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Excel Formula for Percentage in Python In this article, we will explore how to write an Excel formula in Python to calculate the percentage of one value in relation to another value. This formula is useful when you want to determine the proportion of a value compared to a reference value. By using Python, you can easily perform this calculation programmatically and incorporate it into your data analysis workflows. To calculate the percentage in Python, you can use the following formula: This formula takes two inputs, A1 and B1, and returns the percentage as a result. The value in cell A1 is divided by the value in cell B1 using the division operator /. The result of the division is then multiplied by 100 to convert it into a percentage using the multiplication operator . Let's consider an example to understand how this formula works. Suppose we have the values 50 in cell A1 and 200 in cell B1. By applying the formula =(A1/B1)100, we can calculate the percentage of A1 in relation to B1 as follows: Therefore, the result of the formula is 25%, which represents the percentage of 50 in relation to 200. By using this formula in Python, you can easily calculate percentages and incorporate them into your data analysis tasks. Whether you are working with financial data, sales figures, or any other numerical data, this formula can be a valuable tool in your analysis toolkit. In the next sections, we will explore more examples and use cases for calculating percentages in Python. An Excel formula Formula Explanation This formula calculates the percentage of one value in relation to another value. It takes two inputs, A1 and B1, and returns the percentage as a result. Step-by-step explanation 1. The formula starts with an opening parenthesis (. 2. The value in cell A1 is divided by the value in cell B1 using the division operator /. 3. The result of the division is multiplied by 100 to convert it into a percentage using the multiplication operator . 4. The formula ends with a closing parenthesis ). For example, let's say we have the following values in cells A1 and B1: Using the formula =(A1/B1)100, we can calculate the percentage of A1 in relation to B1: So, the result of the formula would be 25%, which represents the percentage of 50 in relation to 200.	{"url":"https://codepal.ai/excel-formula-generator/query/2fFEU2Xj/excel-formula-percentage","timestamp":"2024-11-03T16:12:34Z","content_type":"text/html","content_length":"91206","record_id":"<urn:uuid:d5604576-c94c-4ae5-a4f8-d51dfee8d6e3>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00389.warc.gz"}
7.9 ounces to grams Convert 7.9 Ounces to Grams (oz to gm) with our conversion calculator. 7.9 ounces to grams equals 223.961245065765 oz. Enter ounces to convert to grams. Formula for Converting Ounces to Grams (Oz to Gm): grams = ounces * 28.3495 By multiplying the number of grams by 28.3495, you can easily obtain the equivalent weight in grams from ounces. Understanding the Conversion from Ounces to Grams When it comes to converting ounces to grams, it’s essential to know the conversion factor. One ounce is equivalent to approximately 28.3495 grams. This means that to convert ounces into grams, you simply multiply the number of ounces by this conversion factor. This conversion is particularly important for those who work with both the imperial and metric systems, as it helps bridge the gap between these two measurement systems. Formula for Converting Ounces to Grams The formula to convert ounces (oz) to grams (g) is: grams = ounces × 28.3495 Step-by-Step Calculation: Converting 7.9 Ounces to Grams Let’s take a closer look at how to convert 7.9 ounces to grams using the formula provided: 1. Start with the number of ounces you want to convert: 7.9 ounces. 2. Use the conversion factor: 28.3495 grams per ounce. 3. Multiply the number of ounces by the conversion factor: 7.9 oz × 28.3495 g/oz. 4. Perform the calculation: 7.9 × 28.3495 = 223.96255 grams. 5. Round the result to two decimal places for practical use: 224.00 grams. The Importance of Ounce to Gram Conversion This conversion is crucial for various applications, especially in cooking, where precise measurements can significantly affect the outcome of a recipe. For instance, if a recipe calls for 7.9 ounces of flour, knowing that this is equivalent to 224 grams allows you to measure accurately using a kitchen scale that may only display metric units. In scientific measurements, converting ounces to grams is equally important. Many laboratory protocols require precise mass measurements in grams, making it essential for researchers and students to convert ounces accurately. Additionally, in everyday scenarios, such as when purchasing food items or supplements, understanding this conversion can help consumers make informed choices based on nutritional information that is often provided in grams. In summary, converting 7.9 ounces to grams is a straightforward process that can enhance your cooking, scientific endeavors, and daily life. By mastering this conversion, you can ensure accuracy and consistency in your measurements, bridging the gap between the imperial and metric systems with ease. Here are 10 items that weigh close to 7.9 ounces to grams – • Standard Baseball Shape: Spherical Dimensions: 9 inches in circumference Usage: Used in the sport of baseball for pitching, hitting, and fielding. Fact: A baseball is made of a cork center wrapped in layers of yarn and covered with leather. • Medium-Sized Apple Shape: Round Dimensions: Approximately 3 inches in diameter Usage: Eaten raw, used in cooking, or made into cider. Fact: Apples float in water because 25% of their volume is air. • Standard Coffee Mug Shape: Cylindrical Dimensions: 4 inches tall, 3 inches in diameter Usage: Used for drinking hot beverages like coffee or tea. Fact: The world’s largest coffee mug can hold over 1,000 cups of coffee! • Small Bag of Flour Shape: Rectangular Dimensions: 10 inches tall, 5 inches wide Usage: Used in baking and cooking as a primary ingredient. Fact: Flour is made by grinding raw grains, typically wheat, into a fine powder. • Standard Deck of Playing Cards Shape: Rectangular Dimensions: 2.5 inches by 3.5 inches Usage: Used for various card games and magic tricks. Fact: A standard deck has 52 cards, plus 2 jokers, totaling 54 cards. • Small Water Bottle Shape: Cylindrical Dimensions: 8 inches tall, 2.5 inches in diameter Usage: Used for carrying water or other beverages on the go. Fact: Staying hydrated can improve your mood and cognitive function! • Medium-Sized Dog Toy Shape: Irregular (often shaped like a bone or ball) Dimensions: Approximately 6 inches long Usage: Used for play and exercise for dogs. Fact: Interactive toys can help reduce anxiety in dogs and keep them mentally stimulated. • Small Potted Plant Shape: Cylindrical (pot) with a varied shape for the plant Dimensions: 6 inches in diameter, 8 inches tall Usage: Used for decoration and improving indoor air quality. Fact: Some houseplants can remove toxins from the air, making your home healthier! • Standard Notebook Shape: Rectangular Dimensions: 8.5 inches by 11 inches Usage: Used for writing notes, journaling, or sketching. Fact: The first notebooks were made from papyrus in ancient Egypt! • Small Bag of Rice Shape: Rectangular Dimensions: 12 inches tall, 6 inches wide Usage: Used as a staple food in many cultures around the world. Fact: Rice is the primary food source for more than half of the world’s population. Other Oz <-> Gm Conversions –	{"url":"https://www.gptpromptshub.com/grams-ounce-converter/7-9-ounces-to-grams","timestamp":"2024-11-09T03:15:08Z","content_type":"text/html","content_length":"186495","record_id":"<urn:uuid:6650d340-3ac2-42b6-868a-dc7aead838b3>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00527.warc.gz"}
How to solve a 2x2 using Ortega Method \| INTERMEDIATE This page explains the Ortega Method of solving a 2x2. This method is for intermediate level solving. If you are completely new to 2x2 or if you don't understand notation then you can find the beginner tutorial here: How to solve a 2x2 Beginners Method There are twelve algorithms for this method. You may be familiar with some of the twelve algorithms in this method, as they are also used to solve the 3x3 Rubik’s Cube's final layer. Step 1: Orientate first layer. Solve the white face. You don't need to have the pieces in the correct position. They should only be orientated correctly. Step 2: OLL - Orientate Last Layer You will now have one of eight cases to orientate the Last layer. Many of the algorithms are the same as 3x3 OLL Algorithms, but there are shorter algorithms to use as there are no centers or edges. Hold your cube with the solved face (white) on the bottom and check your top face for the patching pattern. Step 3: PBL - Permutate Both Layers You will now have one of five cases to permutate both layers together. Hold your cube with yellow on top and white at the bottom. Choose from the options below that matched your situation and follow the algorithm to solve the 2x2. 2x2 Algorithm Sets EG-1 \| Erik Gunnar 1 EG-2 \|Erik-Gunnar 2 Ortega \| Intermediate CLL \| Corners of Last Layer	{"url":"https://uk.speedcube.com.au/blogs/speedcubing-solutions/how-to-solve-a-2x2-using-ortega-method-intermediate","timestamp":"2024-11-12T04:16:29Z","content_type":"text/html","content_length":"319443","record_id":"<urn:uuid:60b5a09e-21bf-40d7-babd-ef077680fdf9>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00832.warc.gz"}
This page provides a simple browsing interface for finding entities described by a property and a named value. Other available search interfaces include the page property search, and the ask query List of results • Model:GeoClaw + (Depth, momentum on adaptive grid at specified output times. Time series at specified gauge locations. Maxima observed over full simulation on specified grid.) • Model:HBV + (Discharge) • Model:FuzzyReef + (Dynamic variables: # water energy # depositional (seafloor) slope Final output: # carbonate productivity rate # depositional facies) • Model:BRaKE + (Elevation and slope arrays as well as optional information about bed cover and shear stress distributions, as well as block size distributions and incision rate records.) • Model:Kirwan marsh model + (Elevation, Biomass, Accretion Rate, Erosion Rate, and other characteristics of every cell in domain. Also outputs spatially averaged statistics.) • Model:GOLEM + (Elevation, drainage area, and related gridded information.) • Model:LOADEST + (Estimated constituent loads) • Model:SBEACH + (Estimated post-storm beach profile, cross-shore profile of: maximum wave height; maximum water elevation plus setup; volume change) • Model:Detrital Thermochron + (Estimates of the erosional history and spatial patterns and model diagnostic plots.) • Model:Rescal-snow + (Evolving 3D cellspace and 2D elevation map) • Model:BarrierBMFT + (Extent, and elevation, and cross-shore boundary locations of barrier, marsh (back-barrier and mainland), bay, and forest ecosystems; organic and mineral deposition; shoreline locations; dune elevations; overwash & shoreface fluxes) • Model:Non Local Means Filtering + (Filtered DEM: A new, filtered DEM in .flt binary format. Noise: A .flt binary format grid of the filtered noise.) • Model:GSSHA + (Flow rates, depths, soil moisture, sediment fluxes, erosion/deposition, contaminant/nutrient fluxes and concentrations, groundwater levels, reservoir storages.) • Model:LOGDIST + (Flow velocities at N levels in the vertical, assuming a logarithmic velocity profile) • Model:WASH123D + (Fluid velocity, pressure, temperature, sal … Fluid velocity, pressure, temperature, salinity, concentrations, thermal flexes, and matrial fluxes at all nodes at any desired time. volumetric, energy, and mass balance at all types of boundaries and the entire boundary at any specified time. Br>For details refer to Yeh et al., 2005 Technical Report on WASH123DYeh et al., 2005 Technical Report on WASH123D) • Model:RivMAP + (For a single image: centerlines, widths, c … For a single image: centerlines, widths, channel direction, curvatures</br>For multiple images: (centerline) migration areas, erosion and accretion areas, cutoffs, cutoff statistics, channel belt boundaries, grid generation to map spatial changes, spacetime maps of changes in planform variablestime maps of changes in planform • Model:MPeat2D + (Formation of peatland) • Model:GPM + (Free-surface flow and wave action through time. Erosion and deposition through time. Optionally, compaction, including porosity reduction.) • Model:WAVEWATCH III ^TM + (From wave heights to spectral data, see manual) • Model:TURB + (Gaussian distribution of instantaneous turbulent fluid shear stresses at the bed) • Model:LateralVerticalIncision + (Geometry of river entrenchment thought time) • Model:SEDPAK + (Graphical Display and surface plot) • Model:WSGFAM + (Gravity flow velocity, Depth-integrated sediment load, down-slope sediment flux, flux convergence or divergence, erosion or deposition rate.) • Model:Plume + (Grid of Sediment rate in m/day for specified grain size classes) • Model:AquaTellUs + (Grid of deposition of different grains over time. The model generates postscript files of stratigraphic sections.) • Model:Landlab + (Gridding component provides ASCII and/or netCDF output of grid geometry.) • Model:Sun fan-delta model + (Grids of topography) • Model:DeltaRCM Vegetation + (Grids of water surface elevation, discharge, bed elevation, and vegetation density values for each cell. Additionally, sand fraction of each vertical cell within a grid cell.) • Model:CASCADE + (H, fluxes, discharge, catchment geom, etc, at all time steps, as welle as grid connectivity) • Model:HSPF + (HSPF produces a time history of the runoff … HSPF produces a time history of the runoff flow rate, sediment load, and</br>nutrient and pesticide concentrations, along with a time history of water</br>quantity and quality at any point in a watershed. Simulation results can be</br>processed through a frequency and duration analysis routine that produces</br>output compatible with conventional toxicological measures (e.g., 96-hour</br>LC50).xicological measures (e.g., 96-hour LC50).) • Model:HexWatershed + (Hexagon DEM, flow direction, flow accumulation, stream grid, stream segment, stream order, stream confluence, subbasin, watershed boundary, etc.) • Model:TauDEM + (Hydrologic information derived from DEM) • Model:GSFLOW-GRASS + (Hydrologic model discretization, input files for GSFLOW, output files from GSFLOW (hydrologic model)) • Model:SWMM + (In addition to modeling the generation and … In addition to modeling the generation and transport of runoff flows, SWMM can also estimate the production of pollutant loads associated with this runoff. The following processes can be modeled for any number of user-defined water quality constituents:</br></br>* dry-weather pollutant buildup over different land uses</ br>* pollutant washoff from specific land uses during storm events</br>* direct contribution of rainfall deposition</br>* reduction in dry-weather buildup due to street cleaning</br>* reduction in washoff load due to BMPs</br>* entry of dry weather sanitary flows and user-specified external inflows at any point in the drainage system</br>* routing of water quality constituents through the drainage system</br>* reduction in constituent concentration through treatment in storage units or by natural processes in pipes and channelsby natural processes in pipes and channels) • Model:Spbgc + (It can output local velocity, vorticity, c … It can output local velocity, vorticity, concentration, stream-function, and all derivatives of velocity necessary to calculate dissipation, viscous momentum diffusion, kinetic energy flux, work by pressure forces, and change in kinetic energy. These quantities are written out in a binary file.</br></br> It also has routines for calculating the local height profile and tip position of gravity currents and internal bores, which are outputted every time step and stored as ASCII txt files.y time step and stored as ASCII txt files.) • Model:Cross Shore Sediment Flux + (It outputs all the variables used in the advection-diffusion equation describing bed evolution for both shallow water wave assumptions (all labeled as _s) and linear theory (labeled as _lh).) • Model:Ecopath with Ecosim + (Key indices, Mortalities, Consumption, Respiration, Niche overlap, Electivity, Search rates and Fishery forms.) • Model:NEWTS + (Lake grid) • Model:SLAMM 6.7 + (Land cover and elevation prediction rasters under SLR conditions through 2100.) • Model:GRLP + (Long profile (x, z); output sediment discharge) • Model:FluidMud + (Major quantities: mud floc concentration, flow velocity in longshelf and cross-shelf direction. Other quantities: TKE, turbulent dissipation rate, floc size (if floc dynamics turn on), bottom stress.) • Model:DFMFON + (Mangrove properties and Delft3D-FM output) • Model:ParFlow + (Many: pressure, saturation, temperature, energy fluxes, flow, etc.) • Model:QDSSM + (Maps of geomorphology, discharge, deposition, isopachs, stratigraphic thickness, grain size, contour, subsidence, and environment) • Model:GEOMBEST++ + (Marsh boundary - gives the position of the backbarrier marsh edge through time Shorelines - gives the position of the barrier shoreline through time step number - saves the surface morphology and stratigraphy for the model at each time step) • Model:Wetland3P + (Marsh depth, mudflat depth, mudflat width) • Model:MarshPondModel + (Marsh elevation Pond area and location) • Model:CosmoLand + (Mass, atoms, landslide size, fluvial residence time, mixed mass and atoms fraction) • Model:Lake-Permafrost with Subsidence + (Matlab variables, Matlab graphs) • Model:DeltaRCM + (Matrices of: Water surface elevation; Water unit discharge and velocity field; Delta surface elevation and bathymetry; Stratigraphy (User can choose which time step to output)) • Model:Manningseq-bouldersforpaleohydrology + (Microsoft Excel tables)	{"url":"https://csdms.colorado.edu/wiki/Special:SearchByProperty/:Describe-20output-20parameters-20model/Gridding-20component-20provides-20ASCII-20and-2For-20netCDF-20output-20of-20grid-20geometry.","timestamp":"2024-11-05T07:53:31Z","content_type":"text/html","content_length":"50560","record_id":"<urn:uuid:09821297-2d55-428e-ba3e-5806b6f64fd4>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00144.warc.gz"}
