mark, this is one he's on. When we were talking about our syriza's, they're all going to do with one rectangles, starting with the height at, too. Since we're sort of equals two. That would be a rectangle. Something like this. You have another rectangle here and so on. However, the integral itself is the entire area underneath the curve, which starting at one, means we would include this area in addition to that of the rectangles. So we can see right away that free choice of Interval. There's a little bit, too gets missed bythe Siri's that the integral itself picks up so clearly. The area of the rectangles for Siri's is less in the area of the integral itself. Now, as far as what we can conclude about the serious based on this information, we can see if we actually evaluate that indefinitely. Nero Chrissy, if we take this integral Ah, using, of course. Ah limit. Since this is an improperly, necro will eventually see that this integral does in fact converge. Ah, you know, we do a little bit of magic and we get negative ten over three times, one over T to the one point three plus ten over three, which has t goes to infinity. This term becomes zero and we're left with just ten over three. Since the integral itself converges and thie, Siri's is less than that of the interview has a lower value than the integral. The Syria's itself must converge.
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# EEMIAN DISTRIBUTION OF NEANDERTHALS
Abstract
Aim: We hypothesize that the northern and southern edges of Neanderthals distribution during the Last Interglacial were respectively limited by low winter temperatures, and high temperatures and lack of rainfall during the summer, while high topographic diversity combined with moderate slopes favored presence at the local scale.
Location: Europe (20ºN to 70ºN, 10ºW to 70ºE).
Methods: We used Neanderthal presence, palaeoclimatic, and topographic data to calibrate a species distribution model. We analyzed variable importance at the continental scale with Randon Forest, and at the local scale with local regression and recursive partition trees.
Results: Highest habitat suitability was observed along the Mediterranean coast. Main mountain ranges and continental plains showed low habitat suitability values. Extreme seasonal temperatures and annual rainfall were the most important predictors at the continental scale, while topography defined habitat suitability at the local scale.
Main conclusions: Our results challenge the notion of Neanderthals as a central European species. Therefore, many current interpretations of Neanderthal livelihood during the Last Interglacial may not accurately represent this species preferred habitat.
Keywords: palaeo-species distribution modeling, hominins, habitat suitability, ecological niche
# Introduction
Our knowledge about Neanderthals has greatly increased over the last two decades with more than two thousand research papers published in areas as diverse as chronology (Gaudzinski-Windheuser 2014, Higham 2014), ecology (Finlayson 2007, Henry 2010), population dynamics (Sørensen 2011, Bocquet-Appel 2013), adaptive traits (Sørensen 2009, Rae 2011), diet (Henry 2010, Hardy 2011), technology, cognition and behavior (Shaw 2012, Peris 2012, Riel-Salvatore 2010), genetics (Briggs 2009, Sanchez-Quinto 2014), and their relationship with anatomically modern humans (Hortolà 2013, Sankararaman 2012). Still, we believe there is a lack of tangible information about the factors controlling the Neanderthals’ distribution. For example, Wenzel (2007) describes the distribution of Neanderthals during MIS 5e across central Europe, and briefly discusses the influence of climatic conditions and habitat features over the general occupancy pattern, but does not establish any quantitative link between presence and environment. Banks et al. (2008) analyzed the distribution of Neanderthals at the interphase of the Neanderthal/modern human transition applying palaeo species distribution models (PSDMs hereafter), assessed niche conservatism, and analyzed the importance of climatic predictors, concluding that temperature was the most important driver shaping the distribution Neanderthal at this time.
In this paper we propose a hypothesis about how abiotic drivers may have shaped Neanderthals distribution that is rooted in the Grinnellian niche concept (Grinnell 1917), the hierarchical framework proposed by Pearson et al. (2003), and our current knowledge on the ecology of Neanderthals. According to Pearson et al. (2003), climate influences species distribution at global and continental scales, while the effect of topography is restricted to scales ranging from regional to local. Our hypothesis consist of three main points: 1) the northern edge of the Neanderthals range was limited by low winter temperatures; 2) the southern edge was shaped by a combination of high temperature and low water availability during the summer; 3) high topographic diversity combined with moderate slopes could have favored occupation at the local scale. Winter temperatures at the northern edge could have resulted in a lower availability of small and big game than in temperate areas (Badgley 2000), compromising the high caloric intake required by this species (Steegmann 2002, Sørensen 2009), and an increased cold stress accelerated by a low-caloric diet, that would have lead to reduced fertility (Bocquet-Appel 2013) and a higher mortality rate (Steegmann 2002). Summer temperatures linked to continentality and the higher solar radiation of lower latitudes could have prevented the occupation of southern plains in the Mediterranean peninsulas due to increased heat stress, specially considering the low body surface area/volume ratio of this species (Churchill 2006), which hampers heat dissipation. Also, high evapotranspiration could have reduced the diversity of plants, since under-storey and forest communities are less common under such climatic conditions, hampering the access to plant resources, and making these areas unsuitable during the warm season. Mediterranean coastal areas could have been suitable because of the buffering effect of the sea over temperature and the permanent availability of resources like shellfish (Hardy 2011). At the local scale, high topographic diversity, which fosters biodiversity by an increased availability of ecological niches (Tews 2003), could have provided the required abundance and diversity of prey and shelter (Daujeard 2012), but moderate to low slopes may have been important to reduce the high energetic expenditure of mobility in steep terrain. This limitation, combined with the lower abundance of animals at higher elevations (Brown 2001), could have prevented Neanderthal’s presence in the higher elevations of the European mountain ranges.
To test our hypothesis, we have modeled and analyzed the distribution of Neanderthals during the full Eemian Interglacial (MIS 5e, $$\sim$$130 ka BP) using a PDSM approach (Franklin 2015, Svenning 2011, Varela 2011). PSDMs rely on the same principles of species distribution modeling (Guisan 2000, Guisan 2005), requiring presence coordinates coming from fossil/archaeological remains, a set of predictors (usually palaeo-climatic simulations), and an algorithm to define the relationship between presence and predictors. The result of a PSDM can be defined as a habitat suitability map in which each cell is scored according to how well it resembles the ecological conditions of the localities where the species occurs (Soberon 2005). PSDMs can be used to analyze the influence of particular drivers over the species distribution and gain ecological knowledge, as in Varela et al. (2009) or Rodríguez-Sánchez et al. (2008). PSDMs have been used before to study the distribution of Neanderthals and anatomically modern humans (Banks 2008, Banks 2008a, Beeton 2013, Burke 2014), but in this paper we add to the previous studies by offering a novel insight into the drivers of Neanderthals distribution both at continental and local scales.
We selected the Eemian Interglacial (MIS 5e, $$\sim$$130 ka BP) as modeling period for three different reasons: 1) the distribution of Neanderthals during this period has been previously assessed by i.e. Richter (2005), Richter (2006), Wenzel (2007), and Gaudzinski-Windheuser et al. (2011), providing a good baseline knowledge, but with these studies mainly focus on the core and northern edge of the distribution area across central Europe, and lack a quantitative description of Neanderthals distribution and its limiting factors; 2) since the publication of Wenzel (2007), new Neanderthal remains attributed to the Eemian have been found in Spain (Arsuaga 2012), Italy (Fiorenza 2015), France (Moncel 2007, Daujeard 2010), and these new sites have the potential to change our view about the Eemian distribution of Neanderthals; 3) During the Eemian, the warmer climatic conditions freed Europe from the Saalian ice sheet, offering the Neanderthals a unique opportunity to spread throughout Europe for c. 10000 years, and probably allowing them to reach their maximum range size. The Eemian can therefore be considered the most suitable period to assume a pseudo-equilibrium with climate for Neanderthals distribution, a key assumption for PSDMs (Guisan 2005, Guisan 2000).
In summary, in this paper we propose a hypothesis about how abiotic drivers (climate and topography) could have shaped Neanderthals distribution throughout Europe, and we test it by fitting and analyzing a PSDM describing their distribution during the Last Interglacial period.
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## Description
General purpose section of an Euler-Bernoulli beam in 3D, not assuming homogeneous density or homogeneous elasticity, given basic material properties.
This is the case where one uses a FEA preprocessor to compute the rigidity of a complex beam made with multi-layered reinforcements with different elasticity and different density - in such a case you could not use ChBeamSectionEulerAdvanced because you do not have a single E or single density, but you rather have collective values of bending rigidities, and collective mass per unit length. This class allows using these values directly, bypassing any knowledge of area, density, Izz Iyy, E young modulus, etc. To be used with ChElementBeamEuler. The center of mass of the section can have an offset respect to the centerline. This material can be shared between multiple beams.
#include <ChBeamSectionEuler.h>
## Public Member Functions
ChBeamSectionEulerAdvancedGeneric (const double mAx, const double mTxx, const double mByy, const double mBzz, const double malpha, const double mCy, const double mCz, const double mSy, const double mSz, const double mmu, const double mJxx, const double mMy=0, const double mMz=0)
virtual void SetAxialRigidity (const double mv)
Sets the axial rigidity, usually A*E for uniform elasticity, but for nonuniform elasticity here you can put a value ad-hoc from a preprocessor.
virtual void SetXtorsionRigidity (const double mv)
Sets the torsion rigidity, for torsion about X axis, at elastic center, usually J*G for uniform elasticity, but for nonuniform elasticity here you can put a value ad-hoc from a preprocessor.
virtual void SetYbendingRigidity (const double mv)
Sets the bending rigidity, for bending about Y axis, at elastic center, usually Iyy*E for uniform elasticity, but for nonuniform elasticity here you can put a value ad-hoc from a preprocessor.
virtual void SetZbendingRigidity (const double mv)
Sets the bending rigidity, for bending about Z axis, at elastic center, usually Izz*E for uniform elasticity, but for nonuniform elasticity here you can put a value ad-hoc from a preprocessor.
virtual void SetSectionRotation (const double mv)
Set the rotation in [rad], abour elastic center, of the Y Z axes for which the YbendingRigidity and ZbendingRigidity values are defined.
virtual void SetCentroidY (const double mv)
Sets the Y position of the elastic center respect to centerline.
virtual void SetCentroidZ (const double mv)
Sets the Z position of the elastic center respect to centerline.
virtual void SetShearCenterY (const double mv)
Sets the Y position of the shear center respect to centerline.
virtual void SetShearCenterZ (const double mv)
Sets the Z position of the shear center respect to centerline.
virtual void SetMassPerUnitLength (const double mv)
Set mass per unit length, ex.SI units [kg/m] For uniform density it would be A*density, but for nonuniform density here you can put a value ad-hoc from a preprocessor.
virtual void SetInertiaJxxPerUnitLength (const double mv)
Set the Jxx component of the inertia per unit length (polar inertia), computed at centerline. More...
virtual void SetInertiaJxxPerUnitLengthInMassReference (const double mv)
Set inertia moment per unit length Jxx_massref, as assumed computed in the "mass reference" frame, ie. More...
virtual double GetInertiaJxxPerUnitLengthInMassReference ()
Get inertia moment per unit length Jxx_massref, as assumed computed in the "mass reference" frame, ie. More...
void SetCenterOfMass (double my, double mz)
"mass reference": set the displacement of the center of mass respect to the section centerline reference.
double GetCenterOfMassY ()
double GetCenterOfMassZ ()
virtual double GetAxialRigidity () const override
Gets the axial rigidity, usually A*E, but might be ad hoc.
virtual double GetXtorsionRigidity () const override
Gets the torsion rigidity, for torsion about X axis at elastic center, usually J*G, but might be ad hoc.
virtual double GetYbendingRigidity () const override
Gets the bending rigidity, for bending about Y axis at elastic center, usually Iyy*E, but might be ad hoc.
virtual double GetZbendingRigidity () const override
Gets the bending rigidity, for bending about Z axis at elastic center, usually Izz*E, but might be ad hoc.
virtual double GetSectionRotation () const override
Set the rotation of the Y Z section axes for which the YbendingRigidity and ZbendingRigidity are defined.
virtual double GetCentroidY () const override
Gets the Y position of the elastic center respect to centerline.
virtual double GetCentroidZ () const override
Gets the Z position of the elastic center respect to centerline.
virtual double GetShearCenterY () const override
Gets the Y position of the shear center respect to centerline.
virtual double GetShearCenterZ () const override
Gets the Z position of the shear center respect to centerline.
virtual void ComputeInertiaMatrix (ChMatrixNM< double, 6, 6 > &M) override
Compute the 6x6 sectional inertia matrix, as in {x_momentum,w_momentum}=[Mm]{xvel,wvel}.
virtual void ComputeInertiaDampingMatrix (ChMatrixNM< double, 6, 6 > &Ri, const ChVector<> &mW) override
Compute the 6x6 sectional inertia damping matrix [Ri] (gyroscopic matrix damping) More...
virtual void ComputeInertiaStiffnessMatrix (ChMatrixNM< double, 6, 6 > &Ki, const ChVector<> &mWvel, const ChVector<> &mWacc, const ChVector<> &mXacc) override
Compute the 6x6 sectional inertia stiffness matrix [Ki^]. More...
virtual void ComputeQuadraticTerms (ChVector<> &mF, ChVector<> &mT, const ChVector<> &mW) override
Compute the centrifugal term and gyroscopic term. More...
virtual double GetMassPerUnitLength () const override
Get mass per unit length, ex.SI units [kg/m].
virtual double GetInertiaJxxPerUnitLength () const override
Get the Jxx component of the inertia per unit length (polar inertia), at centerline.
Public Member Functions inherited from chrono::fea::ChBeamSectionEuler
virtual void ComputeInertialForce (ChVector<> &mFi, ChVector<> &mTi, const ChVector<> &mWvel, const ChVector<> &mWacc, const ChVector<> &mXacc)
Compute the total inertial forces (per unit length). More...
void SetArtificialJyyJzzFactor (double mf)
The Euler beam model has no rotational inertia per each section, assuming mass is concentrated on the centerline. More...
double GetArtificialJyyJzzFactor ()
virtual void SetBeamRaleyghDampingAlpha (double malpha)
Set the "alpha" Rayleigh damping ratio,
the mass-proportional structural damping in: R = alpha*M + beta*K
double GetBeamRaleyghDampingAlpha ()
virtual void SetBeamRaleyghDampingBeta (double mbeta)
Set the "beta" Rayleigh damping ratio, the stiffness-proportional structural damping in: R = alpha*M + beta*K
double GetBeamRaleyghDampingBeta ()
virtual void SetBeamRaleyghDamping (double mbeta, double malpha=0)
Set both beta and alpha coefficients in Rayleigh damping model: R = alpha*M + beta*K. More...
Public Member Functions inherited from chrono::fea::ChBeamSection
void SetDrawShape (std::shared_ptr< ChBeamSectionShape > mshape)
Set the graphical representation for this section. More...
std::shared_ptr< ChBeamSectionShapeGetDrawShape () const
Get the drawing shape of this section (i.e.a 2D profile used for drawing 3D tesselation and visualization) By default a thin square section, use SetDrawShape() to change it.
void SetDrawThickness (double thickness_y, double thickness_z)
Shortcut: adds a ChBeamSectionShapeRectangular for visualization as a centered rectangular beam, and sets its width/height. More...
Shortcut: adds a ChBeamSectionShapeCircular for visualization as a centered circular beam, and sets its radius. More...
void SetCircular (bool ic)
OBSOLETE only for backward compability
## Protected Attributes
double Ax
double Txx
double Byy
double Bzz
double alpha
double Cy
double Cz
double Sy
double Sz
double mu
double Jxx
double My
double Mz
Protected Attributes inherited from chrono::fea::ChBeamSectionEuler
double rdamping_beta
double rdamping_alpha
double JzzJyy_factor
Public Attributes inherited from chrono::fea::ChBeamSectionEuler
bool compute_inertia_damping_matrix = true
Flag that turns on/off the computation of the [Ri] 'gyroscopic' inertial damping matrix. More...
bool compute_inertia_stiffness_matrix = true
Flag that turns on/off the computation of the [Ki] inertial stiffness matrix. More...
bool compute_Ri_Ki_by_num_diff = false
Flag for computing the Ri and Ki matrices via numerical differentiation even if an analytical expression is provided. More...
## Constructor & Destructor Documentation
chrono::fea::ChBeamSectionEulerAdvancedGeneric::ChBeamSectionEulerAdvancedGeneric ( const double mAx, const double mTxx, const double mByy, const double mBzz, const double malpha, const double mCy, const double mCz, const double mSy, const double mSz, const double mmu, const double mJxx, const double mMy = 0, const double mMz = 0 )
inline
Parameters
mAx axial rigidity mTxx torsion rigidity mByy bending regidity about yy mBzz bending rigidity about zz malpha section rotation about elastic center [rad] mCy elastic center y displacement respect to centerline mCz elastic center z displacement respect to centerline mSy shear center y displacement respect to centerline mSz shear center z displacement respect to centerline mmu mass per unit length mJxx polar inertia Jxx per unit lenght, measured respect to centerline mMy mass center y displacement respect to centerline mMz mass center z displacement respect to centerline
## Member Function Documentation
void chrono::fea::ChBeamSectionEulerAdvancedGeneric::ComputeInertiaDampingMatrix ( ChMatrixNM< double, 6, 6 > & Ri, const ChVector<> & mW )
overridevirtual
Compute the 6x6 sectional inertia damping matrix [Ri] (gyroscopic matrix damping)
Parameters
Ri 6x6 sectional inertial-damping (gyroscopic damping) matrix values here mW current angular velocity of section, in material frame
Reimplemented from chrono::fea::ChBeamSectionEuler.
## ◆ ComputeInertiaStiffnessMatrix()
void chrono::fea::ChBeamSectionEulerAdvancedGeneric::ComputeInertiaStiffnessMatrix ( ChMatrixNM< double, 6, 6 > & Ki, const ChVector<> & mWvel, const ChVector<> & mWacc, const ChVector<> & mXacc )
overridevirtual
Compute the 6x6 sectional inertia stiffness matrix [Ki^].
Parameters
Ki 6x6 sectional inertial-stiffness matrix [Ki^] values here mWvel current angular velocity of section, in material frame mWacc current angular acceleration of section, in material frame mXacc current acceleration of section, in material frame (not absolute!)
Reimplemented from chrono::fea::ChBeamSectionEuler.
void chrono::fea::ChBeamSectionEulerAdvancedGeneric::ComputeQuadraticTerms ( ChVector<> & mF, ChVector<> & mT, const ChVector<> & mW )
overridevirtual
Compute the centrifugal term and gyroscopic term.
Parameters
mF centrifugal term (if any) returned here mT gyroscopic term returned here mW current angular velocity of section, in material frame
Implements chrono::fea::ChBeamSectionEuler.
## ◆ GetInertiaJxxPerUnitLengthInMassReference()
inlinevirtual
Get inertia moment per unit length Jxx_massref, as assumed computed in the "mass reference" frame, ie.
centered at the center of mass
## ◆ SetInertiaJxxPerUnitLength()
virtual void chrono::fea::ChBeamSectionEulerAdvancedGeneric::SetInertiaJxxPerUnitLength ( const double mv )
inlinevirtual
Set the Jxx component of the inertia per unit length (polar inertia), computed at centerline.
For uniform density it would be Ixx*density or, by polar theorem, (Izz+Iyy)*density, but for nonuniform density here you can put a value ad-hoc from a preprocessor
## ◆ SetInertiaJxxPerUnitLengthInMassReference()
virtual void chrono::fea::ChBeamSectionEulerAdvancedGeneric::SetInertiaJxxPerUnitLengthInMassReference ( const double mv )
inlinevirtual
Set inertia moment per unit length Jxx_massref, as assumed computed in the "mass reference" frame, ie.
centered at the center of mass. Call this after you set SetCenterOfMass() and SetMassPerUnitLength()
The documentation for this class was generated from the following files:
• /builds/uwsbel/chrono/src/chrono/fea/ChBeamSectionEuler.h
• /builds/uwsbel/chrono/src/chrono/fea/ChBeamSectionEuler.cpp
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• #### Quasi-Equilibria and Click Times for a Variant of Muller's Ratchet
[OWP-2022-18] (Mathematisches Forschungsinstitut Oberwolfach, 2022-11-30)
Consider a population of $N$ individuals, each of them carrying a type in $\mathbb N_0$. The population evolves according to a Moran dynamics with selection and mutation, where an individual of type $k$ has the same selective ...
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2 added 93 characters in body
Hi
I have a function F:R^n->R^n $F:\mathbb{R} ^ n\rightarrow \mathbb{R}^n$ for which I know there exist a unique fixed point $x ^ *$ (say). I also know that the Jacobian of F $F$ at each point x $x$ in R^n $\mathbb{R} ^ n$ has all of its eigenvalues in [0,1) $[0,1)$ (but they are different for each x). $x$). Are these facts enough for me to say that the iterative sequence x_{n+1} $x _ {n+1} = F(x_n) F(x_ n)$ converges to $x ^ *$ independently of the initial point x_0$x_ 0$? (I know that if x_0 $x_0$ is close enough to $x ^ *$ then the sequence coverges but my question concerns any x_0 $x_0$ in R^n.$\mathbb{R} ^ n$.) Whatever the answer is, could you give me a reference to some theorem that justifies that?
Thank you
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# fixed point of a particular vector valued function
Hi
I have a function F:R^n->R^n for which I know there exist a unique fixed point x* (say). I also know that the Jacobian of F at each point x in R^n has all of its eigenvalues in [0,1) (but they are different for each x). Are these facts enough for me to say that the iterative sequence x_{n+1} = F(x_n) converges to x* independently of the initial point x_0? (I know that if x_0 is close enough to x* then the sequence coverges but my question concerns any x_0 in R^n.) Whatever the answer is, could you give me a reference to some theorem that justifies that?
Thank you
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# The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.
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The sum of the 2nd and the 7th terms of an AP is 30. If its 15th term is 1 less than twice its 8th term, find the AP.
by (43.8k points)
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Correct Answer - 1, 5, 9, 13,…
T_(2) + T_(7) = 30 rArr (a+d) + (a+6d) = 30 rArr 2a + 17d = 30. " "…(i)
T_(15) = 2T_(8)-1 rArr (a+14d) = 2(a+7d) -1 rArr a = 1.
Putting a = 1 (i), we get d = 4.
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Snake
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I'm going to make a snake game, and I wanted your guys' opinion. It's going to be a grid-based snake game. My plan for movement is to have a vector for the snake pieces. When the player's head is on the same grid point as a piece of food, it will add a new piece to the vector. For movement, each piece would move to the piece in front of it. I know I wasn't very specific, but is this a good way to do it, or how is it usually done?
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Personally, rather than 'move' each segment of the snake. Id save myself a lot of work and use a RingQueue of snake segments, with a running count of how long it is, and which slot contains the head.
Thus rather than 'move' (change tons of parameters for each segment); they individually remain static, with only the memory mapping order changing. As the snake moves forward, correspondingly the head position in the queue changes, and the length counter makes the tail end of the queue expire. Only one segment actually needed to be updated (head) the intermediate body segments remain as is, and the tail is basically ignored(you'll automatically deallocate it as the ringqueue wraps around in the future).
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What's RingQueue, lol?
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It may have several different names depending on what school you went to.
I've known it as RingQueue
but Ring Buffer
Circular Buffer
Circular Queue
etc
may also be valid
wikipedia: http://en.wikipedia.org/wiki/Circular_buffer
basically you make a Big Array, big enough to hold the Max length of your snake.
and you have a pointer to the 'head'(insertion end) of the Queue (also the literal head of the snake) As your snake moves forward, rather than change the x,y coodinates of each body segment, you simply create a new head, and add it to the ringqueue. The old snakehead, stays in the ringqueue, and keeps its old x,y coords, but it gets treated as body segment #1 instead of a head. similarly, old body segment #1 is now treated as #2, and so on down to the end of the queue (and end of the snake).
None of the segments in the Array actually move, you simply draw them to screen in a different order (one slot higher each time)
after a while, the front of the array will be full of 'dead' body segments that are no longer used, and the 'head' will reach the end of the array. To handle this issue, you wrap the array into a 'ring'(hence RingQueue) and reset the 'head' to the front of the array ontop of the dead slots.
(the drawing order of body segments also needs to be able to travel off the end of the array and onto the other side)
The advantage of doing this, instead of a Vector or LinkedList approach, is that memory access is minimal for the snake segments. You are not constantly allocating and deallocating memory as segments are added and expired. And there is no modification to segments after creation.
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Could you give me a little example of how to do it? I'm a little confused...
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I'm 99% sure that the orginal Snake arcade game used a RingQueue, since on that old hardware performance was paramount.
On a modern computer however, Snake is a relatively simple task, and your Vector/LinkList approach should still work.
So if your goal is just to get a game running, Vector is fine.
It is good to know about various data structures though, and this project does sound like a good opportunity to learn RingQueue. So it depends what you want to focus on. If you still want to know, I can write some psuedocode for ringqueue/snake when I get home later...
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I'd like to use Ring Queue since this is just for learning purposes anyway, and I might as well learn as much as I can in the process. So yes, example would be great when you get home.
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Here is an ilistration of a circular array
http://lcm.csa.iisc.ernet.in/dsa/node24.html
This should also help in creating a circular array
http://www.cs.hmc.edu/claremont/keller/webBook/ch07/sec10.html
Hope this helps in creating the circular array datatype.
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Hmmmh
actually, bin11001010's second link is pretty much what I was going to say.
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Well, I guess I understand how it works, but I don't understand how I could implement it into the game.
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You could use a deque and push the new piece on the front, then pop the last piece off the back.
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But, how would I keep track of each ones position etc.?
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Read this pdf it's a snakes clone written in java and will give you all the info you need:
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you will need to create a SnakeSegment Struct or Class, which represents a single segemnt of snake, and contains x and y position.
The RingQueue should then be filled up with SnakeSegments as the snake grows.
Every update, the snake head moves, so you create a new SnakeSegment for the new head position, and enqueue it into the RingQueue. At the same time, the very tip of the snake's tail should be removed (since the snake stay constant length unless it is eating an apple) to remove the tip of the tail, simply do a dequeue on the RingQueue to remove the last segment.
Also, very Important, the links you saw explained how a basic RingQueue works, but for Snake you will need to add an extra feature, to 'View' the queue in order without actually removing anything from it. This will be needed for drawing all the SnakeSegments, and in checking that the head of the snake is not biting its body.
View is easy to add, in additon to a Head(enqueue) and Tail(dequeue) index on the Ring, you should also have a View index, which can be moved forward, backwards, or reset to Tail via a few 'View' functions attached to the RingQueue class.
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Is Queue part of C++ or is it just a way of using an array?
Also, I don't see how it's position in the queue is related to it's position on screen.
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A queue is a data structure that can be created in pretty much any language, and it can be made using array's or like a linked-list where nodes have points to the next node on the queue.
Correct me if i'm wrong, but i think the other people are wanting you to store the positions of each part of the snake in the queue, it's just that the first item in the queue is the head.
Then, when the snake moves, instead of updating every piece of the snake, take the last item, and put it in the front and just change it's position.
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They say I should make a class for each snake "peice" which is what I was planning on doing.
So, when the snake moves, I should move the tail "piece" one place ahead of the head "piece"?
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A queue is a FIFO datastructure. FIFO means First In First Out which means the first element that is inserted in the queue is the first element that is 'popped off'/'dequeued'. I think STL has a class for this.
e.g.
[] - empty queue
[3] - added the element 3 to the queue
[3, 4] - added the element 4 to the queue
[4] - dequeued the queue so the 3 was removed as it was the first element that was inserted.
To relate it to your game, say you had a struct or class to represent the body segment that holds the X,Y position of itself, you would have a queue of this struct/class and to 'move', you create a new segment that holds the new position of the head of the snake and add it to the queue, you then dequeue the queue to remove the last piece of the tail to keep the snake the same length.
[a, b, c, d] - current queue
[a, b, c, d, e] - e represents the new head position, a is the last tail piece
[b, c, d, e] - a is dequeued to keep the length of the snake.
Why would this be better/faster? Say you have a huge snake, 20 odd pieces, using your current method with vectors, you would have to tranverse the entire vector array and update the X/Y positions of the segments and this would get worse performence wise the larger the snake gets. With a queue you just creating and adding a new segment and removing and deleting a tail segment no matter how large the snake gets. Using a ringqueue (which is a fixed size) you don't create or delete any segments, you just overwrite an existing element.
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Ohhhh, okay, I totally understand it now, thanks! Now for the code part. I would just create a single dimensional array that's the maximum size of the snake right? Then make the pointers to the front and the back, then when the snake eats, add a new one to the back?
Or I could just use the Queue class you mentioned right?
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If you use the 'pure' Queue method, then I would just go for the STL Queue class, if the RingQueue, I would create my own class and use a STL Vector to hold the elements and two 'pointers' or iterators to hold the front and rear of the queue in the RingQueue with checks if any of the 'pointers' or iterators reach the end of the vector list.
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Wait, what's the difference between pure "queue" and RingQueue?
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One doesn't have an end, the other does.
So when you transverse through a Ring array, you can keep going indefinitely. A standard array you will hit either the beginning or the end.
Ring:-->--| |--<--Non-ring(?)-->--
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What do you think will be a better aproach? Non-ring looks easier...
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From a performence point of view, RingQueue looks like the way to go. Otherwise it doesn't really matter.
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It warns me that "queue.h is deprecated". What should I use different than that?
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What is the absolute mass and charge of a proton $?$
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Hint :Atoms are made up of subatomic particles like protons, neutrons and electrons. Protons and neutrons are present in the nucleus of the atom whereas electrons are found on its orbitals. Protons are positively charged and denoted as p or ${p^ + }$ .The sum of protons and neutrons in the nucleus gives the mass number of an atom and is collectively called as nucleons.
A proton is a positively charged subatomic particle found in the atomic nucleus. It has an elementary charge of $+ 1e$ . The charge of an electron is the same as that of a proton except for sign.
We know that a hydrogen atom consists of a proton and an electron. When the atom gets ionised and loses the electron, it has only a proton left in its nucleus.
$H\, \to {H^ + } + \mathop e\limits^\_ \,$
Therefore a proton can be referred to as hydrogen nucleus $({H^ + })$ . Thus, the mass of the proton will be equal to the mass of the hydrogen atom.
Mass of $1$ mole of H atom $= \,1.007\,g$
Mass of $1$ proton $= \dfrac{{1.007\,g}}{{6.022 \times {{10}^{23}}}}\,\, = \,1.67 \times {10^{ - 24}}g$
The absolute mass of the proton is $1.67 \times {10^{ - 27}}kg$ .
The absolute charge of a proton is $1.6 \times {10^{ - 19}}C$ . This is the value of an elementary charge (e) and when expressed in atomic units, it takes the value of unity , $+ 1$ .
Note :
The number of protons in a nucleus gives the atomic number (Z) of an atom. The sum of protons and neutrons gives the mass number (A) of an atom.
Protons have the same charge as that of electrons but they differ in sign. But a proton is much heavier than an electron. It is around $1,836$ times heavier than an electron.
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Browse Questions
# Rank each set of ions in order of decreasing size: Ca$^{2+}$, Sr$^{2+}$, Mg$^{2+}$.
$\begin{array}{1 1} Sr^{2+} > Ca^{2+} > Mg^{2+}. \\ Ca^{2+} > Sr^{2+} > Mg^{2+}. \\ Mg^{2+} > Sr^{2+} > Ca^{2+}.\\ Ca^{2+} > Mg^{2+} > Sa^{2+}. \end{array}$
Answer: Sr$^{2+}$ > Ca$^{2+}$ > Mg$^{2+}$.
Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons.
These are members of the same Group 2A(2), and decrease in size going up the group.
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Definition 89.9.1. Let $p : \mathcal{F} \to \mathcal{C}_\Lambda$ be a category cofibered in groupoids. We say $\mathcal{F}$ is smooth or unobstructed if its structure morphism $p$ is smooth in the sense of Definition 89.8.1.
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# VA-11 HALL-A:Cyberpunk Bartender Action
## 游戏机制
• Adelhyde阿德海德
• Bronson Extract布朗森提取物
• Powdered Delta德尔塔粉末
• Flanergide弗拉基德
• Karmotrine卡尔莫丁是含有酒精的配方
# 人物介绍
## VA-11 HALL-A员工
### Julianne Stingray/吉尔·斯汀雷
Jill 有一头乌黑的头发，当玩家有机会看到她的正脸时，可以看到她总是表现得漠不关心。Jill 是贫乳，几乎总是穿着 VA-11 Hall-A 的制服（除了标题界面和购物时，在这些场合可以观察到她穿着皮大衣而不是休闲装，在家中她穿着绿色毛衣，但在 Dana 拜访后，她会改穿灰红条纹的毛衣）。
Jill同样也会隐瞒自己对某件事/某个人的关心，就比如在自己公寓里因在 Dana 面前替 Fore 说话而变得紧张慌乱一样。 显然， Jill在很多方面都和她的祖父很像。 Jill承认过一件事：在大学时的自己表现得很笨拙，还很容易被激怒（至于当上酒保之后嘛，现在已经更能波澜不惊了），平日里总是处于“焦虑不安”的状态。
Jill 不经常外出，而在她不工作的大部分时间里则会做一个家里蹲，窝在被炉里陪着她的宠物猫ｆｏｒｅ——或是直接待在酒吧。
### Dana Zane/达娜·赞恩
Dana 充满活力，无忧无虑，大多时间处于“我才管不了那么多”的态度。她喜欢字面意思上的“一头扎进事物里去”，比如她觉得 Sei 的头盔很有趣，就一把套在自己头上。但她对自己的员工非常关心，不遗余力保证员工的安全。
### Gillian/吉莲
Gil穿着酒保制服，留着半长的黑紫色头发，下巴有稀疏的胡子茬。
Gil给人留下能干可靠的印象，尽管他有时会冷不防被吓一跳，哪怕在工作中，也总是漫不经心的态度。
Gil记得他小时候生活在Arctic，但他不知道自己在哪出生。曾在 Hong Kong riots期间， 他是防暴部队的一份子，但是在窃取补给品后叛逃了. 后来, Robert 死后不久，他衣冠不整，情绪激动地出现在VA-11 Hall-A酒吧的门口。据 Dana所说，他对自己过去的行为非常后悔，尽管Dana并不知道究竟发生了什么。发现他身无分文以后，她不仅提供了他免费的饮料，还提供他一个酒保的职位，他接受了。
Dana在游戏过程的某些环节中尝试发现Gillian的真实身份，比如在真心话大冒险游戏中，她甚至把Gil和一张照片作比较。在过去的某个时候曾参与过克格勃，由于克格勃早就于1991年解散，这也可能暗示他可能比一开始表面上的年龄要更大。值得注意的是，克格勃只是在1991年被正式解散的，而且有可能在过去的某个时候，克格勃被重建了。
### Dorothy Haze/多萝西·海兹
Sei表示这是因为她的修饰手旨在从指尖发射5级子弹。经常看到她穿着一件蓬松的袖子和红色条纹的连衣裙，而其白色织物则穿行。她的头发上还戴着有图案的发箍，脖子下面是大红色蝴蝶结。
## VA-11 HALL-A顾客
### Alma Armas
Alma是职业黑客，是Jill的朋友，也是VA-11 Hall-A的常客。
Alma的双手是机械义肢，能让她有更快的操作速度，并在繁重的操作中避免腕管综合征 (Carpal Tunnel Syndrome)。
Alma tends to be playful and enjoys teasing people
However, she takes family and familial responsibility very seriously. One of the biggest sources of frustration in her life at the moment is that her older sister is not taking the time to actually raise her children.
Alma started coming to VA-11 Hall-A roughly six months before the start of the game. She was served by Gillian the first time she came around but was served by Jill the second time and felt like Jill just "got her."
### Sei Asagiri
Sei 是一位（前）白骑士，Stella 的挚友，VA-11 Hall-A 的顾客
Sei 是一个非常富有同情心和善良的人，总是被人看到亲切地问候陌生人并且很少抱怨。当 Stella 在她们还是孩子时，遭到一个发狂的白骑士攻击，尽管冒着风险，Sei 还是试图拯救她。并在讲述她作为白骑士的职业生涯时表现得勇敢无私。然而，她本人常心不在焉，粗心大意，第一次拜访 VA-11 Hall-A，就将白骑士的头盔留在了酒吧里。
Sei 生在一个非常有问题的家庭中，她的父亲是个酒精中毒者，并有药瘾。当 Stella 要求她的父亲将她带到一个“真正的儿童公园”时，Sei 和 Stella 第一次见面。两人形成了紧密的联系，并开始定期在公园见面，直到有一天，由于一些未公开的原因，一个发狂的白骑士在公园劫持她们作为人质。在被治安当局接近时，他开始挖出 Stella 的眼睛。Sei 试图阻止犯下暴行的白骑士，后者打断她的几根肋骨，并将她的器官踢至破裂，当场倒地。
### Stella Hoshii
Stella Hoshii 是富二代， VA-11 Hall-A 的常客。她是一位 Cat Boomer，也是 Sei 的挚友。
Stella appears to be an average-sized girl with fair skin and red hair, styled into flashy drills. Although her breasts appear to be fairly average-sized for her height, Sei reveals that they're much larger than they appear, and she has no idea how she hides them so well.
Her two most defining traits, however, are her prosthetic right eye and the cat-ear-like protrusions on her head. Her outfit consists of what appears to be a red leather dress and black arm warmers, as well as a golden cross necklace around her neck.
She often acts in a prideful manner, shown when initially meeting her, and when Sei isn't around. However, she's a kind person at heart, as shown during Mega-Christmas when she buys a large amount of gifts for the children in her family.
She also hides her true feelings under her usual haughty demeanor, as evidenced when when Sei was heavily injured at the Apollo Bank incident. It is also evidenced in a more lighthearted manner when she often tries to hide her status as a fan of Kira Miki's music, as shown from her reaction upon seeing Kira Miki's signed glass.