This assignment falls under R Studio which was successfully solved by the assignment writing experts at My Assignment Services AU under assignment help service. R Studio Assignment Assignment Task You will need to download the following data from Quandl — a repository of public data. Tickers for Quandl are in parentheses. • Constant maturity US Treasuries: 3M, 2Y, 10Y, and 30Y (FRED/DGS3MO, FRED/DGS2, FRED/DGS10, FRED/DGS30) • Indices: S&P 500, Russell 2000 (YAHOO/INDEX_GSPC, YAHOO/INDEX_RUT) • Eurodollar futures (CHRIS/CME_ED1, CHRIS/CME_ED24) • Your stock (YAHOO/YOURTICKER) The bond data are yields, so they need to be rescaled to be on the same scale as log-returns (divide them by 100). The index and equity prices need to be corrected for splits and dividends (use the adjusted close field). Finally: Restrict your analysis to 1 September 2013–1 September 2016 (including the start and end dates). Analysis Software To make life easy (really!), we will use R with the Quandl, xts, and PerformanceAnalytics packages for the analysis.1 To get working with R, use the market trading lab: sign up for some of the short tutorials offered there — or download the tutorial slides and go through them yourself. Go to go.uic.edu/cme_lab to sign up and get slides. To install a package: go to the Package Manager in R; get the list of packages from CRAN; and, find the package by name. Highlight the package name, check the box for “Install dependencies,” and then click install. Once PerformanceAnalytics is installed, for example, you can use the SemiDeviation function. Example R Code Below is some R code to get you started. Paste this into a file where you can then modify the code. library(Quandl) library(xts) library(PerformanceAnalytics) # Line that start with a hash mark are comments. Comments are # crucial: They explain what you are doing. That helps you remember # what you did later; lets other people take over your job (when you # get promoted); and, helps you/others fix the analysis quickly when # it breaks. (All analyses break eventually.) # Example of reading in a Quandl dataset to xts. # Name columns so we know what each holds after joining them together ust.tickers <- c("FRED/DGS3MO", "FRED/DGS2", "FRED/DGS10", "FRED/DGS30") ust.raw <- Quandl(ust.tickers, type="xts")/100 colnames(ust.raw) <- c("T3M", "T2Y", "T10Y", "T30Y") # This is a way to get approximate returns for these bonds. # Those of you who take FIN 310 will learn why we can do this. ust.yieldchanges <- diff(ust.raw) ust[,"T3M"] <- -0.25ust.yieldchanges [,"T3M"] ust[,"T2Y"] <- -1.98ust.yieldchanges[,"T2Y"] ust[,"T10Y"] <- -8.72ust.yieldchanges[,"T10Y"] ust[,"T30Y"] <- -19.2ust.yieldchanges[,"T30Y"] # Get Eurodollar futures (settlement) prices and create log-returns. ed1.raw <- Quandl("CHRIS/CME_ED1", type="xts")[,"Settle"] 1You may download R from www.r-project.org. The xts package handles time series data very nicely, taking care of date alignment when doing math or combining data series. The xts package is used worldwide and was written by Jeff Ryan. . . who graduated from UIC in 2004. ed1 <- diff(log(ed1.raw)) colnames(ed1) <- c("ED1") # Get S&P 500 prices (just adjusted close); then create log-returns. # Do similarly for the Russell 2000, and your stock. spx.raw <- Quandl("YAHOO/INDEX_GSPC", type="xts")[,"Adjusted Close"] spx <- diff(log(spx.raw)) colnames(spx) <- c("SPX") # Join all of the datasets together: US Treasuries, Eurodollars, # S&P 500, Russell 2000, and your stock. Then trim them down. alldata.full <- cbind(ust, ed1, ed24, spx, rut, yourticker) alldata <- # Calculate annual volatilities and semideviations like so: apply(alldata, 2, sd)sqrt(250) SemiDeviation(alldata)sqrt(250) # skewness and kurtosis are independent of time; no need to scale them skewness(alldata, method="moment") kurtosis(alldata, method="moment") 1. (Risk-Free Price Risk) When we say US Treasuries (USTs) are risk-free, we mean that their payoff is certain. (Well, as certain as can be for a USD-denominated investment.) However, you can lose money trading USTs since their prices change with interest rates. To get a handle on that capital gains risk, we will calculate some volatilities. (a) (4) We want to get daily yields for some USTs at four tenors (times to maturity): 3M, 2Y, 10Y, and 30Y. Since a 3M T-bill expires in three months, we obviously cannot use the same bill over a two-year period. Therefore, the Fed creates yield series called constant maturity treasuries (CMTs). CMT rates are averages of yields for instruments maturing near a certain amount of time. We use these to infer the yield for a certain maturity. Get daily yields over the past three years for those four instruments. The yields are quoted in percentage points; thus yields of “1.23” and “0.002” are yields of 1.23% and 0.002% (i.e. 0.2 basis points). What is the average yield for each of these instruments over the three years? (b) We cannot calculate daily log-returns for CMTs. Therefore, we must use an approximation. To do this requires two steps: First, compute the changes in yields. Then, multiply those changes by the following numbers: -0.25 (3M), -1.98 (2Y), -8.72 (10Y), and -19.20 (30Y).2 2These numbers are related to the average time of a bond’s cashflows. We’ll get to that later. The result is a percent change for the bond price, on the same scale as the bond yields. Find the average of these approximated log-returns for each of the four maturities (c)Using these approximated daily log-returns, calculate a standard deviation for each matu- rity. These are estimates of daily log-return volatilities. Scale them up to an annual basis (remembering that there are about 250 trading days/year). (d) (8) Again using the approximated daily log-returns, calculate a semi-deviation for each ma- turity. These are estimates of daily log-return semi-deviation. Scale them up to an annual basis (remembering that there are about 250 trading days/year). (Short-term Credit and Price Risk) While we say US Treasuries (USTs) are risk-free, this is not true for money deposited in a bank: That bank can fail. Eurodollar futures can be used to hedge the rate paid for large US dollar deposits in a top-credit London bank. (That rate is called LIBOR, the London Interbank Offered Rate.) Eurodollar futures are some of the most actively-traded instruments in the world. Because banks and finance firms often anticipate cashflows well into the future, Eurodollar futures are not just liquid for a few maturities but for many maturities. For this question, you should look at near Eurodollars (ED1) which are used to hedge three- month rate risk and Eurodollars about two-years out (ED24). (a) Since Eurodollar futures trade at prices (not yields), we can easily calculate daily log- returns for them. Using these daily log-returns, calculate a standard deviation for each maturity. These are estimates of daily log-return volatilities. Scale them up to an annual basis (remembering that there are about 250 trading days/year). Report the scaled-up volatilities. (b) Again using the daily log-returns, calculate a semi-deviation for each maturity. These are estimates of daily log-return semi-deviation. Scale them up to an annual basis (remembering that there are about 250 trading days/year) and report the scaled semi deviations. (c) Compare the volatilities of these Eurodollar contracts to the volatilities of similar-term CMTs. How different are the volatilities? Why would this be? (d) (4) Now we will examine a credit spread. The TED spread is the amount that short-term Eurodollars (the “ED” in TED) yield over a similar-term US Treasury instrument (the “T” in TED). To compute what 3M Eurodollars are yielding, just subtract their price from 100. So if 3M Eurodollars are at 99.735, that implies a yield of 100?99.735 = 0.265 aka 0.265% or 26.5 basis points (bp). The TED spread is then found by subtracting the 3M CMT yield from this number. If the 3M CMT UST is yielding 0.03% (3 bp), then the TED spread is 23.5 bp. Calculate and report the historical average, volatility, and semideviation for the TED spread. (Equity Price Risk) Stocks are not risk-free: they may be rendered worthless (or nearly so) in bankruptcy; and, dividends may be reduced or suspended. All of these possibilities affect the risk of stocks. To get a handle on that risk and how it compares to price risk for USTs, we will calculate more volatilities. For these questions, use the ticker assigned to you. (For example, if my ticker were DAL its Quandl ticker would be YAHOO/DAL.) (a) (3) Get daily prices over the past three years for your stock, the S&P 500, and the Russell 2000. 3 This means you will be working with three equity instruments. For your stock, make sure you get prices that are adjusted for dividends and splits; or, you may get dividends and splits and do the adjustments yourself. What is the average price of each of the equity instruments over the past three years (b) Calculate daily log-returns, differences in logs of daily prices, for all three equity instruments. Find the average log-return for each equity instrument. (c)How do these average daily log-returns (when annualized) compare to average UST yields? Does this make sense? Why? (d)Using the daily log-returns, calculate a standard deviation of log-returns for each equity instrument. These are estimates of daily volatility. Scale them up to an annual basis (remembering that there are about 250 trading days/year). (e) Again using the daily log-returns, calculate a semi-deviation of log-returns for each equity instrument. These are estimates of daily semi-deviation. Scale them up to an annual basis (remembering that there are about 250 trading days/year). (f) How do the volatilities and semi-deviation compare between the equity instruments and USTs? Does this make sense? Why? (4) Save the data and, if you used R, the commands you used to do this homework. Print out your answers AND all the commands — and turn both in. Note that if you do not turn in your code, you will get no credit for this assignment. Also, if the code you turn in does not actually work, you will lose points. You will use the commands and data in Homework 2, so getting these points will make your life easier later. The assignment file was solved by professional R Studio experts and academic professionals at My Assignment Services AU. The solution file, as per the marking rubric, is of high quality and 100% original (as reported by Plagiarism). The assignment help was delivered to the student within the 2-3 days to submission. Looking for a new solution for this exact same question? Our assignment help professionals can help you with that. With a clientele based in top Australian universities, My Assignment Services AU’s assignment writing service is aiding thousands of students to achieve good scores in their academics. Our R Studio assignment experts are proficient with following the marking rubric and adhering to the referencing style guidelines. The R coding output is given below Going through the output, we see that semi deviation of 3 month, 2 year, 30 year, EDI and adjusted close values are 0.0003, 0.0055, 0.0473, 0.1023, 0.00138 and 2705.905 respectively. The table given below shows the workings of bond price for the four maturities Maturity Bond Price 3M -0.0075 2Y -0.01089 10Y -0.4125 30Y -0.201 Going through the output, we see that semi deviation of 3 month, 2 year, 30 year, EDI and adjusted close values are 0.0003, 0.0055, 0.0473, 0.1023, 0.00138 and 2705.905 respectively. The output for Ed1 is given below Going through the output, we see that semi deviation of 3 month, 2 year, 30 year, EDI and adjusted close values are 0.0003, 0.0055, 0.0473, 0.1023, 0.00138 and 2705.905 respectively. On comparing the volatilities of these Eurodollar contracts to the volatilities of similar term CMTs, we see that both are similar in volatilities The average daily prices of three stocks is given below The standard deviation of three log returns for each equity instrument is calculated and is given below The table given below shows the workings of bond price for the four maturities Stocks Bond Price K 68.22 S & P 500 2056 Russel 2000 114.61 The volatilities of these average daily log returns was compared to average UST yields to determine the variation in the daily prices This R Studio assignment sample was powered by the assignment writing experts of My Assignment Services AU. You can free download this R Studio assessment answer for reference. This solved R Studio assignment sample is only for reference purpose and not to be submitted to your university. For a fresh solution to this question, fill the form here and get our professional assignment help.	{"url":"https://www.gradesaviours.com/solutions/finance-310-investments-risk-free-price-risk-r-studio-assessment-answer","timestamp":"2024-11-05T04:05:46Z","content_type":"text/html","content_length":"503409","record_id":"<urn:uuid:0e1ca203-1248-4566-8d29-ce15acad02dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027870.7/warc/CC-MAIN-20241105021014-20241105051014-00072.warc.gz"}
kalman filter prediction The first step uses previous states to predict the current state. The measurement matrix accommodates what you can directly measure and what you can’t. Predict the last estimation to the time of the new measurement using the propagation model, and update the co-variance accordingly. The main goal of this chapter is to explain the Kalman Filter concept in a simple and intuitive way without using math tools that may seem complex and confusing. by David Kohanbash on January 30, 2014 . The second step uses the current measurement, such as object location, to correct the state. I'm new to Kalman filtering, but is it possible to apply kalman filter for prediction and tracking of objects in video frames using MATLAB? If you succeeded to fit your model into Kalman Filter, then the next step is to determine the necessary parameters and your initial values. The Kalman filter is generally credited to Kalman and Bucy. The standard Kalman lter deriv ation is giv When there is a lot of input noise, the Kalman Filter estimate is much more accurate than a direct reading. Its use in the analysis of visual motion has b een do cumen ted frequen tly. The Kalman filter algorithm involves two stages: prediction and measure-ment update. 2. If prediction is enabled, the red line shows the predicted path of your movement (how far the prediction goes is adjustable by the Prediction Amount slider). Kalman Filtering – A Practical Implementation Guide (with code!) To validate the prediction performance of this method, we conduct an empirical study for China’s manufacturing industry. The alpha beta filter is conceptually simpler and works well for slowly evolving systems. Kalman Filter tutorial Part 6. Following a problem definition of state estimation, filtering algorithms will be presented with supporting examples to help readers easily grasp how the Kalman filters work. Kalman filter algorithm can be roughly organized under the following steps: 1. Kalman published his famous paper describing a recursive solution to the discrete-data linear filtering problem [Kalman60]. ie. Subject MI63: Kalman Filter Tank Filling First Option: A Static Model 2. All code is written in Python, and the book itself is written in Ipython Notebook so that you can run and modify the code Kalman Filter (KF) is a well-known algorithm for estimation and prediction especially when data has a lot of noise. Kalman Filter in one dimension. One question, will the Kalman filter get more accurate as more variables are input into it? Let's take the expected value of the observation tomorrow, given our knowledge of the data today: Prior distribution from the Chapman-Kolmogorov equation 2 Department of Computer Science and Biomedical Informatics, University of Thessaly, 35100 Lamia, Greece The ﬁrst is the most basic model, the tank is level (i.e., the true level is constant L= c). The overall errors during prediction will be compared for an analysis of the corrective ability of the filter. The Kalman filter technique allows to capture the temporal dependence as well as the spatial correlation structure through state-space equations, and it is aimed to perform statistical inference in terms of both parameter estimation and prediction at unobserved locations. Yes. I did some image processing on the frames and now i'm able to indicate the bullet as a point. if you have 1 unknown variable and 3 known variables can you use the filter with all 3 known variables to give a better prediction of the unknown variable and can you keep increasing the known inputs as long as you have accurate measurements of the data. Kalman Filter Based Short Term Prediction Model for COVID-19 Spread Suraj Kumar, Koushlendra Kumar Singh,1, Prachi Dixit2, Manish Kumar Bajpai3 1National Institute of Technology, Jamshedpur, India 2Jai Narayan Vyas University, Jodhpur, India 3Indian Institute of Information Technology Design and Manufacturing, Jabalpur, India Corresponding Author … The application of Kalman filter on wind speed prediction is implemented in MATLAB software and results are provided in this paper. The Kalman filter has 2 steps: 1. ... • This is the prediction step of the optimal filter. This discrepancy is given by: ... Time-Varying Kalman Filter. 11.1 In tro duction The Kalman lter [1] has long b een regarded as the optimal solution to man y trac king and data prediction tasks, [2]. Show abstract. The prediction step projects forward the current state and covariance to obtain an a priori estimate. The Kalman filter algorithm involves two steps, prediction and correction (also known as the update step). Further info: I have a sequential set of 20 images of a bullet coming out of a gun (A burst shot of images). The Bayesian approach to the Kalman Filter leads naturally to a mechanism for prediction. Therefore, an Extended Kalman Filter (EKF) is used due to the nonlinear nature of the process and measurements model. Knowledge of the state allows theoretically prediction of the future (and prior) dynamics and outputs of the deterministic system in the absence of noise. Even if I have understood the Bayesian filter concept, and I can efficiently use some of Kalman Filter implementation I'm stucked on understand the math behind it in an easy way. Additionally a final prediction at a later date and another location will serve as an indicator to the usefulness of the prediction capabilities over time. This chapter aims for those who need to teach Kalman filters to others, or for those who do not have a strong background in estimation theory. Kalman filter can predict the worldwide spread of coronavirus (COVID-19) and produce updated predictions based on reported data. In the end, I would like to understand the Extended Kalman Filter in the second half of the tutorial, but first I want to solve any mystery. We provide a tutorial-like description of Kalman filter and extended Kalman filter. • The Kalman filter (KF) uses the observed data to learn about the unobservable state variables, which describe the state of the model. Kalman Filter T on y Lacey. Prediction. One of the typical techniques are filter-based methods which include kalman filter (KF), extended kalman filter (EKF) and unscented kalman filter. Since we have our posterior estimate for the state $\theta_t$, we can predict the next day's values by considering the mean value of the observation. … View. We put in relevance the nugget effect at the observation equation. We have two distinct set of equations : Time Update (prediction) and Measurement Update (correction). Model the state process We will outline several ways to model this simple situation, showing the power of a good Kalman ﬁlter model. derive the Kalman filter equations that allow us to recursively calculate xt t by combining prior knowledge, predictions from systems models, and noisy mea-surements. The principle of Kalman filtering can be roughly summarised as the weighted least square solution of the linearised observation system augmented with a prediction of the estimate as additional equations. I originally wrote this for a Society Of Robot article several years ago. Bayesian Optimal Filter: Prediction Step 16 •Now we have: 1. This chapter describes the Kalman Filter in one dimension. Prediction, estimation, and smoothing are fundamental to signal processing. Using Kalman Filter to Predict Corona Virus Spread (Feb 22) At every point in the time-series, a prediction is made of the next value based a few of the most recent estimates, and on the data-model contained in the Kalman filter equations. Kalman Filter Cycle: The filter equations can be divided in a prediction and a correction step. An adaptive online Kalman filter provides us very good one-day predictions for each region. Fitting time series analysis and statistical algorithms to produce the best short term and long term prediction. 12 STATE SPACE REPRESENTATION State equation: 2. Kalman filters operate on a predict/update cycle. We make a prediction of a state, based on some previous values and model. The predicted estimate and the weighted solution are given as … Chapter 1 Preface Introductory textbook for Kalman lters and Bayesian lters. now let’s consider the covariance xt+1 −x¯t+1 = A(xt −x¯t)+B(ut −u¯t) and so Σx(t+1) = E(A(xt −x¯t)+B(ut −u¯t))(A(xt −x¯t)+B(ut −u¯t)) T = AΣx(t)AT +BΣu(t)BT +AΣxu(t)BT +BΣux(t)AT where Σxu(t) = Σux(t)T = E(xt −x¯t)(ut −u¯t)T thus, the covariance Σx(t) satisﬁes another, Lyapunov-like linear dynamical system, driven by Σxu and Σu The Kalman ﬁlter 8–3 After that the correction step uses the incorporates a new measurement to get an improved a posteriori estimate. in a previous article, we have shown that Kalman filter can produce… In terms of a Kalman Filter, if your state observation system is observable and controllable, you don’t have to directly observe your state. I think we use constant for prediction error, because the new value in a certain k time moment can be different, than the previous. 15. Kalman, Rudolph E., and Richard S. Bucy. The green line represents the Kalman Filter estimate of the true position. 1 Department of Electronic Engineering, Technological Educational Institute of Central Greece, 35100 Lamia, Greece. Welch & Bishop, An Introduction to the Kalman Filter 2 UNC-Chapel Hill, TR 95-041, July 24, 2006 1 T he Discrete Kalman Filter In 1960, R.E. “New results in linear filtering and prediction theory.” (1961): 95-108. The method is now standard in many text books on control and machine learning. The operation of the dynamic prediction is achieved by Kalman filtering algorithm, and a general n-step-ahead prediction algorithm based on Kalman filter is derived for prospective prediction. Links to Medium article can be found here. Hi all Here is a quick tutorial for implementing a Kalman Filter. A Kalman filter tracks a time-series using a two-stage process: 1. The measurement update then adjusts this prediction based on the new measurement y v [n + 1]. 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Kane’s Method in Physics/Mechanics Kane’s Method in Physics/Mechanics¶ sympy.physics.mechanics provides functionality for deriving equations of motion using Kane’s method [Kane1985]. This document will describe Kane’s method as used in this module, but not how the equations are actually derived. Structure of Equations¶ In sympy.physics.mechanics we are assuming there are 5 basic sets of equations needed to describe a system. They are: holonomic constraints, non-holonomic constraints, kinematic differential equations, dynamic equations, and differentiated non-holonomic equations. \[\begin{split}\mathbf{f_h}(q, t) &= 0\\ \mathbf{k_{nh}}(q, t) u + \mathbf{f_{nh}}(q, t) &= 0\\ \mathbf{k_{k\dot{q}}}(q, t) \dot{q} + \mathbf{k_{ku}}(q, t) u + \mathbf{f_k}(q, t) &= 0\\ \mathbf{k_d} (q, t) \dot{u} + \mathbf{f_d}(q, \dot{q}, u, t) &= 0\\ \mathbf{k_{dnh}}(q, t) \dot{u} + \mathbf{f_{dnh}}(q, \dot{q}, u, t) &= 0\\\end{split}\] In sympy.physics.mechanics holonomic constraints are only used for the linearization process; it is assumed that they will be too complicated to solve for the dependent coordinate(s). If you are able to easily solve a holonomic constraint, you should consider redefining your problem in terms of a smaller set of coordinates. Alternatively, the time-differentiated holonomic constraints can be Kane’s method forms two expressions, $F_r$ and $F_r^$, whose sum is zero. In this module, these expressions are rearranged into the following form: $\mathbf{M}(q, t) \dot{u} = \mathbf{f}(q, \dot{q}, u, t)$ For a non-holonomic system with $o$ total speeds and $m$ motion constraints, we will get o - m equations. The mass-matrix/forcing equations are then augmented in the following fashion: \[\begin{split}\mathbf{M}(q, t) &= \begin{bmatrix} \mathbf{k_d}(q, t) \\ \mathbf{k_{dnh}}(q, t) \end{bmatrix}\\ \mathbf{_{(forcing)}}(q, \dot{q}, u, t) &= \begin{bmatrix} - \mathbf{f_d}(q, \dot{q}, u, t) \\ - \mathbf{f_{dnh}}(q, \dot{q}, u, t) \end{bmatrix}\\\end{split}\] Kane’s Method in Physics/Mechanics¶ The formulation of the equations of motion in sympy.physics.mechanics starts with creation of a KanesMethod object. Upon initialization of the KanesMethod object, an inertial reference frame needs to be supplied. along with some basic system information, such as coordinates and speeds >>> from sympy.physics.mechanics import >>> N = ReferenceFrame('N') >>> q1, q2, u1, u2 = dynamicsymbols('q1 q2 u1 u2') >>> q1d, q2d, u1d, u2d = dynamicsymbols('q1 q2 u1 u2', 1) >>> KM = KanesMethod(N, [q1, q2], [u1, u2]) It is also important to supply the order of coordinates and speeds properly if there are dependent coordinates and speeds. They must be supplied after independent coordinates and speeds or as a keyword argument; this is shown later. >>> q1, q2, q3, q4 = dynamicsymbols('q1 q2 q3 q4') >>> u1, u2, u3, u4 = dynamicsymbols('u1 u2 u3 u4') >>> # Here we will assume q2 is dependent, and u2 and u3 are dependent >>> # We need the constraint equations to enter them though >>> KM = KanesMethod(N, [q1, q3, q4], [u1, u4]) Additionally, if there are auxiliary speeds, they need to be identified here. See the examples for more information on this. In this example u4 is the auxiliary speed. >>> KM = KanesMethod(N, [q1, q3, q4], [u1, u2, u3], u_auxiliary=[u4]) Kinematic differential equations must also be supplied; there are to be provided as a list of expressions which are each equal to zero. A trivial example follows: >>> kd = [q1d - u1, q2d - u2] Turning on mechanics_printing() makes the expressions significantly shorter and is recommended. Alternatively, the mprint and mpprint commands can be used. If there are non-holonomic constraints, dependent speeds need to be specified (and so do dependent coordinates, but they only come into play when linearizing the system). The constraints need to be supplied in a list of expressions which are equal to zero, trivial motion and configuration constraints are shown below: >>> N = ReferenceFrame('N') >>> q1, q2, q3, q4 = dynamicsymbols('q1 q2 q3 q4') >>> q1d, q2d, q3d, q4d = dynamicsymbols('q1 q2 q3 q4', 1) >>> u1, u2, u3, u4 = dynamicsymbols('u1 u2 u3 u4') >>> #Here we will assume q2 is dependent, and u2 and u3 are dependent >>> speed_cons = [u2 - u1, u3 - u1 - u4] >>> coord_cons = [q2 - q1] >>> q_ind = [q1, q3, q4] >>> q_dep = [q2] >>> u_ind = [u1, u4] >>> u_dep = [u2, u3] >>> kd = [q1d - u1, q2d - u2, q3d - u3, q4d - u4] >>> KM = KanesMethod(N, q_ind, u_ind, kd, ... q_dependent=q_dep, ... configuration_constraints=coord_cons, ... u_dependent=u_dep, ... velocity_constraints=speed_cons) A dictionary returning the solved $\dot{q}$’s can also be solved for: >>> mechanics_printing(pretty_print=False) >>> KM.kindiffdict() {q1': u1, q2': u2, q3': u3, q4': u4} The final step in forming the equations of motion is supplying a list of bodies and particles, and a list of 2-tuples of the form (Point, Vector) or (ReferenceFrame, Vector) to represent applied forces and torques. >>> N = ReferenceFrame('N') >>> q, u = dynamicsymbols('q u') >>> qd, ud = dynamicsymbols('q u', 1) >>> P = Point('P') >>> P.set_vel(N, u * N.x) >>> Pa = Particle('Pa', P, 5) >>> BL = [Pa] >>> FL = [(P, 7 * N.x)] >>> KM = KanesMethod(N, [q], [u], [qd - u]) >>> (fr, frstar) = KM.kanes_equations(BL, FL) >>> KM.mass_matrix >>> KM.forcing When there are motion constraints, the mass matrix is augmented by the $k_{dnh}(q, t)$ matrix, and the forcing vector by the $f_{dnh}(q, \dot{q}, u, t)$ vector. There are also the “full” mass matrix and “full” forcing vector terms, these include the kinematic differential equations; the mass matrix is of size (n + o) x (n + o), or square and the size of all coordinates and speeds. >>> KM.mass_matrix_full [1, 0], [0, 5]]) >>> KM.forcing_full Exploration of the provided examples is encouraged in order to gain more understanding of the KanesMethod object.	{"url":"https://docs.sympy.org/dev/explanation/modules/physics/mechanics/kane.html","timestamp":"2024-11-11T00:19:18Z","content_type":"text/html","content_length":"103505","record_id":"<urn:uuid:55091472-9d74-42ff-ab98-a387712bf4c9>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00856.warc.gz"}
Energy-Efficiency Model for Residential Buildings Using Supervised Machine Learning Algorithm Intelligent Automation & Soft Computing Energy-Efficiency Model for Residential Buildings Using Supervised Machine Learning Algorithm 1School of Computer Science, National College of Business Administration and Economics, Lahore, 54000, Pakistan 2Center for Cyber Security, Faculty of Information Science and Technology, Universiti Kebansaan Malaysia (UKM), 43600, Bangi, Selangor, Malaysia 3School of Information Technology, Skyline University College, University City Sharjah, 1797, Sharjah, UAE 4Department of Computer Science, Lahore Garrison University, Lahore, 54000, Pakistan 5Canadian University Dubai, Dubai, UAE 6Riphah School of Computing & Innovation, Faculty of Computing, Riphah International University, Lahore, 54000, Pakistan 7Pattern Recognition and Machine Learning Lab, Department of Software Engineering, Gachon University, Seongnam, 13557, Korea 8School of Computer Science, Minhaj University Lahore, Lahore, 54000, Pakistan *Corresponding Author: Muhammad Adnan Khan, Email: adnan.khan@riphah.edu.pk Received: 17 February 2021; Accepted: 15 May 2021 Abstract: The real-time management and control of heating-system networks in residential buildings has tremendous energy-saving potential, and accurate load prediction is the basis for system monitoring. In this regard, selecting the appropriate input parameters is the key to accurate heating-load forecasting. In existing models for forecasting heating loads and selecting input parameters, with an increase in the length of the prediction cycle, the heating-load rate gradually decreases, and the influence of the outside temperature gradually increases. In view of different types of solutions for improving buildings’ energy efficiency, this study proposed a Energy-efficiency model for residential buildings based on gradient descent optimization (E2B-GDO). This model can predict a building’s heating-load conservation based on a building energy performance dataset. The input layer includes area (distribution of the glazing area, wall area, and surface area), relative density, and overall elevation. The proposed E2B-GDO model achieved an accuracy of 99.98% for training and 98.00% for validation. Keywords: Heating-load prediction; machine learning; gradient descent optimization Buildings need to be designed to support people’s health and well-being and to use the least amount of materials and resources. This applies to the construction of new buildings as well as the improvement of existing ones. In performance-driven