Conceived by wealthy parents, before Stella was born she was diagnosed with 纳米机械排异, a potentially life-threatening disease. As such, she was forced to go through genetic treatment by doctors and become a Cat Boomer, called so because of the cat-ear-like protrusions which appear on the heads of those who receive treatment. She did survive because of this, and thus grew up in a rich environment where she received anything she asked for.
At age 5 she asked her father to take her to a "real kids park", where she met her soon to be best friend, Sei. The two formed a close bond and began to meet at the park regularly, until one day while they were there, a rogue 白骑士 took them hostage, on the run for undisclosed reasons. After being approached by the authorities, he began to gouge Stella's eye out and kicked Sei in the chest after she attempted to stop him, breaking several of her ribs and rupturing her organs, as well as knocking her out.
Afterwards, Sei briefly regained consciousness to see the White Knight on the ground, not moving, whilst another had her slumped over his shoulder. He put his helmet over Stella's head and the two were quickly taken to the hospital. Sei was placed into emergency care due to the severity of her injuries and only woke up after Stella began punching her leg, telling her to do so. The two of them recovered eventually, but Stella's right eye had to be removed and replaced with a prosthetic one.
### Beatrice Albert
Betty is a veterinarian who works at Seifar Toy Company, Deal's best friend, and a regular patron at VA-11 Hall-A.
Betty has a somewhat irritable personality and has been said to have a "kicking problem" by Deal. She is also incredibly attached to the Lilim, referring to him as "[her] piece o' scrap", and breaking down in tears when she thought he was going to be taken away from her.
She is a lesbian and has had numerous short-lived relationships with other women, usually ending them because of petty problems she has with her partner. Despite this, she feels she may be attracted to Deal, and asks Jill for advice about it.
Betty also enjoys old romance novels and has an entire shelf dedicated to them at home. Although they are generally heterosexual, she claims that they're written in such a way that it is very easy to put yourself in the place of the protagonist.
According to Deal, Betty has a harsh personality that not many people can stand, but that she's also a sweetheart underneath it. Betty usually softens up when drunk, and has been this way since at least her college days.
According to Betty, her grandfather loved his car more than any of his sons and left all his belongings to it in his will.
Betty has had a life-long hatred of puns, stemming from when she was young. Everyone in her family had a penchant for making puns whenever possible, with the exception of her, who tried but was never able to. This made her feel stupid growing up, and when she finally "got" puns, she hated them for being so unfunny.
When she was in high school Betty was charged with a minor sentence of petty vandalism for painting graffiti outside her school saying "Ms. Thompson's a cunt." Betty claims she did this because said Ms. Thompson was constantly humiliating her students and also constantly bullied two of Betty's classmates.
Betty fosters a dislike of cybernetic enhancements. This is due to her days of studying surgery at medical school, when five of her classmates used shady means to try and get cybernetic enhancements for their hands. Although she managed to talk one of them out of it, two of them lost their hands, one of them became arthritic, and another died on the operating table. According to Jill, the man responsible for hooking them up was later caught, as was covered by a news story. Betty makes a distinction that she doesn't have a problem with cybernetic prostheses, only implants that drastically improve ability rather than serving as a replacement. She feels the same way about Lilim who needlessly upgrade their bodies.
### Deal
Deal 是一个 Lilim , Betty 最好的朋友, 也是 VA-11 Hall-A 的常客之一
Seifar Toy Company 的工作。据Betty所说，他抱着毛绒玩具柯基睡觉，还有一堵贴满了从公司拍摄的照片的墙。 Deal不喜欢喝太多酒，也不是一个酒鬼。
Deal的型号是DT-01d，一种社会发展型Lilim，主要为适应性定制的。他们是“工作用Lilim”。尽管Deal从没升级过自己的肌肉，但他能承受200摄氏度的高温。 他的人类监护人叫Juergen。Juergen背不太好，但不愿意吃药，总说自己“还没那么弱”。Deal一年前通过了人格测试，但仍然和监护人生活在一起。 Deal曾经在人工智能委员会的部门工作过一段时间，负责处理Lilim的收养事宜，这意味着他对收养过程有非常详细的了解。 Deal的一位邻居的狗在Seifar工作。这个邻居最初告诉他公司打算雇用类人机器人。Deal现在有了这份工作，主要是因为法律要求员工名单上至少要有一个类人生物。他觉得自己的工作相当于一个勤杂工，不仅如此，因为他经常得履行他的雇主因为没有拇指而无法完成的工作和职能。
### Jamie
Jamie是一名雇佣兵/杀手，也是 VA-11 Hall-A的常客。通常由 Gillian 提供服务，但他也会时不时向 Jill 点酒
Jamie的外表令人望而生畏，他倾向于隐藏自己的秘密身份。然而一旦被问及自己的职业和过往，他也会表现得十分坦诚。 Jamie是个非常友善的人，这是Jill在聆听他与Ingram交谈时注意到的。 Jill认为Jamie很擅长引出别人友善的一面。 尽管作为雇佣兵，Jamie仍然维持着道德观念。他告诉Jill自己不接受任何杀死无辜人的工作，也不会因为琐碎的理由杀人，而是接受杀死作奸犯科之徒的任务，视自己为一个赏金猎人。
Jamie的过去很模糊，除了他提到自己做过纳米机械清除手术。Jamie形容那种手术的痛苦令人难以忍受，像全身扎满了小针。
## 游戏视频
Last modification：April 10, 2022
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# 怎样让移动图像显示更快一些...
## Hide Controls When Setting Properties to Avoid Multiple Repaints
Every repaint is expensive. The fewer repaints Visual Basic must perform, the faster your application will appear. One way to reduce the number of repaints is to make controls invisible while you are manipulating them. For example, suppose you want to resize several list boxes in the Resize event for the form:
Sub Form_Resize ()
Dim i As Integer, sHeight As Integer
sHeight = ScaleHeight / 4
For i = 0 To 3
lstDisplay(i).Move 0, i * sHeight, _
ScaleWidth, sHeight
Next
End Sub
This creates four separate repaints, one for each list box. You can reduce the number of repaints by placing all the list boxes within a picture box, and hiding the picture box before you move and size the list boxes. Then, when you make the picture box visible again, all of
the list boxes are painted in a single pass:
在vb中用move方法移动图片时,速度有些慢,当图片很大时,这时可以用下面的方法:
Sub Form_Resize ()
Dim i As Integer, sHeight As Integer
picContainer.Visible = False
picContainer.Move 0, 0, ScaleWidth, ScaleHeight
sHeight = ScaleHeight / 4
For i = 0 To 3
lstDisplay(i).Move 0, i * sHeight, _
ScaleWidth, sHeight
Next
picContainer.Visible = True
End Sub
Note that this example uses the Move method instead of setting the Top and Left properties. The Move method sets both properties in a single operation, saving additional repaints.
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# Demystification of the spin-sum for massive spin-1 particles
Tags:
1. Jul 29, 2015
### Dilatino
Assuming that a massive spin-1 particle has momentum only in the z-direction, the polarization vectors are given by
$$\varepsilon_{\mu}(J_z = +1) = (0,-\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$
$$\varepsilon_{\mu}(J_z = 0) = (\frac{p}{m},0,0, \frac{E}{m})$$
$$\varepsilon_{\mu}(J_z = -1) = (0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$
The so-called spin-sum is the claimed to be
$$\sum\limits_{J_z = -1,0,+1} \varepsilon_{\mu}\varepsilon_{\nu}^* = g_{\mu\nu} + \frac{p_{\mu}p_{\nu}}{m^2}$$
I absolutely dont understand how this spin-sum is evaluated.
What does $\varepsilon_{\mu}\varepsilon_{\nu}^*$ even exactly mean? Is it a scalar product between two of the three above polarization vectors, or is it a "tensor-product" between the components of a single polarization vector which results in a 4x4 matrix and one has finally to sum all such matrices for the three possible values of $J_z$?
I would really appreciate it if somebody can explain to me what this spin-sum exactly means and how it is evaluated step-by-step.
2. Jul 29, 2015
### Orodruin
Staff Emeritus
The latter option, it is a tensor product, which results in a rank two tensor (which may be represented by a matrix, it is not a matrix - it is a rank two tensor). The sum is taken over all possible spin states, i.e., you are summing several rank two tensors. The result is a new rank two tensor.
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# Stability of nonsmooth, Lipschitz continuous, autonomous system of differential equations
Consider the following autonomous system of differential equations:
$$\frac{\mathrm d\mathbf x}{\mathrm dt} = \mathbf v(\mathbf x)$$
where $\mathbf x, \mathbf v \in \mathbb R^n$. Assume that $\mathbf v(\mathbf x)$ is Lipschitz continuous (therefore the solution is unique given initial conditions). Suppose that $\mathbf x^*$ is a fixed point, i.e., $\mathbf v(\mathbf x^*) = 0$. I need to characterize the stability of $\mathbf x^*$. However, $\mathbf v(\mathbf x)$ may not be differentiable (it is only Lipschitz continuous). Therefore the straightforward method of looking at the real parts of the eigenvalues of the Jacobian does not apply. I'm trying to find alternative characterizations.
Can someone point me to references in the literature or books about the stability of nonsmooth differential equations, where the right-hand side is Lipschitz continuous (like the system above)? Are there any helpful theorems in this topic?
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# Pamiêci Profesora Andrzeja Krzysztofa Kwa¶niewskiego (1947-2011)
Profesor AKK (2008)
Mathemagics:
Inne:
# Gian Carlo Rota Polish Seminar
## Gian Carlo Rota Polish Seminar *
prowadz±cy: a.k.kwa¶niewski the member of the Institute of Combinatorics and its Applications Editorial Committee Member in Advanced Studies in Contemporary Mathematics Editorial Board Member of JOURNAL OF ANALYSIS AND APPLICATIONS (INDIA)
Fibonacci Number Theorists
### abstrakty
data autor tytu³
April 2008 A.K.Kwasniewski, M. Dziemiañczuk On cobweb posets most relevant codings
Abstract
One considers here orderable acyclic digraphs named KoDAGs which represent the outmost general chains of dibicliques denoting thus the outmost general chains of binary relations. Because of this fact KoDAGs start to become an outstanding concept of nowadays investigation. We propose here examples of codings of KoDAGs looked upon as infinite hyper-boxes as well as chains of rectangular hyper-boxes in $N^\infty$. Neither of KoDAGs codings considered here is a poset isomorphism with $\Pi = \langle P, \leq\rangle$ . Nevertheless every example of coding supplies a new view on possible investigation of KoDAGs properties. The codes proposed here down are by now recognized as most relevant codes for practical purposes including visualization.
More than that. Employing quite arbitrary sequences $F=\{n_F\}_{n\geq 0}$ infinitely many new representations of natural numbers called base of F number system representations are introduced. These constitute mixed radix-type numeral systems. F base nonstandard positional numeral systems in which the numerical base varies from position to position have picturesque interpretation due to KoDAGs graphs and their correspondent posets which in turn are endowed on their own with combinatorial interpretation of uniquely assigned to KoDAGs F-nomial coefficients. The base of F number systems are used for KoDAGs coding and are interpreted as chain coordinatization in KoDAGs pictures as well as systems of infinite number of boxes sequences of F-varying containers capacity of subsequent boxes. Needless to say how crucial is this base of F number system for KoDAGs hence consequently for arbitrary chains of binary relations. New F based numeral systems are umbral base of F number systems in a sense to be explained in what follows.
AMS Classification Numbers: 06A07, 05C78, 11A63; 23 pages, 15 figures,
Bibliography
[1] A. K. Kwa¶niewski, M. Dziemiañczuk On cobweb posets most relevant codings arXiv:math.CO/0804.1728, Thu, 10 Apr 2008
06A07 Combinatorics of partially ordered sets
05C78 Graph labelling
Mixed radix numeral systems are non-standard positional numeral systems in which the numerical base varies from position to position.
March 2008 Wies³aw Bajguz More on Algebraic Structure of the Complete Partition Function for the Zn - Potts Model
Abstract
In this first part of a larger review undertaking the results of the firstauthor and a part of the second author doctor dissertation are presented. Nextwe plan to give a survey of a nowadays situation in the area of investigation.Here we report on what follows. Calculation of the partition function for anyvector potts model is at first reduced to the calculation of traces of productsof the generalized clifford algebra generators. The formula for such traces isderived. This enables one, in principle, to use an explicit calculation algorithm forpartition functions also in other models for which the transfer matrix is anelement from generalized clifford algebra. The method - simple for Z2 case - becomes complicated for Zn, n>2,however everything is controlled , in principle, due to knowledge of thecorresponding algebra properties and those of generalized cosh function. Thediscussion of the content of the in statu nascendi second part is to be foundat the end of this presentation. This constitutes the last section.
Bibliography
[1] A. K. Kwa¶niewski , W. Bajguz, More on Algebraic Structure of the Complete Partition Function for the Zn - Potts Model, Part 1, arXiv: 0803.1305, Sun, 9 Mar 2008, submitted to Rep. Math. Phys.Categories: math.GM math-ph math.MP, Comments: 23 pages, MSC-class: 11E88, 15A66
[2] W. Bajguz; W.Bajguz PhD dissertation
March 2008A.K. Kwa¶niewski Further broadening of incidence algebra horizon - an informal introduction. Part I
Part I
On further Applications if Incidence Algebras description of Posets - QM and all that
1] at first - comments on \bibitem {xx}Vladimir V. Kisil , {\it The Umbral Calculus and Cancellation Semigroup Algebras} Zeitschrift für Analysis und ihre Anwendungen, {\bf 19} no. 2, (2000), 315-338.E-print: arXiv:funct-an/9704001.See page 6 „Our agreement is equivalent to introduction of incidence algebra of …”[knowledge of QM - not needed]then
2] a look on relevant Roman R. Zapatrin's contribution knowledge of QM - not at the first stage obligatory however a bit of QM - recommended,then gradually - more and more
3] goal? how are doing KoDAGs with all this- being so much around?
4] Aristophanes Dimakis? and discrete space time history...
Bibliography
[2] Roman R. Zapatrin
[3] Aristophanes Dimakis
January- February 2008 Ewa Krot-Sieniawska On Incidence Algebras of Cobweb Posets
in Polish
Przedstawiê najnowsze wyniki w³asne dotycz±ce opisu algebrami incydencji ca³ej bez wyj±tków rodziny „cobweb posetów” - rodziny odkrytej i wprowadzonej do matematycznych podstaw informatyki kilka lat temu przez autora [1]. U¿ywam definicji tych digrafów Kwa¶niewskiego tak jak np. w [1] z 2007 roku. Pojawia siê tu w [1] i nowa znamienna charakteryzacja a zatem definicja owych cobweb posetów poprzez tym szczególny typ diagramu Hasse , ¿e stanowi go DAG niejako „matkuj±cy” wszystkim z rodziny grafów ³añcuchów relacji - grafów ³añcuchów wszelkich relacji binarnych. Podkre¶lmy – WSZELKICH! Patrz [1]. Fakt ten stanowi o wyj±tkowej roli rozwa¿anego typu KoDAG grafów [KoDAG = Kwa¶niewski-oDAG] W pierwszej czê¶ci wyprowadzone bêd± wzory jawne na:
1. Funkcjê charakterystyczn± ? porz±dku czê¶ciowego dowolnego cobweb posetu P;
2. Funkcjê Möbiusa µ dowolnego cobweb posetu P
3. Funkcje przeliczaj±ce elementy, multi³añcuchy, ³añcuchy i ³añcuchy maksymalne w dowolnym odcinku P;
4. Liczby Whitney’a I i II rodzaju dowolnego cobweb posetu P
5. Wielomiany charakterystyczne dowolnego skoñczonego cobweb posetu Pn.
Punkty 4 i 5 stanowi± urzeczywistnianie zadania postawionego w [1] str.2 w postaci zagadnienia o tytule:
„The problem to be the next”
W drugiej czê¶ci okre¶lê standardow± zredukowan± algebrê incydencji R(P) dowolnego cobweb posetu P i zajmê siê jej szczegó³owymi w³asno¶ciami. W szczególno¶ci /trzymam nastêpuj±ce Twierdzenie, [2] Niech dany bêdzie dowolny ci±g liczb {Fn}={F0, F1, ...} o warto¶ciach naturalnych, F0=1. Wówczas liczby Fn s± wspó³czynnikami incydencji w standardowej zredukowanej algebrze incydencji R(P) odpowiadaj±cego im cobweb posetu P.
legenda
oDAG = orderable Directed Acyclic Digraph - patrz np. [3] i prace Profesora
Bibliography
[1] A. Krzysztof Kwa¶niewski On cobweb posets and their combinatorially admissible sequences, arXiv:math.CO/0512578 v1 26 Dec 2005 - 21 Oct 2007.
[2] E.Krot-Sieniawska Rozszerzenia Rachunku operatorowego Roty I pokrewne zastosowania analizy kombinatorycznej, przygotowywana praca doktorska
[3] E. Krot The first ascent into the Fibonacci Cob-web Poset, Advanced Studies in Contemporary Mathematics 11 (2005), No. 2, pp.179-184, ArXiv: math.CO/0411007
[4] Spiegel E., ODonnell Ch. J. Incidence algebras, Marcel Dekker, Inc. Basel 1997
[5] Stanley R. P. Enumerative Combinatorics, Volume I, Wadsworth\& Brooks/ColeAdvanced Books \& Software, Monterey California, 1986
[6] Anatoly Plotnikov's Publications http://www.vinnica.ua/~aplot/publicat.htmla
w szczególno¶ci
1. About presentation of a digraph by dim 2 poset Advanced Studies in Contemporary Mathematics volume 12 (1), 2006, p. 55-60.
2. The lower bound of the number of all-or-none DAGs Advanced Studies in Contemporary Mathematics volume 12 (2), 2006, p. 323-326.
January 2008 Maciej Dziemiañczuk On Cobweb admissible sequences- The Production Theorem
In English
Further [1] clue decisive observations on cobweb admissible sequences are shared with the audience. In particular an announced proof of the Theorem 1 (by Dziemiañczuk) from [1] announced in India –Kolkata- December 2007 is delivered here. Namely here and there we claim that any cobweb admissible sequence F is at the point product of primary cobweb admissible sequences taking values one and/or certain power of an appropriate primary number p. Here one delivers also the so much desired [2,3,1] algorithm to produce the family of all cobweb-admissible sequences i.e. the Problem 1 from [1] problems posed in source papers [2,3] is solved using the idea and methods implicitly present already in [4].
Bibliography
[1] A. Krzysztof Kwa¶niewski, M. Dziemiañczuk, Cobweb posets – Recent Results; ISRAMA Congress; December 15-17 2007 Kolcata, INDIA see ArXiv
[2] A. Krzysztof Kwa¶niewski, Cobweb posets as noncommutative prefabs Adv. Stud. Contemp. Math. vol. 14 (1) (2007) 37-47.
[3] A. Krzysztof Kwa¶niewski, On cobweb posets and their combinatorially admissible sequences, arXiv:math.CO/0512578 v1, 26 Dec 2005, 21 Oct 2007.
[4] Maciej Dziemiañczuk On Cobweb posets tiling problem arXiv:0709.4263, 4 Oct 2007
November 2007 Ewa Krot-Sieniawska On Characteristic polynomials of Cobweb Posets
A graded poset, sometimes called a ranked poset, is a partially ordered set (poset) P in which every maximal chain is finite and has the same length. Every graded poset has an integer-valued rank function. Equivalently, a poset P is graded if it admits a partition into maximal antichains An for n = 0, 1, ..., r (where r is a nonnegative integer) such that for each x being an element of An, all of the elements covering x are in An + 1 and all the elements covered by x are in An - 1. The rank of P, written ?(P), is the maximum rank of any element. The rank of a minimal element is 0. A finite graded poset P with a unique minimal element (called 0) has an important invariant called its characteristic polynomial. Let µ be the Möbius function of P. The characteristic polynomial of P is
Examples of graded posets are the so called (Kwa¶niewski’s) Cobweb Posets defined in [3,4]. Following the expression for the Möbius function of Fibonacci and other cobweb posets from [5] the family of characteristic polynomials of these structures will be presented.
in Polish
Zbiór czê¶ciowo uporz±dkowany P nazywamy posetem z rang±, je¶li w dowolnym jego odcinku wszystkie ³añcuchy maksymalne s± tej samej d³ugo¶ci. Rang± ?(x) elementu x z P nazywamy wówczas d³ugo¶æ ³añcucha maksymalnego ³±cz±cego element najmniejszy (ew. minimalny) z x. Dla takiego zbioru P mo¿emy okre¶liæ jego wielomian charakterystyczny nastêpuj±co: Je¶li µ jest funkcj± Möbiusa posetu P, to
Przyk³adami posetów z gradacj± s± zdefiniowane przez Kwa¶niewskiego m.in. w [3,4], tzw. cobweb posety.
Poka¿ê znane [1,2,7] przyk³ady posetów z rang± i ich wielomiany charakterystyczne, a tak¿e przyk³ady zastosowañ tych¿e wielomianów. Korzystaj±c z postaci funkcji Möbiusa dla Fibonacci cobweb posetu z [5], reprezentatywnej dla ca³ej rodziny tych struktur, wyprowadzê wielomiany charakterystyczne dla cobweb posetów
Bibliography
[1] B. Sagan, Why the characteristic polynomial factors, Bull. Amer. Math. Soc. 36 (1999), 113–134.
[2] B. Sagan, M¨obius Functions of Posets IV: Why the Characteristic Polynomial Factors , www.math.msu.edu/~sagan/Slides/mfp4.pdf
[3] A. Krzysztof Kwa¶niewski, Cobweb posets as noncommutative prefabs Adv. Stud. Contemp. Math. vol. 14 (1) (2007) 37-47
[4] A. Krzysztof Kwa¶niewski, On cobweb posets and their combinatorially admissible sequences, arXiv:math.CO/0512578 v1 26 Dec 2005.
[5] E. Krot, The first ascent into the Fibonacci Cob-web Poset, Advanced Studies in Conterporary Mathematics 11 (2005), No. 2, pp.179-184, ArXiv: math.CO/0411007
[7] Joni S. A., Rota. G.-C., Sagan B. From Sets to Functions: Three Elementary Examples, Discreta Mathematics 37 (1981) pp.193-202
December 2007 Maciej Dziemiañczuk On cobweb-admissibility characterization
I shall present some observations on cobweb admissible sequences. Among others a proof is given of the statement: Any cobweb admissible sequence F is at the point product of cobweb admissible sequences primary numbers valued by one and/or a certain primary number p. I shall also disclose family of all such primary cobweb admissible sequences in a form of the corresponding binary tree. As for the background see the source papers [1,2,3].
Abstrakt
a) poka¿ê, i¿ mo¿na ka¿dy cobweb admissible ci±g przedstawiæ za pomoc± iloczynu w punkcie prostych ci±gów sk³adaj±cych siê z warto¶ci {1, p}, gdzie p to liczba pierwsza
b) ujawniê opis wszystkich takich prostych ci±gów za pomoc± nieskoñczonego poddrzewa drzewa binarnego
Bibliography
[1] A. Krzysztof Kwa¶niewski, Cobweb posets as noncommutative prefabs Adv. Stud. Contemp. Math. vol. 14 (1) (2007) 37-47
[2] A. Krzysztof Kwa¶niewski, On cobweb posets and their combinatorially admissible sequences, arXiv:math.CO/0512578 v1 26 Dec 2005.
[3] Maciej Dziemiañczuk On Cobweb posets tiling problem arXiv:0709.4263
November 2007 Maciej Dziemiañczuk On cobweb tilings algorithms
We present some of cobweb tilings algorithms for searching all the partitions of any layer and their computational complexity as rough upper bounds of cobweb tiling problem. We show also that rough upper bound of tiling problem can be smaller than that which was introduced in [2] a month ago. This is a continuation of [6].
Bibliography
[1] A. Krzysztof Kwa¶niewski, Cobweb posets as noncommutative prefabs Adv. Stud. Contemp. Math. vol. 14 (1) (2007) 37-47
[2] A. Krzysztof Kwa¶niewski, On cobweb posets and their combinatorially admissible sequences, arXiv:math.CO/0512578 v1 26 Dec 2005.
[3] Ewa Krot, An Introduction to Finite Fibonomial Calculus, Central European Journal of Mathematics 2(5) (2005) 754-766
[4] A. Krzysztof Kwa¶niewski Main theorems of extended finite operator calculus Integral Transforms and Special Functions Vol. 14, No 6 (2003) 499-516.
[5] A.Krzysztof Kwa¶niewski The logarithmic Fib-binomial formula Advan. Stud. Contemp. Math. v.9 No.1 (2004) 19-26.
[6] Maciej Dziemiañczuk On Cobweb posets tiling problem arXiv:0709.4263
[7] Eduard Lucas Théorie des Fonctions Numériques Simplement Périodiques American Journal of Mathematics Volume 1 (1878), pp. 184-240 (Translated from the French by Sidney Kravitz, Edited by Douglas Lind Fibonacci Association 1969
October 2007 Maciej Dziemiañczuk On Cobweb posets tiling problem
Kwa¶niewskis cobweb posets uniquely represented by directed acyclic graphs are such a generalization of the Fibonacci tree that allows joint combinatorial interpretation for all of them under admissibility condition. This interpretation was derived in the source papers and it entailes natural enquieres already formulated therein. In our note we response to one of those problems. This is a tiling problem. Our observations on tiling problem include proofs of tiling’s existence for some cobweb-admissible sequences. We show also that not all cobwebs admit tiling as defined below.
Bibliography
[1] A. Krzysztof Kwa¶niewski, Cobweb posets as noncommutative prefabs Adv. Stud. Contemp. Math. vol. 14 (1) (2007) 37-47
[2] A. Krzysztof Kwa¶niewski, On cobweb posets and their combinatorially admissible sequences, arXiv:math.CO/0512578 v1 26 Dec 2005.
[3] Ewa Krot, An Introduction to Finite Fibonomial Calculus, Central European Journal of Mathematics 2(5) (2005) 754-766
[4] A.Krzysztof Kwa¶niewski, Main theorems of extended finite operator calculus Integral Transforms and Special Functions Vol. 14, No 6 (2003) 499-516.
[5] A.Krzysztof Kwa¶niewski The logarithmic Fib-binomial formula Advan. Stud. Contemp. Math. v.9 No.1 (2004) 19-26.
[6] Maciej Dziemiañczuk,
[7] Eduard Lucas Théorie des Fonctions Numériques Simplement Périodiques American Journal of Mathematics Volume 1 (1878), pp. 184-240 (Translated from the French by Sidney Kravitz, Edited by Douglas Lind Fibonacci Association 1969
March - April 2007 Ewa Krot-Sieniawska On Combinatorial Species
The central concept of the combinatorial theory is to enumerate discrete structures, i.e.: permutations, graphs, trees, derangements and so on. Each of these has an associated generating function (formal series) which counts how many structures there are of the certain size. We are able to analyse complicated structures by describing them in terms of transformations and combinations of simpler structures. These operations correspond to equivalent manipulations of generating functions. The Category Theory provides the tools to give a proper mathematical foundation to our combinatorial intuition trough this concept - The Theory of Combinatorial Species introduced by Andre Joyal in 1981. It is an abstract, systematic method for analyzing discrete structures (for example) in terms of generating functions. A combinatorial object (species) is a functor from the category B of finite sets and bijections to itself. To each species S one associates its exponential generating series , where sn is the cardinality of set of S- structures on any set with n elements. Certain “natural” operations on species, for example addition, multiplying (product), composition (substitution ), differentiation and so on correspond to some operations on generating functions. I’ll give basic definitions of the theory and some examples of their settings.
Bibliography
[1] André Joyal, Une théorie combinatoire des séries formelles, Advances in Mathematics 42:1-82 (1981).
[2] Francois Bergeron, Gilbert Labelle, Pierre Leroux, Combinatorial Species and Tree-like Structures, Cambridge University Press, Cambridge, 1998.
[3] Francois Bergeron, MI>Species and Variations on the theme of Species, invited talk at Category Theory and Computer Science ’04 ( http://www.itu.dk/research/theory/ctcs2004/ ), Copenhagen (2004). Slides (pdf) ( http://bergeron.math.uqam.ca/Especes_trans.pdf ).
[4] Federico G. Lastaria, An Invitation to Combinatorial Species, http://math.unipa.it/~grim/ELastaria221-230.PDF
[5] http://en.wikipedia.org/wiki/Combinatorial_species
March 2007 Wies³aw Bajguz Introduction to category theory
Categories were first introduced by Samuel Eilenberg and Saunders Mac Lane in 1942-1945 as an important part of the transition from homology (an intuitive and geometric concept) to homology theory. Category theory is applied in most branches of mathematics (see – theory of species in combinatorics). This talk is an introduction to categries. We start with definitions and basic facts from this theory - in the spirit of recollection of the already known to the audience.
Bibliography
[1] Z. Semadeni, A.Wiweger, Wstêp do teorii kategorii i funktorów BM 45, PWN, Warszawa 1978
[2] D. Turi Category Theory Lecture Notes http://www.dcs.ed.ac.uk/home/dt/CT/categories.pdf
[3] M. M. Fokkinga A Gentle Introduction to Category Theory - the calculational approach , http://wwwhome.cs.utwente.nl/~fokkinga/mmf92b.html
[4] Michael Barr and Charles Wells Cathegory theory Lecture Notes http://folli.loria.fr/cds/1999/library/pdf/barrwells.pdf
[5] Federico G. Lastaria An Invitation to Combinatorial Species, http://math.unipa.it/~grim/ELastaria221-230.PDF
February 2007 a. krzysztof
kwa¶niewski
[Tachion]
Cardinals, Ordinals, Non-Archimedian Fields and NSA i.e. Infinitezimal and infinite nonstardard numbers
Nonstandard Analysis is a branch of mathematical logic which introduces hyperreal numbers to allow for the existence of genuine infinitesimals. ANALIZA NIESTANDARDOWA to zia³ matematyki rozwiniêty w latach 60. XX w. przez A. Robinsona, bêd±cy ¶cis³± realizacj± pochodz±cej z XVII w. koncepcji G.W. Leibniza, który chcia³ oprzeæ ca³y rachunek ró¿niczkowy i ca³kowy na pojêciu nieskoñczenie ma³ej; podstawowym obiektem w analizie niestandardowej jest tzw. cia³o liczb hiperrzeczywistych R* (jego podzbiorem s± zwyk³e liczby rzeczywiste), w którym nie zachodzi aksjomat Archimedesa [0!], tzn. istniej± elementy mniejsze od ka¿dej zwyk³ej liczby rzeczywistej dodatniej, a mimo to wiêksze od zera; w jêzyku analizy niestandardowej mo¿na dokonaæ rekonstrukcji w³a¶ciwie ca³ej klasycznej analizy matematycznej- przy czym podanie definicji np. funkcji ci±g³ej, czy pochodnej funkcji, wymaga mniejszej liczby kwantyfikatorów ni¿ w klasycznym rachunku ró¿niczkowym. Newton and Leibniz used infinitesimal methods in their development of the calculus, but were unable to make them precise, and Weierstrass eventually provided the formal epsilon-delta idea of limits. Abraham Robinson developed nonstandard analysis in the 1960's, and the theory has since been investigated for its own sake, and has been applied in areas such as Banach spaces, differential equations, probability theory, microeconomic theory and mathematical physics. Nonstandard analysis is also sometimes referred to as infinitesimal analysis. Basic notions and ideas underlying the presentation are to be found in [1,6].
Bibliography
[0!] J. Browkin Teoria cia³ PWN, Warszawa 1978 [patrz Rozdzia³ IV, §1 ]
[1] E. Nelson Internal set theory: A new approach NonStardard Analysis Bull. Amer. Math. Soc., 83(6):1165–1198, 1977
[2] E. Nelson Confessions of an apostate mathematician debate with Ennio De Giorgi, Forli, Italy, 1995. http://www.math.princeton.edu/~nelson/papers.html
[3] E. Nelson Mathematics and faith in “The Human Search for Truth: Philosophy, Science, Theology– The Outlook for the Third Millennium” International Conference on Science and Faith,The Vatican, 23-25 May 2000, St. Joseph’s University Press, Philadelphia, 2002 http://www.math.princeton.edu/~nelson/papers.html
[4] E. Nelson Completed versus incomplete infinity in arithmetic, presented to STOQ International Con-ference on Infinity in Science, Philosophy, and Theology, Pontifical Lateran University, Vatican City, November 9–11, 2005. http://www.math.princeton.edu/~nelson/papers.html
[5] E. Nelson Syntax and semantics, presented to International Conference: “Foundations and the Onto-logical Quest. Prospects for the New Millennium”, Pontifical Lateran University, Vatican City, January 7–10, 2002. http://www.math.princeton.edu/~nelson/papers.html
[6] H. Jerome Keisler Elementary Calculus An Approach Using Infinitesimals. First edition 1976; 2nd edition 1986 in pdf format available for downloading at http://www.math.wisc.edu/~keisler/calc.html
May 2006 Dominique Foata, Guo Niu Han
przedstawia:
A.K.Kwasniewski, W.Bajguz, E. Krot-Sieniawska
Nombres de Fibonacci et polynomes orthogonaux
RÉSUMÉ : Le calcul des séries de produits de nombres de Fibonacci et des polynômes de Tchebicheff des deux especes est obtenu ici a laide de deux méthodes combinatoires.
Bibliography
[1] Dominique Foata, Guo Niu Han Nombres de Fibonacci et polynômes orthogonaux in Leonardo Fibonacci: il tempo, le opere, l'ereditá scientifica [Pisa. 23-25 Marzo 1994, Marcello Morelli e Marco Tangheroni, ed.], p. 179-200. Pisa, Pacini Editore (Fondazione IBM Italia), 1994. www-irma.u-strasbg.fr/~foata/paper/pub71.html
[2] see references in [1] above
April 2006 Ewa Borak Partition Statistics and q-Bell Numbers
Following Carl G. Wagner [1] tree q-generalization of Stirling numbers of second kind will be presented. These q-Stirling numbers are generating functions for three statistics on the set of partitions of [n] with k blocks. The unified treatment of these numbers will be presented.
Bibliography
[1] Carl G. Wagner: "Partition Statistics and q-Bell Numbers (q=-1)" Journal of Integer Sequences, Vol. 7 (2004), Article 04.1.1
April 2006 Ewa Krot-Sieniawska Generalized Stirling and Lah numbers
Following Carl G. Wagner [1,2] three generalizations of Stirling numbers of the second kind, and of Lah numbers are presented. Combinatorial interpretation of these generalizations will be given by the use of the theory of modular binomial lattices [3].
Bibliography
[1] C.G Wagner: Generalized Stirling and Lah numbers, Discrete Mathematics, 160 (1996), pp.199-218
[2] C.G.Wagner: Suriections, Differences, and Binomial Lattices, Studies In Applied Mathematics, 93 (1994), pp.15-27
[3] P.Doubilet, G.-C. Rota, R.Stanley: On the Foundations of Combinatorial Theory VI:The Idea of Generating Functions, In 6th Berkeley Symp. Math. Stat. Prob. vol. 2 (1972), pp. 267-318
March 13, April 2006 A.K.Kwa¶niewski
przedstawia:
E. Borak
On umbral extensions of Stirling numbers and Dobinski-like formulas
Stirling numbers of the second kind are umbrally extended in a new way and the resulting new type of dobinskian formulae are discovered. These extensions naturally encompass the well known extensions. The fact that the q-umbral extended dobinskian formulae may also be interpreted as the average of powers or polynomials of specific random variable with the extended Poisson like distribution singles out the q-extensions which appear to be a kind of bifurcation point in the domain of umbral extensions. Because of this the Because of this the further consecutive umbral extensions are here realized in a twofold way.
Bibliography
[1] A.K. Kwa¶niewski On umbral extensions of Stirling numbers and Dobinski-like formulas submitted - under correction 2005, ArXiv: math.CO/0503286 2005
[2] see references in [1] above
February 2006 Wies³aw Bajguz Graphs compositions
New results in compositions of unions of graphs shall be presented. The author shall define such a chain of graphs, which is a generalization of ladders and will demonstrate his method of counting of these. Basic notions and ideas underlying the presentation are to be found in [1]; see also [2].
Bibliography
[1] A. Knopfmacher, M. E. Mays, Graph Compositions I: Basic Enumeration Integers: Electronic Journal of Combinatorial Number Theory. 1 #A04 (2001): 1-11. www.integers-ejcnt.org/vol1.html
[2] J. N. Ridley, M. E. Mays Compositions of Unions of Graphs The Fibonacci Quarterly, Vol. 42, No 3 (2004) pp. 222-230.
23 January 2006 Maciej Dziemiañczuk Remarks on cobweb posets Kwasniewski Hypothesis and Related Questions posed by [1]
prime cobweb posets Pn = Cn[F; (F1), (F2),..., (Fn)] introduced by Kwasniewski in [1] are considered. Examples are provided in abundance. Hypothesis concerning their combinatorial interpretation by Kwasniewski is to undergo various attempts of verification along the lines outlined by the author of the hypothesis and the author of the presentation.
Bibliography
[1] A.K. Kwa¶niewski On cobweb posets and their combinatorially admissible sequences http://arxiv.org/PS_cache/math/pdf/0512/0512578.pdf 2006
[2] see references in [1] above
November/December 2005 A.K. Kwa¶niewski Cobweb posets as noncommutative prefabs
A class of new type graded infinite posets with minimal element are considered. These so called cobweb posets introduced recently by the present author provide a wide range of new noncommutative prefab combinatorial schema with characteristic graded subposets as primes. The schema are defined here via relaxing commutativity and associativity requirements imposed on the composition of prefabs by the fathers of this fertile concept. The construction and the very first basic properties of cobweb prefabs are pointed out in what follows. An another single valued commutative amd associative composision is also considered.