building design, simulation is used to understand a building’s energy consumption [1]. This approach is known as Building performance simulation (BPS), and it allows a designer to examine in advance the influence of a building’s shape, materials, and systems on its expected thermal output. The search for an optimal design via simulation has conventionally had a small search space and a narrow region of possible choices. However, optimization, which can improve efficiency, can help to dramatically expand the search space during the design of a product [2]. The advantages of search optimization are realized when the optimization routine can assess thousands of possible options. However, large runs of realistic performance simulations require significant time and resources. Optimization can reduce the amount of specialist labor needed to scan broad additional spaces, but the resulting computational load can be daunting. Surrogate or data-driven models are based on mathematical inputs learned from measured or virtual data related to physical properties. For instance, the thermophysical properties of building materials and environmental-condition parameters can be used to predict indoor environmental conditions. Sufficiently accurate surrogate models, therefore, can provide fast, precise alternatives to performance simulators. It has been suggested that such models are superior to random forest, support vector machines, and decision trees for forecasting the hourly energy demand of buildings [3]. Buildings account for one-third of the world’s overall electricity consumption. Although energy efficiency standards are important instruments for improving buildings’ energy efficiency, the criteria can vary from country to country. The energy standard used in the US has saved more than $56 billion in electricity costs. Meanwhile, china requires new buildings to be 65% more energy efficient than they were in the 1980s. Energy efficiency standards in the EU tend to focus more on existing buildings. In India, the energy conservation building code for large-scale commercial buildings was adopted in 2007 [1,4]. Improvements to global energy efficiency have been slowing down since 2015. Energy Efficiency 2019, which tracks trends in global energy efficiency, has investigated the factors related to this recent slowdown while also emphasizing how digitization can improve energy efficiency [2]. In digitization, the input layer consists of a sphere, which can be described as a flat circle or the space that occupies an object’s surface. This area is measured in units per square centimeter, in square feet, and in square inches [3]. When installing glass in a structure, the area of the wall framing the glass is measured using the standard equation, which is height × length [4]. The surface-to-volume (S/V) ratio (the circumference’s three-dimensional magnitude-area ratio) is a crucial factor for determining heat loss/gain. The larger the surface area, the higher the heat. Low S/V ratios, therefore, reflect minimal heat gain and minimal heat loss [5]. Reducing compactness ratios for heating and cooling buildings can reduce costs and improve efficiency [6]. Most heating and cooling loads are worsened when the building’s thermal load is 25°C, its cooling load temperature is 30°C, and the outputs are below 16°C [7]. Energy included in the UN Environment Program includes climate construction, building envelopes, building services, and energy systems. The temperature created inside a building is estimated at 19°C, according to the researchers [8]. Buildings have large surface areas and high volume. The main thermal envelope area and its volume (ratio A/V) represent the ratio between a building’s compactness and its expression; the thermal envelope separates the indoor and outdoor surroundings. There are various thermal envelope characteristics under different climatic conditions, which have different effects on building management processes [9,10]. The US Environmental Protection Agency has described “low energy consumption as energy quality.” A study in France noted a clear association between energy consumption and size coefficients; if the relationship between building size and energy consumption is adjusted to climate change, its output generates very hot or very cold charges [11]. Development of energy consumptions varies according to temperature rather than the main effects of the glazed region, depending on the orientation of the building and annual energy demands. Various studies have investigated designing for different environments [12,13]. There are three types of relative compactness: window ratios, wall ratios, and glazing, as defined by the solar heat gain coefficient. Other factors, meanwhile, greatly affect the color of the facade of a building. Some structures can be made related to energy generation, the prevention of excessive heat, heat transfer, solar radiation absorption, and compactness [14–16]. Many studies have relied on mathematical models to calculate buildings’ heating efficiency. This approach is suitable for residential buildings but not commercial buildings because of inherent limitations [12,17]. Simulation techniques are mainly used for office buildings where the size of the rectangle L gives the building size at the total annual capacity. Just two building shapes have been studied in previous studies, but instead of heating the space to display the size of the house, its percentage is discussed, along with the relative compactness and glazing styles. The system model needs a low refrigeration load and low total building energy consumption [13,18]. This study investigated the key relationship for most apartments between convex ratio and building power efficiency and discusses the three major levels of convexity. The shape of the building has a direct effect on the building’s energy consumption. This study analyzes the impact of the curved shapes on the final energy demand by studying three different shapes. The differences in shape between the buildings was found to have an impact and on the final energy demand. Both amounts are less convex in the volumes of the curve, higher energy efficiency, and convexity of the direct mean [19,20]. Computational intelligence approaches, such as a fuzzy logic, artificial neural networks, and support vector machines, are robust approaches for future forecasting and prediction purposes [21–23]. 3 Proposed E2B-GDO System Model This study proposes a building energy-efficiency model based on gradient descent optimization (E2B-GDO) to forecast the energy performance of residential buildings. In this model, five layers are pushed toward the cloud if accuracy is achieved to conduct further processing for the validation phase (Fig. 1). The proposed model has 786 samples. Relative compactness, field, and overall height are inputs; heating and cooling loads are outputs. A cloud was used to store data, whether for testing or training. The model collects data from the cloud for the validation process. Trained data or inputs are fed into the cloud, which determines the evaluation system for testing purposes. Data are fed into the cloud and forwarded to the preprocessor layer, which will boost errors and missing values. Forwarding to the final iteration is the last step. The sensory layer is taken as an input layer providing inputs during the training process. The object layer selects the desired data from multiple data and selects the accurate data needed for the missing values in the preprocessing layer and normalizes the result to get an accurate raw data result. The prediction layer used Gradient descent optimization (GDO) to predict the outcome. Consistency is tested in the consistency layer regarding whether its accuracy is adequate, and the miss rate is verified. If the accuracy meets the requirement, it is moved to the cloud, and the process is repeated until all errors have been removed. Once errors are removed, the process moves to the validation phase, where it moves from the sensory layer to preprocessing. If the result is accurate, the final model is obtained; if not, it moves back to the sensory layer and repeats the process. In the proposed model’s input layer, the hidden layer and output layer are used for computation and optimization through the backpropagation algorithm. Different steps are involved in the backpropagation algorithm, such as weight initialization, feedforward, error backpropagation, and weight and bias updating. Each neuron is given the activation function in a hidden layer, such as S (y) = alpha(y). The alpha function was used in the input layer shown in Eq. (2). The sigmoid activation function used in the output layer is given in Eq. (2). We know the equation of the line is αj=a1+∑i=1m⁡(mijxbi), (1) θj=11+exp−αj, (2) where j = 1, 2, 3……n. The second input taken for the output layer is shown in Eq. (3): αk=b2+∑j=1n⁡(njkxθj). (3) The output is shown in Eq. (4): θk=11+exp−αk, (4) where k = 1, 2, 3…r. The error involved in backpropagation is shown in Eq. (5): E=12∑k⁡(Sk−θk), (5) where Sk is the desired output, and θk is an output. The change of weights in the output is shown in Eq. (6): Δwx−δEδw , Δnjk=−εδEδnJk. (6) Apply the chain rule method in Eq. (6): Δnjk=−εδEδθk×δθKδαK×δαkδnjk. (7) After obtaining the value from Eq. (7), the weights are as shown in Eq. (8): Δnjk=ε(sk−θk)×θk(1−θk)×(θj) , Δnjk=εζkθj , ε[∑k⁡(njk)]×θj(1−θj)×yiε[∑k⁡(njk)]×θj(1−θj)×yi, (8) ζk = ( sk−θk ) × θk (1- θk ), Δwijα−[∑kδEδθ×δαk¯xδθk×δαkδθj]×δθjδαj×δαjδwij , Δwij=−ε[∑kδEδθ×δαk¯xδθk×δαkδθj]×δθjδαj×δαjδwij , Δwij=ε[∑k⁡(sk−θk)×θk(1−θk)×(njk)]×θk(1−θk)×yi , Δwij=ε[∑k⁡(sk−θk)×θk(1−θk)×(njk)]×θj(1−θj)×yi , Δwij=ε[∑k⁡ζk(njk)]x×(1−θj)×yi , Δwij=εζjyi, (9) ζj=[∑k⁡ζk(njk)]×θj(1−θj) . Updating the weight and bias between the output and hidden layer is as shown in Eq. (10): njk+=njk+πFΔnjk. (10) At the input layer and hidden layer, updating the weights and bias are as shown in Eq. (11): wij+=wij+πFΔwij. (11) πF is taken as the learning rate of the E2B-GDO model. The convergence of the proposed E2B-GDO model depends on the selection of πF . 3.2 Proposed E2B-GDO System Model Using Fitness Modeling Two layers of a feed-forward network are used in the GDO fitting application. A fitting network is used to train the selected data. Then, those data are divided into training, validation, and testing sets to define the architecture of the network. The Levenberg–Marquardt algorithm was used to train and fit 786 sets of data, randomly divided into 70% for training (538 samples), 15% for validation (115 samples), and 15% for testing (115 samples). 3.3 Proposed E2B-GDO Using Time-Series Modeling The time-series standard was used to solve the three types of nonlinear problems using the dynamic network in GDO. The GDO time series was trained by first selecting and then dividing the data into training, validation, and testing sets to define the architecture of the network. MATLAB was used to train and fit 786 sets of data using a Nonlinear autoregressive exogenous model (NARX). Three layers were used in the time-series model, with a sensory layer as the input, a hidden layer as the prediction layer, and a performance layer that shows the output if the result is accurate. In the case of errors, the process is repeated. MATLAB was used for the clone of the result. Tab. 1 shows the accuracy, missing rate in the training and the validation phase. In Tab. 1, along with the comparison of previous approaches, performance is checked using a fitting model and a time series [14]. Tab. 1 shows a comparison of the accuracy and missing rate in training and validation for two kinds of approaches. Tab. 2 shows the evaluation and prediction of residential buildings’ energy performance using the proposed E2B-GDO model and compares the results with those of other approaches, such as random forest and decision trees. The comparison is based on the factors of heating-load prediction based on building energy performance; the table also shows statistical measures such as accuracy, missing rate, and mean square error. The proposed E2B-GDO model was found to perform better than the other methods with regard to accuracy and MSE. This study investigated buildings’ energy consumption and heat production using gradient descent optimization to predict heating and cooling load. The proposed Energy-efficiency model for residential buildings based on gradient descent optimization (E2B-GDO) can predict a building’s heating-load conservation based on a building energy performance dataset. The proposed E2B-GDO model achieved accuracies of 99.98% in training and 98.00 in validation. Area, relative compactness, and overall height have a significant effect on heating and cooling load. This study’s method can be applied to designing buildings to optimize energy performance for any given input variable based on either experimental or simulated data. Acknowledgement: We thank our families and colleagues for their moral support. Funding Statement: The author(s) received no specific funding for this study. Conflicts of Interest: The authors declare that they have no conflicts of interest to report regarding the present study. 1. L. Pérez-Lombard, J. Ortiz and C. Pout, “A review on buildings energy consumption information,” Energy and Buildings, vol. 40, no. 3, pp. 394–398, 2008. [Google Scholar] 2. C. Huang, A. Zappone, G. C. Alexandropoulos, M. Debbah and C. Yuen, “Reconfigurable intelligent surfaces for energy efficiency in wireless communication,” IEEE Transactions on Wireless Communications, vol. 18, no. 8, pp. 4157–4170, 2019. [Google Scholar] 3. N. Dudley, Guidelines for applying protected area management categories. Iucn, 2008. [Google Scholar] 4. A. Ghaffarian, U. Berardi, A. Hoseini and N. Makaremi, “Intelligent facades in low-energy buildings,” British Journal of Environment and Climate Change, vol. 2, no. 4, pp. 437–448, 2012. [Google 5. I. Danielski, M. Fröling and A. Joelsson, “The impact of the shape factor on final energy demand in residential buildings in nordic climates,” World Renewable Energy Conference, vol. 3, no. 1, pp. 4260–4264, 2012. [Google Scholar] 6. A. A. Anzi, D. Seo and M. Krarti, “Impact of building shape on thermal performance of office buildings in Kuwait,” Energy Conversion and Management, vol. 50, no. 3, pp. 822–828, 2009. [Google 7. S. Abbas, M. A. Khan, L. E. F. Morales, A. Rehman, Y. Saeed et al., “Modeling, simulation and optimization of power plant energy sustainability for iot enabled smart cities empowered with deep extreme learning machine,” IEEE Access, vol. 8, no. 1, pp. 39982–39997, 2020. [Google Scholar] 8. A. J. Khalil, A. M. Barhoomm, B. S. A. Nasser, M. M. Musleh and S. S. A. Naser, “Energy efficiency predicting using artificial neural network,” International Journal of Academic Pedagogical Research (IJAPR), vol. 9, no. 3, pp. 1–7, 2019. [Google Scholar] 9. A. Tsanas and A. Xifara, “Accurate quantitative estimation of energy performance of residential buildings using statistical machine learning tools,” Energy and Buildings, vol. 49, pp. 560–567, 2012. [Google Scholar] 10. Z. Yu, F. Haghighat, B. C. Fung and H. Yoshino, “A decision tree method for building energy demand modeling,” Energy and Buildings, vol. 42, no. 10, pp. 1637–1646, 2010. [Google Scholar] 11. T. Nigitzand and M. Golles, “A generally applicable, simple and adaptive forecasting method for the short-term heat load of consumers,” Applied Energy, vol. 241, pp. 73–81, 2019. [Google Scholar] 12. X. J. Luo, L. O. Oyedele, A. O. Ajayi, C. G. Monyei, O. O. Akinade et al., “Development of an iot-based big data platform for day-ahead prediction of building heating and cooling demands,” Advanced Engineering Informatics, vol. 4, no. 2, pp. 23–36, 2019. [Google Scholar] 13. S. Seyedzadeh, F. Pour Rahimian and P. Rastogi, “Tuning machine learning models for prediction of building energy loads,” Sustainable Cities and Society, vol. 20, no. 4, pp. 1–14, 2019. [Google 14. L. Yakai, T. Zhe and P. Peng, “Gmm clustering for heating load patterns in-depth identification and prediction model accuracy improvement of district heating system,” Energy and Buildings, vol. 190, no. 4, pp. 49–60, 2019. [Google Scholar] 15. R. Niemierko, Töppel, Jannick and Tränkler and Timm, “A divine copula quantile regression approach for the prediction of residential heating energy consumption based on historical data,” Applied Energy, vol. 10, no. 14, pp. 233–234, 2019. [Google Scholar] 16. D. S. Kapetanakis, E. Mangina and D. P. Finn, “Input variable selection for thermal load predictive models of commercial buildings,” Energy and Buildings, vol. 137, no. 17, pp. 13–26, 2017. [ Google Scholar] 17. S. S. Roy, R. Roy and V. E. Balas, “Estimating heating load in buildings using multivariate adaptive regression splines, extreme learning machine, a hybrid model of MARS and ELM,” Renewable & Sustainable Energy Reviews, vol. 17, no. 3, pp. 52–68, 2017. [Google Scholar] 18. P. Potočnik, E. Strmčnik and E. Govekar, “Linear and neural network-based models for short-term heat load forecasting,” Strojniski Vestnik, vol. 9, no. 61, pp. 543–550, 2015. [Google Scholar] 19. K. M. Powell, A. Sriprasad, W. J. Cole and T. F. Edgar, “Heating, cooling, and electrical load forecasting for a large-scale district energy system,” Energy, vol. 13, no. 74, pp. 877–885, 2014. [ Google Scholar] 20. I. Korolija, Y. Zhang, L. M. M. Halburd and V. I. Hanby, “Regression models for predicting UK office building energy consumption from heating and cooling demands,” Energy & Building, vol. 8, no. 59, pp. 214–227, 2013. [Google Scholar] 21. S. Y. Siddiqui, A. Athar, M. A. Khan, S. Abbas, Y. Saeed et al., “Modelling, simulation and optimization of diagnosis cardiovascular disease using computational intelligence approaches,” Journal of Medical Imaging and Health Informatics, vol. 10, no. 5, pp. 1005–1022, 2020. [Google Scholar] 22. F. Khan, M. A. Khan, S. Abbas, A. Athar, S. Y. Siddiqui et al., “Cloud-based breast cancer prediction empowered with soft computing approaches,” Journal of Healthcare Engineering, vol. 2020, no. 1, pp. 1–16, 2020. [Google Scholar] 23. S. Y. Siddiqui, I. Naseer, M. A. Khan, M. F. Mushtaq, R. A. Naqvi et al., “Intelligent breast cancer prediction empowered with fusion and deep learning,” Computers, Materials & Continua, vol. 67, no. 1, pp. 1033–1049, 2021. [Google Scholar] This work is licensed under a Creative Commons Attribution 4.0 International License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.	{"url":"https://www.techscience.com/iasc/v30n3/44089/html","timestamp":"2024-11-07T23:49:32Z","content_type":"application/xhtml+xml","content_length":"89828","record_id":"<urn:uuid:5540d719-c508-4c92-8280-9942eba59c52>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00410.warc.gz"}