Bibliography
[1] A. K. Kwa¶niewski Cobweb posets as noncommutative prefabs submitted 2005 (October). ArXiv: math.CO/0503286 2005
[2] see references in [1] above
November 2005 A.K. Kwa¶niewski
przedstawia E. Krot-Sieniawska
Prefab posets Whitney numbers
We introduce a natural partial order in structurally natural finite subsets the cobweb prefabs sets recently constructed by the present author. Whitney numbers of the second kind of the corresponding subposet which constitute Stirling-like numbers triangular array are then calculated and the explicit formula for them is provided. Next - in the second construction - we endow the set sums of prefabiants with such an another partial order that their their bell like numbers include fibonacci triad sequences introduced recently by the present author in order to extend famous relation between binomial newton coefficients and fibonacci numbers onto the infinity of their relatives among which there are also the fibonacci triad sequences and binomial-like coefficients (incidence coefficients included).
Bibliography
[1] A. K. Kwa¶niewski Prefab posets Whitney numbers in press 2005 (October). ArXiv: math.CO/0510027 Mon, 3 Oct 2005
[2] see references in [1] above
October 2005 Ewa Krot-Sieniawska The first ascent into the incidence algebra of the Fibonacci cobweb poset
The explicite formulas for möbiusien function and some other important elements of the incidence algebra are delivered. For that to do one uses kwa\'sniewski's construction of his fibonacci cobweb poset in the plane grid coordinate system.
Bibliography
[1] E. Krot The first ascent into the incidence algebra of the Fibonacci cobweb poset Advan. Stud. Contemp. Math. The paper is to appear in volume 11 (2), 2005 (October). ArXiv: math.CO/0411007 31 Oct 2004
[2] see references in [1] above
17 October 2005 Wies³aw Bajguz Counting of elements of the composition
The number of compositions of positive integer n is 2n-1 – of course. Let us however try to count various subsets of a composition: compositions with parts each less than N, with m different sizes of parts, palindromic compositions, etc. The knowledge of recurrences and generating functions can help in counting problems. We shall start discussing these and related topics.
Bibliography
[1] George E. Andrews, Sun T. Soh The Theory of Compositions: IV. Compositions with designated summands, (submitted).
[2] Phyllis Chinn, Silvia Heubach, (1,k)-Compositions, Congressus Numerantium 164 (2003), pp. 183-194.
[3] A Knopfmacher and M E Mays The sum of distinct parts in compositions and partitions, Bull. Inst. Comb. & Appl. 25 (1999), 66-78.
[4] Arnold Knopfmacher, M. E. Mays Compositions with m distinct parts Ars Combinatorica 53 (1999), 111-128.
[5] Arnold Knopfmacher, Neville Robbins Compositions with parts constrained by the leading summand, to appear in Ars Combinatoria.
May 2005 Ewa Krot Properties and Applications of Stirling and Bell numbers and their extensions II
Following [1,2] one applies probabilistic argumentation apart from combinatorial one in order to re-derive in that way known results and in order start further applications of Stirling and Bell numbers extensions [3].
Bibliography
[1] D.Branson Stirling numbers and Bell numbers: their role in combinatorics and probability The Mathematical Scientist, Vol. 25, pp.1-31, 2000
[2] D. Branson An extension of Stirling numbers The Fibonacci Quarterly, Vol. 34, pp. 213-223, 1996
[3] A.K.Kwa¶niewski Information on some recent applications of umbral extensions to discrete mathematics to appear in the Proceedings of the International Symposium on Recent Advances in Mathematics and its Applications (ISRAMA 2004) pp. 18-20, December 2004 Kolkata (Calcutta), INDIA ArXiv: math.CO/0411145 7 Nov 2004
May 2005 A. K. Kwa¶niewski,
Wies³aw Bajguz
Whitneys logical expansion and hybrid sets
The formulation of Inclusion – Exclusion Principle by Whitney [1] is being recognized as the one implicitly introducing the algebra of hybrid sets.
Bibliography
[1] Whitney Hassler A logical expansion in mathematics Bull. Amer. Math. Soc. (1932) 38, 572-579
March-April 2005 A. K. Kwa¶niewski Applications diverses de la théorie combinatoire des especes de structures- dapres Jacques Labelle [1] et André Joyal [2]
Le but du présent cycles des presentations est dillustrer ledificacité de la Théorie des especes de structures catégorial par un grand nombres dexamples et de lappliquer a divers problemes de denombrement. Cest sont lapplication simples aussi bien que les plus avancé.
les mots clef : foncteur, ...etc.,...,vertébrés , arborescences, derangeménts, partitions etc.
Bibliography
[1] J. Labelle, Applications diverses de la théorie combinatoire des especes de structures, Annales des Sciences Mathématiques du Québec, (1983) 7 (1), 59-94. (MR 85d: 05018). http://www.labmath.uqam.ca/~annales/volumes/7-1/PDF/59-94.pdf
[2] André Joyal Une théorie combinatoire des séries formelles, Advances in Mathematics, 42, 1981, 1-82.
[3] Federico G. Lastaria : An Invitation to Combinatorial Species http://math.unipa.it/~grim/ELastaria221-230.PDF
March-April 2005 E. Krot
A. K. Kwa¶niewski
Properties and Applications of Stirling and Bell numbers and their extensions I
Following [1,2] one applies probabilistic argumentation apart from combinatorial one in order to re-derive in that way known results and in order start further applications of Stirling and Bell numbers extensions [3].
Bibliography
[1] D.Branson. Stirling numbers and Bell numbers: their role in combinatorics and probability The Mathematical Scientist, Vol. 25, pp.1-31, 2000
[2] D. Branson. An extension of Stirling numbers, The Fibonacci Quarterly, Vol. 34, pp. 213-223, 1996
[3] A.K.Kwa¶niewski Information on some recent applications of umbral extensions to discrete mathematics to appear in the Proceedings of the International Symposium on Recent Advances in Mathematics and its Applications (ISRAMA 2004) pp. 18-20, December 2004 Kolkata (Calcutta), INDIA ArXiv: math.CO/0411145 7 Nov 2004
February-March 2005 Ewa Borak Further Remarks on Duality Triads
The notion of duality triads [1] is presented . Several examples are quoted [1,2] . The new properties of duality triads are revealed . These are presented following Matthias Schork’s generalization of [2] in [3] . His possible way of generalization of duality triads to higher order recurrences is to be discussed also.
Bibliography
[1] A. K. Kwasniewski On duality triads. Bull. Soc. Sci. Lett. Lodz Ser. Rech. Deform. 53, Ser. Rech.Deform. 42 (2003) pp.11-25. math.GM/0402260 [2] A. K. Kwasniewski The second part of on duality triads' paper Bull. Soc. Sci. Lett. Lodz Ser. Rech. Deform. 53, Ser. Rech.Deform. 42 (2003) pp.25-37. math.GM/0402288
[3] Ewa Borak A note on special duality triads and their operator valued counterparts. math.CO/0411041 8 Nov 2004.
[4] Matthias Schork Some remarks on duality triads (draft version- December 2004)
February 2005 A. K. Kwa¶niewski Les especes de structures et les operations combinatoires- lapplications simples
Cest la recapitulation et en meme cas une introduction elementaire aux la theory des especes de structures dapres André Joyal [1] avec quelques application simples; (vertébrés , arborescences, derangeménts, partitions etc.)
Bibliography
[1] André Joyal Une théorie combinatoire des séries formelles, Advances in Mathematics, 42, 1981, 1-82.
[2] Gessel , R. Stanley Algebraic enumerationHandbook of combinatorics (vol. 2) Pages: 1021 - 1061 , 1996 MIT Press Cambridge, MA, USA 10
[3] Mathemagics , Matematyka Dyskretna http://ii.uwb.edu.pl/akk/index.html http://globegate.utm.edu/french/globegate_mirror/vocab.html, http://users.skynet.be/providence/vocabulaire/francais/menu.htm
January 2005 Ewa Borak
Wies³aw Bajguz
Expressing Benoulli numbers (*) and their extensions via Stirling numbers
Following [1,2] one presents a derivation of - among others- the expression of Benoulli numbers in terms of Stirling numbers of the second kind. This is also in order to investigate possible Stirling numbers extensions being involved [3].
Bibliography
[1] M.Kaneko The Akiyama-Tanigawa algorithm for Bernoulli numbers J. Integer Seq. 3 (2000), no. 2, Article 00.2.9, 6 pp. (electronic). Z0982.11009; M2001k:11026
[2] CHEN KWANG-WU Algorithms for Bernoulli numbers and Euler numbers J. Integer Seq. 4 (2001), no. 1, Article 01.1.6, 7 pp. (electronic).Z0973.11021; M2002f:11019
[3] A.K.Kwa¶niewski Information on some recent applications of umbral extensions to discrete mathematics to appear in the Proceedings of the International Symposium on Recent Advances in Mathematics and its Applications (ISRAMA 2004) pp. 18-20 December, 2004 Kolkata (Calcutta), INDIA ArXiv: math.CO/0411145 7 Nov 2004
December 2004 Ewa Borak A generalization and common combinatorial interpretation of binomial & q-binomial Gaussian coefficient and Stirling & q-Stirling numbers coefficients
One presents – following Voigt [1] a simultaneous generalization of binomial, Gaussian and Stirling numbers obtained due to an unified combinatorial interpretation of these as a problem of words. A comparison with the similar extension of binomial, Gaussian and Stirling coefficients due to Konvalina [2,3] is one of the goal of the presentation.
Bibliography
[1] Bernd Voigt A common generalization of binomial coefficients, Stirling numbers and Gaussian coefficients Publ. I.R.M.A. Strasbourg, 1984, 229/S–08 Actes 8e Seminaire Lotharingien, p. 87–89 http://www.mat.univie.ac.at/~slc/opapers/s08voigt.pdf
[2] J. Konvalina Generalized binomial coefficients and the subset-subspace problem Adv. in Appl. Math. Vol. 21 (1998) : 228-240
[3] J. Konvalina A Unified Interpretation of the Binomial Coefficients, the Stirling Numbers, and the Gaussian Coefficients The American Mathematical Monthly vol. 107, No 10 (2000):901-910
December 2004 Wies³aw Bajguz On Kims generalization of sums of powers of consecutive integers formula
Formulas for sum of powers of consecutive numbers in natural way leads to Bernoulli numbers and Bernoulli polynomials. The generalization of this is presented following [1], where natural numbers involved in the problem are replaced by q-integers. Thus sums of powers of consecutive q-numbers are investigated including the corresponding extension of Bernoulli numbers and polynomials.
Bibliography
[1] Taekyun Kim Sums of powers of consecutive q-integers to appear
[2] Yuan-Yuan Shen A note on the sums of powers of consecutive integers Tunghai Science vol. 5, 2003, pp. 101-106
listopad 2004Wies³aw Bajguz Potêgi kolejnych liczb naturalnych, wielomiany i liczby Bernoulliego
Tematem seminarium bêd± sumy postaci 1n+2n+ 3n+... +kn, gdzie k,n >0
Formu³a na sumy sze¶cianów (n=3) by³a znana ju¿ oko³o roku 500 n.e. W XII wieku w Bagdadzie i niezale¿nie w XIII w. W Chinach i Indiach matematycy odkryli zale¿no¶ci pozwalaj±ce wyliczaæ sumy wy¿szych potêg. Jednak skuteczne metody obliczania tych sum znalaz³ dopiero Bernoulli (1654-1705).
Literatura:
[1] D. Bressoud Calculus Before Newton and Leibnitz, Part III, AP Central 7/8/04
[2} Yuan-Yuan Shen A Note on the Sums of Powers of Consecutive Integers, Tunghai Science Vol. 5, July 2003. pp. 101-106
pa¼dziernik 2004Ewa Krot O pewnej klasie krat geometrycznych zwi±zanych z grupami skoñczonymi
Niech X={x1,x2,...,xn}, |X|=n i niech G bêdzie skoñczon± grup± multiplikatywn±. Skonstruowana bêdzie krata czê¶ciowych G-partycji zbioru oraz zbadane bêd± jej w³a¶ciwo¶ci. Poka¿ê, ¿e jest to krata geometryczna (zupe³na, atomowa, semi-modularna). Wyznaczê jej funkcjê Mobiusa, liczby Whitney’a oraz wielomian charakterystyczny, a tak¿e przedstawiê to¿samo¶ci wielomianowe analogiczne do tych spe³nianych przez liczby Stirlinga drugiego rodzaju (tj. liczby Whitney’a w kracie partycji zbioru -elementowego) .
Literatura:
[1] T.A.Dowling: A class of Geometric Lattices Based on Finite Groups, Journal of Combinatorial Theory (B),14(1973)61-86
pa¼dziernik 2004A.K. Kwa¶niewski Umbral extended Stirling and Bell numbers
One introduces explicit formulas for two kinds of Stirling and Bell numbers umbral extensions which include all q-extended ones recalled recently by Wagner [2]. The umbral extended correspondents to the famous combinatorial sequences above proposed by Kwa¶niewski are the following:
a] GHW (*) originated psi-umbral extensions (announced in [1]) and alternative
b] Comtet [3,2] psi-umbral extensions of Stirling and Bell numbers.
The corresponding Dobinski formula for a] was already delivered in [1] based on recent (see: [1]) works of the author where – among others -umbral extended Poisson distributions were introduced.
(*)Graves C. On the principles which regulate the interchange of symbols in certain symbolic equations Proc. Royal Irish Academy vol. 6 (1853-1857), pp. 144-152
S. Pincherle, U. Amaldi Le operazioni distributive e le loro applicazioni all`analisi, N. Zanichelli , Bologna , 1901 (see also [7,8] - below)
Literatura:
[01] A.K. Kwasniewski Psi-Poisson, q-Cigler, psi-Dobinski, psi-Rota and psi-coherent states ArXive: math.CO/0403054 v1 2 March 2004
[02] Carl G. Wagner Partition Statistics and q-Bell Numbers (q = -1) Journal of Integer Sequences, Vol. 7 (2004), Article 04.1.1
[03] Comtet L. Nombres dr Stirling generaux et fonctions symmetriques C.R. Acad. Sci. Paris, Series A, 275 (1972):747-750
[04] M. Wachs, D. White p,q Stirling numbers and set partitions Statistics J. Combin. Theory (A) 56 (1991): 27
[05] M. Schork On the combinatorics of normal ordering bosonic operators and deformations of it J. Phys. A: Math. Gen. 36 (2003) 4651-4665
[06] J.Katriel, M. Kibler Normal ordering for deformed boson operators and operator-valued deformed Stirling numbers J. Phys. A: Math. Gen. 5 (1992): 2683-26-91
[07] A.K.Kwa¶niewski On Simple Characterisations of Sheffer psi-polynomials and Related Propositions of the Calculus of Sequences, Bull. Soc. Sci. Lett. Lodz; 52, Ser. Rech. Deform. 36 (2002) pp.45-65 ArXiv: math.CO/0312397
[08] A.K.Kwa¶niewski Main theorems of extended finite operator calculus Integral Transforms and Special Functions Vol 14, No 6, (2003): 499-516
wrzesieñ 2004Ewa Borak Liczby Whitney’a pierwszego i drugiego rodzaju
Rozwa¿ymy trzy ci±gi wielomianów bazowych przestrzeni R[x]: ci±g potêg xn, ci±g potêg ubywaj±cych i ci±g potêg krocz±cych () oraz wspó³czynniki przej¶cia z jednej bazy do drugiej (tzw. connection constants ). Z ka¿d± par± ci±gów zwi±zany jest odpowiedni poset (zbiór czê¶ciowo uporz±dkowany), a liczby Whitney’a pierwszego i drugiego rodzaju dla tego posetu definiuj± ¿±dane connection constants. Przy pomocy funkcyj odwzorowuj±cych zbiór uniwersum U w pier¶cieñ liczb ca³kowitych wprowadza siê pojêcie kolekcji dopuszczaj±cej ujemn± liczbê swych niektórych elementów. Omawia siê wstêpnie elementarne, kombinatoryczne skutki tego rozszerzenia pojêcia zbioru z powtórzeniami (multi-set) wszechobecnego w kombinatoryce ery nowo¿ytnej.
Literatura:
[1] Joni S.A., Rota G.-C., Sagan B.: From sets to functions: three elementary examples Discrete Mathematics 37 (1981), 193-202.
07-06-2004Wies³aw Bajguz Zbiory hybrydowe – czyli zbiory dopuszczaj±ce ujemn± liczbê elementów
Przy pomocy funkcyj odwzorowuj±cych zbiór uniwersum U w pier¶cieñ liczb ca³kowitych wprowadza siê pojêcie kolekcji dopuszczaj±cej ujemn± liczbê swych niektórych elementów. Omawia siê wstêpnie elementarne, kombinatoryczne skutki tego rozszerzenia pojêcia zbioru z powtórzeniami (multi-set) wszechobecnego w kombinatoryce ery nowo¿ytnej.
Literatura:
[1] Loeb D.E. Sets with Negative Number of Elements Adv. Math. 91 No1 64-74 (1992)
[2] Loeb D.E. A generalization of the binomial coefficients Disrete. Mathemetics 105 (1992) pp.143-156
(*) rodowód: (przed likwidacj± ZMD)
wspomnienie
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You are currently browsing the category archive for the ‘MATH6040’ category.
## Test 1 – Results
Have been emailed to you,
## Week 6
We finished Chapter 2 by looking at Determinants and Cramer’s Rule. Ye did an example in class, video here: Cramer’s Rule Example
The new material was recorded here.
We had about 90 minutes of tutorial time looking at Chapter 2.
## Test 1 – Results
I will have these to you by the end of the week.
## Week 5
After our test, we continued to look at Matrix Inverses — “dividing” for Matrices. This allowed us to solve matrix equations. We will look at linear systems, and spoke again about determinants.
## Test 1
Test 1, worth 15%, takes place from 19:00 to 20:05 sharp, Tuesday 25 February in the usual lecture venue. There is a sample on P.53 of the notes to give you an idea of the length and layout only.
If you want questions answered you have two options:
• ask me questions via email, perhaps with a photo to show your work
• ask me questions via the comment function on this website
## Week 4
We had a Concept MCQ about vectors, and then we started looking at Chapter 2: Matrices. We did some examples of matrix arithmetic and look at Matrix Inverses — “dividing” for Matrices.
Here find a note that answers the question: why do we multiply matrices like we do?
## Test 1
Test 1, worth 15%, takes place from 19:00 to 20:05 sharp, Tuesday 25 February in the usual lecture venue. There is a sample on P.53 of the notes to give you an idea of the length and layout only.
Everything in Chapter 1 is examinable. Additional practise questions may be found by looking at past MATH6040 exam papers (usually vectors are Q. 1, sometimes Q. 2).
You will want to be familiar with all the concepts in the Vector Summary, P. 49-52.
If you want questions answered you have three options:
• hand me up written work next week, which I will correct, scan, and email back to you
• ask me questions via email, perhaps with a photo to show your work
• ask me questions via the comment function on this website
## Week 3
In Week 3 will found out how to calculate a vector product, and we looked at the applications of vectors to work and moments
We managed about 30 minutes of tutorial time.
## Week 2
In Week 2 we finished talking about the scalar product before doing some tutorial work on the p.29 exercises.
We started talking about the vector product, but we did not show how to calculate it.
## Homework/Study
How much time you put into homework is up to you: of course the more time you put in the better but we all have competing interests. Please feel free to ask me questions about the exercises via email, or even better on this webpage.
I would invite you to complete the exercises we started in class:
• P.29, Q.1-11
In addition, you can hand up your work in class and I will have it correct for the week after. You may need the formula that the area of a triangle is equal to $\displaystyle \frac12 ab\sin C$.
## Week 3
In Week 3 we will find out how to calculate a vector product, and we will look at the applications of vectors to work and moments
We might manage a little tutorial time.
## Test 1
If we finish the Vectors chapter in Week 3 the test will be in Week 5 (this is very likely): otherwise we will push this out to Week 6. Official notice will be given in Week 3 (or Week 4 if necessary). There is a sample test in the notes.
## Study
Please feel free to ask me questions about the exercises via email or even better on this webpage.
If you are a little worried about your maths this semester, perhaps after the Diagnostic Test or in general, I would just like to remind you about the Academic Learning Centre. Next week, some students will receive emails detailing areas of maths that they should brush up on. The timetable is here: there is some availability after 17:00 on Mondays and Tuesdays.
## Strike!
There is 24 hour strike this Tuesday, and so we will have our Week 2 class on Wednesday 5 February, 19:00-22:00, in the same room, B188.
## Diagnostic Test
If you haven’t already, you are invited to take the following ‘Diagnostic Test’:
click here
This ‘Test’ does not go towards your grade, but allows me to give you some feedback on where you are in terms of material you have seen before that will be used in this module.
## Manuals
The manuals are available in the Copy Centre and must be purchased before next Wednesday’s class.
• The first piece of advice is to read questions carefully. Don’t glance at a question and go off writing: take a moment to understand what you have been asked to do.
• Don’t use tippex; instead draw a simple line(s) through work that you think is incorrect.
• For equations, check your solution by substituting your solution into the original equation. If your answer is wrong and you know it is wrong: write that on your script.
If you do have time at the end of the exam, go through each of your answers and ask yourself:
2. does my answer make sense?
3. check your answer (e.g. differentiate/antidifferentiate an antiderivative/derivative, substitute your solution into equations, check your answer against a rough estimate, or what a picture is telling you, etc)
## Student Feedback
You are invited to give your feedback on my teaching and this module here.
## Test 2 – Results
Have been emailed to you along with final CA results.
## Week 12
We looked at centroids of laminas and centres of gravity of solids of revolution.
We had one and a half classes of tutorial time.
## Week 13
There is an exam paper at the back of your notes — I will go through this on the board in the lecture times (in the usual venues):
• Monday 16:00
• Tuesday 09:00
• Thursday 09:00
We will also have tutorial time in the tutorial slots. You can come to as many tutorials as you like.
• Monday at 09:00 in B180
• Monday at 17:00 in B189
• Wednesday at 10:00 in F1. 3
The exam is on Friday 13 December. Past exam papers (MATH6040 runs in Semester 1 and Semester 2) may be found here.
## Student Feedback
You are invited to give your feedback on my teaching and this module here.
## Test 2 – Results
Possibly early next week. I am a little behind in my corrections after being sick last week though. Definitely Thursday.
## Week 11
We had our test on Monday, then will looked at completing the square, and work on Tuesday and Thursday.
Here is some video of revision antidifferentiation.
## Week 12
We will look at centroids of laminas and centres of gravity of solids of revolution. Any spare lecture time will be given over to tutorial time.
## Week 13
There is an exam paper at the back of your notes — I will go through this on the board in the lecture times (in the usual venues):
• Monday 16:00
• Tuesday 09:00
• Thursday 09:00
We will also have tutorial time in the tutorial slots. You can come to as many tutorials as you like.
• Monday at 09:00 in B180
• Monday at 17:00 in B189
• Wednesday at 10:00 in F1. 3
The exam is on Friday 13 December.
## Test 2
On Chapter 3, on at 5 pm (not 4 pm) Monday 25 November, Week 11 in Melbourne Hall. You will have one full hour.
The BioEng2B tutorial will take place from 4 pm sharp to 4.45 pm in B263 on that day.
Chapter 3: Differentiation is going to be examined. A Summary of Differentiation (p.147-8): you will want to know this stuff very well. You will be given a copy of these tables
There is a sample test on p. 149 of the notes.
I strongly advise you that attending tutorials alone will not be sufficient preparation for this test and you will have to devote extra time outside classes to study aka do exercises.
Between tutorials and private study you really should aim to have completed as least the following:
• P. 113, Q. 1-6 (not 5c or 6iii)
• P. 119, Q. 1-4
• P. 127, Q. 1-4
• P. 138, Q. 1-3
• P. 146, Q. 1-4
• The Sample Test
There are more questions in most of these exercises.
If you are having any problem, take a photo of your work and email me your question.
## Week 10
We started Chapter 4 on (Further) Integration with a revision of antidifferentiation, and had a look at Integration by Parts. We used implicit differentiation to differentiate inverse sine.
## Week 11
We have our test on Monday, then we will look at completing the square, and work on Tuesday and Thursday.
## Week 12
We will look at centroids of laminas and centres of gravity of solids of revolution. Any spare lecture time will be given over to tutorial time.
## Week 13
There is an exam paper at the back of your notes — I will go through this on the board in the lecture times (in the usual venues):
• Monday 16:00
• Tuesday 09:00
• Thursday 09:00
We will also have tutorial time in the tutorial slots. You can come to as many tutorials as you like.
• Monday at 09:00 in B180
• Monday at 17:00 in B189
• Wednesday at 10:00 in F1. 3
The exam is on Friday 13 December.
## Test 2
On Chapter 3, on at 5 pm (not 4 pm) Monday 25 November, Week 11 in Melbourne Hall
The BioEng2B tutorial will take place from 4 pm sharp to 4.45 pm in B263 on that day.
Chapter 3: Differentiation is going to be examined. A Summary of Differentiation (p.147-8): you will want to know this stuff very well. You will be given a copy of these tables
There is a sample test on p. 149 of the notes.
I strongly advise you that attending tutorials alone will not be sufficient preparation for this test and you will have to devote extra time outside classes to study aka do exercises.
Between tutorials and private study you really should aim to have completed as least the following:
• P. 113, Q. 1-6 (not 5c or 6iii)
• P. 119, Q. 1-4
• P. 127, Q. 1-4
• P. 138, Q. 1-3
• P. 146, Q. 1-4
• The Sample Test
There are more questions in most of these exercises.
If you are having any problem, take a photo of your work and email me your question.
## Week 9
We looked at partial differentiation and its applications to differentials and error analysis.
I had hoped to do a Concept MCQ for Chapter 3 but we kind of ran out of time… I would ask ye to complete this (on your own) before Monday.
## Week 10
We will start Chapter 4 on (Further) Integration. A good revision of integration/antidifferentiation may be found here, but this material will be gone through in Monday’s lecture..
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# Generation of a Turbulent Boundary Layer Inflow for RANS Simulations
Rona, Aldo and Grottadaurea, Marco and Monti, Manuele and Airiau, Christophe and Gandhi, Thangasivam Generation of a Turbulent Boundary Layer Inflow for RANS Simulations. (2009) In: 15th AIAA/CEAS Aeroacoustics Conference (30th AIAA Aeroacoustics Conference), 11 May 2009, Miami, USA .
(Document in English) PDF (Author's version) - Depositor and staff only - Requires a PDF viewer such as GSview, Xpdf or Adobe Acrobat Reader 727kB
Official URL: http://arc.aiaa.org/doi/abs/10.2514/6.2009-3272
## Abstract
The generation of a fully turbulent boundary layer profile is investigated using analytical and numerical methods over the Reynolds number range $300 \leq Re_{\theta} \leq 31000$. The predictions are validated against reference wind tunnel measurements with a zero streamwise pressure gradient. The analytical method is then tested for favourable pressure gradient by modelling the turbulent boundary layer approaching a two-dimensional potential sink. Both methods show a good predictive ability under a zero pressure gradient, with the numerical method providing a complete velocity profile through the laminar sub-layer down to the wall. This work is of practical interest to computational fluid dynamic practitioners for generating an equilibrium thick turbulent boundary layer at the computational domain inflow.
Item Type: Conference or Workshop Item (UNSPECIFIED) International conference proceedings French research institutions > Centre National de la Recherche Scientifique - CNRS (FRANCE)Université de Toulouse > Institut National Polytechnique de Toulouse - Toulouse INP (FRANCE)Université de Toulouse > Université Toulouse III - Paul Sabatier - UT3 (FRANCE)Other partners > University of Leicester (UNITED KINGDOM) download 21 Feb 2013 09:00
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# Math Help - Nested Sequences
1. ## Nested Sequences
Let (In) be a nested sequence of closed bounded intervals. For each natural number n, let xn be in In. Use the Bolzano-Weierstrass Theorem to give a proof of the Nested Intervals property.
2. Originally Posted by frenchguy87
Let (In) be a nested sequence of closed bounded intervals. For each natural number n, let xn be in In. Use the Bolzano-Weierstrass Theorem to give a proof of the Nested Intervals property.
Intervals is $\mathbb{R}$? Do you have to use BW? It's easier to use the fact that since each interval is bounded and the diameter of the sets approaches zero (which I assume is what you want) then we get by the completeness of [tex]\mathbb{R}[/mat] that the intersesection contains a single point. Otherwise, argue that if we have $[a_n,b_n]$ that each must converge to a point by BW. And use the diameter approaching zero to show that it the two points must be equal.
3. ## Boundedness Removed
I do have to use BW...
4. Originally Posted by frenchguy87
I do have to use BW...
Ok. That's fine. But, what have you done as of now?
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# Proving that Aut($S_3$) is isomorphic to $S_3$
I'm doing an exercise were I had to first prove that all automorphisms of $S_3$ induce a permutation in $X= \{ \alpha \in S_3 \, / \,$order$(\alpha) = 2\}$, which was easy enough.
Now I have to prove that $Aut(S_3) \cong S_3$. I noticed that evaluating an automorphism in an element of X and taking the product of that with any other cycle doesnt work, since there's no way you're getting all of $S_3$ out of that. Evaluation in general doesn't seem too useful in general.
I tried googling for a bit and found a pdf that says the following:
Any automorphism of $S_3$ must send elements of order 2 to elements of order 2; in this case the only elements of order 2 are the transpositions, so an element of Aut($S_3$) permutes the transpositions: that is, we obtain a map $\phi$ : Aut($S_3) \rightarrow S_{ \{(1\,2),(2\,3),(1\,3)\}}$ . Since the transpositions generate $S_3$ , this map is injective. On the other hand #Aut($S_3) \geq 6$, since Aut($S_3$) contains the inner automorphisms and Z($S_3$) is trivial. Hence φ is an isomorphism.
I understand that since the 2-cycles generate $S_3$ any automorphism is determined by it's values in them but I'm not sure what $\phi$ would look like nor why it has to exist exactly.
Any help would be greatly appreciated.
Edit: I think I've got it. I've got an isomorphism between Aut($S_3$) and $S_{\{(12),(13),(23)\}}$ by $\phi (f) = (f(1\,2)\,f(1\,3)\,f(2\,3))$, and then I have an isomorphism between $S_{\{(12),(13),(23)\}}$ and $S_3$ by relabeling the elements ie $(12) \rightarrow 1$ , $(23) \rightarrow 2$ and $(23) \rightarrow 3$.
• Are you sure $(23)\rightarrow 3$, not $(13)\rightarrow 3$ ? – Anjan3 Sep 14 '15 at 7:16
• There was a typo, one of the $(23)$ in "$(23) \rightarrow2$ and $(23) \rightarrow 3$" should be a $(13)$. I'm just looking at the problem again for the first time in a while, but I don't see why it would matter if $(23) \rightarrow3$ or $(13) \rightarrow3$. – John Williams Sep 15 '15 at 14:06
$S_3$ is generated by its elements of order $2$, and the elements of $S_3$ of order $2$ are exactly the single transpositions. So all you need to show is that if you send transpositions to transpositions (through the automorphism on $S_3$), then this is equivalent to just permuting the 3 underlying elements. And then you need that permuting the $3$ elements gives an automorphism on $S_3$ always, but this is trivial.
• I've managed to proved that any automorphism of $S_3$ is determined by it's values in the single transpositions. That way I have an isomorphism between Aut($S_3$) and the permutations of the single transpositions, which I think is similar to what you meant in the first part. But now I think I need an isomorphisms from the permutations on the 2-cycles to $S_3$ itself, don't I? – Martin Williams Apr 28 '14 at 19:03
• I think I've got it. I've got an isomorphism between Aut($S_3$) and $S_{\{(12),(13),(23)\}}$ by $\phi (f) = (f(1\,2)\,f(1\,3)\,f(2\,3))$, and then I have an isomorphism between $S_{\{(12),(13),(23)\}}$ and $S_3$ by relabeling the elements ie $(12) \rightarrow 1$ , $(23) \rightarrow 2$ and $(23) \rightarrow 3$. – Martin Williams Apr 28 '14 at 19:29
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# Second derivative seminorm of function composition
Consider the seminorm $\| f \|^2 = \int_{-\infty}^{\infty} dx f''(x)^2$
for $f:\mathbb{R}\rightarrow \mathbb{R}$ in the Sobalev space $W^{k,2}(\mathbb{R})$.
Can we put some upper bound on the composition $\| f\circ g \|$ in terms of $\| f \|$ and $\| g \|$?
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I don't think so. Consider the rescaling $f_\lambda(x) = f(\lambda x)$ and $g_\lambda(x) = g(\lambda x)$. Then the term $\|f_\lambda\circ g_\lambda\|^2$ scales like $\lambda^7$ while $\|f_\lambda\|^2\|g_\lambda\|^2$ scales like $\lambda^6$.
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# Tag Info
3
Yes, this is secure, even though scrypt uses PBKDF2 inside. PBKDF2 has the issue that it the work factor is required $n$ times where $n$ is the number hash outputs concatenated to create the final PBKDF2 output. That means that if you can check the validity of PBKDF2 using only the initial bits (in your case used for the key if the hash was SHA-256, for ...
3
No the IV doesn't get encrypted. The IV is a random vector to make sure that the ciphertext is not identical for identical plaintext. This would leak information to any eavesdropper. It needs to be unique - and in the case of CBC, indistinguishable from random to the eavesdropper ("unpredictable") - but not confidential. As the IV is separate from the ...
1
The authentication tag in GCM is generated by XORing a block cipher output with the Galois field hash (and truncating it for shorter lengths). It is thus assumed to look PRF. So it is effectively just a random nonce that should not collide until a birthday bound of $2^{t/2}$. With a tag length of 96 or more bits, it should be secure. Shorter random IV ...
1
Using the same key and IV for different, independently encrypted cells is a bad idea, regardless of which encryption algorithm you use. It would allow attackers to find either XORs of cells (with CTR) or at least equality of prefixes (with CBC). If you are going to use authenticated encryption, you need to choose whether the MAC applies to each cell ...
Only top voted, non community-wiki answers of a minimum length are eligible
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# nLab co-span
A co-span in a category $V$ is a diagram
$\array{ && S \\ & \nearrow && \nwarrow \\ a &&&& b }$
in $V$, i.e. a span in the opposite category $V^{op}$.
Co-spans in a category $V$ with small co-limits form a bicategory whose objects are the objects of $V$, whose morphisms are co-spans between two objects, and whose 2-morphisms $\eta$ are commuting diagrams of the form
$\array{ && S \\ & {}^{\sigma_{S}}\nearrow && \nwarrow^{\tau_S} \\ a &&\downarrow^\eta&& b \\ & {}_{\sigma_T}\searrow && \swarrow_{\tau_T} \\ && T } \,.$
The category of co-spans from $a$ to $b$ is naturally a category enriched in $V$: for
$\array{ && S \\ & {}^{\sigma_{S}}\nearrow && \nwarrow^{\tau_S} \\ a &&&& b \\ & {}_{\sigma_T}\searrow && \swarrow_{\tau_T} \\ && T }$
two parallel cospans in $V$, the $V$-object ${}_a[S,T]_b$ of morphisms between them is the pullback
$\array{ {}_a[S,T]_b &\to& pt \\ \downarrow && \downarrow^{\sigma_T \times \tau_T} \\ [S,T] &\stackrel{\sigma_S^* \times \sigma_T^*}{\to}& [a \sqcup b, T] }$
formed in analogy to the enriched hom of pointed objects.
If $V$ has a terminal object, $pt$, then co-spans from $pt$ to itself are bi-pointed objects in $V$.
# References
Topological cospans and their role as models for cobordisms are discussed in
• Marco Grandis, Collared cospans, cohomotopy and TQFT (Cospans in algebraic topology, II) (pdf)
Revised on May 17, 2013 23:56:36 by Urs Schreiber (89.204.154.16)
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Adjective. Adjective vs. Adverb. Additionally, not all adverbs end in -ly: here, there, together, yesterday, aboard, very,almost, … (Adverb)Sue has an early start in the morning. Adverbs … What's the adjective for difference? learn the difference between adverb and adjective phrases. “She seems unhappy today.” Here, unhappy is an adjective that modifies the pronoun, she. The main difference between adverb and adjective is that adverbs modify verbs, adjectives, and other adverbs whereas adjectives modify nouns.. What is an Adverb. The main difference between an adverb and an adjective is in what they describe: adjectives describe a noun or pronoun, while adverbs are used to describe verbs and other non-nouns/pronouns. Asking for a thing. “I feel badly” means that you are bad at feeling things. It can sometimes be difficult to tell the difference between adjectives and adverbs in a sentence. (Adjective)Sue starts early. An adjective is a word used to describe a noun. When comparing the difference between adverb and adjective, we already know that it is the questions that each part of speech answers that ultimately help us to summarize their varied natures. Also includes a practice activity where students identify adverbs and adjectives as used in sentences. Here are some special situations that give even native English speakers problems: Deciding when to use good vs well: Good is an adjective, and well is an adverb, except when you’re talking about your health: I am good. (Adjective)Billy speaks fast. Features of Adjectives in English First of all, adjectives in English describe nouns or pronouns. 3. This leads to a common type of error: incorrectly substituting an adverb in place of a predicate adjective. Adjectives can come before the word they describe. An adjective is one of the eight parts of speech which define a noun. We can find them everywhere and are generally present when the language is used, however, at no time should be confused; since their functions are totally different. Adjective clauses begin with a relative pronoun, while adverb clauses start with a subordinating conjunction. As an adjective modifies the noun, an adverb modifies the verb. Included below are past participle and present participle forms for the verbs differ , difference , differentiate , diff and differentialize which may be used as adjectives within certain contexts. A cleverly developed idea. She walks slowly (tells how)She walks very slowly (very – tells how slowly)My cousin arrived today (tells when)My cousin will arrive in an hour (this adverb phrase tells when)We looked for her toy in the attic (tells where). What is an adjective? If you would like to read some Grammar Notes about Adverbs and Adjectives and the difference between them, visit this page: Adverbs vs. Adjectives. An adjective performs the function of qualifying a noun. Adjectives describe nouns and pronouns, whereas, adverbs describe verbs, adjectives and other adverbs. In this instance, we merely need to take a look at the example sentence outlined previously. You will learn about the main difference between adjectives and adverbs as well as the different questions they answer. Good is an We have listed the main differences in usage below, along with some tricky adverbs. Learn the difference between adverbs and adjectives in English grammar with Lingolia’s simple grammar rules and explanations. It can be said that an adverb modifies everything in a sentence except the noun and pronoun; it brings explains the verbs, adjectives and other adverbs used in the sentence. Complete the sentence with the appropriate adverb or adjective. The main difference is that: -Adjectives (French: les adjectives ) aim to describe nouns. It, therefore depends on the sentence context as to whether it is an adjective or an adverb. Sometimes, though, we use well as an adjective when we are talking about health and well-being. - We describe a thing. Modifies a verb, adjective, preposition or conjunction. It can sometimes be difficult to tell the difference between adjectives and adverbs in a sentence. Read each sentence and identify the word as either an adjective or adverb. Meanwhile, adverbials act like adverbs to modify a verb or a clause. The adverbs fairly, really and pretty can generally be used with both gradable and non-gradable adjectives.. When referring to health, we often use well rather than good. This English grammar game is to help you learn about the difference between adverbs and adjectives.. 2. An adjective is a describing word (or set of words) that modify or describe a noun or a pronoun. Using adverbs in sentences. An adverb describes a verb or anything apart from a noun and pronoun: "That boy speaks so loudly!" Not all words that end in –ly are adverbs: lovely, costly, friendly, etc. In the free exercises, you can test yourself. The difference between adjectives and adverbs is that; adjectives describe nouns or pronouns, and adverbs describe verbs and other things not related to nouns or pronouns. Adjectives change the nouns, whereas Adverbs change the verbs. Generally is an adverb, and general is an adjective. Examples of verbs that relate a state of being are: am, are, is, will, was, were. An adjective is one of the eight parts of speech which define a noun. Good and well are two words that tend to create confusion. Learn more about what sets them apart from each other with this guide. An adverb is a word, which changes or describes a verb, adjective or another adverb, which it precedes or succeeds. The difference lies in what they describe – a noun or pronoun, or a verb, adjective or other adverbs. Adjectives and adverbs - English Grammar Today - a reference to written and spoken English grammar and usage - Cambridge Dictionary http://www.engvid.com/ Many beginners get confused between adjectives and adverbs. are also non-gradable. Sources: In the table “Adverbs and Adjectives” you will learn how to form adverbs in Russian. We can find them everywhere and are generally present when the language is used, however, at no time should be … The carpenter hit the nail with a hammer. An adverb tells the way how the verb is functioning. Well, you can learn certain rules around the differences between adjectives and adverbs, as long as you know what are adjectives and adverbs. - We describe a person. The difference between adjectives and adverbs can be difficult to spot. (Adverb). Key difference: A noun is a word that is used for identifying people, places, objects, events, happenings, etc.A verb is a word used for expressing any action in a sentence. The difference lies in what they describe – a noun or pronoun, or a verb, adjective or other adverbs. Pin. The other key difference is that adjectives always agree with the gender and number of the noun they are describing, while adverbs do not change. Adverbs are used to answer how questions and adverbs could end in “ly” ex; quickly, or beautifully where as with adjectives it would be quick or beautiful. After reading through today’s post, you should finally feel confident enough to be able to make use of these helpful parts of speech going forward when creating sentence structures. Difference Between Adjective and Adverb: Conclusion. A verb is a word for an action or a state of being. If the adjective ends in -y, change -y to -i.Then add -ly:. What is an Adverb Adverbs are words that express manner, time, place, frequency, degree, etc. Nouns are one of the elementary rules of English grammar. Your email address will not be published. For example:How are you today?I'm well, thanks. An adjective usually, but not all the time, comes before the noun or the pronoun which it describes. Adverb and adjective (advanced): adjectives and adverbs with the same form (He works hard, A hard worker) and two forms of the adverb having different meanings (He hardly worked) exercise 1: choose the adjective or adverb (in combination with linking verbs) While –ly is helpful, it’s not a universal rule. person, place, animal or thing. Adverbs are words in the English language that can modify a verb, adjective, or even other adverbs. Understanding the difference between an adjective and an adverb can be confusing. As against, adverbs will answer the questions like how, when, where, how much, how often, to what extent etc. Lovely is a... gotcha! … What is the difference between Adjective and Verb? We have listed the main differences in usage below, along with some tricky adverbs. Her friend Zoe is a clever girl. Like adjectives, they tell which one, what kind, how much, or how many.. Features of Adjectives in English. Adverbs and Adjectives are both used to modify different words in a sentence. An adverb would describe how you perform the action of feeling—an adjective describes what you feel. (Adverb). Adverb: A word that modifies an adjective, a verb, or even another adverb. He bought her a necklace which was horribly expensive. Many adverbs end in –ly: Strictly is an adverb, and strict is an adjective. A lot of the time this difference can be seen in the structure of the words: 1. Knowing the ways adjective and adverb clauses differ from one another is the key to identifying them correctly. AD40 - Different Meanings of Adjectives and Adverbs Gap-fill exercise. In the free exercises, you can test yourself. Examples of Differences between Adjectives and Adverbs. To help support students in learning to make this distinction, we have created a series of grammar worksheets. The adjectives / adverbs that take the same form include:fast, hard, early, late, high, low, right, wrong, straight and long.For example: Billy has a fast car. Terms in this set (25) adverb. Today, I will explain to you in simple high school tongue the real difference between an adjective and an adverb and I will make sure that you won’t forget again ever in your life. A clevernew idea. In grade 2, students learn to distinguish between adjectives and adverbs. Adverbs are words that express manner, time, place, frequency, degree, etc. The show \on television tonight is about snow leopards \in Asia. Read on to learn more! There are three forms of adjectives and adverbs used to show varying degrees of comparison: the positive, the comparative, and the superlative. What can you remember about the difference between adjectives and adverbs? The second rule applies when writing or in some cases spelling. The word well can be an adjective, too. If you would like to read some Grammar Notes about Adverbs and Adjectives and the difference between them, visit this page: Adverbs vs. Adjectives. Difference between Adjective and Adverb: – In our oral and written speech we always use adverbs and adjectives. Adjectives and Adverbs – What’s the Difference. If you apply this ending to an adjective, it will most likely transform into an adverb. Today, I will explain to you in simple high school tongue the real difference between an adjective and an adverb and I will make sure that you won’t forget again ever in your life. : If you're not sure whether to use adjectives or adverbs, pay close attention to the following rules and explanations. After you watch this lesson, the difference will be clear. You must ask yourself what word the modifier is modifying. Examples: He bought an expensive car last week. Adjective or adverb? The main difference is that adjectives always describe a noun, whereas adverbs can describe a verb, adjective or another adverb. Clever is an adjective, and cleverly is an adverb. A thing is either annual, digital, domestic etc., or it is not.. When? An adjective is a word or a set of words that modify or describe a noun or pronoun. The main difference is that adjectives always describe a noun, whereas adverbs can describe a verb, adjective or another adverb. AD040 - Different meanings of adjectives and adverbs - English Grammar exercise. 2. An adjective and adverb phrase differ in that an adverb modifies verbs, adjectives, and other adverbs. 1. Fill in all the gaps, then press "Check" to check your answers. The woman who lives next door is a doctor. Adjectives usually come before the noun they modify although they can be used after the noun as well. Adjective or Adverb The difference between adjective and adverb. A cleverlydeveloped idea. You must ask yourself what word the modifier is modifying. An … They also have to identify if the selection is an adjective or an adverb. What’s the Difference Between Adverbs and Adjectives? Many times, adverbs are formed by adding “ly” to an adjective. Here students are given a word bank and are asked to complete the sentences with the correct adverbs. This is the main difference between adjectives and adverbs. How to Form Adverbs from Adjectives – Grammar Rules for Different Word Ending . • Both are parts of speech but very different … An adverb is a word used to modify a verb, an adjective, or another adverb. The K5 Learning Blog urges parents to be pro-active in helping their children reach their full academic potential. An adjective is a word or set of words that describes a noun or pronoun. Adjectives can also follow the word they describe. She likes a professional hockey player. Adverbs modify verbs, adjectives, and other adverbs. How, when, where, how often, to what extent and how much. Here's a quick reminder: An adjective describes a noun or pronoun: "That boy is so loud!" You are now very well informed on the difference between these two parts of speech. If you want to know if a word is an adverb, you can see if that word asks how, when, where, why, to what extent, how often, or how much as a linking verb. Adverbs modify verbs, adjectives, or other adverbs. In grammar, the adjective is among the eight parts of speech which identifies and describes a noun or a pronoun, i.e. Adjectives can come before the word they describe. Learn the difference between adverbs and adjectives in English grammar with Lingolia’s simple grammar rules and explanations. • Adjective tells us more about the noun, whereas verb tells us about the condition, experience, or the state of mind of the subject. This adjective + lyconstruction is a short-cut to identifying adverbs. adjective. She was terribly sorry for being late again. To help remember the difference, the word itself has “verb” inside it, and adverbs tend to end in “-ly.” “Slowly,” “loudly,” and “happily” are all adverbs. A lot of the time this difference can be seen in the structure of the words: A clever new idea. What is the Difference between an Adjective and an Adverb? First of all, adjectives in English describe nouns or pronouns. Adjective VS Adverb Examples | Difference Between Adjectives and Adverbs An adjective is the property of the noun and it modifies the noun which it describes adding detail to it; whereas adverb is the quality or the property of the verb. The hard and fast rule is:Good is an adjective.Well is an adverb.For example: Sarah is a good singer. As we’ve learned, adjectives and adverbs act in similar but different roles. Adjective vs. Adverb. - How is the house? There were strange green creatures with bobbing heads. To help remember the difference, the word itself has “verb” inside it, and adverbs tend to end in “-ly.” “Slowly,” “loudly,” and “happily” are all adverbs. Adverbs and Adjectives are both used to modify different words in a sentence. Many of us learned in school that adjectives modify nouns and that adverbs modify verbs. The first rule is a general rule that if a word ends in -ly then it is an adverb. There are different types and different forms of adverbs, and they can be used almost anywhere in a sentence. Adjectives typically come before the noun or pronoun that they modify. Where? Adjective provides answers to questions such as which, how many, what kind, etc. MILTON’S COMUS|JOHN MILTON; Summary. VERB Examples of action verbs are: walk, talk, think, see, eat, find, believe, sit. • Verb is an action word whereas adjective is a describing word. An adjective and adverb phrase differ in that an adverb modifies verbs, adjectives, and other adverbs. These rules will help you properly use both types of … There are a number of adjectives / adverbs that take the same form. Adverb Examples: The children were playing happily with their toys. Be careful. In the table “Adverbs and Adjectives” you will learn how to form adverbs in Russian. What are Adjectives and Adverbs. Though they, adjectives and adverb, are more connected with nouns and verbs, there is an interesting … Use the "Hint" button to get a free letter if an answer is giving you trouble. An adjective tells us more about a noun. Adjective prepositional phrases follow the nouns they modify, unlike adjectives which generally go immediately before the nouns they modify. person, place, animal or thing. This English grammar game is to help you learn about the difference between adverbs and adjectives.. If … An adjective is a word or set of words that describes a noun or pronoun. safe – safe ly Not all words ending in -ly are adverbs:. Differences Between Adjectives and Adverbs. Difference between Adjective and Adverb: – In our oral and written speech we always use adverbs and adjectives. If you’re trying to read Braille through thick leather gloves, then it might make sense for you to say “I feel badly.” tell us how something happens or is done.They are usually placed either after the main verb or after the object. Adverbs and Adjectives Lesson – A slide show lesson teaching students the differences between adverbs and adjectives. Adverbs are words that describe verbs and adjectives are words that describe nouns. Adjective vs Adverb The difference between Adjective and Adverb can be better understood by their types and how they are used. Example. Here are some example: That is an adorable kitten. If the modified word is a noun, the modifier is an adjective. Adverbs of Manner. We must finish our project before the holidays. You will learn about the main difference between adjectives and adverbs as well as the different questions they answer. Adjective is a word that describes, qualifies and identifies a noun or pronoun, whereas an adverb describes a verb, adjective or other adverbs. Adjectives describe nouns and pronouns, whereas, adverbs describe verbs, adjectives and other adverbs. A word that qualifies, identifies and describes a noun or pronoun is known as Adjective, whereas an Adverb delineates the adjective, verb, or other adverbs. If you wrote “Sharon’s cough sounds badly,” it would not make sense, because badly would be an adverb modifying the verb, sounds,meaning that her cough isn't very good at sounding. The difference between adjectives and adverbs can be difficult to spot. Adjective is a word that qualifies a noun whereas adverb is a word that describes a verb. Clever is an adjective, and cleverly is an adverb. Remember, adjectives modify nouns. Is there any difference between an adverb and an adverbial? The following examples explain the differences between adjectives and adverbs: “Sharon's cough sounds bad.” In this case, bad is an adjective that modifies the noun, cough. An adverb, on the other hand, modifies or describes a verb, adjective or another adverb. adjectives ending in -ly: friendly, silly, lonely, ugly; nouns, ending in -ly: ally, bully, Italy, melancholy Adjectives and adverbs - English Grammar Today - uma referência à Gramática e uso do Inglês escrito e falado - Cambridge Dictionary As we’ve learned, adjectives and adverbs act in similar but different roles. adverb. Adjective Clause: This is a subordinate clause that modifies either a noun or a pronoun, and it usually begins with words such as who, whom, whose, that, or which. As against, an adverb is also one of the parts of speech, which gives you further information about a verb, adjective or any other adverb. Here's the word you're looking for. Nicely is an adverb, and nice is an adjective. Complete the sentence with the appropriate adverb or adjective. So, what is the difference between an adjective vs adverb? Adjectives describe nouns or pronouns, and adverbs describe verbs, adjectives, or other adverbs. Adverbs of manner (such as cheerfully, efficiently, painfully, secretly, quietly, peacefully, carefully, slowly, badly, closely, easily, well, fast, quickly, etc. ) Adjective – a word used to modify or describe a noun or pronoun. If you wrote “She seems unhappily today,” unhappily w… For example:That is a cute hamster.She likes a handsome man. While an adjective qualifies a noun or pronoun, the adverb is used to modify the verb, clause, phrase, adjective, preposition and conjunction. (Adjective)Sarah sings well. The main difference between adverb and adjective is that adverbs modify verbs, adjectives, and other adverbs whereas adjectives modify nouns. They answer the questions: How? Main Difference – Adverb vs Adjective. The main difference between an adverb and an adjective is in what they describe: adjectives describe a noun or pronoun, while adverbs are used to describe verbs and other non-nouns/pronouns. If you are forming an answer to a "how" question, adverbs in that frame often cause grammar issues. An adjective performs the function of qualifying a noun. An adverb tells the way how the verb is functioning. Privacy. An adverb is a word or set of words that describes verbs, adjectives, or other adverbs. An adjective is a word, which adds to the meaning of the noun or pronoun it precedes or succeeds. Key Differences Between Adjective and Adverb The difference between adjective and adverb can be drawn clearly on the following grounds: In grammar, the adjective is among the eight parts of speech which identifies and describes a noun or a pronoun, i.e. Adjective describes a noun or pronoun adjectives typically come before the noun or pronoun adverb, which changes describes. Universal rule, or other adverbs time, place, frequency, degree, etc, was were... Are one of the time, place, frequency, degree, etc Blog urges parents to be pro-active helping... Choose from to complete the sentence context as to whether it is not lesson, difference! Close attention to the meaning of the time this difference can be seen in the English language that can a... How '' question, adverbs describe verbs, adjectives, or a pronoun verb! Verb examples of verbs that relate a state of being are: walk, talk, think, see eat. How, when, where, how many, what is the main between... We have listed the main difference between adverbs and adjectives answer to a common of! And general is an adorable kitten to complete the sentences with the appropriate adverb adjective! This English grammar when adverb and adjective difference to health, we use well rather good! To identifying them correctly speaks so loudly! pay close attention to the meaning of the time this can... 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# Negative neutral real interest rate
What is the intuition behind having a negative neutral real interest rate?
The $$y$$ axis measures interest rates both real and nominal and $$x$$ axis output. The IS curve gives all the combinations of interest rates and output at which investment equals to savings. Whereas the LM curve gives you all combinations of interest rates and output at which money supply equals money demand (see Blanchard et al. Macroeconomics an European perspective for more detail behind IS-LM). Furthermore, the LM curve becomes flat at zero lower bound as, if nominal interest rates become negative, peoples preference for holding money as cash becomes infinite. Furthermore, the $$Y^*$$ line here would represent the natural level of output economy can attain.
If there is too much savings in the market but too little worthwhile investment opportunities (i.e. there are excess savings) the IS curve will shift to the left. The intuition for that is that if there is low demand for saving and huge amount of saving then any equilibrium interest rate (which can be viewed as the price for saving) at which $$I=S$$ will be lower for a given level of output (where $$S$$ is saving and $$I$$ investment). Since nominal interest rate cannot be negative it will stay at zero but the real interest rate will be given by the intersection of the IS curve with the natural level of output which will be below zero.
The intuition is similar to why oil futures price became negative recently - too many people wanting to sell the oil while too little people being interested to buy it. With the oil price, there was also an additional problem that people did not have any space to store the oil and so they were willing to pay people to take it off them. Here the problem is not so much with storage space but the fact that people simply often have to save - for example, due to precautionary savings or other reasons and even though they could hold cash at nominal 0 rate the problem is that it will be eaten by inflation as from Fisher equation we know that nominal interest rate $$i$$ is approximately equal to the real interest rate $$r$$ plus inflation $$\pi$$ - which in turn implies that if the nominal interest rate is above real we get inflation $$i \approx r+\pi \implies \pi \approx i-r$$. So people might desperately supply savings even when the real interest rate is negative because they have no other choice (except for stopping to save altogether- which again might not be reasonable for many people who might need to save due to various reasons such as to form precautionary savings).
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# --- Find the determinant of a 5 X 5 matrix ---
by VinnyCee
Tags: determinant, matrix
P: 492 1. The problem statement, all variables and given/known data Find the determinant of the matrix. A = $$\left[\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ 3 & 0 & 4 & 5 & 6 \\ 2 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 6 & 5 \\ 0 & 0 & 0 & 5 & 6 \end{array}\right]$$ 2. Relevant equations Laplace Expansion forumla For an Expansion across the $$i^{th}$$ row of an n x n matrix: det(A) = $$\sum_{j=1}^{n}\,(-1)^{i\,+\,j}\,a_{i\,j}\,det\left(A_{i\,j}\right)$$ (for a fixed i) For an Expansion across the $$j^{th}$$ column of an n x n matrix: det(A) = $$\sum_{i=1}^{n}\,(-1)^{i\,+\,j}\,a_{i\,j}\,det\left(A_{i\,j}\right)$$ (for a fixed j) 3. The attempt at a solution So, I start by doing a Laplace Expansion across the first row and down the second column. So i = 1 and j = 2. det(A) = $$(-1)^{1\,+\,2}\,a_{1\,2}\,det\left(A_{1\,2}\right)$$ det(A) = $$(-1)\,(2)\,det\left(\left[\begin{array}{cccc} 3 & 4 & 5 & 6 \\ 2 & 2 & 3 & 4 \\ 0 & 0 & 6 & 5 \\ 0 & 0 & 5 & 6 \end{array}\right]\right)$$ I continue by doing another Laplace Expansion, this time across the first row and down the first column. So i = 1 and j = 1. det(A) = $$(-2)\,(-1)^{1\,+\,1}\,a_{1\,1}\,det\left(\left[\begin{array}{ccc} 2 & 3 & 4 \\ 0 & 6 & 5 \\ 0 & 5 & 6 \end{array}\right]\right)$$ det(A) = $$(-2)\,(1)\,(3)\,det\left(\left[\begin{array}{ccc} 2 & 3 & 4 \\ 0 & 6 & 5 \\ 0 & 5 & 6 \end{array}\right]\right)$$ For the 3 x 3 matrix, I use Sarrus's Rule to get a determinant of 22. det(A) = $$(-6)\,(22)\,=-132$$ However, when I plug the original matrix into my TI-92, I get det(A) = 99! I tried a second time with i = 1 and j = 1 for the original matrix and then i = 1 and j = 2 for the 4 x 4 matrix. Here I get det(A) = -44. Neither are right! What am I doing wrong here?
Sci Advisor HW Helper P: 9,398 You haven't summed. Fix a row and you need to do the Laplace thing for every entry in that row, not just one that takes your fancy, and take the signed sum. In short, you've not used the formula properly. It is easier to use row operations, anyway.
P: 492 I found my error, it was exactly as you said. I was not doing the sum for the second non-zero term in the column! The right equation is: det(A) = $$(-1)^{1\,+\,2}\,a_{1\,2}\,det\left(A_{1\,2}\right)\,+\,(-1)^{3\,+\,2}\,a_{3\,2}\,det\left(A_{3\,2}\right)$$
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# American Institute of Mathematical Sciences
June 2020, 28(2): 1031-1036. doi: 10.3934/era.2020055
## A family of $q$-congruences modulo the square of a cyclotomic polynomial
School of Mathematics and Statistics, Huaiyin Normal University, Huai'an 223300, Jiangsu, China
Received January 2020 Revised April 2020 Published June 2020
Fund Project: The author was partially supported by the National Natural Science Foundation of China (grant 11771175)
Using Watson's terminating $_8\phi_7$ transformation formula, we prove a family of $q$-congruences modulo the square of a cyclotomic polynomial, which were originally conjectured by the author and Zudilin [J. Math. Anal. Appl. 475 (2019), 1636-646]. As an application, we deduce two supercongruences modulo $p^4$ ($p$ is an odd prime) and their $q$-analogues. This also partially confirms a special case of Swisher's (H.3) conjecture.
Citation: Victor J. W. Guo. A family of $q$-congruences modulo the square of a cyclotomic polynomial. Electronic Research Archive, 2020, 28 (2) : 1031-1036. doi: 10.3934/era.2020055
##### References:
[1] G. Gasper and M. Rahman, Basic Hypergeometric Series, Encyclopedia of Mathematics and Its Applications, 96, Cambridge University Press, Cambridge, 2004. doi: 10.1017/CBO9780511526251. Google Scholar [2] C.-Y. Gu and V. J. W. Guo, $q$-Analogues of two supercongruences of Z.-W. Sun, Czechoslovak Math. J., in press. doi: 10.21136/CMJ.2020.0516-18. Google Scholar [3] V. J. W. Guo, Common $q$-analogues of some different supercongruences, Results Math., 74 (2019), 15pp. doi: 10.1007/s00025-019-1056-1. Google Scholar [4] V. J. W. Guo, Proof of a generalization of the (B.2) supercongruence of Van Hamme through a $q$-microscope, Adv. in Appl. Math., 116 (2020), 19pp. doi: 10.1016/j.aam.2020.102016. Google Scholar [5] V. J. W. Guo, $q$-Analogues of Dwork-type supercongruences, J. Math. Anal. Appl., 487 (2020), 9pp. doi: 10.1016/j.jmaa.2020.124022. Google Scholar [6] V. J. W. Guo and J.-C. Liu, $q$-Analogues of two Ramanujan-type formulas for $1/\pi$, J. Difference Equ. Appl., 24 (2018), 1368-1373. doi: 10.1080/10236198.2018.1485669. Google Scholar [7] V. J. W. Guo and M. J. Schlosser, Some new $q$-congruences for truncated basic hypergeometric series: Even powers, Results Math., 75 (2020), 15pp. doi: 10.1007/s00025-019-1126-4. Google Scholar [8] V. J. W. Guo and J. Zeng, Some $q$-supercongruences for truncated basic hypergeometric series, Acta Arith., 171 (2015), 309-326. doi: 10.4064/aa171-4-2. Google Scholar [9] V. J. W. Guo and W. Zudilin, A $q$-microscope for supercongruences, Adv. Math., 346 (2019), 329-358. doi: 10.1016/j.aim.2019.02.008. Google Scholar [10] V. J. W. Guo and W. Zudilin, On a $q$-deformation of modular forms, J. Math. Anal. Appl., 475 (2019), 1636-1646. doi: 10.1016/j.jmaa.2019.03.035. Google Scholar [11] V. J. W. Guo and W. Zudilin, A common $q$-analogue of two supercongruences, Results Math., 75 (2020), 11pp. doi: 10.1007/s00025-020-1168-7. Google Scholar [12] J.-C. Liu, Some supercongruences on truncated $_3F_2$ hypergeometric series, J. Difference Equ. Appl., 24 (2018), 438-451. doi: 10.1080/10236198.2017.1418863. Google Scholar [13] J.-C. Liu, On Van Hamme's (A.2) and (H.2) supercongruences, J. Math. Anal. Appl., 471 (2019), 613-622. doi: 10.1016/j.jmaa.2018.10.095. Google Scholar [14] L. Long and R. Ramakrishna, Some supercongruences occurring in truncated hypergeometric series, Adv. Math., 290 (2016), 773-808. doi: 10.1016/j.aim.2015.11.043. Google Scholar [15] H.-X. Ni and H. Pan, Some symmetric $q$-congruences modulo the square of a cyclotomic polynomial, J. Math. Anal. Appl., 481 (2020), 12pp. doi: 10.1016/j.jmaa.2019.07.062. Google Scholar [16] Z.-H. Sun, Generalized Legendre polynomials and related supercongruences, J. Number Theory, 143 (2014), 293-319. doi: 10.1016/j.jnt.2014.04.012. Google Scholar [17] Z.-W. Sun, On sums of Apéry polynomials and related congruences, J. Number Theory, 132 (2012), 2673-2699. doi: 10.1016/j.jnt.2012.05.014. Google Scholar [18] H. Swisher, On the supercongruence conjectures of van Hamme, Res. Math. Sci., 2 (2015), 21pp. doi: 10.1186/s40687-015-0037-6. Google Scholar [19] L. Van Hamme, Some conjectures concerning partial sums of generalized hypergeometric series, in $p$-Adic Functional Analysis, Lecture Notes in Pure and Appl. Math., 192, Dekker, New York, 1997,223–236. Google Scholar [20] C. Wang and H. Pan, On a conjectural congruence of Guo, preprint, arXiv: 2001.08347. Google Scholar
show all references
##### References:
[1] G. Gasper and M. Rahman, Basic Hypergeometric Series, Encyclopedia of Mathematics and Its Applications, 96, Cambridge University Press, Cambridge, 2004. doi: 10.1017/CBO9780511526251. Google Scholar [2] C.-Y. Gu and V. J. W. Guo, $q$-Analogues of two supercongruences of Z.-W. Sun, Czechoslovak Math. J., in press. doi: 10.21136/CMJ.2020.0516-18. Google Scholar [3] V. J. W. Guo, Common $q$-analogues of some different supercongruences, Results Math., 74 (2019), 15pp. doi: 10.1007/s00025-019-1056-1. Google Scholar [4] V. J. W. Guo, Proof of a generalization of the (B.2) supercongruence of Van Hamme through a $q$-microscope, Adv. in Appl. Math., 116 (2020), 19pp. doi: 10.1016/j.aam.2020.102016. Google Scholar [5] V. J. W. Guo, $q$-Analogues of Dwork-type supercongruences, J. Math. Anal. Appl., 487 (2020), 9pp. doi: 10.1016/j.jmaa.2020.124022. Google Scholar [6] V. J. W. Guo and J.-C. Liu, $q$-Analogues of two Ramanujan-type formulas for $1/\pi$, J. Difference Equ. Appl., 24 (2018), 1368-1373. doi: 10.1080/10236198.2018.1485669. Google Scholar [7] V. J. W. Guo and M. J. Schlosser, Some new $q$-congruences for truncated basic hypergeometric series: Even powers, Results Math., 75 (2020), 15pp. doi: 10.1007/s00025-019-1126-4. Google Scholar [8] V. J. W. Guo and J. Zeng, Some $q$-supercongruences for truncated basic hypergeometric series, Acta Arith., 171 (2015), 309-326. doi: 10.4064/aa171-4-2. Google Scholar [9] V. J. W. Guo and W. Zudilin, A $q$-microscope for supercongruences, Adv. Math., 346 (2019), 329-358. doi: 10.1016/j.aim.2019.02.008. Google Scholar [10] V. J. W. Guo and W. Zudilin, On a $q$-deformation of modular forms, J. Math. Anal. Appl., 475 (2019), 1636-1646. doi: 10.1016/j.jmaa.2019.03.035. Google Scholar [11] V. J. W. Guo and W. Zudilin, A common $q$-analogue of two supercongruences, Results Math., 75 (2020), 11pp. doi: 10.1007/s00025-020-1168-7. Google Scholar [12] J.-C. Liu, Some supercongruences on truncated $_3F_2$ hypergeometric series, J. Difference Equ. Appl., 24 (2018), 438-451. doi: 10.1080/10236198.2017.1418863. Google Scholar [13] J.-C. Liu, On Van Hamme's (A.2) and (H.2) supercongruences, J. Math. Anal. Appl., 471 (2019), 613-622. doi: 10.1016/j.jmaa.2018.10.095. Google Scholar [14] L. Long and R. Ramakrishna, Some supercongruences occurring in truncated hypergeometric series, Adv. Math., 290 (2016), 773-808. doi: 10.1016/j.aim.2015.11.043. Google Scholar [15] H.-X. Ni and H. Pan, Some symmetric $q$-congruences modulo the square of a cyclotomic polynomial, J. Math. Anal. Appl., 481 (2020), 12pp. doi: 10.1016/j.jmaa.2019.07.062. Google Scholar [16] Z.-H. Sun, Generalized Legendre polynomials and related supercongruences, J. Number Theory, 143 (2014), 293-319. doi: 10.1016/j.jnt.2014.04.012. Google Scholar [17] Z.-W. Sun, On sums of Apéry polynomials and related congruences, J. Number Theory, 132 (2012), 2673-2699. doi: 10.1016/j.jnt.2012.05.014. Google Scholar [18] H. Swisher, On the supercongruence conjectures of van Hamme, Res. Math. Sci., 2 (2015), 21pp. doi: 10.1186/s40687-015-0037-6. Google Scholar [19] L. Van Hamme, Some conjectures concerning partial sums of generalized hypergeometric series, in $p$-Adic Functional Analysis, Lecture Notes in Pure and Appl. Math., 192, Dekker, New York, 1997,223–236. Google Scholar [20] C. Wang and H. Pan, On a conjectural congruence of Guo, preprint, arXiv: 2001.08347. Google Scholar
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Multistable mechanical metamaterials are materials that have multiple stable configurations. The geometrical changes caused by the transition of the metamaterial from one stable state to another, could be exploited to obtain multifunctional and programmable materials. As the stimulus amplitude is varied, a multistable metamaterial goes through a sequence of stable configurations. However, this sequence (which we will call the deformation sequence) is unpredictable if the metamaterial consists of identical unit cells. This paper proposes to use small variations in the unit cell geometry to obtain a deterministic deformation sequence for one type of multistable metamaterial that consists of bistable unit cells. Based on an analytical model for a single unit cell and on the minimization of the total strain energy, a rigorous theoretical model is proposed to analyze the nonlinear mechanics of this type of metamaterials and to inform the designs. The proposed theoretical model is able to accurately predict the deformation sequence and the stress–strain curves that are observed in the finite-element simulations with an elastic constitutive model. A deterministic deformation sequence that matches the sequence predicted by the theory and finite-element simulations is obtained in experiments with 3D-printed samples. Furthermore, an excellent quantitative agreement between simulations and experiments is obtained once a viscoelastic constitutive model is introduced in the finite-element model.
## Introduction
Mechanical metamaterials are materials whose effective properties arise from the underlying architecture, rather than from the bulk behavior of their constituents [1]. For example, acoustic metamaterials can exhibit unusual acoustic behavior, such as the ability to guide or stop elastic wave propagation along a desired path [24]. Other interesting and unconventional properties of metamaterials include negative bulk modulus [58], negative Poisson's ratio [9,10], high specific energy absorption [11,12], or negative (dynamic) mass [5,8,13,14].
However, the functionality of a mechanical metamaterial is limited if its properties cannot be tuned after fabrication of the metamaterial. Recent studies have shown great interest in developing tunable metamaterials by exploiting elastic instabilities that can cause configurational changes in the microarchitecture of these metamaterials [15]. For example, reversible changes in the microarchitecture geometry due to elastic instabilities have been used to tune the bandgaps in acoustic [1620] and photonic [21,22] metamaterials.
A snap-through instability is a kind of elastic instability in which a structure instantaneously jumps from one configuration to another configuration when an applied stimulus reaches a critical level [23]. The design of mechanical metamaterials with snap-through instabilities has been the focus of active research in recent years [2428]. For example, Correa et al. [25] and Restrepo et al. [26] developed metamaterials that are made up of multiple unit cells that each includes an initially curved beam. Through the elastic snap-through of high number of unit cells arranged in series, this kind of metamaterial can exhibit large hysteresis loops in response to cyclic loads. Because of these hysteresis loops, these metamaterials dissipate a large amount of mechanical energy. In contrast to traditional honeycombs that rely on plastic energy dissipation, the deformation can be fully recovered in these metamaterials. While some of the proposed metamaterials with snap-through instabilities are monostable, i.e., they return to their undeformed configuration after removal of the load (for example, the negative stiffness honeycombs designed by Correa et al. [25]), other designs are multistable, i.e., they have multiple stable configurations. For example, Shan et al. [24] developed multistable metamaterials using bistable tilted beams. These metamaterials can achieve large deformations and offer significant energy absorption capacity due to the ability of the material to trap elastic energy in a stable deformed configuration in response to an impact. Haghpanah and Salari-Sharif [28] investigated another design for multistable metamaterials, based on bistable structural elements designed to be considerably stronger than the previous designs. While these previous studies focus on multistable metamaterials in response to compressive loads, Rafsanjani et al. [27] showed that the design of architectures that exhibit multistability in response to tensile loads is possible.
In multistable metamaterials, a sequence of stable configurations is obtained as the stimulus amplitude is increased. However, this sequence (which we will call the deformation sequence) is unpredictable if, as in most previous works, the metamaterials consist of identical unit cells [24,26]. Imperfections in the geometry, material properties [2931] or in the boundary conditions determine the deformation sequence in the case of a multistable metamaterial with identical unit cells. Although previous papers about multistable metamaterials give important insight into the design and mechanics of these metamaterials, these previous works have not investigated strategies that would make it possible to control their deformation sequence.
The objective of this paper is to develop and analyze approaches to tune the deformation sequence of one kind of multistable metamaterial. To achieve this objective, a rigorous theoretical model, based on an analytical model for a single unit cell and on the minimization of the total strain energy, is developed to analyze the nonlinear mechanics and deformation sequence of multistable metamaterials. Experiments with 3D-printed samples are used to test the theory. One possible application of metamaterials with a deterministic deformation sequence could be the control of elastic waves. In a multistable metamaterial with a deterministic deformation sequence, a preload could be applied to cause a predictable switch to another stable configuration; this switch in the geometry of the microarchitecture would affect wave propagation due to a predictable change in the frequency bandgaps of the metamaterials. In contrast to previous tunable metamaterials which requires to maintain the stimulus [16,19], these metamaterials would remain in a deformed configuration after removal of the preload due to multistability. Thus, although tuning the deformation sequence only requires snap-through buckling, multistable metamaterials are the focus of this paper because multistability could have advantages in applications such as the development of tunable metamaterials with switchable properties.
The organization of this article is as follows. First, the theoretical model and the experimental methods are presented. Then two methods, both based on small variations in the unit cell geometry, are proposed to obtain a deterministic deformation sequence. These methods are validated using theoretical analysis, finite-element simulations, and experiments performed on 3D-printed samples.
## Methods
### Theoretical Model for Multistable Metamaterials.
The mechanics of a specific kind of multistable metamaterial, shown in Fig. 1(a), is investigated using a theoretical model. The metamaterial is made up of multiple bistable unit cells (that are similar to [26]), whose geometrical parameters are shown in Fig. 1(b). The unit cell consists of thick horizontal and vertical elements and a thin curved part. As in the previous studies [24,26], because the loading is intended to be only in the y-direction, these metamaterials are called one-dimensional (1D) metamaterials.
#### Overview of Single Unit Cell Model.
Analyzing the mechanics of multistable metamaterials first requires a model of the unit cell. The model of Qiu et al. [32], developed for a single initially curved beam without any mode imperfection, was extended to allow the presence of a mode shape imperfection. The curved part of the unit cell is modeled as an initially curved clamped–clamped Euler–Bernoulli beam, while other parts of the unit cell are assumed to be rigid (Fig. 1(c)). A transverse force, f, is applied at the midpoint of the beam. The governing equation for this beam is
$EI(d4wdx4−d4w0dx4)+pd2wdx2=−fδ(x−l2)$
(1)
where w is the lateral deflection of the beam, $w0(x)$ is the initial shape of the beam, E is the Young's modulus, I is the area moment of inertia of the beam, p is the compressive force, l is the beam span (Fig. 1(b)), and δ is the Dirac delta function. In the case of the model without any mode imperfection, $w0(x)$ is given by
$w0(x)=h2W1(x)$
(2)
where h is the initial apex height of the beam, and $W1(x)=1−cos(2πx/l)$ is the first buckling mode of a clamped–clamped Euler–Bernoulli beam. A model with an imperfection proportional to the third-buckling mode was also considered in order to obtain a deterministic deformation sequence. In this model, $w0(x)$ is given by
$w0(x)=h2W1(x)+δW(x)$
(3)
where $δW(x)$ is the mode shape imperfection, which is prescribed to be given by
$δW(x)=a3h2W3(x)$
(4)
where $W3(x)=1−cos(4πx/l)$ is the third-buckling mode of a clamped–clamped beam, and a3 is the mode imperfection size. The case of a model without imperfection corresponds to a3 = 0, such that the model with the mode imperfection is a generalization of the model without imperfection. For a single unit cell, the compressive force, p, is caused by the shortening of the beam total length and is given by
$p=−EAΔss0$
(5)
where A is the cross section area of the beam, Δs is the change in the beam length, and s0 is the initial beam length. The beam length, s, is given by
$s=∫0l1+(dwdx)2dx$
(6)
such that the following approximations, valid for $(dw/dx)≪1$ and $(dw0/dx)≪1$, are used
$Δs=∫0l12[(dwdx)2−(dw0dx)2]dxs0=l$
(7)
The lateral deflection of the beam at the beam midpoint, d, is given by
$d=w0(l/2)−w(l/2)$
(8)
Since d corresponds to the change in the height of the unit cell when the transverse load is applied, we will refer to d as the deformation of the unit cell. To make the results scale-independent, the following normalizations [32] are used:
$f¯=l3EIhf, d¯=dh, U¯=l3EIh2U$
(9)
where U is the strain energy, and an overbar denotes a normalized quantity. The procedure and equations needed to obtain the relationship between the normalized force, normalized strain energy, and normalized deformation are given in Appendix A.