Generating a Bitcoin Private Key In this article, we learn how to generate a bitcoin private key, that is later used to generate a bitcoin address. Table of contents. 1. Introduction. 2. Private keys. 3. Generating a Private Key. 4. Summary. 5. References. Cryptography, more specifically public-key cryptography is at the heart of blockchain technology. Hashing is also used to secure transactions and prevent blockchain mutability. In public-key cryptography also known as asymmetric cryptography, we generate two key pairs, a private key which we keep private and a public key which we share, this key will be used by the recipient together with his/her private key to decrypt the encrypted message. In cryptography, a cryptographic key is randomized data used to scramble data so that it looks random. Plaintext data goes through an encryption algorithm whose result is randomized data(encrypted). Unlike symmetric cryptography such s ceases cipher where a single key is used to encrypt and decrypt information, asymmetric ensure all challenges encounter using the former such as sharing of keys, and reversal of the encrypted message was solved. Asymmetric cryptography is considered a trapdoor function, whereby it is easy to compute in one direction but impossible to compute in the opposite direction without some key information, in our case this key piece of information is the private key. In this article, we will learn how to generate a bitcoin private key, we will later use this generated private key to generate a bitcoin address. Usually, private keys are generated by third parties, for example, crypto exchanges and other bitcoin wallets. Private keys. This is just a variable used by an encryption algorithm to encrypt and decrypt data. In Bitcoin specifically, it is a 32-byte series of characters that can be converted into any format such as binary, hexadecimal, base64, etc. As we learned in previous articles, Bitcoin uses the ECDSA algorithm which uses a specific elliptic curve referred to as secp256k1. In this case, we generate a 32-byte key to satisfy the curve parameters, that is, since the curve is in the order of 256 bits, it takes 256 bits as input and returns 256-bit integers. 256 bits is equivalent to 32-bytes(32 * 8 = 256). The secp256k1 curve also has a specific rule about the size of the key, it should be less than the curve order, also the key should be a positive number. Generating a private key. As mentioned, a private key is 32-bytes and it is randomized. To generate a randomized 32-byte string using python, we will use the python random library. import random bits = random.getrandbits(256) key_hex = hex(bits) private_key = key_hex[2:] The output for the above code is as follows; Each time we run this code we will generate a new randomized private key. This is good, however, the python random generator library was not built for cryptography. The private key generated in this way uses the current time as the seed meaning that if a malicious user knows the exact time we generated this key then a brute force algorithm would crack this key in a very short time. Remember once coins on the blockchain are stolen, there is no way of recovering them so we should be very careful. Now let's use a stronger random number generator library referred to as secrets, it is a python module used for generating stronger randomized numbers which we can then use to generate a private key and it was specifically designed for cryptographic operations. For this we write; import secrets bits = secrets.randbits(256) key_hex = hex(bits) private_key = key_hex[2:] We have the following output; An entropy is randomness generated by an operating system that is used in algorithms that need randomized data. Using the above way to generate a private key is much more secure than the previous since it gets its entropy from the operating system meaning a malicious user would have a hard time finding such since he/she is on a different machine and even with access to the original machine, the task is still very difficult. In this case, the seed is created by the algorithm itself. We can also use the python PyCryptodome library which implements RSA, the most used algorithm for public-key cryptography. To generate a private key we write; from Crypto.PublicKey import RSA key = RSA.generate(2048) private_key = key.exportKey("PEM") The output is shown below; Above we use a 2048 key length which is the recommended key length for utmost security. This is also very difficult to crack because of some of its features such as the use of symmetric ciphers, authenticated encryption, and strong one-way hashing functions. Randomization is very essential to cryptography, especially public-key cryptography. We have used three python libraries one superior to its predecessor for cryptographic work, however, we can also use websites such as random.org and bitaddress.org, these and many more to generate public and private keys. Each has its own way of implementing randomness. The latter is preferred since the code can be downloaded and run locally, this means that no one can know your private key except you. Private keys should be kept hidden, this means that the owner should be the only one in possession of the key. However, in the case of third parties who generate the private key for its users, it leaves users vulnerable since they are not the only ones with the knowledge of the private key. With this article at OpenGenus, you must have the complete idea of how to generate a Bitcoin Private Key.	{"url":"https://iq.opengenus.org/generating-bitcoin-private-key/","timestamp":"2024-11-14T12:25:25Z","content_type":"text/html","content_length":"42718","record_id":"<urn:uuid:a9ec8a4b-cb81-41ad-b9b8-bbaa38aaa51f>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00377.warc.gz"}
Valid Triangle Number - DSA and Algorithm - Javascript Problem and Solution Given an array consists of non-negative integers, your task is to count the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle. Example 1: Input: [2,2,3,4] Output: 3 Explanation: Valid combinations are: 2,3,4 (using the first 2) 2,3,4 (using the second 2) Example 2: Input: nums = [4,2,3,4] Output: 4 The length of the given array won't exceed 1000. The integers in the given array are in the range of [0, 1000]. Let's start with the algorithm, const triangleNumber = nums => { // Count of triangles let count = 0; // The three loops select three // different values from array for (let i = 0; i < nums.length; i++) for (let j = i + 1; j < nums.length; j++) // The innermost loop checks for // the triangle property for (let k = j + 1; k < nums.length; k++) // Sum of two sides is greater // than the third if (nums[i] + nums[j] > nums[k] && nums[i] + nums[k] > nums[j] && nums[k] + nums[j] > nums[i]) return count; Here we are using the brute force method, to run three loops and keep track of the number of triangles possible so far. The three loops select three different values from an array. The innermost loop checks for the triangle property which specifies the sum of any two sides must be greater than the value of the third side). Given steps to solve the problem: • Run three nested loops each loop starting from the index of the previous loop to the end of the array i.e run first loop from 0 to n, loop j from i to n, and k from j to n • Check if array[i] + array[j] > array[k], i.e. sum of two sides is greater than the third • Check condition 2 for all combinations of sides by interchanging i, j, k • If all three conditions match, then increase the count • Print the count I hope you understood the algorithm, if you have any doubts let us know in the comments. Post a Comment	{"url":"https://www.thecodingdev.com/2023/01/valid-triangle-number-dsa-and-algorithm.html","timestamp":"2024-11-09T15:37:42Z","content_type":"application/xhtml+xml","content_length":"373179","record_id":"<urn:uuid:e0614c3c-6c81-44a9-9c95-0d8233f6245d>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00332.warc.gz"}
Plant Breeding Methods (CS, HS 541) Exercise 3 – Quantitative Genetics Review Give complete answers, but be brief and specific. Show your work. 20 points total 1. What is the mean and variance for a population (where only 1 locus is segregating) consisting of the F[2] of the cross BB x bb where BB yields 24, Bb yields 18 and bb yields 9 fruits per plot? 2. How could you measure directly all components of the epistatic variance (s^2[I]) of a population? 3. A population with a mean yield of 19 fruits per plant was intercrossed, and 5 S[0] plants were sampled from it with yields of 28, 21, 29, 16 and 19 fruits per plant each, respectively. The half-sib families derived from the 5 S[0] plants had yields of 25, 21, 24, 15, and 20 fruits per plant, respectively. What is the breeding value (A) of the highest yielding S[0]? 4. A quantitative genetic study was run for lodging resistance in a population of sweet corn that had been wind-intercrossed several times during development. Given the following analysis of variance, estimate s^2A, s^2D, and h^2N. Does the population have a high enough heritability to warrant single-plant selection for the trait? • 2 df Sets (S) • 9 df Blocks / S • 297 df Males / S MS=205 • 1200 df Females/M/S MS=45 • 4491 df Error MS=5 5. If the population in question 4 had a mean of 7 before selection, what would it be after 2 cycles of selection where the best 5% of the FS families in the population were intercrossed to form the next generation (assuming the variances are not changed by selection)? Given – k for selection percentage: 1%=2.64, 2%=2.42, 5%=2.06, 10%=1.76, 20%=1.40, 50%=0.80.	{"url":"https://cucurbitbreeding.wordpress.ncsu.edu/courses-training-opportunities-and-admissions/plant-breeding-methods-cs-hs-541/course-schedule/exercise-3/","timestamp":"2024-11-02T23:24:55Z","content_type":"text/html","content_length":"28180","record_id":"<urn:uuid:0f2ceb00-6412-485a-8627-b4010900d661>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00160.warc.gz"}
Stress Strain Equations Formulas Calculator Young's Modulus - Stress Online Web Apps, Rich Internet Application, Technical Tools, Specifications, How to Guides, Training, Applications, Examples, Tutorials, Reviews, Answers, Test Review Resources, Analysis, Homework Solutions, Worksheets, Help, Data and Information for Engineers, Technicians, Teachers, Tutors, Researchers, K-12 Education, College and High School Students, Science Fair Projects and Scientists By Jimmy Raymond Contact: aj@ajdesigner.com Privacy Policy, Disclaimer and Terms Copyright 2002-2015	{"url":"https://www.ajdesigner.com/phpstress/stress_strain_equation_young_modulus_stress.php","timestamp":"2024-11-12T23:59:10Z","content_type":"text/html","content_length":"25899","record_id":"<urn:uuid:91ad188b-cbc1-42af-ba21-51fe221013ae>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00505.warc.gz"}
Continuum Thermodynamics (SISSA Springer Series) 1st ed. 2019 Edition by Paolo Podio-Guidugli (Author) This book deals with an important topic in rational continuum physics, thermodynamics.Although slim, it is fairly well self-contained; some basic notions in continuum mechanics, which a well-intentioned reader should but may not be familiar with, are collected in a final appendix. Modern continuum thermodynamics is a field theory devised to handle a large class of processes that typically are neither spatially homogeneous nor sequences of equilibrium states. The most basic chapter addresses the continuum theory of heat conduction, in which the constitutive laws furnish a mathematical characterization of the macroscopic manifestations of those fluctuations in position and velocity of the microscopic matter constituents that statistical thermodynamics considers collectively. In addition to a nonstandard exposition of the conceptual steps leading to the classical heat equation, the crucial assumption that energy and entropy inflows should be proportional is discussed and a hyperbolic version of that prototypical parabolic PDE is presented. Thermomechanics comes next, a slightly more complex paradigmatic example of a field theory where microscopic and macroscopic manifestations of motion become intertwined. Finally, a virtual power format for thermomechanics is proposed, whose formulation requires that temperature is regarded formally as the time derivative of thermal displacement. It is shown that this format permits an alternative formulation of the theory of heat conduction, and a physical interpretation of the notion of thermal displacement is given. It is addressed to mathematical modelers – or mathematical modelers to be – of continuous material bodies, be they mathematicians, physicists, or mathematically versed engineers.	{"url":"https://ketabdownload.com/continuum-thermodynamics","timestamp":"2024-11-03T10:47:11Z","content_type":"text/html","content_length":"39959","record_id":"<urn:uuid:b16b49e9-0a75-4f6f-8d0b-c326a40a027b>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00787.warc.gz"}