The theoretical model was validated using nonlinear finite-element analysis (FEA) in abaqus/standard using the static/general nonlinear procedure. The FEA model is built with four-node bilinear quadrilateral plane-stress element with reduced integration (CPS4R in abaqus) and an elastic, isotropic constitutive model. The model is stimulated using displacement control, such that the vertical (y) displacement on the top edge of the unit cell is prescribed. The other boundary conditions are the following: the bottom edge of the unit cell is fixed, and the top edge of the unit cell has a fixed x-direction displacement.
Results of Figs. 1(d)1(g) show that there is excellent agreement between the theoretical model and the FEA simulation, both in the case a3 = 0 which corresponds to the model developed by Qiu et al. [32] (Figs. 1(d) and 1(e)), and in the case a3 ≠ 0 which corresponds to the model extension of this paper (Figs. 1(f) and 1(g)). It can be seen in the normalized force versus normalized deformation curve that there is one critical point whose normalized force value is $f¯cr$, from which point the beam begins to exhibit a negative stiffness. In a force control simulation, this point would correspond to the point at which the unit cell would snap through. In the normalized strain energy versus normalized deformation curve, there is a global minimum (point I in Figs. 1(e) and 1(g)) when $d¯=0$, and a local minimum (point III in Figs. 1(e) and 1(g)) when $d¯≈2$. These are the two stable equilibrium positions that correspond to points I and III in Figs. 1(d) and 1(f). Moreover, there is one local maximum (point II in Figs. 1(e) and 1(g)) in the strain energy versus deformation curve, which is an unstable equilibrium position corresponding to point II in Figs. 1(d) and 1(f).
#### Multiple Cells in Series: Equations and Algorithm.
The mechanics of a multistable metamaterial that is comprised of a series of bistable unit cells can be derived from the model for a single unit cell. Consider the metamaterial with n unit cells in series (Fig. 2(a)) and the corresponding theoretical model shown in Fig. 2(b). In the theoretical model, the beams are connected by a rigid link. The bottom beam (beam 1) has clamped–clamped boundary conditions; the left and right sections of the other beams (beams 2, 3,…, n) are only allowed vertical (y-direction) displacement. The thicknesses and mode imperfection sizes of the curved part of each unit cell are ti and a3,i, respectively. For all unit cells, the initial height of the curved parts is h. The displacements of the midpoint of the curved part of unit cell i is ui (see Fig. 2(b)); the deformations of the unit cells can be expressed as d1 = u1, $di=ui−ui−1$ for i = 2,…, n, and the total deformation is $dtot=∑i=1ndi$. The total deformation, dtot, the total strain energy, Utot, and the loading force, f, are normalized with respect to the parameters of unit cell 1, i.e., the following definitions are used for the normalized force, $f¯$, the normalized deformation, $d¯tot$, and the normalized strain energy, $U¯tot$:
$f¯=l3EI1hf, d¯tot=dtoth, U¯tot=l3EI1h2Utot$
(10)
where I1 is the area moment of inertia of beam 1. When each of the unit cells is isolated, a subscript i refers to the properties of the ith unit cell. For single unit cells, the applied force fi, the deformation, di and the strain energy, Ui, are normalized with respect to the properties of unit cell 1
$fi¯=l3EI1hfi, d¯i=dih, U¯i=l3EI1h2Ui$
(11)
where $f¯i, d¯i$, and $U¯i$ are the normalized force, deformation, and strain energy for unit cell i, respectively.
The deformations of each unit cell can be expressed as a vector $(d¯1,d¯2,…,d¯n)$. In order to determine the values of the components in this vector when the total normalized deformation $d¯tot$ is increased, an algorithm similar to the algorithm proposed by Oh and Kota [33] and Overvelde et al. [34] for multistable mechanisms can be used. This algorithm requires calculating the total normalized strain energy of the system, $U¯tot(d¯1,d¯2,…,d¯n)$, which can be computed by
$U¯tot(d¯1,d¯2,…,d¯n)=∑i=1nU¯i(di¯)$
(12)
where $d¯n=d¯tot−∑i=1n−1d¯i$. The normalized strain energy for each unit cell $U¯i$ is determined using the theoretical model of Sec. 2.1.1. In this algorithm, the loading from $d¯tot=0$ to $d¯tot,max$ is discretized into small steps of amplitude $δd¯$. Assuming that at step k of this algorithm, the values of $(d¯1,d¯2,…,d¯n)$ are $(d¯1k,d¯2k,…,d¯nk)$ ; at step k + 1, the values are found by solving the following optimization problem:
$min U¯tot(d¯1,d¯2,…,d¯n)with respect to (d¯1,d¯2,…,d¯n−1)subject to d¯n=d¯tot−∑i=1n−1d¯i$
(13)
This optimization problem is solved using a local optimization algorithm (the interior-point algorithm using the fmincon function in matlab). Using a local minimization algorithm is critical; during loading the metamaterial does not switch to a state that corresponds to the global minimum but remains in the neighborhood of the current configuration.
Once the values of $d¯i$ are obtained at all steps for the loading (or unloading), the values of $d¯i$ can be plotted as a function of $d¯tot$ to analyze the sequence of deformed configurations the metamaterial goes through when the loading parameter is increased. In the case of a metamaterial with 2 unit cells in series, the curve of $d¯1$ as a function of $d¯tot$ completely characterizes how the material deforms during loading. This curve will be referred to as the deformation path of the metamaterial.
In addition to the theoretical model, a corresponding FEA model with elastic material is built using the boundary conditions shown in Fig. 2(a), i.e., the bottom edge of unit cell 1 is fixed, the left and right sides of these unit cells have a fixed x-direction displacement, and the top edge of unit cell n also has a fixed x-direction displacement. The deformation path and the force versus deformation curves predicted by a static nonlinear FEA simulation with displacement control are used to validate the theoretical model.
### Experimental Methods.
Samples were fabricated by a multimaterial 3D printer (Objet Connex 260, Stratasys, Edina, MN). The printing material is DM9895, which is a digital material derived by mixing two base materials. One of the two base materials is TangoblackPlus, which is a rubbery material at room temperature; the other one is Verowhite, which is a rigid plastic at room temperature. The in-plane dimensions of the printed samples are about 10 cm × 10 cm; the out-of-plane thickness is 1 cm.
In order to determine the deformation sequences of printed samples, compression tests were conducted on a universal Material Testing System (MTS, Model Insight 10, Eden Prairie, MN) in a displacement control manner with a 10 kN load cell. The samples were compressed (at room temperature) using a customized compression fixture (Fig. 6(b)) at a testing velocity of 10 mm/min until all unit cells collapse. A digital camera was used to record the whole testing process.
## Results
In this section, the theoretical model for elastic multistable metamaterials is first applied to a metamaterial that consists of two unit cells in series to demonstrate the validity of the model. The same algorithm is then applied to design metamaterials with a larger number of unit cells.
### Analysis of the Strain Energy Landscape for Metamaterials With Identical Unit Cells.
Using the theoretical method from Sec. 2.1, the total strain energy contours of the metamaterial with two unit cells in series can be obtained as a function of $d¯tot$ and $d¯1$, as shown in Fig. 3. The strain energy contours show that there are four local minima, A, B, C, and D (that correspond to stable equilibria) and one local maximum of the total strain energy (that corresponds to an unstable equilibrium). The presence of multiple minima indicates the multistability of this metamaterial. The global minimum A is the initial undeformed configuration. The local minima B and C correspond to cases when only unit cell 2 has collapsed and only unit cell 1 has collapsed, respectively. The local minimum D is when both unit cell 1 and unit cell 2 have collapsed. Because the two unit cells are identical, the local minima B and C have the same strain energy value, which means the deformation path (the curve $d¯1(d¯tot)$) during loading can follow the sequence A–B–D or A–C–D. In an experiment, the presence of imperfections, for example, caused by the manufacturing process or in the experimental setup (for instance in the application of the boundary conditions), would determine which of the deformation paths is preferred.
Based on the analysis of the metamaterial with identical unit cells, it can be seen that the sequence of stable configurations the metamaterial switches in response to a given stimulus is unpredictable if it consists of identical unit cells. In order to get a deterministic deformation sequence, small variations in the unit cell geometry are utilized, specifically varying the thicknesses or initial shapes of the curved parts of the unit cells. These variations will give different values of the critical forces (as defined in Sec. 2.1.1) for each unit cell. When the unit cells are assembled together, the unit cell with smallest critical force will deform first, followed by the unit cell with next smaller critical force. These two methods are analyzed next.
### Using Thickness Variation to Obtain a Deterministic Deformation Sequence.
One method to obtain a deterministic deformation sequence is to use thickness variation, i.e., varying the thickness value of the curved part of the unit cell ti from row to row; for example, t2 = 1.5t1 for a metamaterial architecture with two unit cells (Fig. 2(a)). The results for single unit cells show that the critical force value of unit cell 2, $f¯cr,2$, is higher than the critical force value of unit cell 1, $f¯cr,1$, which means unit cell 2 requires a larger loading force to collapse. Moreover, the strain energy of unit cell 2 is also greater than that of unit cell 1.
If the unit cells are combined in series, the theoretical results of the total strain energy contour, deformation path and normalized force versus deformation curves can be determined using the algorithm of Sec. 2.1.2 as the solid lines in Figs. 4(c)4(e), respectively. As shown in the contour plot of this metamaterial's total strain energy (Fig. 4(c)), the four local minima of the strain energy are different, such that a unique deformation path is obtained. In order to validate the theoretical results, the simulation results for the corresponding FEA model are given in Figs. 4(d)4(e), as the blue dashed lines. Note the excellent agreement for the deformation path between the theoretical model and the FEA simulation (Fig. 4(d)). Both the theoretical model and the FEA simulation show that the deformation sequence is A–C–D for loading and D–B–A for unloading. As shown in Figs. 4(c) and 4(d), when the normalized total deformation $d¯tot$ is about 3.4, there will be one snap-back of unit cell 1 (i.e., the value of $d¯1$ drops instantaneously) during loading when unit cell 2 snaps through. In the normalized force versus normalized deformation curve 4e, the first peak has a smaller amplitude than the second peak due to the lower stiffness of unit cell 1; futhermore, the force goes below the horizontal axis (which means a tensile force is observed) twice during loading.
### Using Mode Shape Imperfection to Obtain a Deterministic Deformation Sequence.
Besides the method of varying the thickness, the deformation sequence can also be tuned by varying the mode shape imperfection size a3 (where the imperfection in the initial shape of the beam, δW(x), is given by Eq. (4)). In order to prove that the deformation sequence can be tuned by mode shape imperfection, the force-deformation (Fig. 5(b)) and strain energy-deformation (Fig. 5(c)) curves are first compared for a unit cell without a mode shape imperfection (i.e., a3 = 0) and a unit cell with a 10% third-mode imperfection (a3 = 0.1). The results of both the theoretical model and FEA simulations show that the critical loading force and the strain energy of a single unit cell decrease when the model has a third-mode imperfection.
After obtaining the results for a single unit cell, the results for the metamaterial with two cells with different mode imperfection sizes can be determined using the same procedure as was used in Sec. 3.2. Consider the metamaterial with two unit cells in series, where unit cell 2 is perfect (a3,2 = 0) and unit cell 1 has a 10% third-mode imperfection (a3,1 = 0.1). In contrast to the case of identical unit cells, the local minima of the strain energy are different, which makes the deformation path unique (Fig. 5(d)). The deformation sequence is A–C–D for loading, and D–B–A for unloading. As in the case of the metamaterial with two cells using thickness variation, there will be one snap-back of the unit cell 1 during the initial snap-through of unit cell 2 (Fig. 5(d)), when the normalized total deformation $d¯tot$ is about 3.3. As in the case of thickness variation, the first peak in the normalized force versus normalized deformation curve has a smaller amplitude due to the smaller critical force in unit cell 1 (Fig. 5(f)); furthermore, a tensile force is observed twice during the loading. However, in contrast to the case of the thickness variation, the first peak has a smoother shape than the second peak because adding an imperfection significantly affects the shape of the normalized deflection versus normalized deformation curve for a single unit cell (see Fig. 5(a)).
### Experimental Validation of the Two Methods
#### Qualitative Validation: Deformation Sequence.
Several multistable architectures were fabricated using a 3D printing technique. These samples were tested using a customized compression fixture (shown in Fig. 6(b)) to experimentally validate that the methods of thickness variation and mode shape imperfection can be used to tune the deformation sequence. Moreover, considering the viscoelasticity of the samples, FEA models with a viscoelastic constitutive model were built for comparisons with the experiments (see Appendix B). In these models, the bottom edge is fixed and the top edge has a fixed x-direction displacement, while its y-displacement is prescribed to move at a constant velocity; the left and right edges are left free. The response to stimulation of the metamaterial at a constant strain rate was simulated using nonlinear quasi-static FEA simulations (VISCO procedure in abaqus) with displacement control.
One of the printed samples is the multistable metamaterial shown in Fig. 6(a) (metamaterial A), which consists of five rows, each with 5 unit cells (in the rest of the paper, we will call a metamaterial with N rows, each with M unit cells as an N × M metamaterial). For this 5 × 5 multistable metamaterial, the thickness is constant within each row, while the thickness varies from row to row (t1 < t5 < t2 < t3 < t4, where the unit cells are ordered from bottom to top). Figures 6(b)6(c) shows snapshots of the response of the multistable architecture at different effective strains ϵ for both experiment and finite-element simulation. These snapshots demonstrate that the deformation sequence matches what is expected from the previous analysis (i.e., row 1 collapses first, then row 5, row 2, row 3, and row 4).
Another printed sample is the 5 × 5 multistable metamaterial shown in Fig. 7(a) (metamaterial B). The unit cells of this 5 × 5 metamaterial have a uniform mode imperfection size within each row, but the imperfection size varies from row to row. The thickness of each unit cell's curved part is 1 mm. Snapshots of the response at different strains ϵ for both experiment and finite-element simulation (Figs. 7(b)7(c)) demonstrate that the deformation sequence matches the expected sequence (i.e., row 1 collapses first, then row 5, row 2, row 3, and row 4).
#### Quantitative Comparisons.
The experiments were also analyzed more quantitatively by plotting the value of the normalized deformation of each unit cell, $d¯i$, as a function of the total deformation, $d¯$, in Figs. 8(a) and 8(b). The normalized deformation was computing using frames of the movies recorded using the digital camera during the compression test. For the experiments, FEA simulations (with the same boundary conditions as in Sec. 3.4.1) and theoretical model, the same clear deformation sequence can be seen by inspecting the results (i.e., row 1 collapses first, then row 5, row 2, row 3, and row 4 in both Figs. 8(a) and 8(b)). While both the FEA simulations and the theoretical model are able to qualitatively capture the deformation sequence, only the FEA simulations are able to quantitatively match the experiments. In particular, the unit cells in the theoretical model deform earlier than in the FEA models and experiments. Moreover, there is an instantaneous snap-back in theoretical model; however, a more gradual and limited snap-back is observed in the FEA model and in experiments (for example, see ϵ = 18% for $d¯1$ in Fig. 8(a) and ϵ = 15% for $d¯1$ in Fig. 8(b)). The inaccuracies of the theoretical model are due to two sources: (1) the viscoelasticity of the printed materials (see Appendix B for the relaxation modulus of the material) and (2) the assumption regarding the kinematics in the theoretical model (a small rotation assumption is used and parts of the unit cell are assumed to be rigid). The smoothing of the curves seen in the experiments and FEA simulations (with a viscoelastic model) are primarily due to the viscoelasticity of the printed samples. Simulations with finite element with an elastic constitutive model (not shown in the manuscript) do show a discontinuous instability jump similar to what is seen in the theoretical model. In spite of the quantitative differences between the theory and experiments, the theoretical model does capture the important characteristics of the deformation paths of this multistable metamaterial and can be used to inform the design of metamaterials with a deterministic deformation sequence.
The effective stress versus strain curve was also obtained during the compression test and compared with the FEA model with viscoelastic material as shown in Figs. 9(a) and 9(c). The effective stress is calculated from the compression force divided by the bottom area of the model, and the effective strain is calculated from the total deformation divided by the initial height of the model. For both metamaterials, there is a good quantitative agreement between the experiments and FEA simulation results. This excellent quantitative agreement is interesting because of the difference in the boundary conditions applied in the FEA and in the experiments: the top edge of the sample was prescribed to move at a constant velocity in the FEA simulations; in the experiments, the top edge of the compression fixture was prescribed to move at a constant velocity and the top edge of the sample was not attached to the compression fixture. In the case of an elastic sample, a separation of the top edge of the sample from the compression fixture would have been expected when one the row of the samples snap-through. However, the presence of viscoelasticity significantly affects the kinetics of the snap-through process, such that the sample remains in contact with the compression fixture. The experiments and FEA simulations with a viscoelastic constitutive model do not show that the force crosses the horizontal axis, which could be wrongly interpreted as the absence of multistability. However, the multistability was experimentally verified by letting the metamaterial relax to any of the expected stable configurations. Furthermore, the first peak has a higher amplitude than the second peak, in contrast to what has been observed in Figs. 4 and 5.
Because of the viscoelasticity of the 3D-printed sample, the theoretical model cannot be directly compared to the experiments or the FEA simulations with a viscoelastic material; hence, the theoretical model was compared to FEA simulations with an elastic constitutive model (Figs. 9(b) and 9(d)). For both metamaterials, the theoretical result has excellent agreement with the FEA with elastic material and captures the major characteristics of this metamaterial, showing its multistability (i.e., crossings of the horizontal axis) and growing stress peaks. This implies that the absence of a tensile force and the fact that the second stress peak has a lower amplitude than the first peak are due to the viscoelasticity of the 3D-printed samples (note, however, that obtaining a tensile force would also require to use an adhesive between the compression fixture and the sample, as done, for example, in Ref. [24]). Because of the assumption of small rotation, some parts being rigid in the theoretical model, the values of the strain corresponding to maxima and minima are somewhat shifted in the theoretical model compared to FEA with an elastic constitutive model.
## Summary and Conclusions
In this paper, we propose to use the methods of small geometric variations to tune the deformation sequence of one kind of metamaterial that consists of a quasi-periodic microarchitecture with multiple stable states. This work overcomes the limit shown in the previous studies that when the unit cells used in a multistable metamaterial are identical, the stable configuration that the metamaterial switches to is unpredictable due to the effects of imperfections in the manufacturing process and in the boundary conditions. Two different methods of thickness variation and mode shape imperfection are analyzed and used to obtain a deterministic deformation sequence. In order to obtain a deterministic deformation sequence, a rigorous theoretical model is developed to analyze the mechanics and help to design this metamaterial. The theoretical model is validated by comparison to finite-element simulations of a model that consists of multiple bistable unit cells in series. Both the loading and unloading response of the theoretical model are in excellent agreement with finite-element simulations.
Moreover, the results of finite-element simulations and experiments on 3D-printed samples demonstrate that the deformation sequence of multiple unit cells can be tuned using either varied thickness or higher mode shape imperfection of the unit cell curved parts. Despite viscoelasticity, which is not taken into account in the theoretical model, the theory does predict the experimentally observed deformation sequence. Excellent quantitative match between experiments and FEA simulations is observed when viscoelasticity is taken into account. Despite the presence of manufacturing imperfections, a deterministic deformation sequence is obtained for thickness variations or mode shape imperfections of moderate sizes. The validity of the proposed model was demonstrated for systems with a small number of unit cells and with simple variations in the model parameters. However, the proposed model is general and could be applied to more complex metamaterials with a large number of unit cells with spatial variations in multiple geometrical and material parameters and might be particularly useful for the analysis of multistable materials of larger sizes, which could be useful for the design multifunctional metamaterials with programmable and switchable properties.
## Acknowledgment
This research was supported in parts by start-up funds from Georgia Tech, by NSF Award No. CMMI-1462894 to H. Jerry Qi and by the China Scholarship Council (Chao Yuan).
### Appendix A: Theoretical Model for a Single Unit Cell With a Mode Shape Imperfection
In order to determine the normalized force versus normalized deformation curve, the deflection of the beam is decomposed into the sum of the buckling mode shapes for an initially straight clamped–clamped beam
$w(x)=∑i=1∞AiWi(x)$
(A1)
where $Wi(x)$ are the mode shapes, and Ai are the modal amplitudes (for the numerical results, only the first 13 modes were taken into account). As in Ref. [32], the expressions for $Wi(x)$ are
$Wi(x)=1−cos(Nixl)Ni=(i+1)π}i=1,3,5,…$
$Wi(x)=1−2xl−cos(Nixl)+2Nisin(Nixl)Ni=2.86π,4.92π,…}i=2,4,6,…$
and system of nonlinear equations for the modal amplitudes, Ai, can be obtained by a variational method. The variation of the total potential energy ∂π is given by
$∂π=∂Ub+∂Uc−∂Uf$
(A2)
where ∂Ub, ∂Uc, and ∂Uf are the variations in the bending strain energy, compression strain energy, and potential energy of the external force, respectively. These variations are given by,
$∂Ub=∂[∫0lEI2(d2wdx2−d2w0dx2)2dx]∂Uc=−p∂Δs∂Uf=f∂d$
(A3)
The following normalizations from Ref. [32] are introduced
$x¯=xl, w¯(x¯)=w(x)h, f¯=fl3EIh, d¯=dh, Bi=Aih, p¯=pl2EI, U¯b=Ubl3EIh2, U¯c=Ucl3EIh2, π¯c=πl3EIh2$
(A4)
Using Eqs. (5), (A1), (8), and (A4), the normalized deformation, $d¯$, and the normalized compression force, $p¯$, can be expressed as a function of the normalized modal amplitudes, Bi
$d¯=1−2∑i=1,5,9,13.,…Bi$
(A5)
$p¯=12Q2[N12+a32N3216−∑i=1∞Bi2Ni24]$
(A6)
where $Q=h/t$. Moreover, the variation in the total potential energy, $∂π¯$, can be expressed as
$∂π¯=(N14−p¯N122B1−N144+2f¯)∂B1+(N24−p¯N222B2)∂B2+(N34−p¯N322B3−a3N344)∂B3+∑i=4,6,7,…(Ni4−p¯Ni24)∂Bi2+∑i=5,9,13,…(Ni4−p¯Ni22Bi+2f¯)∂Bi$
(A7)
The theorem of minimum potential energy, $∂π¯=0$, leads to the following system of nonlinear equations:
$(N14−p¯N122B1−N144+2f¯)∂B1=0(N24−p¯N222B2)∂B2=0(N34−p¯N322B3−a3N344)∂B3=0(Ni4−p¯Ni22)Bi∂Bi=0 for i=4,6,7,…(Ni4−p¯Ni22Bi+2f¯)∂Bi=0 for i=5,9,13,…$
(A8)
For the theoretical model of the initially curved beam, the second mode is constrained, i.e., only symmetric solutions were considered (because it is observed that the deformation mode of the unit cell is almost symmetric in FEA simulations), such that B2 = 0.
#### Model With a Third-Mode Imperfection
With a third-mode imperfection (a3 ≠ 0), there is only one form of solutions. The normalized modal amplitudes are given by
$B1=−12N12p¯−N12+4f¯N12(p¯−N12)B3=−12a3N32p¯−N32Bi=4f¯Ni2(p¯−Ni2), i=5,9,13,…Bi=0, for other i values$
(A9)
From Eqs. (A6) and (A9), we can get a quadratic equation for the normalized force, $f¯$
$αf¯2+βf¯+γ=0$
(A10)
where
$α=∑i=1,5,9,13…4(p¯−N12)2Ni2(p¯−Ni2)2β=−N12γ=p¯(p¯−N12)212Q2−N12p¯(p¯−2N12)16+a32p¯(2N32−p¯)N32(p¯−N12)216(p¯−N32)2$
(A11)
In order to obtain the normalized force versus normalized deformation curve, a vector is first formed for the normalized compression force, $p¯$. Equation (A10) is solved for each entry of this vector. If the discriminant of Eq. (A10), Δ, is negative, it means that the value of the normalized compression force is impossible. If Δ is positive, then two roots are obtained for the normalized force, $f¯$. For each of these roots, the mode coefficients Bi can be obtained from Eq. (A9), and the normalized deformation can be derived from Eq. (A5). Thus, the value of normalized force can be plotted as a function of the normalized deformation for each root. Connecting the points obtained for each entry of the vector for $p¯$ that corresponds to Δ > 0, a curve $f¯$ versus $d¯$ is obtained for each of the two families of roots; these two curves converge to the point that corresponds to Δ = 0. A single normalized force versus normalized deformation curve is obtained by connecting the two curves.
#### Model Without Imperfections
If there is no third-mode imperfection (a3 = 0), then, the solutions are comprised of two forms of solutions.
• Solution 1: if $p¯, a solution of the form as in the case a3 ≠ 0 is obtained. The equations obtained in the general case can be applied by setting a3 = 0.
• Solution 2: If $p¯=N32$, then another form of solution is obtained. In that case, the value of B3 cannot be found directly using Eq. (A8), while the values of all other Bi are the same as in the case a3 ≠ 0. Using Eq. (A5) and the equation $p¯=N32$, a linear equation that relates the normalized force, $f¯$, and the normalized deformation, $d¯$, is obtained
$f¯=−−N12N32−N12+d¯−1∑i=1,5,9,13…1Ni2(N32−Ni2)$
(A12)
In the model without imperfections, the normalized force versus normalized deformation curve is obtained by connecting the two forms of solutions, as described in Ref. [32].
### Appendix B: Viscoelastic Material Model
In order to more accurately characterize the deformation path, finite-element simulations with a viscoelastic, isotropic constitutive model were considered. The material was modeled in abaqus finite deformation viscoelastic model using a generalized standard linear solid with N = 4 viscoelastic branches, where the relaxation shear modulus, G(t), is given by
$G(t)=G∞+∑i=1NGie−tτi$
(B1)
where G is the long-term shear modulus, Gi is the shear modulus of branch i, and τi is the relaxation time constant. These parameters were fit to relaxation data of one printed sample (3 mm × 1 mm × 10 mm) using the printed material (DM9895) at room temperature, as shown in Fig. 10. The Poisson's ratio was assumed to be equal to 0.495 and to be time-independent. The value of G is set to 0.0167 MPa, and the values of Gi and τi are shown in Table 1.
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|
## Nostalgia corner: John Riordan’s referee report of my first paper
In 1971/1972 academic year, I was an undergraduate student at the Hebrew University of Jerusalem and toward the end of the year I wrote a paper about Abel’s sums. I sent it to John Riordan the author of the books “Combinatorial Identities” and “Combinatorial Analysis”.
I received this letter shortly after my 17th birthday, and I was surely very happy to read the sentence “I think you have had a splendid idea”. Here is part of Riordan’s remarks. The full report is here.
It took me some time to revise the paper and get it printed. And here is the report for the second version.
And here is part of Riordan’s second round of remarks. The full report is here.
I was certainly happy to read the following sentence: “I would remark that the result for p = -1 is new and perhaps the simplest derivation of Abel’s result.”
In 1978 I actually visited John Riordan in his office at Rockefeller University, NYC. I remember him as very cheerful and he told me that when his first book appeared he was working at Bell Labs and his managers wanted to talk to him. He was a bit worried that they would not approve of him spending time and effort to write a book in pure mathematics. But actually, they gave him a salary raise!
(If you have a picture of John Riordan, please send me.)
In 1979 the paper appeared.
Posted in Combinatorics, personal | | 7 Comments
## At the Movies III: Picture a Scientist
A few days ago I saw the great, emotionally steering, movie Picture a Scientist. I strongly recommend it. Here is the link to the hompage trailer, and IMBd page.
### SYNOPSIS
PICTURE A SCIENTIST chronicles the groundswell of researchers who are writing a new chapter for women scientists. Biologist Nancy Hopkins, chemist Raychelle Burks, and geologist Jane Willenbring lead viewers on a journey deep into their own experiences in the sciences, ranging from brutal harassment to years of subtle slights. Along the way, from cramped laboratories to spectacular field stations, we encounter scientific luminaries – including social scientists, neuroscientists, and psychologists – who provide new perspectives on how to make science itself more diverse, equitable, and open to all.
Raychelle Burks, Nancy Hopkins, and Jane Willenbring
Posted in Movies, Women in science | | 2 Comments
## At the Movies II: Kobi Mizrahi’s short movie White Eye makes it to the Oscar’s short list.
Update: White eye have made it to the list of five Oscar candidates! Congratulations!
My nephew Kobi Mizrahi is a well known movie producer and it was just announced that his short film “White eye” (עין לבנה) made it to the short list of ten Oscar candidates in the live-action short category. The director is Tomer Shushan.
Congratulations, Kobi!
I saw the movie and I highly recommend it! Let me use use the opportunity to recommend a full-length action movie the Dive that Kobi produced two years ago.
Kobi Mizrahi
Our previous post: And the Oscar goes to: Meir Feder, Zvi Reznic, Guy Dorman, and Ron Yogev.
## And the Oscar goes to: Meir Feder, Zvi Reznic, Guy Dorman, and Ron Yogev
My mother Carmela Kalai often said that if there was something she is thankful for it was that she was born in the era of movies. Indeed, she loved movies from a very early age throughout her life. So, I was especially excited to hear that my long-time friend Meir Feder and his colleagues at the Amimon company won the 2021 Academy Award for scientific contributions to the movie industry. It is very nice to see deep ideas from information theory, computer science and engineering come to play in the movies. Congratulations!
From right to left: Ron Yogev, Meir Feder, Zvi Reznic and Guy Dorman
Here is Meir’s thank-you speech.
| | 1 Comment
## Thomas Vidick: What it is that we do
A wonderful post by Thomas Vidick to cheer you up in difficult times with a lot of food for thought and for discussion. What is it that we (mathematicians) do?
It goes back to ancient Greece and also mention the legendary historian, poet, and philosopher Reviel Nets (whose two wonderful talks in Jerusalem we mentioned here and here).
This post is a follow-up on some somewhat off-hand comments that I made earlier regarding the notion of truth in a “proof-based” discipline such as pure mathematics or theoretical computer science. Since the former is easier to circumscribe and also has a larger literature available on it, for the purposes of the post I will discuss my emerging take on truth in mathematics; what I say applies to TCS as well. (I wasn’t able to find satisfactory writings on the practice of computer science, even more broadly interpreted than just “theory”; any pointers on this are very welcome.) I obviously don’t claim any originality here; I suspect that to some the points I make might be interesting while to others they could feel passé–in the latter case, please help me make progress in the comments!
View original post 2,277 more words
Posted in What is Mathematics | Tagged | 1 Comment
## New lower bounds for van der Waerden numbers by Ben Green
Abstract: We show that there is a red-blue colouring of [N] with no blue 3-term arithmetic progression and no red arithmetic progression of length $e^{C(\log N)^{3/4}(\log \log N)^{1/4}}.$ Consequently, the two-colour van der Waerden number w(3,k) is bounded below by $k^{b(k)}$, where $b(k)=c(\frac{\log k}{\log \log k})^{1/3}$. Previously it had been speculated, supported by data, that $w(3,k)=O(k^2)$.
The left side of the picture shows the world record holders for W(3,k). On the left Ben Green (LB) and in the centre Tomasz Schoen (UB). The pictures on the right shows protective mittens for people who make bold mathematical conjectures (see Igor Pak’s post
The two-colour van der Waerden number $w(m,k)$ is the smallest N such that however [N] = {1, . . . , N} is coloured blue and red, there is either a blue m-term arithmetic progression or a red k-term arithmetic progression. The celebrated theorem of van der Waerden implies that $w(m, k)$ is finite.
The van der Waerden number $w(m,k)$ is analogous to the Ramsey number $R(m,k)$. Finding the behaviors of $R(m,k)$ is an important problem in Ramsey theory and much attention is given to $R(k,k)$ and of $R(3,k)$. Similarly, understanding the values of $W(m,k)$, and especially of $W(k,k)$ and $W(3,k)$ are also central problems in Ramsey theory. A big difference between van der Waerden number and Ramsey numbers is that there are density theorems for the existence of $k$-terms arithmetic progressions. (Roth’s theorem for $k=3$ and Szemeredi’s theorem for general $k$.) There are several important methods to derive those density theorems (including Fourier methods, ergodic methods, and Szemeredi-type regularity) and these methods, as far as I know, do not apply for ordinary Ramsey numbers. (But correct me if I am wrong here, and if I am right and you have some insights as to why ergodic methods or Fourier methods do not apply to “ordinary” Ramsey, please share.)
Green’s paper studies the values of $w(3,k)$. The best known upper bound is of Tomasz Schoen, $w(3, k) for some constant $c > 0.$ The best known lower bound until the new paper, was by Li and Shu: $w(3, k) \gg (k/ log k)^2$. (This result, as the earlier bound by Robertson, used a probabilistic argument and relied on Lovasz’s local lemma.)
Several people conjectured, also based on empirical data, that $w(3,k) = O(k^2)$ but now Green proved a super-polynomial lower bound! This is amazing! Congratulations, Ben!
It is largely conjectured that the Behrend-type bounds give the correct quantitative behaviour for Roth’s theorem (and Szemeredi theorem). In rough terms what we see from Green’s example is that this might be true also for van der Waerden numbers.
The proof is rather involved and long, so, naturally, there is little I can say about it, which only slightly exceeds the little I actually know about it. The overview and other fragments of the paper I looked at are very illuminating. Here are a few things that caught my eyes.
1) A word about Tomasz Schoen’s upper bound and important paper: A subexponential upper bound for van der Waerden numbers W(3,k). Among other things Schoen’s proof relies on a lemma developed by Schoen for improved Roth bound. This relies on a structure theory of Bateman and Katz. The paper gives a nice description of the state of the art regarding the diagonal values $w(k,k)$. (Schoen’s upper bound on $w(3,k)$ follows also from the more recent bound for Roth’s theorem by Bloom and Sisask.)
2) Among the people that speculated that $w(3,k)$ behaves like $k^2$ is Ben Green himself. This is recorded in reference [9] of the paper. However, Green’s first reaction to this possibility was that it must be false. But he realized that some ideas for showing that it is false are themselves false.
3) Reference [9] in the paper is: B. J. Green, 100 open problems, manuscript, available on request. If you are curious about the list, request it!
4) New lower bounds in Ramsey theory are nor frequent. Thirteen years ago I described Elkin’s improvement to Behrend’s bound and a few days ago I mentioned Linial and Shraibman’s new lower bounds for the corner problem. Green’s study started by looking at complements of 3-AP free sets. An example by Green and Julia Wolf (that followed Elkin’s result) turned out to be important for reaching some parts of Green’s strategy.
5) In some sense, something about the $k^2$ prediction is not entirely lost. The construction gives a sort of a multi-scale behaviour where in the $r$th scale the example’s cardinality is $k^r$. (So all the empirical data comes from the $r=2$ regime.) Ben Green boldly suggests that the true values of $w(3,k)$ might exhibit such multi-scale behaviour. He conjectures that the true value of $w(3, k)$ is quasi-polynomial, namely it lies somewhere in between the bound given by his construction and something like $k^{c\log k}$ (which is Behrend-bound behaviour on the nose) .
6) Until Ben’s list of 100 problems becomes available to you, you may find interest in Francis Su’s 100 questions about mathematics for discussion and reflection.
7) In connection with Linial and Shraibman’s new lower bounds for the corner problem, let me mention that the best upper bound is by I.D. Shkredov. The paper is: On a two-dimensional analog of Szemeredi’s Theorem in Abelian groups, Izvestiya of Russian Academy of Sciences, 73 (2009), 455–505.
Posted in Combinatorics, Number theory | Tagged , | 2 Comments
## To cheer you up in difficult times 19: Nati Linial and Adi Shraibman construct larger corner-free sets from better numbers-on-the-forehead protocols
What will be the next polymath project? click here for our previous post.
Larger Corner-Free Sets from Better NOF Exactly-N Protocols, by Nati Linial and Adi Shraibman
Abstract: A subset of the integer planar grid [N]×[N] is called corner-free if it contains no triple of the form (x,y),(x+δ,y),(x,y+δ). It is known that such a set has a vanishingly small density, but how large this density can be remains unknown. The best previous construction was based on Behrend’s large subset of [N] with no 3-term arithmetic progression. Here we provide the first substantial improvement to this lower bound in decades. Our approach to the problem is based on the theory of communication complexity.
In the 3-players exactly-N problem the players need to decide whether x+y+z=N for inputs x,y,z and fixed N. This is the first problem considered in the multiplayer Number On the Forehead (NOF) model. Despite the basic nature of this problem, no progress has been made on it throughout the years. Only recently have explicit protocols been found for the first time, yet no improvement in complexity has been achieved to date. The present paper offers the first improved protocol for the exactly-N problem. This is also the first significant example where algorithmic ideas in communication complexity bear fruit in additive combinatorics.
This is remarkable for various reasons. On the additive combinatorics side, improved constructions are rare. For example, we reported here in 2008 Elkin’s (small) improvements of Behrend’s bound. For the corner-free problem the paper of Nati and Adi goes beyond the Behrend’s (and Elkin’s) constructions. On the communication complexity side this is significant progress on a classical 1983 problem of Chandra, Furst and Lipton. The connection that goes from improved result on communication complexity to additive combinatorics is exciting — certainly a new frontier for Nati and Adi. On the blogging side, I cannot compete with the beautifully written introduction. Click here to read the paper.
Remark 1: The number of the forehead problem is related to Levine’s hat problem that we discussed in this post.
Remark 2: Ryan Alweiss just told me about Ben Green’s new paper New lower bounds for van der Waerden numbers. It gives a construction of a red blue colouring of {1,2,…,N} with no 3 term blue or a k-term red arithmetic progression, where N is super-polynomial! Stay tune for a fuller report.
| Tagged , | 3 Comments
## Possible future Polymath projects (2009, 2021)
What will be our next polymath project?
A polymath project (Wikipedia) is a collaboration among mathematicians to solve important and difficult mathematical problems by coordinating many mathematicians to communicate with each other on finding the best route to the solution. The project began in January 2009 on Timothy Gowers’s blog when he posted a problem and asked his readers to post partial ideas and partial progress toward a solution. This experiment resulted in a new answer to a difficult problem, and since then the Polymath Project has grown to describe a particular process of using an online collaboration to solve any math problem.
After the success of Polymath1 and the launching of Polymath3 and Polymath4, Tim Gowers wrote a blog post “Possible future Polymath projects” for planning the next polymath project on his blog. The post mentioned 9 possible projects. (Three of them later turned to polymath projects, and one turned into a project of a different nature.) Following the post and separate posts describing some of the proposed projects, a few polls were taken and a problem – the Erdős discrepancy problem, was selected for the next project polymath5.
One of our next posts will have the same title and similar purpose as Tim’s 2009 post. I will describe several possibilities for my next polymath project. (A quick rather vague and tentative preview can be found at the end of this post.) Today we go back to Tim’s 2009 post and the problems posed there.
Comments on the 2009 proposed projects, the new proposed projects, other proposed projects, and on the polymath endeavor, are most welcome. (At the end of the post I also mention a few “meta” questions.)
Let me also mention The PolyTCS Project aimed for proposing projects in theoretical computer science. There are so far three very interesting proposals there, and the first proposal is the Friedgut-Kalai Entropy/Influence conjecture. For various proposals, see also the polymath blog administered by Tim Gowers, Michael Nielsen, Terry Tao, and me, and this MO question.
## The proposed projects in Gowers’s 2009 post and updates regarding these projects.
1. Littlewood’s conjecture and related problems.
The conjecture states that if $\alpha$ and $\beta$ are any two real numbers, and $\epsilon>0$, then there exists a positive integer $n$ such that $||\alpha n||\,||\beta n||\leq\epsilon/n$. Famously, Einsiedler, Katok and Lindenstrauss proved that the Hausdorff dimension of the set of counterexamples to the conjecture is zero. Gowers had ideas for an elementary approach, and his ideas are described in this later post. This project was not launched and I am also not aware of progress related to the problem (but I am not an expert).
2. A DHJ-related project.
DHJ stands for “density Hales Jewett” which was the topic of polymath1. The second proposed project was to build on the success of polymath1 and at a later post the following problem was proposed.
Conjecture: For every $\delta>0$ and every positive integer $k$ there exists $n$ such that if $A$ is any subset of $[k]^{[n]^2}$ of density at least $\delta$, then $A$ contains a combinatorial line such that the wildcard set is of the form $X\times X$ for some subset $X\subset[n]$.
As far as I know, this conjecture is still open.
Both questions “what kind of forbidden patters in $[k]^n$ force exponentially small density” and “what kind of forbidden patters in $[k]^n$ force vanishing density” are fascinating. Let me recommend again the Frankl-Rodl theorem and its proof as a role model.
3. Four Erdős-style combinatorial problems.
3a. Erdős’s discrepancy problem
This was the problem that was eventually chosen for polymath5. This was a very nice story. The problem was presented in this blog post and selected as polymath5 after some polls among readers. The polymath project was the longest in polymath history. There were six preliminary discussion posts with more than 600 comments followed by 21 official posts EDP1-EDP21. There was some short revival of the project (EDP22-EDP27) in 2012 where I contributed three posts. Famously, in 2015 Terry Tao solved the problem. The paper appeared in the journal Discrete Analysis. Tao’s solution relies on some insights from the polymath project including a crucial reduction that Terry himself contributed. It also crucially relied on a (then) new theory by Kaisa Matomaki and Maksym Radziwill. The solution was triggered by a blog comment by Uwe Stroinski who pointed to a possible connection to EDP, and a subsequent one by Kodlu who seconded Uwe’s suggestion. This is reported in EDP28, and here on my blog we celebrated the solution in this post.
3b. The Erdős-Hajnal conjecture.
Let $k$ be a positive integer and let $H$ be a graph. Erdös and Hajnal conjectured that there is a constant $C=C(H)$ such that if $G$ is any graph with at least $k^C$ vertices that does not contain any induced copy of $H$, then either $G$ or $G^c$ contain a clique of size $k$.
Tim asserted that the simplest open case is where $H$ is a pentagon. This special case was recently settled by Maria Chudnovsky, Alex Scott, Paul Seymour and Sophie Spirkl. They rely on recent results by Janos Pach and Istvan Tomon. See this videotaped lecture by Paul Seymour at IBS Discrete Mathematics Group, South Korea.
3c. Frankl’s union-closed conjecture.
## Peter Cameron: Doing research
To cheer you up in difficult times, here is a wonderful post by Peter Cameron about doing research. And see also this entertaining post by Peter on mathematics and religion.
Probably every research mathematician has been asked the question, “How do you do mathematical research?” Some lay people think we simply figure out ways of doing bigger and bigger long multiplications. Many more people think that all the mathematics must have been discovered by now, so what are we doing?
These misunderstandings are about what we do research on rather than how we do it, the question I want to focus on here.
In 1945, Jacques Hadamard published a remarkable book, The Psychology of Invention in the Mathematical Field. He was a great mathematician himself, one of whose achievements was the proof of the Prime Number Theorem fifty years earlier. He examined his own mental processes, searched the writings of his predecessors, and sent questionnaires to many leading mathematicians and scientists of his time, including Albert Einstein.
Hadamard’s conclusion was that the process of mathematical discovery can be broken…
View original post 1,204 more words
## To cheer you up in difficult times 18: Beautiful drawings by Neta Kalai for my book: “Gina Says”
In 2009 I wrote a book “Gina Says” that appeared here on the blog, about the adventures of “Gina” in the blogsphere. In 2017 the book (edited and shortened a little) appeared in “world scientific.” The most important additions were beautiful drawings contributed by my daughter Neta. Here are three of them. More to come.
“My own heart goes to chemistry” (Chapter 5)
“Octopus music” (Chapter 25)
anti anti anti missile missile missile missle (Chapter 17, see also the post Chomskian Linguistics)
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# How to count the total number of sections within a chapter?
How can one count the total number of sections within each chapter?
I am trying to use the totcount package, but it returns the total number of sections of the last chapter, not total number of sections of the current chapter.
In the following MWE, the desired behaviour would be to report 3 sections for the first Chapter, and 1 section for the second and last chapter.
totcount reports only 1 section (the counter value for the last chapter).
\documentclass[12pt]{book}
\usepackage{totcount}
\regtotcounter{section}
\begin{document}
\chapter*{First Chapter}
The total number of sections in this chapter should be 3.
The totcount package reports: \total{section}
\section{First Section}
First text
\section{Second Section}
Second text
\section{Third Section}
Third text
\chapter*{Last Chapter}
The total number of sections in this chapter should be 1.
The totcount package reports: \total{section}
\section{First and only section}
Section text
\end{document}
How can one count the total number of sections within each chapter?
Note: totcount reports 0 sections if using \section*, see Sigur's comment.
• \section*{} does not increase the counter section so it is not counted. – Sigur May 1 '15 at 12:54
• I see, thank you. I guess, if needed,, one could create and manage his own counter. That answer the second issue, thank you. – gsl May 1 '15 at 12:59
• @gsl: assoccnt might help – user31729 May 1 '15 at 13:02
• @christian-hupfer: Thank you, I believe that would be a solution. Would you care to submit a brief answer (perhaps with example) , so that I can accept it? – gsl May 1 '15 at 13:04
• @gsl: I just uploaded a pre-version of a new package named cntpersec to CTAN. It provides the means of storing counter values on a per chapter base and displaying the values in advance after two compilation runs have been made. If you send me an email to my address (see my profile page), I'll send you the package files with some preliminary documentation as long it's not fully available on CTAN – user31729 May 3 '15 at 0:40
Edit The look-ahead - version of knowing section numbers per chapter in advance is at the very end of this post. It requires two runs of compilation.
This question lead to a new package cntperchap which is available in its version 0.2 on CTAN since 2015/9/5
This uses the assoccnt package (whose author I accidentally know quite well ;-))
It associates a counter totalsections to the section counter -- each time the section counter is increased, the totalsections counter is stepped as well.
However, there is no automatic resetting for \chapter* usage. In this case, it can be done automatically by prepending some code to \chapter using \xpretocmd from xpatch package
Note The author of assoccnt should really incorporate this into his package ;-)
\documentclass[12pt]{book}
\usepackage{assoccnt}
\usepackage{xpatch}
\newcounter{totalsections}[chapter] % Does not work with chapter*
% Automatically provide for resetting of the section counter each time
%\chapter or \chapter* is used -- in this setup this is requested.
\xpretocmd{\chapter}{\setcounter{totalsections}{0}}{}{}
\DeclareAssociatedCounters{section}{totalsections}
\begin{document}
\chapter*{First Chapter}
The total number of sections in this chapter should be 3.
\section{First Section}
First text
\section{Second Section}
Second text
\section{Third Section}
Third text
There \number\value{totalsections} sections in this chapter
\chapter*{Last Chapter}
\section{First and only section}
Section text
There \number\value{totalsections} sections in this chapter
\end{document}
Edit Some new version, which uses a per chapter count of sections (requires two runs to be successful).
Explanation: Each time a new chapter is used, the accumulated number of sections is written to external file, \jobname.seccnt, say foo.seccnt for short. This file is read again at the next latex compile run and the values are stored to a etoolbox list. The macro \GetTotalSectionCounter will advance through this list until it's at the right position and will then print out the number of sections in this chapter, even ahead. (The macro should be expandable, I think this is the case)
At the moment, it's necessary to manually remove the foo.seccnt file if there have been changes to number of chapters/sections.
I'll try to get around this drawback.
\documentclass{book}
\usepackage{ifthen}
\usepackage{assoccnt}
\usepackage{xpatch}
\usepackage{pgffor} % Only for quick usage of a lot of sections, demo only
\newwrite\seccountfile%
\newcounter{currentchapter}
\newcounter{totalsections}
\newcounter{togglecounter}
\DeclareAssociatedCounters{section}{totalsections}
\newcommand{\getsectioncountnumbers}{%
\setcounter{togglecounter}{0}%
\whiledo {\value{togglecounter} < 1}{%
\ifeof\seccountinfile%
\stepcounter{togglecounter}%
\else%
\fi%
}%
}
\xpretocmd{\chapter}{%
\stepcounter{currentchapter}%
\immediate\write\seccountfile{%
\number\value{totalsections}%
}%
\setcounter{totalsections}{0}
}{}{}
\makeatletter
\newcounter{tempcount@a}
\newcommand{\@getsectiontotalcounter}[1]{%
\setcounter{tempcount@a}{0}%
\renewcommand*{\do}[1]{%
\ifnumequal{\value{#1}}{\value{tempcount@a}}{%
##1\listbreak%
}{%
\stepcounter{tempcount@a}%
}%
}%
\dolistloop{\seccountlist}%
}
\newcommand{\GetSectionTotalCounter}[1][currentchapter]{%
\ifdef{\seccountlist}{%
\@getsectiontotalcounter{#1}%
}{}%
}
\makeatother
\AtBeginDocument{%
\IfFileExists{\jobname.seccnt}{%
\immediate\openin\seccountinfile=\jobname.seccnt%
\getsectioncountnumbers%
}{%
% Open for writing
\immediate\openout\seccountfile=\jobname.seccnt%
}%
}
\AtEndDocument{%
\immediate\write\seccountfile{%
\number\value{totalsections}%
}%
\immediate\closeout\seccountfile%
}
\begin{document}
\chapter*{First}
This chapter will have \GetSectionTotalCounter sections
\section{First}
\section{Second}
\chapter*{Second}
This chapter will have \GetSectionTotalCounter sections
\section{First}
\section{Second}
\section{Third}
\section{Fourth}
% Now a really large chapter
\chapter*{Third}
This chapter will have \GetSectionTotalCounter sections
\foreach \x in {1,...,100} {%
\section{\x}
}
\end{document}
Edit This is the version without explicit deleting the foo.seccnt file
I used the addtocontents approach to let LaTeX write the section numbers to a separate file just as it is done with the toc related stuff. The foo.seccnt is then treated as a faked toc, read in before (and the values temporarily stored) rewritten in the run.
\documentclass{book}
\usepackage{ifthen}
\usepackage{assoccnt}
\usepackage{xpatch}
\usepackage{pgffor}
\newcounter{currentchapter}
\newcounter{totalsections}
\newcounter{togglecounter}
\DeclareAssociatedCounters{section}{totalsections}
\makeatletter
\newcommand{\getsectioncountnumbers}{%
\setcounter{togglecounter}{0}%
\whiledo {\value{togglecounter} < 1}{%
\ifeof\tf@seccnt
\stepcounter{togglecounter}%
\else%
\fi%
}%
}
\xpretocmd{\chapter}{%
\stepcounter{currentchapter}%
\number\value{totalsections}%
}%
\setcounter{totalsections}{0}
}{}{}
\newcounter{tempcount@a}
\newcommand{\@getsectiontotalcounter}[1]{%
\setcounter{tempcount@a}{0}%
\renewcommand*{\do}[1]{%
\ifnumequal{\value{#1}}{\value{tempcount@a}}{%
##1\listbreak%
}{%
\stepcounter{tempcount@a}%
}%
}%
\dolistloop{\seccountlist}%
}
\newcommand{\GetSectionTotalCounter}[1][currentchapter]{%
\ifdef{\seccountlist}{%
\@getsectiontotalcounter{#1}%
}{}%
}
% This is a modified version from \@starttoc, being defined latex.ltx
\def\@startfaketoc#1{%
\begingroup
% Generate the file handle first
\expandafter\newwrite\csname tf@#1\endcsname%
\makeatletter
% Read first before deleting it
\ifcsdef{tf@#1}{%
\IfFileExists{\jobname.#1}{%
\immediate\openin\csname tf@#1\endcsname \jobname.#1\relax
\getsectioncountnumbers%
}{}
}{%
\typeout{No section count numbers so far}
}%
\if@filesw
% Write only if not \nofiles is specified
\immediate\openout \csname tf@#1\endcsname \jobname.#1\relax
\fi
\@nobreakfalse
\endgroup%
}
\AtBeginDocument{%
\@startfaketoc{seccnt}
}
\AtEndDocument{%
% Write the last section count to the file
\number\value{totalsections}%
}%
}
\makeatother
\begin{document}
\tableofcontents
\chapter*{First}
This chapter will have \GetSectionTotalCounter sections
\section{First}
\section{Second}
\chapter*{Second}
This chapter will have \GetSectionTotalCounter sections
\section{First}
\section{Second}
\section{Third}
\section{Fourth}
% Now a really large chapter
\chapter*{Third}
This chapter will have \GetSectionTotalCounter sections
\foreach \x in {1,...,100} {%
\section{\x}
}
\chapter{Fourth}
This chapter will have \GetSectionTotalCounter sections
\section{A single section}
\end{document}
Edit The OP gsl noted some error in this code. I could track it down to the fact that \@startfaketoc tries to read in the external foo.seccnt file in the first run already. This fails of course, since there is no such file if it was deleted or the documentat is compiled for the very first time.
• @gsl: Yes, it has some advantages, but it has to be used with care! It's no swiss-knife ;-) Happy texing – user31729 May 1 '15 at 13:22
• If one needs to know the value of totalsections at any place withing a chapter, like for example at the beginning, like in the MWE, it seems that system is not working. Just move the line There are \number\value{totalsections} sections in this chapter right after \chapter to see what I mean, or see this MWE: paste.lisp.org/display/147711 – gsl May 1 '15 at 14:13
• @gls: Yes, I see the point. I will think about it. It will probably need more than one run to achieve since assoccnt can't ahead how many sections there will be per chapter in advance. – user31729 May 1 '15 at 14:32
• @gls: How many chapters do you intend to use in your document? I try to use a new setup – user31729 May 1 '15 at 15:09
• @gsl: I am still working on it, but the first version will be a hack, nothing nice – user31729 May 1 '15 at 17:42
|
Publication
Title
Tuning the Fermi level of $SiO_{2}$-supported single-layer graphene by thermal annealing
Author
Abstract
The effects of thermal annealing in inert Ar gas atmosphere of SiO2-supported, exfoliated single-layer graphene are investigated in this work. A systematic, reproducible change in the electronic properties of graphene is observed after annealing. The most prominent Raman features in graphene, the G and 2D peaks, change in accord to what is expected in the case of hole doping. The results of electrical characterization performed on annealed, back-gated field-effect graphene devices show that the neutrality point voltage VNP increases monotonically with the annealing temperature, confirming the occurrence of excess hole accumulation. No degradation of the structural properties of graphene is observed after annealing at temperatures as high as 400 °C. Thermal annealing of single-layer graphene in controlled Ar atmosphere can therefore be considered a technique to reproducibly modify the electronic structure of graphene by tuning its Fermi level.
Language
English
Source (journal)
The journal of physical chemistry : C : nanomaterials and interfaces. - Washington, D.C., 2007, currens
Publication
Washington, D.C. : 2010
ISSN
1932-7447 [print]
1932-7455 [online]
Volume/pages
114:5(2010), p. 6894-6900
ISI
000276562500002
Full text (Publisher's DOI)
UAntwerpen
Faculty/Department Publication type Subject
|
For some time now, I have been using Git to keep track of my work - both source code and otherwise. Even when you’re not using it for code, Git is perfect - easy backups to a Git server in the cloud along with the possibility of maintaining multiple working versions of my papers and writing projects is just great.
Except - Git treats my Word files as binaries. Different versions can be accessed just fine, but I have no access to diff functionality because Git doesn’t understand what has changed between two versions. Not to mention, most programmers will flinch uncontrollably at my mention of storing binaries in version control - even if it’s only the stupid play I’m writing.
The solution is to use something that Git can understand out of the box - like Markdown - or find a way to help Git interpret .docx files, which you know deep (deep) down are text files. Choosing not to give up the flexibility of Word (and mostly out of habit), I started looking at ways to implement the latter.
Martin Fenner has a very informative post on one way of doing this. He uses pandoc to convert docx files to markdown that git can then use to compare two versions of the same file. However, I ran into problems when using pandoc to convert files, and had to use a different implementation. I’m including both methods below, so feel free to try either.
## Using Pandoc to interpret Word Documents
Martin recommends the following steps:
Before you begin, install pandoc. On a Mac, if you have HomeBrew or MacPorts, this is as simple as typing brew install pandoc or port install pandoc. For a PC, you can find an installer on the downloads page. If you use LaTex, there’s a good chance you already have it.
• Create a .gitattributes file in your project directory (if you don’t have one already) and add the following:
*.docx diff=pandoc
This will let git know what to do when it encounters files with a .docx extension.
• Append the following to the .gitconfig file (On Mac, this should be ~/.gitconfig. For Windows, this will depend on your installation, but for default settings, it should be in your home folder - or %HOMEDRIVE%%HOMEPATH%) -
[diff "pandoc"]
textconv=pandoc --to=markdown
prompt = false
[alias]
wdiff = diff --word-diff=color --unified=1
We’re adding pandoc to the config, along with creating an alias for a specific diff call.
• Once this is done, you can make sure everything works by commiting a Word file, and running the following command:
git wdiff <your_file>.docx
Unfortunately, this didn’t work for me. For every file I tried, pandoc failed with a ‘UTF-8’ error. Some Googling showed that this was a common error with pandoc and some versions of Word. There were a few fixes, but none solved my problem. Not wanting to trawl the manpage for answers, I moved on to door number two.
## Using a custom converter for Word in Git
If pandoc worked for you, then this is really unnecessary and you can move on to actual work. But it didn’t work for me, so I started looking for docx to markdown converters. There are a few online that are pretty good. There’s a Ruby Gem that does it pretty well. There’s even a Visual Basic Script that does it, if you’re looking for integration into Word itself. For Mac, the included textutil does docx conversions pretty easily. It comes pre-installed, and is pretty light-weight. Unfortunately, it doesn’t support markdown, but if you really need to track stylistic changes, there is a Ruby Gem that uses textutil for docx to markdown conversion that shouldn’t be hard to use. Just install it and replace the direct calls to textutil in the directions. Converting to plain text was simple enough for me. I’m mostly using Terminal to track changes to my work, so textutil’s txt conversion seemed enough (at least for the time being). It also meant that I didn’t have to install new software on all my workstations.
To implement textutil-based tracking, do the following:
• Create and add the following to .gitattributes in your project directory (same as before):
*.docx diff=textutil
• Add these lines to ~/.gitconfig:
[diff "textutil"]
textconv=textutil -convert txt -stdout
prompt = false
[alias]
wdiff = diff --word-diff=color --unified=1
The -convert txt switch lets us convert the file to plain text, and the -stdout switch makes sure we’re printing to standard out instead of to file.
• Once this is done, you can run git wdiff <filename>.docx to see the results.
It should now be easy to modify Git to work with complex file formats, now that you know how.
|
• ### Announcements
#### Archived
This topic is now archived and is closed to further replies.
# possible to use d3dfvf_diffuse and d3dfvf_normal at same time for vertices?
## Recommended Posts
johnnyBravo 100
Hi with my vertices i want to use lighting aswell as being able to set the diffuse for my vertices. that is use d3dfvf_normal |d3dfvf_diffuse at the same time. When ive tried it the diffuse doesn''t show up heres some code showing my use of the fvf in settings (ignore this if u want , just to show what im doing)
void create(LPDIRECT3DDEVICE9 d3dDevice, int bufferArraySize)
{
int vertexByteSize = bufferArraySize * sizeof(Vertex);
d3dDevice->CreateVertexBuffer(vertexByteSize, 0,
(D3DFVF_XYZ|D3DFVF_DIFFUSE|D3DFVF_NORMAL|D3DFVF_TEX1) ,
D3DPOOL_DEFAULT, &lp_VB, NULL ) ;
}
void select()
{
LPDIRECT3DDEVICE9 d3dDevice;
lp_VB->GetDevice(&d3dDevice);
d3dDevice->SetStreamSource( 0, lp_VB, 0, sizeof(Vertex));
d3dDevice->SetFVF((D3DFVF_XYZ|D3DFVF_DIFFUSE|D3DFVF_NORMAL|D3DFVF_TEX1) );
}
//and part of my vertice class
class Vertex {
private:
D3DXVECTOR3 point;
DWORD diffuse;
D3DXVECTOR3 normals;
D3DXVECTOR2 textureCoords;
...
if its possibly to use the normal and diffuse fvf flags together, could you maybe hint towards how thanks,
##### Share on other sites
don 431
Swap the order of your normal vector and the diffuse color in your vertex class declaration, then read the description of D3DRS_COLORVERTEX in the SDK docs.
##### Share on other sites
johnnyBravo 100
Ok igot it all working like that now, but now i want to try and make the diffuse work with lighting.
is there a way to actually use diffuse with lighting?,
Thanks,
[edited by - johnnyBravo on March 27, 2004 10:44:55 PM]
##### Share on other sites
psykr 295
Hm, now that I think about it, you can put diffuse colors into the material. However, this way you can''t control the diffuse color as finely.
|
# Expedition
Time Limit: 1000MS
Memory Limit: 65536K
[显示标签]
## Description
A group of cows grabbed a truck and ventured on an expedition deep into the jungle. Being rather poor drivers, the cows unfortunately managed to run over a rock and puncture the truck's fuel tank. The truck now leaks one unit of fuel every unit of distance it travels. To repair the truck, the cows need to drive to the nearest town (no more than 1,000,000 units distant) down a long, winding road. On this road, between the town and the current location of the truck, there are N (1 <= N <= 10,000) fuel stops where the cows can stop to acquire additional fuel (1..100 units at each stop). The jungle is a dangerous place for humans and is especially dangerous for cows. Therefore, the cows want to make the minimum possible number of stops for fuel on the way to the town. Fortunately, the capacity of the fuel tank on their truck is so large that there is effectively no limit to the amount of fuel it can hold. The truck is currently L units away from the town and has P units of fuel (1 <= P <= 1,000,000). Determine the minimum number of stops needed to reach the town, or if the cows cannot reach the town at all.
## Input
* Line 1: A single integer, N * Lines 2..N+1: Each line contains two space-separated integers describing a fuel stop: The first integer is the distance from the town to the stop; the second is the amount of fuel available at that stop. * Line N+2: Two space-separated integers, L and P
## Output
* Line 1: A single integer giving the minimum number of fuel stops necessary to reach the town. If it is not possible to reach the town, output -1.
## Sample Input
4
4 4
5 2
11 5
15 10
25 10
## Sample Output
2
## Hint
INPUT DETAILS:
The truck is 25 units away from the town; the truck has 10 units of fuel. Along the road, there are 4 fuel stops at distances 4, 5, 11, and 15 from the town (so these are initially at distances 21, 20, 14, and 10 from the truck). These fuel stops can supply up to 4, 2, 5, and 10 units of fuel, respectively.
OUTPUT DETAILS:
Drive 10 units, stop to acquire 10 more units of fuel, drive 4 more units, stop to acquire 5 more units of fuel, then drive to the town.
## Source
USACO 2005 U S Open Gold
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# Troubleshooting IBM Installations¶
## Opening Port 12345¶
For default IBM Power9 systems with RHEL 7 installed, be sure to open port 12345 in the firewall. For example:
firewall-cmd --zone=public --add-port=12345/tcp —permanent
firewall-cmd --reload
## Growing the Disk¶
Some users may find it necessary to grow their disk. An example describing how to add disk space to a virtual machine is available at https://www.geoffstratton.com/expand-hard-disk-ubuntu-lvm. The steps for an IBM Power9 system with RHEL 7 would be similar.
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# Math Help - cofinality, infinite cardinal
1. ## cofinality, infinite cardinal
Show that for each infinite cardinal $\kappa$, $\kappa <_c \kappa^{\text{cf}(\kappa)}$.
Notation: $\text{cf}$ denotes the cofinality. I know some properties of $\text{cf}(\kappa)$. They may be helpful.
$\text{cf}(\kappa) \leq_c \kappa$
For each infinite cardinal number $\kappa$, $\text{cf}(2^{\kappa}) >_c \kappa$.
However, I do not see how to prove this. Any hints would be great. Thanks.
2. Here one of Koenig's lemmas helps:
Consider $(\mu_i),\ (\lambda_i)$ two families of cardinals indexed by $I,$ such that for any $i\in I,\ \mu_i<\lambda_i,$ then:
$\sum_I\mu_i<\prod_I\lambda_i$
Use this result with the fact that given a cardinal $\kappa:$
$\text{cf}(\kappa)$ is the lowest cardinal such that there exists a family $(\mu_{\xi})_{\xi\in\text{cf}(\kappa)}$ with for all $\xi\in\text{cf}(\kappa),\ \mu_{\xi}<\kappa$ and $\sum_{\xi\in\text{cf}(\kappa)}\mu_{\xi}=\kappa$
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# Import Once in Julia
### Background
I’m reading Think Julia to repick the language again.
• To import a package/module, we using Foo.
• To run a script within the REPL, we include("foo.jl").
This is useful for user-defined functions, which are usually multi-line.
### Problem
How to avoid importing a module/including a file twice?
### Solution
Here’s a modified code from the book to give an example.
1. polygon.jl depends on the module :ThinkJulia, and it defines a function which moves the object t along a polygon-shaped closed loop.
2. circle.jl depends on the script polygon.jl, and it attempts to move the object t along a “circle”.
# polygon.jl
if !isdefined(Main, :ThinkJulia)
using ThinkJulia
end
function polygon(t, n, len)
angle = 360 / n
for i in 1:n
forward(t, len)
turn(t, -angle)
end
end
# circle.jl
if !(@isdefined polygon) || !(typeof(polygon) <: Function)
include("polygon.jl")
end
function circle(t, r)
circumference = 2 * π * r
n = 50
len = circumference / n
polygon(t, n, len)
end
### Notes
1. <:(T1, T2) -> Bool is a function that returns true if T1 is a subtype of T2.
2. isa(x, type) -> Bool is a function, e.g. :foo isa Symbol.
3. eval() evaluates the expression inside. This can be used to get an object of type Module. Say typeof(:Main) is Symbol, but typeof(eval (:Main)) is Module.
4. filter((x) -> f(x), arr) filters element from arr using the predicate f.
Reference: https://stackoverflow.com/q/37823591
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## General developer forum
### Open external links in the pop-up window using lightbox | fancybox
Open external links in the pop-up window using lightbox | fancybox
Hello!
Moodle 3.4. There is a task - to open links to extra resources with js libraries like "lightbox" | "fancybox" and so on.
The task is to ensure that the page on which the link is located:
1. A script from the "lightbox" | "fancybox" library supposed to be connected.
2. The <body> tag must have a specific class.
3. A certain <a> tag must have a certain attribute.
I'm trying to modify the mod / url module.
I change the file lib.php, view.php. But all this is a roulette game guess which line passes the desired parameter to course / view.php.
I understand that somewhere supposed to be
$PAGE-> requires-> js$ PAGE-> requires-> css
\$ PAGE-> add_body_class
but, all my attempts did not give the desired result, and much time has already been spent. This can continue indefinitely.
Help me please. How can I pass additional attributes to tags and connect libraries through the mod / url plugin?
Thank you in advance!
Average of ratings: -
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# Nukes with cup-sized mushroom clouds
A few sequential questions:
• Is it possible for a nuclear explosion to be small enough to produce a 250-ml (one cup) mushroom cloud?
• If so, how much uranium would that take?
• How close to the explosion could one be (in normal clothing) not to suffer from burns or excessive radiation exposure?
• Also: mushroom clouds happen because of high temperatures in a big explosion. They're not specific to nuclear explosions. – Dan Aug 9 '13 at 6:09
Mushroom clouds are formed in explosions (not necessarily nuclear - see picture) as a result of the Rayleigh-Taylor instability. For given density contrast $\frac{\rho_{cold}-\rho_{hot}}{\rho_{cold}+\rho_{hot}}$ between the hot cloud and the cold atmosphere, the timescale $t_{RT}$ for this instability scales with the length scale $L_{RT}$ according to:
$$t_{RT} \approx \sqrt{\frac{L_{RT}}{g}\frac{\rho_{cold}+\rho_{hot}}{\rho_{cold}-\rho_{hot}}}$$
With $g$ the gravitational acceleration. For $\frac{\rho_{cold}-\rho_{hot}}{\rho_{cold}+\rho_{hot}}\approx 0.1$ and $L_{RT} = 1 km$ we find $t_{RT} \approx 30 s$. For a cup sized ($L_{RT} = 0.05 m$) explosion with the same density contrast we find $t_{RT} \approx 0.2 s$.
The 'mushroom' has disappeared before it really takes shape. Bottom line is that you need fairly large explosions to observe a mushroom.
• ... or you need a very fast explosion that's observed at high time resolution! – Emilio Pisanty Sep 4 '13 at 11:44
Let's try to answer your question in terms of fluid mechanics. A mushroom cloud is quite interesting, because of the very large scale phenomena, down to the vary small scales, which are called Kolmogorov microscales.
The mushroom cloud is quite complicated, as there are probably supersonic velocities, buoyancy effects and of course combustion playing important roles. Lets ignore all these effects, and assume ideal isotropic turbulence as a first order approximation.
The large scale we call $L$, and the small scale, the Kolmogorov scale $\eta$. Then we are interested in the fraction of these scales, as this is determines whether you'll qualify it as a mushroom or not.
$\eta$ can be approximated by
$$\eta=\left(\frac{\nu^3}{\epsilon}\right)^{\frac{1}{4}}.$$
Here, $\nu$ is the kinematic viscosity and $\epsilon$ the dissipation rate of turbulent kinetic energy. Under our assumptions, we can take the dissipation rate at the small scale equal to the large scale production, thus
$$\epsilon=\frac{U^3}{L}$$
Here, $U$ is the large scale velocity. And thus
$$\frac{L}{\eta}=\frac{U^3 L^3}{\nu^3 }^{\frac{1}{4}}=Re^{\frac{3}{4}},$$
where $Re=\frac{UL}{\nu}$ is the Reynolds number.
For the large mushroom, with assuming $U=100 m/s$, $L=1km$ and $\nu=10^-5 m^2/s$ (air viscosity), I would get $L/\eta=3\times 10^7$, while for a small mushroom, with $L=0.1m$, and probably a lower velocity, $U=1m/s$, the range of scales would be $L/\eta=10^3$, which would look completely different. You could also argue the other way around: What velocity $U$ do I need to get the same scale separation? This means Reynolds similarity and you would end up with $U=10^6 m/s$ in the small cup! In all cases it would be gone instantly.
In addition it is worth to know that to make a nuclear explosion, you need enough fissile material. Sufficient mass is called "the critical mass".
http://en.wikipedia.org/wiki/Critical_mass
As you can see here, you need at least about 15 kg of uranium. This costs a lot. Moreover you would need plenty of gear to initiate explosion (because you firstly have to assemble everything and then, using explosives rise the temperature to lower the critical mass parameter).
Such nuclear device would be bigger than explosion itself.