CS代考计算机代写 algorithm THE SIGN-RANK OF AC0 ∗ THE SIGN-RANK OF AC0 ∗ ALEXANDER A. RAZBOROV† AND ALEXANDER A. SHERSTOV‡ Abstract. The sign-rank of a matrix A = [Aij] with ±1 entries is the least rank of a real matrix B = [Bij] with AijBij > 0 for all i,j. We obtain the first exponential lower bound on the sign-rank of a function in AC0. Namely, let f(x,y) = 􏰓i=1,…,m 􏰔j=1,…,m2(xij ∧ yij). We show that the matrix [f(x, y)]x,y has sign-rank exp(Ω(m)). This in particular implies that Σcc ̸⊆ UPPcc, 2 which solves a longstanding open problem in communication complexity posed by Babai, Frankl, and Simon (1986). Our result additionally implies a lower bound in learning theory. Specifically, let φ1, . . . , φr : {0, 1}n → R be functions such that every DNF formula f : {0, 1}n → {−1, +1} of polynomial size has the representation f ≡ sgn(a1φ1 + ··· + arφr) for some reals a1,…,ar. We prove that then r 􏰟 exp(Ω(n1/3)), which essentially matches an upper bound of exp(O ̃(n1/3)) due to Klivans and Servedio Finally, our work yields the first exponential lower bound on the size of threshold-of-majority circuits computing a function in AC0. This substantially generalizes and strengthens the results of Krause and Pudl ́ak (1997). Key words. Sign-rank; communication complexity; complexity classes Σcc , Πcc , and UPPcc ; 22 constant-depth AND/OR/NOT circuits. AMS subject classifications. 03D15, 68Q15, 68Q17 1. Introduction. The sign-rank of a real matrix A = [Aij] with nonzero entries is the least rank of a matrix B = [Bij] with AijBij > 0 for all i,j. In other words, sign-rank measures the stability of the rank of A as its entries undergo arbitrary sign-preserving perturbations. This fundamental notion has been studied in contexts as diverse as matrix analysis, communication complexity, circuit complexity, and learning theory [40, 2, 4, 13, 14, 26, 32, 48, 51]. We will give a detailed overview of these applications shortly as they pertain to our work. Despite its importance, progress in understanding sign-rank has been slow and difficult. Indeed, we are aware of only a few nontrivial results on this subject. Alon et al. [2] obtained strong lower bounds on the sign-rank of random matrices. No nontrivial results were available for any explicit matrices until the breakthrough work of Forster [13], who proved strong lower bounds on the sign-rank of Hadamard matri- ces and, more generally, all sign matrices with small spectral norm. Several extensions and refinements of Forster’s method were proposed in subsequent work [14, 15, 32]. Near-tight estimates of the sign-rank were obtained in [51] for all symmetric prob- lems,i.e.,matricesoftheform[D(􏰂xiyi)]x,y whereD:{0,1,…,n}→{−1,+1}is a given predicate and x, y range over {0, 1}n . This paper focuses on AC0, a prominent class whose sign-rank has seen no progress in previous work. It will henceforth be convenient to view Boolean functions as mappings of the form {0, 1}n → {−1, +1}, where the elements −1 and +1 of the range represent “true” and “false,” respectively. The central objective of our study is to estimate the maximum sign-rank of a matrix [f(x,y)]x,y, where f : {0,1}n×{0,1}n → ∗An extended abstract of this article appeared in Proceedings of the 49th IEEE Symposium on Foundations of Computer Science (FOCS), pages 57–66, 2008. †Department of Computer Science, University of Chicago and Steklov Mathematical Institute (razborov@cs.uchicago.edu). Supported by NSF grant ITR-0324906 and by the Russian Foundation for Basic ‡Department of Computer Science, The University of Texas at Austin (sherstov@cs.utexas.edu). 2 A.A. RAZBOROV, A.A. SHERSTOV {−1,+1} is a function in AC0. An obvious upper bound is 2n, while the best lower bound prior to this paper was quasipolynomial. (The quasipolynomial lower bound is immediate from Forster’s work [13] and the fact that AC0 can compute inner product modulo 2 on logc n variables, for every constant c > 1.) Our main result considerably tightens the gap by improving the lower bound to 2Ω(n1/3). 􏰓m 􏰔m2 Theorem 1.1 (Main result). Let fm(x,y) = i=1 j=1(xij ∧ yij). Then the matrix [fm(x, y)]x,y has sign-rank 2Ω(m). It is not difficult to show that the matrix in Theorem 1.1 has sign-rank 2O(m log m), i.e., the lower bound that we prove is almost tight. (See Remark 6.1 for details.) More- over, Theorem 1.1 is optimal with respect to circuit depth: AC0 circuits of depth 1 and 2 lead to at most polynomial sign-rank. Indeed, the function 􏰔mi=1(xi ∧yi) which is universal in this context possesses the matrix representation [sgn(1/2 − 􏰂i xiyi)]x,y and thus has sign-rank at most m + 1. Our main result states that AC0 contains matrices whose rank is rather stable in that it cannot be reduced below 2Θ(n1/3) by any sign-preserving changes to the matrix entries. We proceed to discuss applications of this fact to communication complexity, learning theory, and circuits. 1.1. Communication complexity. The study of sign-rank is synonymous with the study of unbounded-error communication complexity, a rich model introduced by Paturi and Simon [40]. Fix a function f : X × Y → {−1, +1}, where X and Y are some finite sets. Alice receives an input x ∈ X, Bob receives y ∈ Y, and their objective is to compute f(x,y) with minimal communication. The two parties each have an unlimited private source of random bits which they can use in deciding what messages to send. Their protocol is said to compute f if on every input (x,y), the output is correct with probability greater than 1/2. The cost of a protocol is the worst-case number of bits exchanged on any input (x,y). The unbounded-error communication complexity of f, denoted U(f), is the least cost of a protocol that computes f. The unbounded-error model occupies a special place in the study of communica- tion because it is more powerful than almost any other standard model (deterministic, nondeterministic, randomized, quantum with or without entanglement). More pre- cisely, the unbounded-error complexity U(f) can be only negligibly greater than the complexity of f in any of these models—and often, U(f) is exponentially smaller. We defer exact quantitative statements to Appendix A. The power of the unbounded- error model resides in its very liberal acceptance criterion: it suffices to produce the correct output with probability even slightly greater than 1/2 (say, by an exponen- tially small amount). This contrasts with the more familiar, bounded-error models, where the correct output is expected with probability at least 2/3. Another compelling aspect of the unbounded-error model is that it has an exact matrix-analytic formulation. Let f : X × Y → {−1, +1} be a given function and M = [f (x, y)]x∈X, y∈Y its communication matrix. Paturi and Simon [40] showed that U(f) = log2(sign-rank(M)) ± O(1). In other words, unbounded-error complexity and sign-rank are essentially equivalent notions. In this light, our main result gives the first polynomial lower bound on the unbounded-error complexity of Corollary 1.2 (Unbounded-error communication complexity of AC0). Let 􏰓m 􏰔m2 fm(x,y)= i=1 j=1(xij ∧yij). Then U(fm)=Ω(m). THE SIGN-RANK OF AC0 3 Corollary 1.2 solves a longstanding problem in communication complexity. Specif- ically, the only models for which efficient simulations in the unbounded-error model were unknown had been developed in the seminal paper by Babai, Frankl, and Si- mon [3]. These models are the communication analogues of the classes PH and PSPACE. Babai et al. asked [3, p. 345] whether Σcc ⊆ UPPcc. Forster [13] made sub- 2 stantial progress on this question, proving that PSPACEcc ̸⊆ UPPcc. We resolve the original question completely: Corollary 1.2 implies that Πcc ̸⊆ UPPcc and hence (since UPPcc is closed under complementation) Σcc ̸⊆ UPPcc. See Section 7 for detailed back- 2 ground on these various complexity classes. 1.2. Learning theory. In a seminal paper [55], Valiant formulated the probably approximately correct (PAC) model of learning, now a central model in computational learning theory. Research has shown that PAC learning is surprisingly difficult. (By “PAC learning,” we shall always mean PAC learning under arbitrary distributions.) Indeed, the learning problem remains unsolved for such natural concept classes as DNF formulas of polynomial size and intersections of two halfspaces, whereas hardness results and lower bounds are abundant [22, 24, 28, 12, 29, 27]. One concept class for which efficient PAC learning algorithms are available is the class of halfspaces, i.e., functions f : Rn → {−1, +1} representable as f(x)≡sgn(a1x1 +···+anxn −θ) for some reals a1,…,an,θ. Halfspaces constitute one of the most studied classes in computational learning theory [46, 37, 34, 5] and a major success story of the field. Indeed, a significant part of computational learning theory attempts to learn rich concept classes by reducing them to halfspaces. The reduction works as follows. Let C be a given concept class, i.e., a set of Boolean functions {0, 1}n → {−1, +1}. One seeks functions φ1, . . . , φr : {0, 1}n → R such that every f ∈ C has a representation f (x) ≡ sgn(a1 φ1 (x) + · · · + ar φr (x)) for some reals a1, . . . , ar. This process is technically described as embedding C in half- spaces of dimension r. Once this is accomplished, C can be learned in time polynomial in n and r by any halfspace-learning algorithm. For this approach to be practical, the number r of real functions needs to be reasonable (ideally, polynomial in n). It is therefore of interest to determine what natural concept classes can be embedded in halfspaces of low dimension [4, 27]. For brevity, we refer to the smallest dimension of such a representation as the dimension complexity of a given class. Formally, the dimension complexity dc(C) of a given class C of functions {0, 1}n → {−1, +1} is the least r for which there exist real functions φ1,…,φr : {0,1}n → R such that every f ∈ C is expressible as f (x) ≡ sgn(a1 (f )φ1 (x) + · · · + ar (f )φr (x)) (1.1) for some reals a1(f), . . . , ar(f) depending on f only. To relate this discussion to sign-rank, let MC = [f(x)]f∈C, x∈{0,1}n be the characteristic matrix of C. A moment’s reflection reveals that the identity (1.1) is yet another way of saying that MC has the same sign pattern as the matrix [􏰂ri=1 ai(f)φi(x)]f,x of rank at most r, whence the dimension complexity of a concept class is precisely the sign-rank of its characteristic matrix. Indeed, the term “dimension complexity” has been used interchangeably with sign-rank in the recent literature [53, 48], which does not lead to confusion since concept classes are naturally identified with their characteristic matrices. 4 A.A. RAZBOROV, A.A. SHERSTOV Thus, the study of sign-rank yields nontrivial PAC learning algorithms. In par- ticular, the current fastest algorithm for learning polynomial-size DNF formulas, due to Klivans and Servedio [26], was obtained precisely by placing an upper bound of 2O ̃(n1/3) on the dimension complexity of that concept class, with the functions φi corresponding to the monomials of degree up to O ̃(n1/3). Klivans and Servedio also observed that their 2O ̃(n1/3) upper bound is best possible when the functions φi are taken to be the monomials up to a given degree. Our work gives a far-reaching generalization of the latter observation: we prove the same lower bound without assuming anything whatsoever about the embedding functions φi. That is, we have: Corollary 1.3 (Dimension complexity of DNF). Let C be the set of all read-once (hence, linear-size) DNF formulas f : {0,1}n → {−1,+1}. Then C has dimension complexity 2Ω(n1/3). Proof. Let fm(x, y) be the function from Theorem 1.1, where m = ⌊n1/3⌋. Then for any fixed y, the function fy(x) = ¬fm(x,y) is a read-once DNF formula. Learning polynomial-size DNF formulas was the original challenge posed in Valiant’s paper [55]. More than twenty years later, this challenge remains a central open problem in computational learning theory despite active research, e.g., [7, 54, 26]. To account for this lack of progress, several hardness results have been obtained based on complexity-theoretic assumptions [24, 1]. Corollary 1.3 complements that line of work by exhibiting an unconditional, structural barrier to the efficient learning of DNF formulas. In particular, it rules out a 2o(n1/3)-time learning algorithm based on dimension complexity. While restricted, the dimension-complexity paradigm is quite rich and captures many PAC learning algorithms designed to date, with the notable exception [18, 6] of learning low-degree polynomials over GF(p). Furthermore, it is known [23, p. 124] that an unconditional superpolynomial lower bound for learning polynomial-size DNF formulas in the standard PAC model would imply that P ̸= NP; thus, such a result is well beyond the reach of the current techniques. The lower bound in this work also points to the importance of generalizing the dimension complexity framework while maintaining algorithmic efficiency. Unfortu- nately, natural attempts at such a generalization do not work, including the idea of sign-representations that err on a small fraction of the inputs [12]. 1.3. Threshold circuits. Recall that a threshold gate g with Boolean inputs x1,…,xn is a function of the form g(x) = sgn(a1x1 +···+anxn −θ), for some fixed reals a1,…,an,θ. Thus, a threshold gate generalizes the familiar majority gate. A major unsolved problem in computational complexity is to exhibit a Boolean function that requires a depth-2 threshold circuit of superpolynomial size, where by the size of a circuit we mean the number of gates. Communication complexity has been crucial to the progress on this problem. Through randomized communication complexity, many explicit functions have been found [17, 16, 36, 47, 49] that require majority-of-threshold circuits of exponential size. This solves an important case of the general problem. Lower bounds for the unbounded-error model (or, equivalently, on the sign-rank) cover another important case, that of threshold-of-majority circuits. More precisely, Forster et al. [14] proved the following: Lemma 1.4 ([14, Lemma 5]). Let f : {0, 1}n × {0, 1}n → {−1, +1} be a Boolean function computed by a depth-2 threshold circuit with arbitrary weights at the top gate THE SIGN-RANK OF AC0 5 and integer weights of absolute value at most w at the bottom. Then the sign-rank of F = [f (x, y)]x,y is O(snw), where s is the number of gates. Combined with our main result, this immediately gives the following. Corollary 1.5 (Threshold circuits). In every depth-2 threshold circuit that 􏰓m 􏰔m2 computes the function fm(x,y) = i=1 j=1(xij ∧yij), the sum of the absolute values of the (integer) weights at the bottom level must be of magnitude 2Ω(m). This is the first exponential lower bound for threshold-of-majority circuits com- puting a function in AC0. It substantially generalizes and strengthens an earlier result of Krause and Pudl ́ak [30, Thm. 2], who proved an exponential lower bound for threshold-of-MODr circuits (for any constant r 􏰟 2) computing a function in AC0. Our work also complements exponential lower bounds for majority-of-threshold cir- cuits computing functions in AC0, obtained by Buhrman et al. [8] and independently in [49, 50]. 1.4. Our proof and techniques. At a high level, we adopt the approach in- troduced in [51] in the context of determining the sign-rank of symmetric functions. That approach is based on Forster’s method [13] for proving lower bounds on the sign-rank in combination with the pattern matrix method [50]. In more detail, Forster [13] showed that a matrix with entries ±1 has high sign- rank if it has low spectral norm. Follow-up papers [14, 15] relaxed the assumptions on the entries of the matrix. We begin with a simple generalization of these results (Theorem 5.1), proving that in order to ensure high sign-rank, it suffices to require, along with low spectral norm, that most of the entries are not too small in absolute value. In [50, 51] it was shown how to construct such matrices from a function g : {0, 1}n → R with the following properties: (1) low-order Fourier coefficients of g are zero, i.e., g is orthogonal to all low- degree polynomials; (2) g is not too small in absolute value on most inputs x ∈ {0, 1}n. The entries of the matrix are the values of g repeated in certain patterns; the resulting matrix is referred to as the pattern matrix for g. If the original function g is Boolean, then (2) is immediate, but (1) rarely holds. A way to fix the problem is to find a probability distribution μ on {0, 1}n which is orthogonalizing in the sense that the combined function g′(x) = g(x)μ(x) is orthogonal to all low-degree polynomials. But then care must be taken to ensure property (2) for the new function g′. In other words, μ must be nonnegligible on most inputs (smooth). In [51], this program was carried out to obtain lower bounds on the sign-rank of symmetric functions. The existence of the required smooth orthogonalizing distribu- tion was established using a linear-programming dual interpretation of Paturi’s lower bounds [39] for the uniform approximation of symmetric functions. In this paper we study the sign-rank of AC0 functions, and specifically the sign- rank of the matrix derived from the communication version of the Minsky-Papert function : m 4m2 MPm(x) = 􏰗 􏰘 xi,j. i=1 j=1 Our proof relies on a linear-programming dual interpretation of the Minsky-Papert lower bound for the sign-representation of MPm. The construction of the smooth or- 6 A.A. RAZBOROV, A.A. SHERSTOV thogonalizing distribution in this paper, which is the crux of the program, is unrelated to the corresponding step in [51] and requires considerable new ideas. Having described our proof at a high level, we will now examine it in more detail, from the bottom up. Figure 1.1 illustrates the main components of our proof. A starting point in our study is an elegant result due to Minsky and Papert [34], who constructed a linear-size DNF formula that cannot be sign-represented by polynomials of low degree. Second, we revisit a fundamental technique from approximation theory, the in- terpolation bound, which bounds a degree-d univariate polynomial p on an interval based on the values of p at d + 1 distinct points. By combining the interpolation bound with an adapted version of Minsky and Papert’s argument, we establish a key intermediate result (Lemma 3.4). This result concerns multivariate polynomials that have nonnegligible agreement with the Minsky-Papert function and constrains their behavior on a large fraction of the inputs. We proceed by deriving a Fourier-theoretic property common to all low-degree multivariate polynomials on {0, 1}n: we show that their values on {0, 1}n can be con- veniently bounded in terms of their behavior on certain small subcubes (Lemma 3.2). In light of this Fourier-theoretic observation, our intermediate result on multivariate polynomials takes on a much stronger form. Namely, we prove that multivariate poly- nomials with any nontrivial agreement with the Minsky-Papert function are highly constrained throughout the hypercube (Theorem 3.6). With some additional work in Section 4, we are able to deduce the existence of a smooth distribution on {0,1}n with respect to which the Minsky-Papert function is orthogonal to all low-degree polynomials. This completes step 1 of the above program, as desired. The techniques of our proof seem to be of independent interest. Multivariate polynomials on {0,1}n arise frequently in the complexity literature and pose a con- siderable analytic challenge. A solution that we introduce is to project a multivariate polynomial in several ways to univariate polynomials, study the latter objects, and recombine the results using Fourier analysis (see Section 3). To our knowledge, this Main result Pattern matrix Forster’s method generalized bound STEP 1 Smooth orthogonalizing distribution Multivariate approximation Intermediate result Subcube lemma Minsky & Papert Interpolation bound Fig. 1.1: Proof outline. THE SIGN-RANK OF AC0 7 approach is novel and shows promise in more general contexts. 2. Preliminaries. All Boolean functions in this paper are represented as map- pings {0, 1}n → {−1, +1}, where −1 corresponds to “true.” For x ∈ {0, 1}n, we define \|x\| = x1 +x2 +···+xn. The symbol Pd stands for the set of all univariate real poly- nomials of degree up to d. By the degree of a multivariate polynomial, we will always mean its total degree, i.e., the largest total degree of any monomial. The notation [n] refers to the set {1,2,…,n}. Set membership notation, when used in the subscript of an expectation operator, means that the expectation is taken with respect to the uniformly random choice of an element from the indicated set. 2.1. Matrix analysis. The symbol Rm×n refers to the family of all m × n matrices with real entries. The (i,j)th entry of a matrix A is denoted by Aij. We frequently use “generic-entry” notation to specify a matrix succinctly: we write A = [F(i,j)]i,j to mean that the (i,j)th entry of A is given by the expression F(i,j). In most matrices that arise in this work, the exact ordering of the columns (and rows) is irrelevant. In such cases we describe a matrix by the notation [F (i, j)]i∈I, j∈J , where I and J are some index sets. Let A = [Aij] ∈ Rm×n be given. We let ∥A∥∞ = maxi,j \|Aij\| and denote the singular values of A by σ1(A) 􏰟 σ2(A) 􏰟 · · · 􏰟 σmin{m,n}(A) 􏰟 0. The notation ∥ · ∥2 refers to the Euclidean norm on vectors. Recall that the spectral norm, trace norm, and Frobenius norm of A are given by ∥A∥ = max ∥Ax∥2 = σ1(A), x∈Rn, ∥x∥2=1 ∥A∥Σ = 􏰕 σi(A), ∥A∥F =􏰜􏰕A2ij =􏰜􏰕σi(A)2. An essential property of these norms is their invariance under orthogonal transforma- tions on the left and on the right, which incidentally explains the alternative expres- sions for the spectral and Frobenius norms given above. The following relationship follows at once by the Cauchy-Schwarz inequality: ∥A∥Σ 􏰞 ∥A∥F 􏰛rank(A) (A ∈ Rm×n). (2.1) For A,B ∈ Rm×n, we write ⟨A,B⟩ = 􏰂i,j AijBij. A useful consequence of the singular value decomposition is: ⟨A, B⟩ 􏰞 ∥A∥ ∥B∥Σ (A, B ∈ Rm×n). (2.2) The Hadamard product of A and B is the matrix A ◦ B = [Aij Bij ]. Recall that rank(A ◦ B) 􏰞 rank(A) rank(B). (2.3) The symbol J stands for the all-ones matrix, whose dimensions will be apparent from the context. The notation A 􏰟 0 means that all the entries in A are nonnegative. The shorthand A ̸= 0 means as usual that A is not the zero matrix. 2.2. Fourier transform over Zn2. Consider the vector space of functions {0, 1}n → R, equipped with the inner product ⟨f,g⟩ = 2−n 􏰕 f(x)g(x). x∈{0,1}n 8 A.A. RAZBOROV, A.A. SHERSTOV For S ⊆ [n], define χS : {0,1}n → {−1,+1} by χS(x) = (−1)􏰂i∈S xi. Then {χS}S⊆[n] is an orthonormal basis for the inner product space in question. As a result, every function f : {0, 1}n → R has a unique representation of the form f(x) = ˆˆ 􏰕ˆ f(S)χS(x), where f(S) = ⟨f,χS⟩. The reals f(S) are called the Fourier coefficients of f. The ˆ following fact is immediate from the definition of f(S): Proposition 2.1. Let f : {0, 1}n → R be given. Then ˆ −n􏰕 max \|f(S)\| 􏰞 2 2.3. Symmetric functions. Let Sn denote the symmetric group on n elements. For σ ∈ Sn and x ∈ {0,1}n, we denote by σx the string (xσ(1),…,xσ(n)) ∈ {0,1}n. A function φ : {0, 1}n → R is called symmetric if φ(x) = φ(σx) for every x ∈ {0, 1}n and every σ ∈ Sn. Equivalently, φ is symmetric if φ(x) is uniquely determined by \|x\|. Observe that for every φ : {0, 1}n → R (symmetric or not), the derived function 􏰙􏰚 φsym(x) = E φ(σx) σ∈Sn is symmetric. Symmetric functions on {0,1}n are intimately related to univariate polynomials, as demonstrated by Minsky and Papert’s symmetrization argument: Proposition 2.2 (Minsky & Papert [34]). Let φ : {0, 1}n → R be representable by a real n-variate polynomial of degree r. Then there is a polynomial p ∈ Pr with E 􏰙φ(σx)􏰚 = p(\|x\|) ∀x ∈ {0, 1}n. σ∈Sn We will need the following straightforward generalization. Proposition 2.3. Let n1,…,nk be positive integers, n = n1 + ··· + nk. Let φ : {0,1}n → R be representable by a real n-variate polynomial of degree r. Write x ∈ {0, 1}n as x = (x(1), . . . , x(k)), where x(i) = (xn1+···+ni−1+1, . . . , xn1+···+ni ). Then there is a polynomial p on Rk of degree at most r such that E 􏰙φ(σ1x(1), . . . , σkx(k))􏰚 = p􏰆\|x(1)\|, . . . , \|x(k)\|􏰇 ∀x ∈ {0, 1}n. σ1 ∈Sn1 ,…,σk ∈Snk 2.4. Sign-rank. The sign-rank of a real matrix A = [Aij] is the least rank of a matrix B = [Bij] such that AijBij > 0 for all i,j with Aij ̸= 0. (Note that this definition generalizes the one given above in the abstract and introduction, which only applied to matrices A with nonzero entries.) In general, the sign-rank of a matrix can be vastly smaller than its rank. For example, consider the following nonsingular matrices of order n 􏰟 3, representing well-known problems greater-than and equality in communication complexity: 1 −1 1 1−1 1 1 −1 .., ... .. −1 1 1 −1 1 −1 THE SIGN-RANK OF AC0 9 These matrices have sign-rank at most 2 and 3, respectively. Indeed, the first matrix has the same sign pattern as the matrix [2(j − i) + 1]i,j . The second has the same sign pattern as the matrix [(1−ε)−⟨vi,vj⟩]i,j, where v1,v2,…,vn ∈ R2 are arbitrary pairwise distinct unit vectors and ε is a suitably small positive real, cf. [40, §5]. Bounding the sign-rank from below is a considerable challenge. In a breakthrough result, Forster [13] proved the first nontrivial lower bound on the sign-rank of an explicit ±1 matrix. The centerpiece of Forster’s argument is the following theorem, which is a crucial starting point for our work. Theorem 2.4 (Forster [13], implicit). Let X,Y be finite sets and M = [Mxy]x∈X,y∈Y a real matrix (M ̸= 0). Put r = sign-rank(M). Then there is a matrix R = [Rxy]x∈X,y∈Y such that: rank(R) = r, M ◦ R 􏰟 0, ∥R∥∞ 􏰞 1, ∥R∥F =􏰛\|X\|\|Y\|/r. Appendix B provides a detailed explanation of how Theorem 2.4 is implicit in Forster’s work. 2.5. Pattern matrices. Pattern matrices were introduced in [49, 50] and proved useful in obtaining strong lower bounds on communication complexity. Rele- vant definitions and results from [50] Let n and N be positive integers with n \| N. Split [N] into n contiguous blocks, with N/n elements each: 􏰌 N􏰍 􏰌N 2N􏰍 􏰌(n−1)N 􏰍 [N]= 1,2,…, n ∪ n +1,…, n ∪···∪ n +1,…,N . Let V(N,n) denote the family of subsets V ⊆ [N] that have exactly one element from each of these blocks (in particular, \|V \| = n). Clearly, \|V(N, n)\| = (N/n)n. For a bit string x ∈ {0, 1}N and a set V ∈ V(N, n), define the projection of x onto V by x\|V = (xi1,xi2,…,xin) ∈ {0,1}n, wherei1 k. Then 􏰅 xk−xi􏰅dkd 􏰅k=0 i=0 􏰅 k=0 We now establish another auxiliary fact. It provides a convenient means to bound a function whose Fourier transform is supported on low-order characters, in terms of its behavior on low-weight Lemma3.2. Letkbeaninteger,0􏰞k􏰞n−1.Letf:{0,1}n →Rbegiven ˆ n k􏰈n􏰉 \|f(1 )\| 􏰞 2 max\|f(x)\|. k \|x\|􏰞k Proof. Define the symmetric function g : {0,1}n → R by g(x) = χ[n](x)p(\|x\|), The following properties of g are immediate: g(1n) = (−1)n, g(x) = 0 The degree of every monomial in g is between k + 1 and n, so that p(t)= 􏰖 t−i. gˆ(S) = 0 (\|S\| 􏰞 k). 􏰕k 􏰈 n 􏰉 􏰈 n − t − 1 􏰉 􏰈 n 􏰉 􏰕k 􏰕 􏰕k 􏰈 n 􏰉 \|g(x)\|= t \|p(t)\|= n − k 􏰈 k 􏰉 n−t t t n−k−1 = k We are now prepared to analyze f. By (3.3), On the other hand, (3.1) and (3.2) show that n−i k k. We will not need this observation, however. We are now in a position to study the approximation problem of interest to us. Define the sets Z = {0,1,2,…,4m2}m, Z+ = {1,2,…,4m2}m. Looking ahead, Z+ and Z Z+ correspond to the sets on which the Minsky-Papert 􏰓m 􏰔4m2 function i=1 j=1 xij is true and false, respectively. Accordingly, we define F : Z → {−1, +1} by 􏰀−1 ifx∈Z+, 1 otherwise. Foru,z∈Z,let∆(u,z)=\|{i:ui ̸=zi}\|betheordinaryHammingdistance. Weshall prove the following intermediate result, inspired by Minsky and Papert’s analysis [34] of the threshold degree of CNF formulas. Lemma 3.4. Let Q be a degree-d real polynomial in m variables, where d 􏰞 m/3. Assume that F (z)Q(z) 􏰟 −1 (z ∈ Z). (3.7) Then \|Q(z)\| 􏰞 4m+d at every point z ∈ Z+ with ∆(u,z) < m/3, where u = (12, 32, 52, . . . , (2m − 1)2) ∈ Z+. F(z) = Proof. Fix z ∈ Z+ with ∆(u,z) < m/3. Define p ∈ P2d by p(t) = Q(p1(t), p2(t), . . . , pm(t)), where 􏰀(t − 2i + 1)2 if zi = ui (equivalently, zi = (2i − 1)2), pi(t) = zi otherwise. Letting S = {i : ui = zi}, inequality (3.7) implies that p(2i − 1) 􏰟 −1 (i ∈ S), p(2i) 􏰞 1 (i = 0,1,...,m). Claim3.5. Leti∈S.Then\|p(ξ)\|􏰞1forsomeξ∈[2i−2,2i−1]. (3.8) (3.9) Proof. The claim is trivial if p vanishes at some point in [2i − 2, 2i − 1]. In the contrary case, p maintains the same sign throughout this interval. As a result, (3.8) and (3.9) show that min{\|p(2i − 2)\|, \|p(2i − 1)\|} 􏰞 1. Claim 3.5 provides \|S\| > 2m/3 􏰟 2d 􏰟 deg(p) points in [0,2m], with pairwise distances at least 1, at which p is bounded in absolute value by 1. By Lemma 3.1, max \|p(t)\|􏰞2 0􏰞t􏰞2m deg(p) 􏰈 2m 􏰉 m+d This completes the proof since Q(z) = p(0). Finally, we remove the restriction on ∆(u, z), thereby establishing the main result of this section. THE SIGN-RANK OF AC0 13 Theorem 3.6. Let Q be a degree-d real polynomial in m variables, where d < m/3. Assume that F(z)Q(z) 􏰟 −1 (z ∈ Z). Then \|Q(z)\| 􏰞 16m (z ∈ Z+). Proof. As before, put u = (12,32,52,...,(2m − 1)2). Fix z ∈ Z+ and define the “interpolating” function f : {0, 1}m → R by f(x)= Q(x1z1 +(1−x1)u1,...,xmzm +(1−xm)um). In this notation, we know from Lemma 3.4 that \|f (x)\| 􏰞 4m+d for every x ∈ {0, 1}m with \|x\| < m/3, and our goal is to show that \|f(1m)\| 􏰞 16m. Since Q has degree d, the Fourier transform of f is supported on characters of order up to d. As a result, \|f(1 )\| 􏰞 2 􏰞 2 􏰞 16m. m d􏰈m􏰉 d \|x\|􏰞d max \|f(x)\| 2m+3d 􏰈m􏰉 by Lemma 3.2 by Lemma 3.4 d 4. A smooth orthogonalizing distribution. An important concept in our work is that of an orthogonalizing distribution [51]. Let f : {0,1}n → {−1,+1} be given. A distribution μ on {0, 1}n is d-orthogonalizing for f if E 􏰙f(x)χS(x)􏰚 = 0 (\|S\| < d). x∼μ In words, a distribution μ is d-orthogonalizing for f if with respect to μ, the function f is orthogonal to every character of order less than d. This section focuses on the following function from {0, 1}4m3 to {−1, +1}: m 4m2 MPm(x) = 􏰗 􏰘 xi,j. i=1 j=1 (Recall that we interpret −1 as “true”.) This function was originally studied by Min- sky and Papert [34] and has played an important role in later works [30, 38, 49, 50]. An explicit m-orthogonalizing distribution for MPm is known [49]. However, our main result requires a Θ(m)-orthogonalizing distribution for MPm that is addition- ally smooth, i.e., places substantial weight on all but a tiny fraction of the points, and the distribution given in [49] severely violates the latter property. Proving the existence of a distribution that is simultaneously Θ(m)-orthogonalizing and smooth is the goal of this section (Theorem 4.1). We will view an input x ∈ {0, 1}n = {0, 1}4m3 to MPm as composed of blocks: x = (x(1),...,x(m)), where the ith block is x(i) = (xi,1,xi,2,...,xi,4m2). The proof 14 A.A. RAZBOROV, A.A. SHERSTOV that is about to start refers to the sets Z,Z+ and the function F as defined in Section 3. Theorem 4.1. There is a 1 m-orthogonalizing distribution μ for MPm such that 3 μ(x) 􏰟 1 16−m 2−n for all inputs x ∈ {0, 1}n with MPm(x) = −1. 2 Proof. Let X be the set of all inputs with MPm(x) = −1, i.e., X ={x∈{0,1}n :x(1) ̸=0, ..., x(m) ̸=0}. It suffices to show that the following linear program has optimum at least 116−m: 2 (LP1) The optimum being nonzero, it will follow by a scaling argument that any optimal solution has 􏰂 μ(x) = 1. As a result, μ will be the sought probability distribution. For x ∈ {0, 1}n, we let z(x) = (\|x(1)\|, . . . , \|x(m)\|); note that MPm(x) = F (z(x)). Since the function MPm is invariant under the action of the group S4m2 × · · · × S4m2 , in view of Proposition 2.3, the dual of (LP1) can be simplified as follows: (LP2) The programs are both feasible and therefore have the same finite optimum. Fix an optimal solution η,Q,δz to (LP2). For the sake of contradiction, assume that variables: maximize: subject to: μ(x) 􏰟 0 for x ∈ {0,1}n μ(x)MPm(x)χS(x) = 0 μ(x) 􏰞 1, x∈{0,1}n μ(x) 􏰟 ε2−n ε 􏰟 0; ε 􏰕 x∈{0,1}n 􏰕 for \|S\| < m/3, for x ∈ X. variables: a polynomial Q on Rm of degree < m/3; η􏰟0; δz 􏰟0forz∈Z+ minimize: η subject to: 􏰕 δz(x) 􏰟 2n, x∈X F(z)Q(z) 􏰟 −η F(z)Q(z) 􏰟 −η + δz for z ∈ Z, for z ∈ Z+. η 􏰞 1 16−m. Then \|Q(z)\| 􏰞 1 for each z ∈ Z+, by Theorem 3.6. From the constraints 22 of the third type in (LP2) we conclude that δz 􏰞 1 +η < 1 (z ∈ Z+). This contradicts 2 the first constraint. Thus, the optimum of (LP1) and (LP2) is at least 1 16−m . 2 5. A generalization of Forster’s bound. Using Theorem 2.4, Forster gave a simple proof of the following fundamental result [13, Thm. 2.2]: for any matrix A = [Axy ]x∈X, y∈Y with ±1 entries, 􏰛\|X\|\|Y\| sign-rank(A) 􏰟 ∥A∥ . Forster et al. [14, Thm. 3] generalized this bound to arbitrary real matrices A ̸= 0: 􏰛\|X\|\|Y\| sign-rank(A) 􏰟 · min \|Axy \|. (5.1) ∥A∥ x,y THE SIGN-RANK OF AC0 15 Forster and Simon [15, §5] considered a different generalization, inspired by the notion of matrix rigidity (see, e.g., [43]). Let A be a given ±1 matrix, and let A ̃be obtained from A by changing some h entries in an arbitrary fashion (h < \|X \| \|Y \|). Forster and Simon showed that 􏰛\|X\|\|Y\| sign-rank(A ̃) 􏰟 √ . (5.2) ∥A∥+2 h The above generalizations are not sufficient for our purposes. Before we can proceed, we need to prove the following “hybrid” bound, which combines the ideas of (5.1) and (5.2). Theorem 5.1. Let A = [Axy]x∈X,y∈Y be a real matrix with s = \|X\|\|Y\| entries (A ̸= 0). Assume that all but h of the entries of A satisfy \|Axy\| 􏰟 γ, where h and γ > 0 are arbitrary parameters. Then γs sign-rank(A) 􏰟 ∥A∥√s + γh. Proof. Let r denote the sign-rank of A. Theorem 2.4 supplies a matrix R = [Rxy] with On the other hand, 􏰟 γ∥R∥2F − γh = γs − γh rank(R) = r, A ◦ R 􏰟 0, ∥R∥∞ 􏰞 1, ∥R∥F = 􏰛s/r. (5.3) (5.4) (5.5) (5.6) The crux of the proof is to estimate ⟨A, R⟩ from below and above. On the one hand, ⟨A,R⟩ 􏰟 􏰕 AxyRxy x,y: \|Axy \|􏰟γ 􏰎􏰏 􏰟γ 􏰕\|Rxy\|−h by (5.4) by (5.5) by (5.6). ⟨A, R⟩ 􏰞 ∥A∥ · ∥R∥Σ 􏰞 ∥A∥ · ∥R∥F by (2.2) by (2.1), (5.3) by (5.6). √ √ = ∥A∥ s Comparing these lower and upper bounds on ⟨A,R⟩ yields the claimed estimate of r = sign-rank(A). 6. Main result. At last, we are in a position to prove our main result. 􏰓m 􏰔m2 Theorem 1.1 (Restated from p. 2). Define fm(x, y) = i=1 j=1(xij Then the matrix [fm(x, y)]x,y has sign-rank 2Ω(m). ∧ yij ). 16 A.A. RAZBOROV, A.A. SHERSTOV Proof. Let M be the (N, n, MPm)-pattern matrix, where n = 4m3 and N = 176n. Let P be the (N, n, μ)-pattern matrix, where μ is the distribution from Theorem 4.1. We are going to estimate the sign-rank of M ◦ P. By Theorem 4.1, all but a 2−Ω(m2) fraction of the inputs x ∈ {0,1}n satisfy μ(x) 􏰟 1 16−m 2−n. As a result, all but a 2−Ω(m2) fraction of the entries of M ◦ P are at least 1 16−m 2−n in absolute value. Theorem 5.1 at once implies that 􏰌16−m 2−n√s Ω(m2)􏰍 sign-rank(M)􏰟sign-rank(M◦P)􏰟min 4∥M◦P∥ , 2 where s = 2N+n 􏰃N 􏰄n denotes the number of entries in M ◦ P. n , (6.1) We now bound the spectral norm of M ◦ P precisely as in [51, §6]. Note first that M ◦P is the (N,n,φ)-pattern matrix, where φ : {0,1}n → R is given by φ(x) = MPm(x)μ(x). Since μ is a 1m-orthogonalizing distribution for MPm, we have 3 φˆ(S) = 0 for \|S\| < 1 m. 3 Since 􏰂x∈{0,1}n \|φ(x)\| = 1, Proposition 2.1 shows that \|φˆ(S)\| 􏰞 2−n for each S ⊆ [n]. Theorem 2.6 implies, in view of (6.2) and (6.3), that √ 􏰈N􏰉−m/6 √ (6.2) (6.3) ∥M◦P∥􏰞 s·2−n n =17−m2−n s. Along with (6.1), this estimate shows that M has sign-rank at least 2Ω(m). It remains to verify that M is a submatrix of [fcm(x,y)]x,y, where c = ⌈􏰛8N/n⌉ = Θ(1). The set V ∈ V (N,n) in Definition 2.5 is naturally represented by a tuple (...,vij,...)