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= P O(semicircle). 14 * (a * a) / 8. The area of compound shapes worksheets consist of a combination of two or more geometric shapes, find the area of the shaded parts by adding or subtracting the indicated areas, calculate the area of rectilinear shapes (irregular figures) and rectangular paths as well. Serving the east coast (Pennsylvania (PA), New Jersey (NJ), New York (NY), Florida (FL), Georgia (GA), South Carolina (SC), North Carolina (NC), West Virginia (WV), Virginia (VA), and Maryland. 1597 is the area to the right. When I use trapz keyword it gives me the whole area. Final Answer. The writers are reliable, honest, extremely knowledgeable, and the results are always top of the class! - Pam, 3rd Year Art Visual Studies. For rectangles and squares, multiply width by height. Area to Z Score Formula. Area of a circle = π * r 2. Middle: Shade the area of the values that you entered. Area of an arch given angle. If you want to total surface area remember to add on the area of the base of the cone. The Are is a unit of area equal to 100 square metres (10m x 10m). Calculate the shaded area. Considering my options are a small stretch that gets 1. This represents the population that does not fall within this z score range. 14 ∗ 2) = 3. Note the number of square units it takes to fill it. Area of shaded region = area of bigger shape - area of smaller shape. One Step Complete for Dense Shade Mulch Areas with Smart Seed, Mulch, Fertilizer Mix. An intermediate step in this learning process is combining the two shapes. Calculate the shaded area in cm 2 between the two circles shown below. If you know radius and angle, you may use the following formulas to calculate the remaining segment parameters:. The Area of the shaded region = (Area of the largest circle) - (Area of the circle with radius 3) - (Area of the circle with radius 2). The area of shaded region is 13. Diameter is just 2x radius, so to get the circle's area from the diameter: a = π ( d / 2) 2. Share 0 Shaded area = area of circle - area of inner rectangle pi * (13/2)^2 - 12*5 sq cm = 3. we're asked to find the area of the shaded region so the area of this red shaded region so this is interesting this is almost a 10 by 10 square except we have these quarter circles that are cut out so the area of this would be the area of what a 10 by 10 square would be minus the area of these quarter circles and each of these quarter circles it's a quarter of a circle with a radius 3 I think. The surface area of a solid is a measure of the total area occupied by the surface of an object. There are 8 triangles in an octagon, so Area of a one triangle x 8. Calculate the area of the shaded region if the length of the side of the square is 10 cm. Our mission is to provide a free, world-class education to anyone, anywhere. m = f (x) & m = g (x) m = f (x) & m = g (x) Where. Use inequalities to shade. In terms of 2b and x, r2 = b2 + x, so height = e−(b2 + x 2 ) = e−b2 e−x 2 = ce−x 2 3. In our case, the number 4 is less, the rounding is therefore 53 600. By following this way a circumference. Calculate the area of the shaded region, if the diameter of the semi circle is equal to 14 cm. Looks good. Substitute the values in area of sector formula, Area = πr 2 × (θ / 360). 52 The area of the shaded region is X 3 Part 2 of 4 0. This means that the region we're interested in must have one of the two curves on every boundary of the region. Try this Drag one of the orange dots that define the endpoints of the segment. The only improvement I wish for is that it would show more exactly how many hours of sun each area got, Instead of increments of "full shade, part shade, part sun, and full sun" it would be better if it was "-1 hour, 1-2 hours, 2-3 hours, 3-4 hours, 4-5 hours, 5-6 hours, 6 hours+ (of sun)". The curve y = x —x2 — 7x + 5 and the line y = 2x — 4 are shown opposite. The area of the square is 16(r)(r). Select unit: Diameter of the Whole Area {{selectedunit. First of all, just what do we mean by "area enclosed by". It can be visualized as the amount of paint that would be necessary to cover a surface, and is the two-dimensional counterpart of the one-dimensional length of a curve, and three-dimensional volume of a solid. Area = radius 2 2 * ( π 180 * θ - sin(α) ). Microsoft Excel does not. Example problem 2 •For this problem, the lower boundary is 15 and the upper boundary is theoretically infinity. Apr 05, 2012 · The Area of the shaded region = (Area of the largest circle) – (Area of the circle with radius 3) – (Area of the circle with radius 2). Tag: area of the shaded region calculator Find The Area Of The Shaded Region. There are 8 triangles in an octagon, so Area of a one triangle x 8. We select and ship only the finest quality bulbs and perennials. The radius can be any measurement of length. Kevin Wilda. r = C ( ∗ 2) r = 20 ( 3. Begin by sketching the distribution and labeling the relevant information. 52 - 84375 P O. Simple step-by-step tutorial by PreMat. To find the curved surface area of any cone, multiply the base radius of the cone by pi. How calculate the shaded area in this picture? Stack Exchange Network. Coverage Area 1. 66 - 20) (23. For example, if the angle is 45° and the radius 10 inches, the area is (45 / 360) x 3. 13 x 36 = 113. the radius of the circle is 3 inches. (a) Find the coordinates of P and Q. By following this way a circumference. I am using maxent in R to have an idea of suitable area of certain crop for whole country. So, the area between two curves calculator computes the area where two curves. r = radius of the circle. The following Java program to print the area of a circle has been written in five simple different ways, static method, using constructor, Interface, inheritance with sample outputs for each program. In our case, the number 4 is less, the rounding is therefore 53 600. Since each of the 4 sides of a square are exactly the same size by definition then you can measure the size of any side of the square and enter it into the field below and press calculate to calculate the area of. We can find the areas between curves by using its standard formula if we have two different curves. = P O(semicircle). Calculate the moment of inertia of the shaded area about the x-axis. 1415926535898 √ = square root Calculator Use. It is important to use the "Length A", the long measurement in the box with the Length A label. The calculator will then determine the length of the arc. By following this way a circumference. Put the value into the equation. Dividing the sum of the area moments by the total area we calculate the x-centroid 1 1 n ii i n i i xA x A = = = ∑ ∑ ID Area x i x i*Area (in2) (in) (in3) A 1 2 0. Area of a regular polygon. You can calculate the area of an oval by following a simple mathematical equation. Moment of shaded area of web about neutral axis calculator uses mi_of_the_area_of_section = (( Thickness of web *((( Inner length of I-section ^2)/4)-( Distance b/w considered and neutral layer ^2)))/2) to calculate the M. How to find the shaded area of a circle with a triangle inside calculator A common beginning geometry problem is calculating the area of standard shapes such as squares and circles. The idea might seem stupid at first, as the area of interest is usually over a continuous curve, but if we take a large enough number of vertices the approximation will be very reasonable. R = 5 ⇒ Shaded Area = 16 Loading. When I use trapz keyword it gives me the whole area. The lengths are measured in inches, so the area is measured in square inches. Step #4: Click on "CALCULATE" button. Shade Tool Calculator 1. To get an area of the plane curve depicted in figure, one needs to calculate definite integral of the form: Functions and as a rule are known from a problem situation, abscisses of their cross points and need to be calculated. We can also calculate probabilities of the form P (a < X ≤ b)--in such cases, the shaded region would be more limited. D = Diameter of circle. Area of the square = (15 x 15) cm 2. Add together all of the y coordinates and then divide the answer by 3. About 63% of the area is drained by marine-terminating glaciers and ice shelf tributary glaciers cover 35% of the area. Calculate the shaded area. A shaded region = A composite figure - A inner circle A shaded region = 1875 - 452. To calculate missing value in circle, based on one known value, you need to remember just three formulas. Thus, Area of annulus = πr₁² - πr₂². The lengths are measured in inches, so the area is measured in square inches. I of the area of section, The Moment of shaded area of web about neutral axis formula is defined as a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. So, area of rectangle is 2x 2 + 17x+ 21. to risk their money and their reputation in college. Step 4: Calculate the area of each shape using our Square footage calculator. The calculator will then determine the length of the arc. Calculated out this gives an area of 28. Substitute the values in area of sector formula, Area = πr 2 × (θ / 360). Download the data for the current chart or look up a random popular name. Plane curves area calculation is one of the main applications of definite integral. How to use the calculator Enter the outer and inner radii R1 and R2 (with R1 > R2) as positive real numbers and press "enter". Drag the selection handles to resize (outside red and blue handles for boxes with rotation performed using the middle. 13 and z 2. Basic Drapery Panels; To figure yardage for basic drapery panels, you will need the cut length (finished length + allowance for your desired hem and rod pocket size NOTE: Our workroom adds 16") x the number of panels you are making. So, the formula for. Shade Ball Calculator. For negative infinity enter. hope someone can help 2 Comments. Calculate the moment of inertia of the shaded area about the x-axis. Area of triangle part: ½ x base x height. Since the total area under the curve is 1, whatever the area to the left is, the area to the right is 1 - area to the left. Area to Z Score Formula. Convert between 0-255 RGB values and HTML hex code dialup color calculator. In order to calculate the area between these two scores, or the probability that a score would fall between X1 and X2, calculate the difference between F(Z2) and F(Z1) in cell I17. We can also calculate probabilities of the form P (a < X ≤ b)--in such cases, the shaded region would be more limited. Area of a Segment Calculator. Choose the shape, then enter the values. Then tap or click the Calculate button. Area of shaded region = Area of ΔABC - Area of ΔADB. In order to calculate S, we first need to know the total area. Find The Area Of The Shaded Region The location of the shaded region is usually seen in typical geometry inquiries. #AreaofShadedRegionofacircle Find the area of a shaded region where you first determine the area of a bigger circle minus area of smaller circle. Using this calculator, we will understand methods of how to find the perimeter and area of a circle. So, area of shaded region is 4x 2 + 15x. One way to make sure your outdoor living spaces are shaded on sunny days is to set up your seating areas near existing structures that provide shade, such as fences or walls. Calculate F(Z1) and F(Z2) using NORMSDIST, you should get areas of 0. View the source. May 12, 2005 · The trick now consists of expressing our region of interest as a polygon. The only improvement I wish for is that it would show more exactly how many hours of sun each area got, Instead of increments of "full shade, part shade, part sun, and full sun" it would be better if it was "-1 hour, 1-2 hours, 2-3 hours, 3-4 hours, 4-5 hours, 5-6 hours, 6 hours+ (of sun)". Cumulative Area Under the Standard Normal Curve Calculator. Question 1183224: Find the area of the shaded region. To find the area of a shaded sector: Get the radius and central angle. Add together all of the x coordinates and then divide the answer by 3. Try this Drag one of the orange dots that define the endpoints of the segment. FindMyShadow. Shade Sail Billing Rates. Substitute 6 for s in P = 4 s. Area between upper curve and x- axis. Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Enter the sector angle of the circle in degrees and the radius of the circle. If you were new to java or at the beginning stage then, Check - 500+ simple Java programs for beginners. Online calculators and formulas for a surface area and other geometry problems. \therefore P_O = 551797. We've created this handy grass seed calculator so you can work out how much you need to purchase. Area of a Octagon: : first calculate the area of one triangle. The surface area of a solid is a measure of the total area occupied by the surface of an object. A lake on the map is 6. This is a composite shape; therefore, we subdivide the diagram into shapes with area formulas. The base is 4r and the height is 4r. The area of the square is 16(r)(r). Enter one or more probabilities or x-values that define the endpoints. May 12, 2005 · The trick now consists of expressing our region of interest as a polygon. Irregularly shaped areas are often divided into several rectangles when one needs to calculate their area, but can't to a precise calculation. According to the figure that is following (in our case the ten, or 53 6 4 3) we must validate if the latter is less than 5 (0,1,3 or 4) or is equal to or greater than 5 (5,6,7,8 or 9). Area of a parallelogram given sides and angle. Use our calculator to find out how much grass seed you'll need for your project. Calculate the area of the shaded region common between two quadrants of circles of radius 7 cm each (as shown in Figure-8). 32 best Geometry Worksheets images on Pinterest from Area Of Shaded Region Worksheet, source:pinterest. Area of a rectangle calculations have a vast array of practical applications: construction, landscaping, internal decoration, architecture, engineering, physics, and so on and so forth. Note: you must enter your Depth Key Code Value. Area = (½ x 40 x 20) cm 2. Click "Add box" or "Add triangle" to add an object. and shaded the area to the right of 15. Side of polygon given area. There is a triangle inside of a circle. Coverage Area 1. Calculate the y-coordinate of the centroid of the shaded area. There are 3 parts shaded in out of 4 parts in total. r = C ( ∗ 2) r = 20 ( 3. Circle area calculator and circumference calculator provide you with a simple way of determining the square footage area for various shapes. In this area worksheet, students calculate area using shaded grids, writing total area in square units. Example Calculate the area shaded between the graphs y= x+2 and y = x2. Due to significant price increases we can no longer offer all functionally. Calculate the area of the shaded region? Solution: As per the given details, A semi-circle is inscribed in a square. This means that you have to. Area of a circle in terms of circumference: Area = C2 4π = 6. Quickly and easily find out how many Shade Balls are needed to complete your project. Step 2: Enter length of horizontal sides into Length 1 and Length 2. Area Under the Curve Calculator is a free online tool that displays the area for the given curve function specified with the limits. Then, use that area to answer probability questions. The lengths are measured in inches, so the area is measured in square inches. Shaded area: # people from that year alive w/ that name as of Jan. Home >> Area. Just like how the shaded area was a fraction of the total area, the arc is a fraction of. Calculate area of Cylinder from radius. The area moment of inertia is the second moment of area around a given axis. And Width of the vertical sides into Width 1 and Width 2. The formula to find the area of the segment is given below. Calculate the shaded area. Example: Simple Compound Shapes. CurveMin =Zmin*StdDev+Mean. The simple tool can provide you with results in minutes. This is known as 'semi-major axis. So the area of the sector is this fraction multiplied by the total area of the circle. Considering my options are a small stretch that gets 1. 14159 x 100 = 39. Area of a parallelogram given sides and angle. Area of a cyclic quadrilateral. Middle: Shade the area of the values that you entered. This geometry video tutorial explains how to calculate the area of the shaded region of circles, rectangles, triangles, and squares. Calculate the shaded area. The ring-shaped object is called as the annulus (i. Area of a circle diameter. Please enter the necessary parameter values, and then click 'Calculate'. For example, in the table above, 80% would be equal to a decimal amount of. Step #3: Select Length Unit. Calculate the azimuth and altitude of the sun. Online calculator to calculate the enclosed area (in blue) of a circular ring when outer and inner radii are known. For triangles, multiply the base by the height and then divide by 2. The area of the shaded region is 6. Measure the length of the ellipse from 'A' to 'B' and divide it by 2. Using this calculator, we will understand methods of how to find the perimeter and area of a circle. Area of a circular sector. a=\pi r^2 a = πr2. Normal distribution or Gaussian distribution (named after Carl Friedrich Gauss) is one of the most important probability distributions of a continuous random variable. The fraction of the shaded shape shown is 1 / 3 , which means one out of 3. Plane curves area calculation is one of the main applications of definite integral. 2744 Square Inches. (b) Hence calculate the shaded area. The area of a circle is the number of square units inside the circle. If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below. Your input: find the area between the following curves $$y = x^{2}$$$, $$y = \sqrt{x}$$$ on the interval $$\left(-\infty, \infty\right)$$$. How to calculate a centroid? Gather both the x and y coordinate points of each vertex. To find the area of the square, write it in terms of the radius of the circles. Add restrictions to shade only part of a region. The area of the shaded region is 6. Since we know that a semicircle is half of a circle, we can simply divide that equation by two to calculate the area of a semicircle. If the angle is θ, then this is θ/2π the fraction of the full angle for a circle. South is often preferred because it will receive the most sunlight if there's no shading, but West can be better in some cases. One Step Complete for Dense Shade Mulch Areas with Smart Seed, Mulch, Fertilizer Mix. 1415926535898 √ = square root Calculator Use. Area of shaded region is: Area of Sector COD - Area of triangle COD Since the shaded region is part of sector O so it is not needed to calculate area of circle. Learn more about the effect of shading. Simply enter the two z-scores below and then click the "Calculate" button. As such, this calculator will focus on the equations for calculating the surface area of the. All of the objects addressed in this calculator are described in more detail on the Volume Calculator and Area Calculator pages. we're asked to find the area of the shaded region so the area of this red shaded region so this is interesting this is almost a 10 by 10 square except we have these quarter circles that are cut out so the area of this would be the area of what a 10 by 10 square would be minus the area of these quarter circles and each of these quarter circles it's a quarter of a circle with a radius 3 I think. The diameter of a circle is the length of a straight line drawn between two points on a circle where the line also passes through the centre of a circle, or. (a) Find the coordinates of P and Q. Although we didn't make a separate calculator for the equilateral triangle area, you can quickly calculate it in this triangle area calculator. It composes the required formula in the run time after gathering the values and that is why it is safe to say that it is also a surface area of a triangular pyramid formula calculator. S and T are the centers of each of the semi-circles. 14159 x 100 = 39. I want to kill the grass in this ~ 616 sq. Here is an example of dientifying the fraction shown. •Calculate the area under the standard normal curve between -2 and 1. Area between upper curve and x- axis Area between lower curve and x- axis Altogether: Example Calculate the shaded area enclosed between the parabolas with equations y = 1 + 10x – 2x2 and y = 1 + 5x – x2. Often times this p-value is also considered to. The total area of a circle is πR 2 corresponding to an angle of 2π radians for the full circle. Add restrictions to shade only part of a region. That's why we have entry tests for all applicants who want to work for us. Name : Printable Math Worksheets @ www. How to use the inside car temperature calculator: Select the unit of measurement. d = diameter. Area of a shaded region part 2 (algebra style) by. Enter the diameter of a circle. The output is the area of the circular ring. F1: A = area. To calculate acreage, start by determining the width and length of the area in yards using a surveyor's wheel. Area of the shaded region = 400 - 154 = 246 cm². Figure 5 shows a top view of the slice whose area we're calculating. Area of a cyclic quadrilateral. 1415926535898 √ = square root Calculator Use. Area is a quantity that describes the size or extent of a two-dimensional figure or shape in a plane. You'll need: Tape measurer or ruler Calculator You may also be interested in: How to Calculate the Volume of a Cube - Formula and Examples. Moment of shaded area of web about neutral axis calculator uses mi_of_the_area_of_section = (( Thickness of web *((( Inner length of I-section ^2)/4)-( Distance b/w considered and neutral layer ^2)))/2) to calculate the M. If the diameter of the largest is 14 cm and of the smallest is 3. Calculate the shaded area. Example Calculate the area shaded between the graphs y= x+2 and y = x2. Whatever is left over is the shaded region. Let’s say we want to shade the region represented by P(-3. Area of a circle = π * r 2. Serving the east coast (Pennsylvania (PA), New Jersey (NJ), New York (NY), Florida (FL), Georgia (GA), South Carolina (SC), North Carolina (NC), West Virginia (WV), Virginia (VA), and Maryland. 32 best Geometry Worksheets images on Pinterest from Area Of Shaded Region Worksheet, source:pinterest. Area = radius 2 2 * ( π 180 * θ - sin(α) ). There is a triangle inside of a circle. The area represents probability and percentile values. Irregularly shaped areas are often divided into several rectangles when one needs to calculate their area, but can't to a precise calculation. Area of inner circle = π x 8 2 = 64π. • Free trial • Search map for an address or longitude and latitude coordinates • Upload any. The area of a circle calculator helps you compute the surface of a circle given a diameter or radius. The shaded area of a circle is a fraction of the area of the entire circle. A = Circle area. The following Java program to print the area of a circle has been written in five simple different ways, static method, using constructor, Interface, inheritance with sample outputs for each program. The calculator allows area look up with out the use of tables or charts. 14 x 62 = 3. How to use the calculator Enter the outer and inner radii R1 and R2 (with R1 > R2) as positive real numbers and press "enter". In some cases, you want to determine how shaded exterior walls, windows, or planting will be. A trapezoid, also known as a trapezium, is a 4-sided shape with two parallel bases that are different lengths. The fraction of the shaded shape shown is 1 / 3 , which means one out of 3. In right triangle PSQ,PQ2 = PS2 + QS2 | By Pythagoras Theorem= (12)2 + (16)2= 144 + 256 = 400 New, for a = 20 cm b = 48 cm c = 52 cm ∴ Area of the shaded portion= Area of ΔPQR - Area of ΔPSQ= 480 - 96 = 384 cm2. •Calculate the area under the standard normal curve between -2 and 1. You can choose between degrees Celsius or degrees Fahrenheit. 52 - 84375 P O. To identify the fraction of a shaded shape, use these steps:. Evaluate the Amount of Sunlight. r = C ( ∗ 2) r = 20 ( 3. Step #3: Select Length Unit. This is a composite shape; therefore, we subdivide the diagram into shapes with area formulas. and that of the smallest is , calculate the area of the shaded region. Use the "Hint" button to get a free letter if an answer is giving you trouble. 13 x 36 = 113. The circumference – the length of the outside boundaries of the circle. Just like how the shaded area was a fraction of the total area, the arc is a fraction of. The average person knows that area is usually calculated by multiplying the length of an object by its width, but for a sloped surface one of those measurements is difficult to determine exactly, even with the help of a ruler or measuring tape. Irregular area's containing angles or curves are therefore easily calculated without complex geometry math. Apr 19, 2021 · Since all semicircles are of same radius, therefore, area of all semicircles will be equal. 6m square sizes. 1) The area formula for a circle: Area of a circle = πr². This calculator determines the area under the standard normal curve given z-Score values. Area of the annulus between r 1 and r 2, the shaded area, is the area contained by the outer circle minus the area contained by the inner circle or A 0 = A 1 - A 2. Side of polygon given area. 14159 x 10 2 = 0. Get the free "Calculate the Area of a Polar curve" widget for your website, blog, Wordpress, Blogger, or iGoogle. Area of a circle diameter. Quickly and easily find out how many Shade Balls are needed to complete your project. The scores are normally distributed with a known mean and standard deviation. The diameter of a circle is the length of a straight line drawn between two points on a circle where the line also passes through the centre of a circle, or. 66 - 20) (23. Shaded area = 9 - 2. Attend to precision. Apr 05, 2012 · The Area of the shaded region = (Area of the largest circle) – (Area of the circle with radius 3) – (Area of the circle with radius 2). Enter the Function = Lower Limit = Upper Limit = Calculate Area. Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Circumference = 2 π r. We have coloured in 1 of the 3 parts of this shape. 9541 1in 1 in 1 in 3 in 1 in A 2 A 3 A 1 4 29 Centroids by Composite Areas Monday, November 12. Calculate the azimuth and altitude of the sun. Shade Tool Calculator 1. The shaded area in the complex figure given is 452. If you only know the side lengths of a regular trapezoid, you can break the trapezoid into simple shapes to find the height and finish your calculation. If you want to total surface area remember to add on the area of the base of the cone. The area of a circle is the number of square units inside the circle. (Note: This is essentially the way you calculate the area of any shaded region on the GMAT). What is the area of a circle with a DIAMETER of 8 inches (radius of 4 inches)? How big is a 8 inch circle? Use this easy and mobile-friendly calculator to compute the area of a circle given its diameter. Now area of the circle " A" = pi x radius x radius = 3. Second Moment of Area is defined as the capacity of a cross-section to resist bending. Area of a cyclic quadrilateral. The writers are reliable, honest, extremely knowledgeable, and the results are always top of the class! - Pam, 3rd Year Art Visual Studies. I found the area of the circle. Calculate the area shaded between the graphs y= x+2 and y = x 2. Moment of shaded area of web about neutral axis calculator uses mi_of_the_area_of_section = (( Thickness of web *((( Inner length of I-section ^2)/4)-( Distance b/w considered and neutral layer ^2)))/2) to calculate the M. : Area of Greenhouse - This number is used. Choose type: Enter a function: y=f(x)=, Rotate around the -axis. (i) (ii) 17. Area and Perimeter of a Ellipse Shape Calculator Calculate Area and Perimeter of an Ellipse Shape. Area of shaded region = 320 - 36 = 284 cm 2. 16 cm2 Area of sector = 452. #AreaofShadedRegionofacircle Find the area of a shaded region where you first determine the area of a bigger circle minus area of smaller circle. Easily Calculate Irregular Area drawn over Google Maps, Images, Photographs or CAD files. 6 square units. AB^2=12^2+16^2. CurveMin =Zmin*StdDev+Mean. Simply enter the length and width of your lawn to calculate how many kilograms you need. Diameter of Circle. 04 in2 Calculate the white area to the nearest hundredth. Please enter the necessary parameter values, and then click 'Calculate'. Despite the fact that clouds do technically block out the sun and cast shade, you shouldn't worry about solar production on cloudy days. : Height of Sidewalls (Feet) Measurement S on the diagram. Free area under the curve calculator - find functions area under the curve step-by-step This website uses cookies to ensure you get the best experience. Area of the shaded region = area of the square - area of the hexagon. Planting at the vertices of a polygon inscribed inside a circle is the best use of this area. Here's it is very easy - the 4 irregular shapes are all the same size (from symmetry). Set a background image* (eg. Online calculator to calculate the enclosed area (in blue) of a circular ring when outer and inner radii are known. Area of a regular polygon. a=\pi r^2 a = πr2. Evaluate the Amount of Sunlight. What we should think about is that this area is the same as subtracting the area to the left of the negative z score from the area to the left of the positive z-score. We pack them carefully to ensure they arrive safely. So, area of shaded region is 4x 2 + 15x square meter. Simple step-by-step tutorial by PreMat. Area of shaded region = Area of ΔABC - Area of ΔADB. Enter the diameter of a circle. SketchAndCalc™ is an irregular area calculator app for all manner of images containing irregular shapes. At first glance, calculating the area of a triangular, sloped surface seems like an extremely tricky task. 16667 x bar 1. (b) Hence calculate the shaded area. This expression, which calculates the area under the curve from the extreme left (negative infinity) to x = c, refers to the shaded region shown below. Cumulative Area Under the Standard Normal Curve Calculator. 52 mm^4 ∴ P O. Get the free "Calculate the Area of a Polar curve" widget for your website, blog, Wordpress, Blogger, or iGoogle. Circle area calculator and circumference calculator provide you with a simple way of determining the square footage area for various shapes. To find the area of the square, write it in terms of the radius of the circles. Area of shaded region = area of outer shape – area of the unshaded inner shape Area of a triangle = ½ bh. Area of sector of circle is the area of the portion of a circle that is enclosed between its two radii and the arc adjoining them is calculated using area_of_sector = (Radius * Arc Length)/2. This process is called finding the definite integral. 61 The area of the shaded region is Part 3 of 4 TO Part 3 of 4 -2. 00 per hour based on an 8 hour work day. The pyramid area calculator automatically utilizes the relevant formula and calculates the area, volume, base, and other terms accordingly. R = 5 ⇒ Shaded Area = 16 Loading. The working area for calculating the potential vegetation and shading consists of 120 foot buffers around all the streams in the basin. Sometimes, you may be required to calculate the area of shaded regions. Circular segment - is an area of a "cut off" circle from the rest of the circle by a secant (chord). Related Volume Calculator | Area Calculator | Body Surface Area Calculator. Plane curves area calculation is one of the main applications of definite integral. You can calculate the area of an oval by following a simple mathematical equation. By following this way a circumference. How To Calculate The Shaded Area Essay essay writing service: get custom How To Calculate The Shaded Area Essay papers created by academic experts. Reasoning 6. This tool performs annulus (annular) area calculation based on the inputs of the radius of the. Find the shaded area A in terms of x for : So, area of shaded region is 10x2. Area of an arch given angle. Area of outer shape = (½ x 15 x 10) cm 2. If your problem area gets less than that, prune back some of the nearby trees to let in more light. Free area under the curve calculator - find functions area under the curve step-by-step This website uses cookies to ensure you get the best experience. If you have ever seen a hoop shade, it was probably in a field protecting crops. Enter a lower limit: If you need -oo, type. Step 3: Finally, the area between the two curves will be displayed in the new window. Find the area of the shaded shape. What is Area?. To find the area of a shaded region in a rectangle, find the total area of the rectangle and the area of the white region. Our tool works both ways - no matter if you're looking for an area to radius calculator or a radius to the area one, you've found the right place. Dec 9, 2016 #9 Ray Vickson. = 22 7 × 14 2 cm 2. If we know the radius, then we can calculate the area of a circle using the formula: A=πr² (Here A is the area of the circle and r is radius). Area between curves online calculator. To identify the fraction of a shaded shape, use these steps:. I of the area of section, The Moment of shaded area of web about neutral axis formula is defined as a geometrical property of an area which reflects how its points are. Therefore, the Area of the shaded region = 25π - 9π - 4π = 12π. Please select an object, enter parameter such as radius or lenght of side or base and/or height and click 'Calculate'. How do you find the area of a shaded sector? The area of a shaded sector can be calculated by the same method we calculate the area of a sector. Step #3: Select Length Unit. INV(PctShade)*StdDev+Mean. The lengths are measured in inches, so the area is measured in square inches. 1415* (13/2)^2 - 60 sq cm = 72. The fraction of the circle that is shaded can be determined by dividing the angle that the shaded region covers by 360 degrees. Simple step-by-step tutorial by PreMat. 72 mm -x 72 mm 44 mm. Tag: area of the shaded region calculator Find The Area Of The Shaded Region. The formula used to calculate circle area is: A = π x (ø/ 2) 2. Laura Schultz Always start by drawing a sketch of the normal distribution that you are working with. C = circumference. Vertical, complementary, and supplementary angles. Begin by sketching the distribution and labeling the relevant information. = 616 cm 2. Please select an object, enter parameter such as radius or lenght of side or base and/or height and click 'Calculate'. •Calculate the area under the standard normal curve between -2 and 1. Calculate the percentage of the shaded region for each !gure. Here is an example of dientifying the fraction shown. If the angle is θ, then this is θ/2π the fraction of the full angle for a circle. ) Lets refer back to a figure that we used earlier. y = x^ {2}$. Leave your answer in terms of π. Circumference = 2 π r. It can be visualized as the amount of paint that would be necessary to cover a surface, and is the two-dimensional counterpart of the one-dimensional length of a curve, and three-dimensional volume of a solid. Shade Tool Calculator 1. All of the objects addressed in this calculator are described in more detail on the Volume Calculator and Area Calculator pages. Learn how to calculate the area of a shaded region inside of a triangle when the areas of the other regions are given. 14 x 62 = 3. 14 square meters (*) (*) 3. On this page, you can calculate area of a Segment. The graphs intersect at (-1 ,1) and (2,4). Providing commercial and residential pool resurfacing, repair, and conversions as well as deck, patio, and plumbing work. Let's find the percentage of adults who score between 90 and 110 on the Weschler IQ test. 14 * (a * a) / 8. On this page, you can calculate area of a Segment. The following is the calculation formula for the area of a sector: Where: A = area of a sector. The area of the outer square is 9 square inches. Area of a cyclic quadrilateral. How to Calculate the Area of a Sector and the Length of an Arc. If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below. According to the figure that is following (in our case the ten, or 53 6 4 3) we must validate if the latter is less than 5 (0,1,3 or 4) or is equal to or greater than 5 (5,6,7,8 or 9). As such, this calculator will focus on the equations for calculating the surface area of the objects and the use of. Available in 3m and 3. The calculator will find the area of the surface of revolution (around the given axis) of the explicit, polar, or parametric curve on the given interval, with steps shown. Then subtract the white area from the rectangle's area. Show transcribed image text Venkatesh Chinna answered this 592 answers Question: Calculate the y-coordinate of the centroid of the shaded area. Although we didn't make a separate calculator for the equilateral triangle area, you can quickly calculate it in this triangle area calculator. And to find area from circumference, it's easiest to first conve. The pyramid area calculator automatically utilizes the relevant formula and calculates the area, volume, base, and other terms accordingly. 1416 x 5 2 x 10. Students will need to multiply polynomials first and then subtract polynomials last to get an answer. Related Surface Area Calculator | Volume Calculator. The side of the square is 25cm. You are also able to save your areas for use later on. Question: In the figure below, RU is a straight line. Area of triangle part: ½ x base x height. Area of the regular hexagon = 6 × area of the equilateral ∆OPQ. Java program to calculate or to print area of a circle in a simple method. Free area under the curve calculator - find functions area under the curve step-by-step This website uses cookies to ensure you get the best experience. Length of RS is the same as length of TU. geometry - Calculating area of a shaded region inside … Details: The area of the outer squ a re is indeed $16$. Area of the Shaded Region: Computation of areas of combinations of geometric shapes is very common when we need to establish the area of a figure composed of several figures, this occurs when the. How to calculate a centroid? Gather both the x and y coordinate points of each vertex. AB^2=12^2+16^2. Although we didn't make a separate calculator for the equilateral triangle area, you can quickly calculate it in this triangle area calculator. Second Moment of Area is defined as the capacity of a cross-section to resist bending. The formula to find the area of the segment is given below. 39 square inches. Dearly Missed. The calculator will generate a step by step explanation along with the graphic representation of the area you want to find and standard normal tables you need to use. Shade in the area (probability) that you are given or trying to find, and label the mean, standard deviation, lower. As such, this calculator will focus on the equations for calculating the surface area of the. ,) the area surrounded by two concentric circles. Calculate F(Z1) and F(Z2) using NORMSDIST, you should get areas of 0. You will not have a single worry if assists you on your schoolwork. The following Java program to print the area of a circle has been written in five simple different ways, static method, using constructor, Interface, inheritance with sample outputs for each program. The calculator does not work outside of these. Calculated out this gives an area of 28. Basic Drapery Panels; To figure yardage for basic drapery panels, you will need the cut length (finished length + allowance for your desired hem and rod pocket size NOTE: Our workroom adds 16") x the number of panels you are making. Enter mean (average), standard deviation, cutoff points, and this normal distribution calculator will calculate the area (=probability) under the normal distribution curve. 14159 x 10 2 = 0. Calculate the shaded area of the square below if the side length of the hexagon is 6 cm. Students will need to multiply polynomials first and then subtract polynomials last to get an answer. hope someone can help 2 Comments. Here is a handy little tool you can use to find the area of plane shapes. Area between upper curve and x- axis Area between lower curve and x- axis Altogether: Example Calculate the shaded area enclosed between the parabolas with equations y = 1 + 10x – 2x2 and y = 1 + 5x – x2. Online calculator to calculate the enclosed area (in blue) of a circular ring when outer and inner radii are known. Area and circumference of circles challenge. Lesson: Circles…. Circumference and rotations calculator also computes the area of a circumference. So, area of rectangle is 2x 2 + 17x+ 21. Free area under between curves calculator - find area between functions step-by-step This website uses cookies to ensure you get the best experience. Since each of the 4 sides of a square are exactly the same size by definition then you can measure the size of any side of the square and enter it into the field below and press calculate to calculate the area of. INV(PctShade)*StdDev+Mean. The graphs intersect at (-1 ,1) and (2,4). This represents the population that does not fall within this z score range. does everything it says it will do and on time. One part is not shaded in. With a = 15, b = 14, c = 13, we have s = 21, so Δ = 84 and hence r = 168 / 42 = 4. Area = π r 2. Not only is this sheet great for both Higher and Foundation GCSE revision, but it can also be used with younger students (I have used this sheet very successfully with Y8. Expert Answer My Textbook Solutions Microecono. The probability of the shaded region. Free area under the curve calculator - find functions area under the curve step-by-step This website uses cookies to ensure you get the best experience. Students will need to multiply polynomials first and then subtract polynomials last to get an answer. In order to calculate S, we first need to know the total area. Knowledge of area of a rectangle required. 95, the boundary of the shaded region consists of four semi-circular arcs, the smallest two being equal. Second Moment of Area Calculator for I beam, T section, rectangle, c channel, hollow rectangle, round bar and unequal angle. The lengths are measured in inches, so the area is measured in square inches. In our example, a 900-square-foot building requires a 33,700-BTU heating system if it is located in an area where it receives equal amounts. The Square Area Calculator will instantly calculate the area of a square if you enter in the size of one of the sides of the square. Do this by finding the area to the left of the number, and multiplying the answer by 100. Then subtract the white area from the rectangle's area. The side of the square is 25cm. we're asked to find the area of the shaded region so the area of this red shaded region so this is interesting this is almost a 10 by 10 square except we have these quarter circles that are cut out so the area of this would be the area of what a 10 by 10 square would be minus the area of these quarter circles and each of these quarter circles it's a quarter of a circle with a radius 3 I think. by the two smaller triangles has an area of 16 cm2. Area required for turfing job [8] 2020/12/16 04:45 60 years old level or over / A retired person / Very / To improve this 'Scalene triangle Calculator', please. () square 2 circle shaded 10 10 100 5 25. SECOND MOMENT OF AREA (AREA MOMENT OF INERTIA) CALCULATOR. The diameter of a circle calculator uses the following equation: Area of a circle = π * (d/2) 2. Can calculate area, arc length,chord length, height and perimeter of circular segment by radius and angle. This process is called finding the definite integral. 1 ; even i wanna know. Created Date:. The diagram shows a circle inside a square calculate the shaded area take pi to be 3. Khan Academy is a 501(c)(3) nonprofit organization. Use the following formulas and sets of equations below to calculate measures of an annulus. AB^2=12^2+16^2. = 4,800 cm 2. Approach 1. A square with side a is inscribed in a circle. Example: find the area of a sector. Homework Helper. Geometry Home: Shaded Area: Radius of an Inscribed Circle: Radius of a Circle: Common Beam Cross. The area of the shaded area is the large rectangle minus the small rectangle. Simple step-by-step tutorial by PreMat. Microsoft Excel does not. Volume & surface area of cylinder calculator uses base radius length and height of a cylinder and calculates the surface area and volume of the cylinder. Easily Calculate Irregular Area drawn over Google Maps, Images, Photographs or CAD files. D = Diameter of circle. Therefore, the Area of the shaded region = 25π - 9π - 4π = 12π. Formula for area of circular ring. Just enter the measurement you know. Then tap or click the Calculate button. Calculate the moment of inertia of the shaded area about the x-axis. calculate the length of the boundary and the area of the shaded region. If each of the smaller triangle has a base of 14 cm and a height of 5 cm, find the area of the unshaded region. Second Moment of Area Calculator for I beam, T section, rectangle, c channel, hollow rectangle, round bar and unequal angle. Example: there is a Circular shaped pool 20 feet wide diameter the owner wants to put in a 2 foot wide border (or walkway) of stone around the pool and would like it to be 3 inches deep (see Depth Key Code ). How To Calculate The Shaded Area Essay, essay topic about changing one think, college application essay pay, greek life on college campuses essay. Usually, we would subtract the area of a smaller inner shape from the area of a larger outer shape in order to find the area of the shaded region. By following this way a circumference. You can also use the normal distribution calculator to find the percentile rank of a number. (b) Hence calculate the shaded area. The area of a circle is the number of square units inside the circle. If you are not pleased with your purchase, please call us at 1-800-552-9916 or email us at [email protected] Weekend work or Holiday’s are billed out at time and ½ or $156. To find the area between a negative z-score and a positive z-score is perhaps the most difficult scenario to deal with due to how our z-score table is arranged. 5 cm, find (FIGURE) (i) the. You will not have a single worry if assists you on your schoolwork. Then subtract the white area from the rectangle's area. 52 - 84375 P O. It can be visualized as the amount of paint that would be necessary to cover a surface, and is the two-dimensional counterpart of the one-dimensional length of a curve, and three-dimensional volume of a solid. Calculate the area of the shaded region common between two quadrants of circles of radius 7 cm each (as shown in Figure-8). - Lauren, 4th Year Education. This calculator can be used to find area under standard normal curve$ ( \mu=0 , \sigma=1 )\$. Areas under the x-axis will come out negative and areas above the x-axis will be positive. Correct option is. Find Surface Area of different objects like Cone, Cube, Cylinder, Pipe, Prism, Pyramid, Rectangular Box, Sphere, Torus (Ring), Tube and more. You will need to know 2 elements in order to calculate the remaining variables. The writers are reliable, honest, extremely knowledgeable, and the results are always top of the class! - Pam, 3rd Year Art Visual Studies. Area of sector of circle is the area of the portion of a circle that is enclosed between its two radii and the arc adjoining them is calculated using area_of_sector = (Radius * Arc Length)/2.
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# Quark Matter 2017
Feb 5 – 11, 2017
Hyatt Regency Chicago
America/Chicago timezone
## D-tagged jet measurements in p-Pb collisions in ALICE
Not scheduled
2h 30m
Hyatt Regency Chicago
#### Hyatt Regency Chicago
151 East Wacker Drive Chicago, Illinois, USA, 60601
Board: H13
Poster
### Speaker
Barbara Antonina Trzeciak (Utrecht University)
### Description
Heavy charm quarks are produced dominantly in the initial stage of high-energy heavy-ion collisions. They take part in the whole evolution of the medium, interact with the produced Quark-Gluon Plasma (QGP) and lose energy due to collisional and radiative processes. Therefore, they serve as unique probes of the QGP transport properties. Charm quarks can be studied by measuring D mesons that originate from the charm-quark fragmentation. Furthermore, identifying the charm-jets (with D-meson tagging) gives additional information about the energy-loss distribution in the medium. In order to disentangle effects related to the presence of the hot and dense medium, measurements in elementary pp and in p--A collisions are needed. In addition to the charm-jet transverse momentum spectrum, an important observable is the charm fragmentation function (fraction of the jet momentum carried by D mesons), which is still not fully understood even in pp collisions, and provides important constraint for Monte-Carlo event generators. The study of possible charm-jet modifications in p--Pb collisions is an important intermediate step between pp and Pb--Pb collisions that complements the simple pp vacuum case and the heavy-ion case with interplay of many different effects.
D mesons are identified via their hadronic decay channels using information from the ALICE Inner Tracking System to reconstruct decay topologies displaced with respect to the collision primary vertex. The ALICE Time Projection Chamber is used for the tracking and the hadron identification, the hadron selection being improved with the Time Of Flight detector. Charged jets containing D mesons are reconstructed with the anti-$k_{\rm T}$ algorithm.
In this poster, we will show the status of the D-tagged jets analysis in p--Pb minimum-bias collisions at $\sqrt{s_{\rm NN}}$ = 5.02 TeV. We will also present the prospects of the analysis with the new dataset to be gathered in the upcoming 2016 p--Pb run at $\sqrt{s_{\rm NN}}$ = 5.02 and 8 TeV. Both the charm-jet transverse momentum spectrum and the charm fragmentation function are studied.
Preferred Track Open Heavy Flavors ALICE
### Primary author
Barbara Antonina Trzeciak (Utrecht University)
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Report the patches or turtles among agents2 within given distances of each of the agents. Currently, this function multiplies radius by 1.0000001 so that the response of inRadius is inclusive.
inRadius(agents, radius, agents2, world, torus = FALSE)
# S4 method for matrix,numeric,matrix
torus = FALSE)
## Arguments
agents Matrix (ncol = 2) with the first column pxcor and the second column pycor representing the patches coordinates, or AgentMatrix object representing the moving agents. Numeric. Vector of distances from agents to locate agents2. Must be of length 1 or of length agents. Matrix (ncol = 2) with the first column pxcor and the second column pycor representing the patches coordinates, or AgentMatrix object representing the moving agents. WorldMatrix or worldArray object. Logical to determine if the world is wrapped. Default is torus = FALSE.
## Value
Matrix (ncol = 3) with the first column pxcor and the second column pycor representing the coordinates of the patches among agents2 within radius distances for each agents which are represented by the id column, if agents2 are patches, or
Matrix (ncol = 2) with the first column who representing the who numbers of the turtles among agents2 within radius distances for each agents which are represented by the id column, if agents2 are turtles.
## Details
Distances from/to patches are calculated from/to their center.
If torus = FALSE, world does not need to be provided.
If torus = TRUE, the radius distances are calculated around the sides of the world to select agents2.
## References
Wilensky, U. 1999. NetLogo. http://ccl.northwestern.edu/netlogo/. Center for Connected Learning and Computer-Based Modeling, Northwestern University. Evanston, IL.
w1 <- createWorld(minPxcor = 0, maxPxcor = 4, minPycor = 0, maxPycor = 4)
t3 <- inRadius(agents = turtle(t1, who = 0), radius = 2, agents2 = t1)
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# Young's Modulus
Written by Jerry Ratzlaff on . Posted in Thermodynamics
Young's modulus ( $$Yo$$ ) (also known as elastic modulus, modulus of elasticity and tension modulus) measures the stiffness of an elastic material. The ratio of the longitudinal stress applied to a body or substance to the resulting longitudinal strain within the elastic limits.
## formula
$$E = \frac { \sigma } { \epsilon }$$
Where:
$$E$$ = Young's modulus
$$\sigma$$ (Greek symbol sigma) = stress
$$\epsilon$$ (Greek symbol epsilon) = strain
Solve for:
$$\sigma = E \epsilon$$
$$\epsilon = \frac { \sigma } { E }$$
Tags: Equations for Stress
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Creates a unified directory structure of all namespace packages, symlinking to the actual contents, in order to ease navigation.
## Change history
### 0.5 (2008-05-29)
• Added uninstall entry point so that the omelette can be uninstalled on Windows without clobbering things outside the omelette path. [optilude]
• Support Windows using NTFS junctions (see http://www.microsoft.com/technet/sysinternals/FileAndDisk/Junction.mspx) [optilude]
• Ignore zipped eggs and fakezope2eggs-created links. [davisagli]
• Added ‘packages’ option to allow merging non-eggified Python packages to any directory in the omelette (so that, for instance, the contents of Zope’s lib/python can be merged flexibly). [davisagli]
### 0.4 (2008-04-07)
• Added option to include Products directories. [davisagli]
• Fixed ignore-develop option. [davisagli]
### 0.3 (2008-03-30)
• Fixed test infrastructure. [davisagli]
• Added option to ignore develop eggs [claytron]
• Added option to ignore eggs [claytron]
• Added option to override the default omelette location. [davisagli]
### 0.2 (2008-03-16)
• Fixed so created directories are not normalized to lowercase. [davisagli]
### 0.1 (2008-03-10)
• Initial basic implementation. [davisagli]
• Created recipe with ZopeSkel. [davisagli]
## Detailed Documentation
### Introduction
Namespace packages offer the huge benefit of being able to distribute parts of a large system in small, self-contained pieces. However, they can be somewhat clunky to navigate, since you end up with a large list of eggs in your egg cache, and then a seemingly endless series of directories you need to open to actually find the contents of your egg.
This recipe sets up a directory structure that mirrors the actual python namespaces, with symlinks to the egg contents. So, instead of this…:
egg-cache/
my.egg.one-1.0-py2.4.egg/
my/
egg/
one/
(contents of first egg)
my.egg.two-1.0-py2.4.egg/
my/
egg/
two/
(contents of second egg)
…you get this:
omelette/
my/
egg/
one/
(contents of first egg)
two/
(contents of second egg)
You can also include non-eggified python packages in the omelette. This makes it simple to get a single path that you can add to your PYTHONPATH for use with specialized python environments like when running under mod_wsgi or PyDev.
### Typical usage with Zope and Plone
For a typical Plone buildout, with a part named “instance” that uses the plone.recipe.zope2instance recipe and a part named “zope2” that uses the plone.recipe.zope2install recipe, the following additions to buildout.cfg will result in an omelette including all eggs and old-style Products used by the Zope instance as well as all of the packages from Zope’s lib/python. It is important that omelette come last if you want it to find everything:
[buildout]
parts =
...(other parts)...
omelette
...
[omelette]
recipe = collective.recipe.omelette
eggs = ${instance:eggs} products =${instance:products}
modules = ${zope2:location}/lib/python ./ ### Supported options The recipe supports the following options: eggs List of eggs which should be included in the omelette. location (optional) Override the directory in which the omelette is created (default is parts/[name of buildout part]) ignore-develop (optional) Ignore eggs that you are currently developing (listed in${buildout:develop}). Default is False
ignores
(optional) List of eggs to ignore when preparing your omelette.
packages
List of Python packages whose contents should be included in the omelette. Each line should be in the format [package_location] [target_directory], where package_location is the real location of the package, and target_directory is the (relative) location where the package should be inserted into the omelette (defaults to top level).
products
(optional) List of old Zope 2-style products directories whose contents should be included in the omelette, one per line. (For backwards-compatibility – equivalent to using packages with Products as the target directory.)
### Windows support
Using omelette on Windows requires the junction utility to make links. Junction.exe must be present in your PATH when you run omelette.
### Using omelette with eggtractor
Mustapha Benali’s buildout.eggtractor provides a handy way for buildout to automatically find development eggs without having to edit buildout.cfg. However, if you use it, the omelette recipe won’t be aware of your eggs unless you a) manually add them to the omelette part’s eggs option, or b) add the name of the omelette part to the builout part’s tractor-target-parts option.
### Running the tests
Just grab the recipe from svn and run:
python2.4 setup.py test
Known issue: The tests run buildout in a separate process, so it’s currently impossible to put a pdb breakpoint in the recipe and debug during the test. If you need to do this, set up another buildout which installs an omelette part and includes collective.recipe.omelette as a development egg.
## Contributors
• David Glick [davisagli], Author
• Clayton Parker [claytron], Code Monkey
• Martin Aspeli [optilude], Windows Hacker
## Project details
### Source Distribution
collective.recipe.omelette-0.5.tar.gz (11.9 kB view hashes)
Uploaded source
### Built Distribution
collective.recipe.omelette-0.5-py2.4.egg (25.7 kB view hashes)
Uploaded 2 4
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Web service architecture is the first phase before building every project, it’s like you prepare to build a house and start by creating the architecture plan.
This article will present how I structure my projects when I need to create a simple web service in Golang. It’s very important for you to keep a simple but intuitive architecture, because as you know, in golang you can call methods by referencing the package name.
In the following lines I’ll present a simple, but traditional model of web service architecture used by me in most of the projects that I’m involved in, treating each individual web service’s component.
## /api
The API package is the folder where all the API endpoints are grouped into sub-packages by the purpose they serve. That means, I prefer to have a special package with it’s main scope to solve a specific problem.
For example all the login, register, forgot password, reset password handlers, I prefer to be defined into a package named registration.
The registration package can look like below:
.
├── api
│ ├── auth
│ │ ├── principal.middleware.go
│ │ └── jwt.helper.go
│ ├── cmd
│ │ └── main.go
│ ├── registration
│ │ ├── register.handler.go
│ │ ├── social_register.handler.go
│ │ ├── reset.handler.go
│ │ ├── helper.go
├── cmd
│ └── main.go
├── config
│ ├── config.dev.json
│ ├── config.local.json
│ ├── config.prod.json
│ ├── config.test.json
│ └── config.go
├── db
│ ├── handlers
│ ├── models
│ ├── tests
│ ├── db.go
│ └── service.go
├── locales
│ ├── en.json
│ └── fr.json
├── public
├── vendor
├── Makefile
..........................
#### handler.go
As you can see, there is a handler.go suffix in the name of the files. In these you can effectively write the code, which will handle the request, where the data requested will be retrieved from the database, processed and in the end the response will be composed.
A simple example which explain better can be shown below:
http.HandleFunc("/bar", func(w http.ResponseWriter, r *http.Request) {
...
// handle the request
})
#### helper.go
Sometimes, before sending the response, you need to collect data from multiple places to process them, and after that, when all the details are collected, the response can be sent to the client app. But the code must be kept as simple as possible in the handler, so all that extra code which is part of the process can be put here.
In the interaction between a client and a web service, they are sending and receiving data, but at the same time, there is probably a third party API involved, another application, or the database. Having this in mind, before transferring the data from an application to another one, we need to convert the format, before being accepted by the new app. This conversion function can be written here, in this adapter.go file.
For example, if I need to convert a struct A to a struct B, I need an adapter function which looks like:
type A struct {
FirstName string
LastName string
Email string
}
type B struct {
Name string
Email string
}
func ConvertAToB(obj A) B {
return B{
Name: obj.FirstName + obj.LastName,
Email: obj.Email,
}
}
### /api/auth
Most web services must have at least one authorization method implemented, like:
• OAuth — Open Authentication
• Basic Authentication
• Token Authentication (I prefer this one with JWT — JSON Web Token)
• OpenID
Personally, I use JWT, because I write web services for our clients (ATNM), mostly for mobile apps or CMS. If you’d like to read more about the Web Authentication API, Mozilla has a great article that explains it very well.
##### What is JWT ?
JSON Web Tokens are an open, industry standard RFC 7519 method for representing claims securely between two parties.
##### Why you should use JWT ?
• Authorization: This is the most common scenario for using JWT. Once the user is logged in, each subsequent request will include the JWT, allowing the user to access routes, services, and resources that are permitted with that token. Single Sign On is a feature that widely uses JWT nowadays, because of its small overhead and its ability to be easily used across different domains.
• Information Exchange: JSON Web Tokens are a good way of securely transmitting information between parties. Because JWTs can be signed—for example, using public/private key pairs—you can be sure the senders are who they say they are. Additionally, as the signature is calculated using the header and the payload, you can also verify that the content hasn’t been tampered with.
So, you have to verify the signature, to encode or decode the body, or to compose the JWT body. For this kind of processes I created the file jwt.helper.go, to keep a consistency and to find all the code related to JWT in a single place under the package auth.
Let’s discuss about the other file, from the auth package, principal.middleware.go. The file has this name because it is the first middleware in the interaction with any API, so all the requests are coming through it. In this file I write a function which serves as a blocker for any requests, and if the rules are not passed, a 401 status code will be sent as the response. Now, if you’re asking which are those rules, we already talked about JWT, so attached to any request (except endpoints like login, register, which don’t need authorization) the client must send an HTTP header, Authorization, which must contain the JWT token.
As a sum up, if the client app doesn’t send a token, or the token is corrupted or invalid, the web service will invalidate the request.
##### Where to get that token from ?
Probably this is a question that you thought of while reading the previous paragraph, so let’s make this clear. I mentioned that on login or register (and yes, probably other routes as well don’t require authentication) you don’t need to send a token, because you will actually get the token from these requests. So you fill in your credentials, and if they are right, you’ll get a token in the response, on login, which will be sent for each request that imposes it.
## /cmd
I always prefer to put the main.go file in this package, which contains all the sub packages from a project. It’s like a wrapper which encapsulate all the submodules, to work all together.
Why is name like that ? It’s simple, because cmd is short for command.
What to understand through command? A command represents a task which is part of something, call other tasks, or run independently. The main.go file is a command which usually wraps all the functions and packages of a web service in a single file and calls just the main functions of any package. At any moment, if you want to remove a functionality, you can simply remove it just by commenting the instance from the main file.
## /config
This package is very important in my opinion, because I found it very useful to keep all the configs in a single place and not search around in all the files of the project. In this package, I usually write a file called config.go which contains the model for the configuration. This model is nothing more than a structure, for example:
type JWT struct {
Secret string required:"true"
}
type Database struct {
Dialect string default:"postgres"
Debug bool default:"false"
Username string required:"true"
Password string required:"true"
Host string required:"true"
Port int
SSLMode bool
}
type Configuration struct {
Database Database required:"true"
JWT JWT required:"true"
}
But this, is just the structure definitions and we still need the real data to be placed somewhere. For that part I prefer to have multiple JSON files, depends by environment and to name them like config.ENV.json. For the structs defined before a dummy JSON example is the following:
{
"Database": {
"Dialect": "postgres",
"Debug": true,
"Host": "example.com",
"Port": 5432,
"SSLMode": true,
},
"JWT": {
"Secret": "abcdefghijklmnopqrstuvwxyz"
}
}
Let’s talk about business, because this part is very special for me and very important for the time invested in finding the best answer. I don’t know if you faced this problem or not, or for you, maybe it is not a problem, but I really encountered some problems trying to import the config in a good way. There are many possibilities, but I had to face the dilemma to choose between two:
• passing the config object as a variable from main.go to the final function, where I need to use it. This for sure is a good idea, because I pass that variable just for those instances which need it, so in this way I don’t compromise speed quality. But this is very time consuming for development or refactoring, because I need to pass the config from one function to another one all the time, so in the end, you want to kill yourself, meeh.., maybe not, but I still don’t like it.
• declaring a global variable and using that instance everywhere I need. But this is not the best option at all in my opinion, because I have to declare a variable, for example in main.go file, and later in the main function I need to Unmarshal() the JSON file, to put that content into the variable object declared as global. But guess what, maybe I’m trying to call that object before it’s initialization is ready, so I’ll have an empty object, with no real values, so in this case my app will crash.
• inject the config object directly where I need, and yes, this is my best option which fits perfectly with me. In config.go file, at the end of it, I declare the following lines:
var Main = (func() Configuration {
var conf Configuration
if err := configor.Load(&conf, "PATH_TO_CONFIG_FILE"); err != nil {
panic(err.Error())
}
return conf
})()
What you need to know for this implementation is that I use a library called Configor which unmarshals a file, in our case a JSON, and loads it into a variable conf, which is returned.
Any time when you need to use something from the config, it’s enough to type the package name, which is config, and to call the variable Main, as the following example which retrieves the configuration for the database:
var myDBConf = config.Main.Database
!!!Tips: As you can see, you must insert there the path to your config file, but because you want to have a different file for different environments, maybe you can set an environment variable called CONFIG_PATH. Define that as an env variable, or put it before you run your go like:
### How to manage packages ?
Probably your question now it’s “Ok, but how can I keep all dependencies together and to install them with a simple command, instead to run multiple commands, if I need to change the environment for example ?” and the answer it’s very simple, use a management dependency tool. With a dependency tool you can achieve basic tasks and you can save some time.
I prefer DEP, the default one from Golang.
It can be install simple with brew for MAC
$brew install dep$ brew upgrade dep
or with CURL
$curl https://raw.githubusercontent.com/golang/dep/master/install.sh | sh ## Makefile I use make file, because it’s simple and can automatize some tasks that I have to repeat from time to time and because I have to make some steps before to create for example a build and I need to do this process after few months or maybe years, probably I need to spend some time figuring out how to do that build. But instead to spend all that time discovering again how should I build my project, I can do it when that information is hot and after that I just need to look inside the make file and to choose what task I need to run. I want to share with you some simple and basic tasks that I use in the most of my projects: IP = "XXX.XXX.XXX.XXX" PEM_FILE = "...PATH_TO_PEM/key.pem" # Run the server run: CONFIG=config/config.local.json go run cmd/main.go --port 8000 # Remove generated files under gen/ folder clean: rm -r gen serve: realize start # Build build: go build cmd/main.go # Build for linux build-linux: env GOOS=linux go build cmd/main.go # Deploy just the code to dev deploy-dev-code: build-linux copy-project # Full deploy to dev # With tests and swagger generation deploy-dev: gen deploy-dev-code # Copy the project build and dependencies to server copy-project: ssh -i$(PEM_FILE) ubuntu@$(IP) 'sudo service api stop' scp -i$(PEM_FILE) -r locales/ ubuntu@$(IP):/home/ubuntu/project_name scp -i$(PEM_FILE) main ubuntu@$(IP):/home/ubuntu/project_name ssh -i$(PEM_FILE) ubuntu@\$(IP) 'sudo service api start'
rm main
You can find a great article about makefile and how to use it from GNU.org.
## Summary
In this article you’ve learn about APIs and how to build an architecture, how to interact with a database from your web service, how to make a config file, take care of security and permissions between client and server with JWT and how to make your life easier using other packages, in the end you learned how to run multiple tasks using a make file.
If you have any question or I you consider that I missed some informations, please leave me a comment.
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# Birthday paradox for large numbers
I'm attempting to test the claim: "Every card deck shuffled is unique. A shuffled deck of cards will exist once and never again."
Assumption: A perfect world where a deck of cards is perfectly random all of the time. 52 cards are used in the deck.
The birthday paradox and it's easy enough to calculate for small numbers. 23 people and 365 birthdays as used in the 50% examples. But how do you approach (or approximate) the birthday paradox for values like 52!?
I understand 52! is a large (~226 bit) number but I would like to get a feel of the order of magnitude of the claim. Is it 1% or 0.00001%?
To calculate the probability of a shuffle collision would be: (52!)!/(52!^n*(52!-n)!)
I understand the formula. But 52!! is incomputable so where do I go from here? How to approach a problem like this?
This is just for my own curiosity and not homework. If it can be done for a deck of cards I'd want to give it a try on collisions in crypto keys. (RSA, and AES256 etc.)
• I have no idea what you are asking. What does shuffling a deck of cards have to do with the birthday paradox? – 5xum Mar 6 at 11:23
• @5xum The probability of a collision. – JWRM22 Mar 6 at 11:24
• By the way, $52!! \neq (52!)!$. – Deepak Mar 6 at 11:42
• @Deepak, I learn something new every day. :) – JWRM22 Mar 6 at 11:51
A simple estimate: given $$n$$ random variables, independently and identically distributed evenly over $$M$$ states, the probability that some two of them are equal is at most $$\frac{n(n-1)}{2M}$$. Why? Because that's the expected value for the number of pairs that are equal; there are $$\binom{n}{2}$$ pairs, each with a probability of $$\frac1M$$ of being equal.
So then, if $$n$$ is significantly less than $$\sqrt{M}$$, the probability of some two being equal is small.
Now, we focus on the shuffling problem. How big is $$52!$$, really? For that, we have Stirling's formula: $$n!\approx \frac{n^n}{e^n}\cdot\sqrt{2\pi n}$$ Take logarithms, and we get $$n(\log n-\log e)+\frac12\log(2\pi n)$$. Calculating that in the spreadsheet I've got open already, I get a base 10 log of about $$67.9$$, for about $$8\cdot 10^{67}$$ possible shuffled decks.
So, then, if we take $$10^{30}$$ shuffles, that's a probability of less than $$\frac{5\cdot 10^{59}}{8\cdot 10^{67}}< 10^{-8}$$ (one in a hundred million) that any two of them match exactly. That $$10^{30}$$ - if you had a computer for every person on earth, each generating shuffled decks a billion times per second, the planet would be swallowed up by the sun before you got there.
• And yet these sneaky magicians seem to always get the deck shuffled as they intended to! – Christoph Mar 6 at 11:57
• Thank you! There is a chance but too tiny to care even when talking about the 10^30 shuffles. Good to know! My curiosity is satisfied. – JWRM22 Mar 6 at 12:02
If you shuffle two decks, the probability of them being in the same order is $$\frac{1}{52!} \approx 1.24\times 10^{-66} \%.$$
If you shuffle $$n$$ decks, the probability of a collision is $$1 - \left( 1- \frac{1}{52!}\right)\left(1-\frac{2}{52!}\right)\cdots\left(1-\frac{(n-1)}{52!}\right).$$ Are you looking for the smallest $$n$$ such that this number is at least $$50\%$$?
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# Blindstore
Blindstore is a private information retrieval data store. It provides an online lookup service which returns the correct answer without ever knowing what the question was.
## Centralised
Unlike a distributed hash table, Blindstore is not prone to a Sybil attack. The data is not inherently public, since the server is free to restrict access with authentication and authorization.
## Privacy
The protocol guarantees that the server can't know which record the client requested. The client still only receives the record they requested.
## Here be dragons
Blindstore is very much work in progress, made using experimental cryptographic software. For the moment, it is for research purposes only.
### Introduction
Blindstore is a cryptographic product that implements a Private Information Retrieval (PIR) protocol. PIR protocols allow a user to retrieve the $$i$$-th record of a database without letting the database server know the content of $$i$$. Otherwise said, the user accesses a database (which provides all of its content through a mapping index → value) and upon requesting an index he is served with the correct value, although the server cannot know which index (and therefore which value) it served. The same result can also by obtained trivially by having the user retrieve the whole database and perform the search locally, but this solution is often very costly. The protection is effective even if the server is not to be trusted. Thus Blindstore safeguards the privacy of users accessing the database.
The implementation of Blindstore has its theoretical foundation in the protocol described in the paper “Single-Database Private Information Retrieval from Fully Homomorphic Encryption” published by Xun Yi, Mohammed Golam Kaosar, Russell Paulet, and Elisa Bertino in May 2013. The protocol illustrated in the paper uses Somewhat Homomorphic Encryption (SHE), a simplification of Fully Homomorphic Encryption (FHE) which allows homomorphic computation of operations on plaintexts. I.e., in FHE schemes encryption and decryption (which take place using a public/private key pair) have the following property: performing an operation on the ciphertext of two encrypted messages, by adding or multiplying them, will result in a ciphertext which, when decrypted, will be equivalent to the addition or multiplication of the two plaintext messages.
### Theory
The theoretical protocol behind Blindstore describes a database accessible by index → value pairs, with the index being $$l$$-bit long. The user generates a public/private key pair and sends the public key to the database server. Then he chooses an index $$i$$, encrypts it with the public key, and sends it to the server as a query. The server computes an encryption of $$i$$-th record based on the database content, the query, and the public key, and sends the response back to the user. The user decrypts the response to obtain the requested value. Since the server cannot decrypt $$i$$ alone, and uses all the values in the database for the computation of the response, the server is unable to know which index (and therefore which value from the database) was requested by the user.
### Implementation
The cryptographic library is written in C++. The whole application is work in progress. FHE schemes being computationally expensive, the current (alpha, non-optimized) implementation of Blindstore has a response time of 2.88 seconds for a query to a database with 10-bit indexes and 1-Kb records (benchmarks measured on an i5 @ 2.50 GHz machine).
### Practical applications
A Blindstore server is an added value to a service provider, as it can guarantee to clients that it has no way to know which piece of information they requested to the service provider. There are several possible practical applications of Blindstore.
#### People directory
As Blindstore protects the privacy of the query, it can be used to create databases storing contact information for human right activists. This would allow citizens in totalitarian countries to access the database without disclosing the identity of the person they've requested information about.
#### Online encyclopedia
Another use case for Blindstore is a shared online encyclopedia such as Wikipedia, with the content indexed by the article's title. In this case the user could access any article without the server knowing which article has been requested.
#### Service location
Blindstore can be used to anonymize a service that provides the user with a list of points of interest (restaurants, bars, hotels...) near his location, as pinpointed by the GPS in the user's smartphone. Usually the user would need to provide his location to the server in order to obtain the relevant information, at the detriment of the user's privacy. With Blindstore, information about points of interest can be organized in geographical quadrants, with the user sending a query containing the coordinates of the quadrant he's in at that moment. The server is unable to decrypt the request (the user location quadrant) but can nonetheless provide the relevant data (points of interests in the user quadrant).
Many commodities (real estate, stock exchange, currencies, plane tickets, hotel stays) are subject to price fluctuations depending on demand. The public demand or interest on a particular commodity (for instance, a house) can be evaluated by the number of information queries about it (for instance, hits on a particular page of the real estate agency website). Being unable to evaluate the number of requests would make financial speculations impossible. The same reasoning can be applied to request of information from a large multinational about the status or health of a small company: this kind of request can have an impact on the stock value of the company.
Let's imagine a scenario in which a mining company knows that there is gold (or oil or diamonds or any other valuable resource) in a particular area. To extract this gold, the company must request a license to exploit that area before anyone else does. Asking the state for an exploitation permit for that particular area informs them that there is gold in that area, independently of whom asked for it; subsequently the state could use this information, either by selling it to some other company or by exploiting the area themselves. In this case, the mining company is the only party in possess of a valuable piece of information (the presence of gold in that area); using the Blindstore protocol to request the exploitation permit makes sure that the mining company remains the only holder of that information.
# Contribute
Blindstore has been developed in occasion of two CERN hackathons in Summer and Fall 2014. It is, however, not affiliated to CERN in any way.
The main communication channel is the Blindstore repository Gitter:
gitter.im/blindstore/blindstore
Check out the current issues:
github.com/blindstore/blindstore/issues
Source code can be found on our GitHub organisation:
github.com/blindstore
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## Sunday, April 26, 2015
### "Open Data in Science"
Recently, I got invited to a meeting of Eindhoven's Social Media Club, which has interesting meetings in the knowledge city capital of The Netherlands [ref]. This months topic was Open Data and I was asked to present Open Data in research, which I eagerly accepted. The quite liked the title too: The great wide Open Data.
I very much enjoyed the other presentation too, mostly by Allard Couwenberg, whom gave an excellent introduction into Open Data, which simplified my presentation, allowing me to focus on the role of Open Data in research and possible at universities. For example, I discussed that I think we can improve the quality of our education of we improve the access to knowledge for our students. I got great questions from the audience, mostly consisting of people outside the scholarly community, and including a few people working with Open Data a lot. A full storify is available.
I have uploaded my slides to SpeakerDeck:
But I only today sent the slides around today, because I just spent (for the first time ever) annotation my slides with source information (the last two slides).
Also, for the first time, I really felt I could have spoken for much longer. While I was able to mention a number of Open Data initiatives, like the Open Knowledge Foundation and its Dutch Open Science working group, WikiData and Wikidata4Research, the Blue Obelisk movement, the Open Notebook Science Challenge, Open Source Malaria, and crowdsourcing initatives like Mark2Cure, I realized there is so much around nowadays, that this can no longer be covered in a single presentation.
Congrats to the scholarly Open Data community!
## Saturday, April 18, 2015
### Bioclipse 2.6.2 with recent hacks #1: Wikidata & Linked Data Fragments
Bioclipse dialog to upload chemicalstructures to an OpenTox repository.
Us chem- and bioinformaticians have it easy when it comes to Open Science. Sure, writing documentation, doing unit testing, etc, takes a lot of time, but testing some new idea is done easily. Yes, people got used to that, so trying to explain that doing it properly actually takes long (documentation, unit testing) can be rather hard.
Important for this is a platform that allows you to easy experiment. For many biologists this environment is R or Python. To me, with most of the libraries important to me written in Java, this is Groovy (e.g. see my Groovy Cheminformatics book) and Bioclipse (doi:10.1186/1471-2105-8-59). Sometimes these hacks grow to be full papers, like with what started with OpenTox support (doi:10.1186/1756-0500-4-487) which even paved (for me) the way to the eNanoMapper project!
But often these hacks are just for me personal, or at least initially. However, I have no excuse to not make this available to a wider audience too. Of course, the source code is easy, and I normally have even the smallest Bioclipse hack available somewhere on GitHub (look for bioclipse.* repositories). But it is getting even better, now that Arvid Berg (Bioclipse team) gave me the pointers to ensure you can install those hacks, taking advantage from Uppsala's build system.
So, from now on, I will blog how to install Bioclipse hacks I deem useful for a wider audience, starting with this post on my Wikidata/Linked Data Fragments hack I used to get more CAS registry number mappings to other identifiers.
Install Bioclipse 2.6.2
The first thing you need is Bioclipse 2.6.2. That's the beta release of Bioclipse, and required for my hacks. From this link you can download binary nightly builds for GNU/Linux, MS-Windows, and OS/X. For the first two 32 and 64 bit build are available. You may need to install Java and version 1.7 should do fine. Unpack the archive, and then start the Bioclipse executable. For example, on GNU/Linux:
$tar zxvf Bioclipse.2.6.2-beta.linux.gtk.x86.tar.gz$ cd Bioclipse.2.6.2-beta/
$./bioclipse Install the Linked Data Fragments manager The default update site already has a lot of goodies you can play with. Just go to Install → New Feature.... That will give you a nice dialog like this one (which allows you to install the aforementioned Bioclipse-OpenTox feature): But that update site doesn't normally have my quick hacks. This is where Arvid's pointers come in, which I hope to carefully reproduce here so that my readers can install other Bioclipse extensions too. Step 1: enable the 'Advanced Mode' The first step is to enable the 'Advanced Mode'; that is, unless you are advanced, forget about this. Fortunately, the fact that you haven't given up on reading my blog yet is a good indicated you are advanced. Go to the Window → Preferences menu and enable the 'Advanced Mode' in the dialog, as shown here: When done, click Apply and close the dialog with OK. Step 2: add an update site from the Uppsala build system The first step enables you to add arbitrary new update sites, like update sites available from the Uppsala build system, by adding a new menu option. To add new update sites, use this new menu option and select Install → Software from update site...: By clicking the Add button, you go this dialog where you should enter the update site information: This dialog will become a recurrent thing in this series, though the content may change from time to time. The information you need to enter is (the name is not too important and can be something else that makes sense to you): 1. Name: Bioclipse RDF Update Site 2. Location: http://pele.farmbio.uu.se/jenkins/job/Bioclipse.rdf/lastSuccessfulBuild/artifact/site.p2/ After clicking OK in the above dialog, you will return to the Available Software dialog (shown earlier). Step 3: installing the Linked Data Fragments Feature The Available Software dialog will now show a list of features available from the just added update site: You can see the Linked Data Fragments Feature is now listed which you can select with the checkbox in front of the name (as shown above). The Next button will walk you through a few more pages in this dialog, providing information about dependencies and a page that requires you to accept the Open Source licenses involved. And at the end of these steps, it may require you to reboot Bioclipse. Step 4: opening the JavaScript Console and verify the new extension is installed Because the Linked Data Fragments Feature extends Bioclipse with a new, so-called manager (see doi:10.1186/1471-2105-10-397), we need to use the JavaScript Console (or Groovy Console, or Python Console, if you prefer those languages). Make sure the JavaScript Console is open, or do this via the menu Windows → Show View → JavaScript Console and type in the console view man ldf which should result in something like this: You can also type man ldf.createStore to get a brief description of the method I used to get a Linked Data Fragments wrapper for Wikidata in my previous post, which is what you should reread next. Have fun and looking forward to hear how you use Linked Data Fragments with Bioclipse! ### Chemistry Central and the ORCID identifier If you are a scientist you have heard about the ORCID identifier by now. If not, you have been focusing on groundbreaking research and isolated yourself from the rest of the world, just to make it perfect and get that Nobel prize next year. If you have been working on impactful research, Nobel prize-worthy, and have been blogging and tweeting about your progress, as a good Open Scholar, you know ORCID is the DOI for "research contributors" and you already have one yourself, and probably also that T-shirt with your own identifier. Mine is 0000-0001-7542-0286, and almost 1.3M other authors got one too. The list of ORCIDs on Wikipedia is growing (and Wikidata), thanks to Andy Mabbett, whom also made it possible to add your ORCID on WikiPathways. Anyway, what I was pleased to see today that you can now log in with your ORCID identifier with the Chemistry Central article submission system (notice the green icon): Many other publishers allow logging in with your ORCID too, which benefits many: 1. authors who just enter a list of ORCID identifiers, instead of a long list of author names and affiliations 2. publishers, which have a simpler submission system and get more accurate information about submitters 3. funding agencies which can more easily track what is done with the research funding 4. research institutes which can more easily track what their employees are studying Don't have one yet? Get your very own ORCID here. ### CC-BY with the ACS Author Choice: CDK and Blue Obelisk papers liberated Screenshot of an old CDK-basedJChemPaint, from the first CDK paper.CC-BY :) Already a while ago, the American Chemical Society (ACS) decided to allow the Creative Commons Attribution license (version 4.0) to be used on their papers, via their Author Choice program. ACS members pay$1500, which is low for a traditional publisher. While I even rather seem them move to a gold Open Access journal, it is a very welcome option! For the ACS business model it means a guaranteed sell of some 40 copies of this paper (at about \$35 dollar each), because it will not immediately affect the sale of the full journal (much). Some papers may sell more than that had the paper remained closed access, but many for papers that sounds like a smart move money wise. Of course, they also buy themselves some goodwill and green Open Access is just around the corner anyway.
Better, perhaps, is that you can also use this option to make a past paper Open Access under a CC-BY license! And that is exactly what Christoph Steinbeck did with five of his papers, including two on which I am co-author. And these are not the least papers either. The first is the first CDK paper from 2003 (doi:10.1021/ci050400b), which featured a screenshot of JChemPaint shown above. Note that in those days, the print journal was still the target, so the screenshot is in gray scale :) BTW, given that this paper is cited 329 times (according to ImpactStory), maybe the ACS could have sold more than 40 copies. But for me, it means that finally people can read this paper about Open Science in chemistry, even after so many years. BTW, there is little chance the second CDK paper will be freed in a similar way.
The second paper that was liberated this way, is the first Blue Obelisk paper (doi:10.1021/ci050400b), which was cited 276 times (see ImpactStory):
This screenshot nicely shows how readers can see the CC-BY license for this paper. Note that it also lists that the copyright is with the ACS, which is correct, because in those days you commonly gave away your copyright to the publisher (I have stopped doing this, bar some unfortunate recent exceptions).
So, head over to your email client and email [email protected] and let them know you also want your JCICS/JCIM paper available under a CC-BY license! No excuse anymore to make your seminal work in cheminformatics not available as gold Open Access!
Of course, submitting your new work to the Journal of Cheminformatics is cheaper and has the advantage that all papers are Open Access!
## Tuesday, April 14, 2015
### Ambit.js 0.0.2 release: a scatterplot
Two weeks ago I made a second release of ambit.js, a small project to show the power of the eNanoMapper API (doi:10.5281/zenodo.16517). It still is a client library to use the API from JavaScript and a few weeks ago I posted a few screenshots. This post is aimed at announcing the 0.0.2 release which doesn't change a lot since the previous version, but now features online documentation:
It also has a number of online examples, which include the code behind the screenshots in the earlier post, but also a new scatterplot example (still using d3.js):
Now, this scatter plot shows basically that the data we are looking at does not show a particular correlation, but that was not really the hypothesis here anyway. However, it does show how hypotheses can be tested with the API and scatter plots.
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Math Help - Qualitative Analysis
1. Qualitative Analysis
Show that there is at least one solution in each of the second and fourth quadrants that tend to the origin as t goes to infinity.
Similarly, show that there is at least one solution in each of the first and third quadrants that tend to the origin as t goes to negative infinity.
No idea how I do this. Some help would be most appreciated!
Guess it would help if I posted the system :P
$dx/dt=y$
$dy/dt=x-x^2$
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# If $f:[2,3] \to R$ is defined by $f(x) =x^3+3x-2,$ then the range $f(x)$ is contained in the interval:
$\begin {array} {1 1} (1)\;[1,12] & \quad (2)\;[12,34] \\ (3)\;[35,50] & \quad (4)\;[-12,12] \end {array}$
## 1 Answer
(2) [12,34]
answered Nov 7, 2013 by
1 answer
1 answer
1 answer
1 answer
1 answer
2 answers
1 answer
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#### Vol. 3, No. 1, 2020
Poincaré duality is central to the understanding of manifold topology. Dimension 3 is critical in various respects, being between the known territory of surfaces and the wilderness manifest in dimensions $\ge$ 4. The main thrust of 3-manifold topology for the past half century has been to show that aspherical closed 3-manifolds are determined by their fundamental groups. Relatively little attention has been given to the question of which groups arise. This book is the first comprehensive account of what is known about PD${}_{3}$-complexes, which model the homotopy types of closed 3-manifolds, and PD${}_{3}$-groups, which correspond to aspherical 3-manifolds. In the first half we show that every P${}^{2}$-irreducible PD${}_{3}$-complex is a connected sum of indecomposables, which are either aspherical or have virtually free fundamental group, and largely determine the latter class. The picture is much less complete in the aspherical case. We sketch several possible aproaches for tackling the central question, whether every PD${}_{3}$-group is a 3-manifold group, and then explore properties of subgroups of PD${}_{3}$-groups, unifying many results of 3-manifold topology. We conclude with an appendix listing over 60 questions. Our general approach is to prove most assertions which are specifically about Poincaré duality in dimension 3, but otherwise to cite standard references for the major supporting results.
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