∈[N/n]n.Thenforallx∈{0,1}N and(V,w)∈V(N,n)×{0,1}n, m 4m2 m 4m2 N/n 􏰗 􏰘(x\|V ⊕w)ij = 􏰗 􏰘 􏰘 􏰘 ((xijk =ε)∧(wij ̸=ε)∧(vij =k)). i=1 j=1 i=1 j=1 k=1 ε∈{0,1} Merging the three groups of OR gates gives bottom fan-in 8m2N/n 􏰞 (cm)2. Remark 6.1. The lower bound in Theorem 1.1 is essentially optimal. To see this, note that the matrix [fm(x,y)]x,y has the same sign pattern as mm2 2 R= 1−􏰖 􏰕xijyij . i=1 j=1 By property (2.3) of the Hadamard product, the sign-rank of [fm(x,y)]x,y does not exceed m2m + 1 = 2O(m log m). 7. On communication complexity classes. We proceed to explore the con- sequences of our main result in the study of communication complexity classes. Throughout this section, the symbol {fn} shall stand for a family of functions f1,f2,...,fn,...,wherefn :{0,1}n×{0,1}n →{−1,+1}. The first class that we consider, UPPcc, corresponds to functions with efficient unbounded-error protocols. Definition 7.1 (Babai et al. [3, §4]). A family {fn} is in UPPcc iff for some constant c and all natural n > c, there exists a probabilistic communication protocol with private coins such that: THE SIGN-RANK OF AC0 17 (1) for every input (x,y), the protocol outputs the correct value fn(x,y) with probability greater than 1/2; (2) the number of bits exchanged is at most logc n. Note that in this model the number of coin flips is not included in the complexity measure. Requiring the same bound of logc n on the number of coin flips results in another extensively studied class [3], called PPcc. For our purposes, however, an equivalent matrix-analytic characterization [40] is more convenient. By the sign-rank of a function f : X × Y → {−1, +1}, where X, Y are finite sets, we shall mean the sign-rank of the matrix [f (x, y)]x∈X, y∈Y . Theorem 7.2 (Paturi & Simon [40]). A function family {fn} is in the class UPPcc iff for some constant c > 1 and all n > c, the function fn has sign-rank at most 2logc n. We now turn to the communication-complexity analogue of the polynomial hier- archy, also defined by Babai et al. [3]. A function fn : {0, 1}n × {0, 1}n → {−1, +1} is called a rectangle if there exist subsets A, B ⊆ {0, 1}n such that fn(x,y)=−1 ⇔ x∈A, y∈B. We call fn the complement of a rectangle if the negated function ¬fn = −fn is a rectangle. Definition 7.3 (Babai et al. [3, §4]). (1) Afamily{f }isinΠcc iffeachf isarectangle. Afamily{f }isinΣcc iff n0n n0 {¬f } is in Πcc. (2) Fix an integer k = 1,2,3,4,…. A family {fn} is in Σcc iff for some constant c>1 and all n>c, f = 􏰘 􏰗 􏰘 ··· 􏰑 gi1,i2,…,ik, 2logc n 2logc n 2logc n 2logc n nn i1=1 i2=1 i3=1 ik=1 where 􏰐 = 􏰔 (resp., 􏰐 = 􏰓) for k odd (resp., even); and each gi1,i2,…,ik is a rectangle (resp., the complement of a rectangle) for k odd (resp., even). A family {fn} is in Πcc iff {¬fn} is in Σcc. (3) The polynomial hierarchy is given by PHcc = 􏰒 Σcc = 􏰒 Πcc, where k = kk kk 0, 1, 2, 3, . . . ranges over all constants. (4) A family {fn} is in PSPACEcc iff for some constant c > 1 and all n > c, 2logc n 2logc n 2logc n 2logc n f = 􏰘 􏰗 􏰘 ··· 􏰘 gi1,i2,…,ik, i1=1 i2=1 i3=1 ik=1 where k < logc n is odd and each gi1,i2,...,ik is a rectangle. n Thus, the zeroth level (Σcc and Πcc) of the polynomial hierarchy consists of 00 rectangles and complements of rectangles, the simplest functions in communication complexity. The first level is easily seen to correspond to functions with efficient nondeterministic or co-nondeterministic protocols: Σcc = NPcc and Πcc = coNPcc. 11 This level is well understood, and there exist powerful methods to show that {fn} ̸∈ Σcc for a host of explicit functions {f }. Finding an explicit sequence {f } ̸∈ Σcc, on 1nn2 the other hand, is a longstanding open problem. The circuit class AC0 is related to the polynomial hierarchy PHcc in communica- tion complexity in the obvious way. Specifically, if fn : {0, 1}n × {0, 1}n → {−1, +1}, k n 18 A.A. RAZBOROV, A.A. SHERSTOV n = 1,2,3,4,..., is an AC0 (or even quasi-AC0) circuit family of depth k with an or gate at the top (resp., and gate), then {fn} ∈ Σcc (resp., {fn} ∈ Πcc ). In k−1 k−1 particular, the depth-3 circuit family {f } in Theorem 1.1 is in Πcc, whereas {¬f } n2n is in Σcc. In this light, Theorem 1.1 establishes the following separations: 2 Theorem 7.4. Σcc ̸⊆ UPPcc, Πcc ̸⊆ UPPcc. 22 Several years prior to our work, Forster [13] proved that the familiar inner prod- uct function ipn(x,y) = (−1)􏰂xiyi has sign-rank 2Θ(n). Since {ipn} ∈ PSPACEcc, Forster’s result yields the separation PSPACEcc ̸⊆ UPPcc. Theorem 7.4 in this pa- per substantially strengthens it, showing that even the second level (Σcc,Πcc) of the 22 polynomial hierarchy is not contained in UPPcc. This settles the open problem due to Babai et al. [3, p. 345], who asked whether Σcc ⊆ UPPcc. Observe that Theorem 7.4 is best possible in that UPPcc trivially contains Σcc, Σcc, Πcc, and Πcc. 010 1 The classes PPcc and UPPcc both correspond to small-bias communication and, in fact, were both inspired by the class PP in computational complexity. It is well-known and straightforward to show that PPcc ⊆ UPPcc. It turns out that UPPcc is strictly more powerful than PPcc, as shown by Buhrman et al. [8] and independently in [48]. In this light, Theorem 7.4 in this paper substantially strengthens earlier separations of Σcc and Πcc from PPcc, obtained independently in [8] and [49]: 22 Theorem 7.5 (Buhrman et al. [8], Sherstov [49]). Σcc ̸⊆ PPcc, Πcc ̸⊆ PPcc. 22 Sign-rank is a much more challenging quantity to analyze than discrepancy, a combinatorial complexity measure that is known [25] to characterize membership in PPcc. Indeed, exponentially small upper bounds on the discrepancy were known more than twenty years ago [56, 10], whereas the first exponential lower bound on the sign- rank for an explicit function was only obtained recently by Forster [13]. It is not surprising, then, that this paper has required ideas that are quite different from the methods of both [8] and [48, 49]. 8. Open problems. Our work is closely related to several natural and impor- tant problems. The first is a well-known and challenging open problem in complexity theory. Are there matrices computable in AC0 that have low spectral norm? More pre- cisely, does one have ∥[f(x, y)]x∈X,y∈Y ∥ 􏰞 2−nΩ(1) 􏰛\|X\| \|Y \| for some choice of an AC0 function f : {0, 1}n × {0, 1}n → {−1, +1} and some multisets X, Y of n-bit Boolean strings? An affirmative answer to this question would subsume our results and ad- ditionally imply that AC0 is not learnable in Kearns’ statistical query model [21]. A suitable lower bound on the spectral norm of every such matrix, on the other hand, would result in a breakthrough separation of PHcc and PSPACEcc. See [3, 43, 33, 48] for relevant background. The second problem concerns the sign-rank of arbitrary pattern matrices. For a Boolean function f : {0,1}n → {−1,+1}, its threshold degree deg±(f) is the least degree of a multivariate polynomial p(x1,...,xn) such that f(x) ≡ sgnp(x). Let Mf denote the (nc , n, f )-pattern matrix, where c 􏰟 1 is a sufficiently large constant. It is straightforward to verify that the sign-rank of Mf does not exceed nO(deg±(f)). Is that upper bound close to optimal? Specifically, does Mf have sign-rank exp(deg± (f)Ω(1)) for every f? Evidence in this paper and prior work suggests an answer in the affir- mative. For example, our main result confirms this hypothesis for the Minsky-Papert function, f = MP. For f = parity the hypothesis follows immediately the seminal work of Forster [13]. More generally, it was proven in [51] that the hypothesis holds for all symmetric functions. In the field of communication complexity, we were able to resolve the main ques- tion left open by Babai et al. [3], but only in one direction: PHcc ̸⊆ UPPcc. As we 2 THE SIGN-RANK OF AC0 19 already noted, the other direction remains wide open despite much research: no lower bounds are known for PHcc or even Σcc. The latter question is in turn equivalent 2 to lower bounds for bounded-depth circuits in the context of graph complexity (e.g., see [41, 42, 19] and the literature cited therein). It is also known [43, Thm. 1] that sufficiently rigid matrices do not belong to PHcc, which provides further motivation to be looking for lower bounds on matrix rigidity. Acknowledgments. The authors would like to thank Scott Aaronson, Adam Klivans, Dieter van Melkebeek, Yaoyun Shi, and Ronald de Wolf for helpful feedback on an earlier version of this manuscript. We are also grateful to our anonymous reviewers for their useful comments. REFERENCES [1] Michael Alekhnovich, Mark Braverman, Vitaly Feldman, Adam R. Klivans, and Toni- ann Pitassi, Learnability and automatizability, in Proc. of the 45th Symposium on Foun- dations of Computer Science (FOCS), 2004, pp. 621–630. [2] Noga Alon, Peter Frankl, and Vojtech Ro ̈dl, Geometrical realization of set systems and probabilistic communication complexity, in Proc. of the 26th Symposium on Foundations of Computer Science (FOCS), 1985, pp. 277–280. [3] La ́szlo ́Babai, Peter Frankl, and Janos Simon, Complexity classes in communication complexity theory, in Proc. of the 27th Symposium on Foundations of Computer Science (FOCS), 1986, pp. 337–347. [4] Shai Ben-David, Nadav Eiron, and Hans Ulrich Simon, Limitations of learning via embed- dings in Euclidean half spaces, J. Mach. Learn. Res., 3 (2003), pp. 441–461. [5] Avrim Blum, Alan M. Frieze, Ravi Kannan, and Santosh Vempala, A polynomial-time algorithm for learning noisy linear threshold functions, Algorithmica, 22 (1998), pp. 35–52. [6] Avrim Blum, Adam Kalai, and Hal Wasserman, Noise-tolerant learning, the parity problem, and the statistical query model, J. ACM, 50 (2003), pp. 506–519. [7] Nader H. Bshouty, A subexponential exact learning algorithm for DNF using equivalence queries, Inf. Process. Lett., 59 (1996), pp. 37–39. [8] Harry Buhrman, Nikolai K. Vereshchagin, and Ronald de Wolf, On computation and communication with small bias, in Proc. of the 22nd Conf. on Computational Complexity (CCC), 2007, pp. 24–32. [9] E.W.Cheney,IntroductiontoApproximationTheory,ChelseaPublishing,NewYork,2nded., 1982. [10] Benny Chor and Oded Goldreich, Unbiased bits from sources of weak randomness and probabilistic communication complexity, SIAM J. Comput., 17 (1988), pp. 230–261. [11] Ronald de Wolf, Quantum Computing and Communication Complexity, PhD thesis, Uni- versity of Amsterdam, 2001. [12] Vitaly Feldman, Parikshit Gopalan, Subhash Khot, and Ashok Kumar Ponnuswami, New results for learning noisy parities and halfspaces, in Proceedings of the 47th Annual Symposium on Foundations of Computer Science (FOCS), 2006, pp. 563–574. [13] Ju ̈rgen Forster, A linear lower bound on the unbounded error probabilistic communication complexity, J. Comput. Syst. Sci., 65 (2002), pp. 612–625. [14] Ju ̈rgen Forster, Matthias Krause, Satyanarayana V. Lokam, Rustam Mubarakzjanov, Niels Schmitt, and Hans-Ulrich Simon, Relations between communication complex- ity, linear arrangements, and computational complexity, in Proc. of the 21st Conf. on Foundations of Software Technology and Theoretical Computer Science (FST TCS), 2001, pp. 171–182. [15] Ju ̈rgen Forster and Hans Ulrich Simon, On the smallest possible dimension and the largest possible margin of linear arrangements representing given concept classes, Theor. Comput. Sci., 350 (2006), pp. 40–48. [16] Mikael Goldmann, Johan H ̊astad, and Alexander A. Razborov, Majority gates vs. gen- eral weighted threshold gates, Computational Complexity, 2 (1992), pp. 277–300. [17] Andra ́s Hajnal, Wolfgang Maass, Pavel Pudla ́k, Mario Szegedy, and Gyo ̈rgy Tura ́n, Threshold circuits of bounded depth, J. Comput. Syst. Sci., 46 (1993), pp. 129–154. [18] David P. Helmbold, Robert H. Sloan, and Manfred K. Warmuth, Learning integer lat- tices, SIAM J. Comput., 21 (1992), pp. 240–266. 20 A.A. RAZBOROV, A.A. SHERSTOV [19] Stasys Jukna, On graph complexity, Combinatorics, Probability and Computing, 15 (2006), pp. 1–22. [20] BalaKalyanasundaramandGeorgSchnitger,Theprobabilisticcommunicationcomplexity of set intersection, SIAM J. Discrete Math., 5 (1992), pp. 545–557. [21] Michael Kearns, Efficient noise-tolerant learning from statistical queries, in Proc. of the 25th Symposium on Theory of Computing (STOC), 1993, pp. 392–401. [22] Michael Kearns and Leslie Valiant, Cryptographic limitations on learning Boolean formu- lae and finite automata, J. ACM, 41 (1994), pp. 67–95. [23] Michael J. Kearns and Umesh V. Vazirani, An Introduction to Computational Learning Theory, MIT Press, Cambridge, 1994. [24] Michael Kharitonov, Cryptographic hardness of distribution-specific learning, in Proc. of the 25th Symposium on Theory of Computing, 1993, pp. 372–381. [25] Hartmut Klauck, Lower bounds for quantum communication complexity, SIAM J. Comput., 37 (2007), pp. 20–46. [26] Adam R. Klivans and Rocco A. Servedio, Learning DNF in time 2O ̃(n1/3), J. Comput. Syst. Sci., 68 (2004), pp. 303–318. [27] Adam R. Klivans and Alexander A. Sherstov, A lower bound for agnostically learning disjunctions, in Proc. of the 20th Conf. on Learning Theory (COLT), 2007, pp. 409–423. [28] , Unconditional lower bounds for learning intersections of halfspaces, Machine Learning, 69 (2007), pp. 97–114. [29] , Cryptographic hardness for learning intersections of halfspaces, J. Comput. Syst. Sci., 75 (2009), pp. 2–12. [30] Matthias Krause and Pavel Pudla ́k, On the computational power of depth-2 circuits with threshold and modulo gates, Theor. Comput. Sci., 174 (1997), pp. 137–156. [31] EyalKushilevitzandNoamNisan,Communicationcomplexity,CambridgeUniversityPress, New York, 1997. [32] Nati Linial, Shahar Mendelson, Gideon Schechtman, and Adi Shraibman, Complexity measures of sign matrices, Combinatorica, 27 (2007), pp. 439–463. [33] Satyanarayana V. Lokam, Spectral methods for matrix rigidity with applications to size-depth trade-offs and communication complexity, J. Comput. Syst. Sci., 63 (2001), pp. 449–473. [34] Marvin L. Minsky and Seymour A. Papert, Perceptrons: Expanded edition, MIT Press, Cambridge, Mass., 1988. [35] Ilan Newman, Private vs. common random bits in communication complexity, Inf. Process. Lett., 39 (1991), pp. 67–71. [36] Noam Nisan, The communication complexity of threshold gates, in Combinatorics, Paul Erdo ̋s is Eighty, 1993, pp. 301–315. [37] A. B. J. Novikoff, On convergence proofs on perceptrons, in Proc. of the Symposium on the Mathematical Theory of Automata, vol. XII, 1962, pp. 615–622. [38] Ryan O’Donnell and Rocco A. Servedio, New degree bounds for polynomial threshold func- tions, in Proc. of the 35th Symposium on Theory of Computing (STOC), 2003, pp. 325–334. [39] Ramamohan Paturi, On the degree of polynomials that approximate symmetric Boolean func- tions, in Proc. of the 24th Symposium on Theory of Computing, 1992, pp. 468–474. [40] Ramamohan Paturi and Janos Simon, Probabilistic communication complexity, J. Comput. Syst. Sci., 33 (1986), pp. 106–123. [41] Pavel Pudla ́k, Vojtech Ro ̈dl, and Petr Savicky ́, Graph complexity, Acta Inf., 25 (1988), pp. 515–535. [42] Alexander A. Razborov, Bounded-depth formulae over the basis {&,⊕} and some combi- natorial problems, Complexity Theory and Applied Mathematical Logic, vol. “Problems of Cybernetics” (1988), pp. 146–166. In Russian, available at http://www.mi.ras.ru/ ~razborov/graph.pdf. [43] , On rigid matrices. Manuscript in Russian, available at http://www.mi.ras.ru/ ~razborov/rigid.pdf, June 1989. [44] , On the distributional complexity of disjointness, Theor. Comput. Sci., 106 (1992), pp. 385–390. [45] Theodore J. Rivlin, An Introduction to the Approximation of Functions, Dover Publications, New York, 1981. [46] Frank Rosenblatt, The perceptron: A probabilistic model for information storage and orga- nization in the brain, Psychological Review, 65 (1958), pp. 386–408. [47] Alexander A. Sherstov, Powering requires threshold depth 3, Inf. Process. Lett., 102 (2007), pp. 104–107. [48] , Halfspace matrices, Comput. Complex., 17 (2008), pp. 149–178. Preliminary version in 22nd CCC, 2007. [49] [50] [51] [52] [53] [54] [55] [56] THE SIGN-RANK OF AC0 21 , Separating AC0 from depth-2 majority circuits, SIAM J. Comput., 38 (2009), pp. 2113– 2129. Preliminary version in 39th STOC, 2007. , The pattern matrix method for lower bounds on quantum communication, in Proc. of the 40th Symposium on Theory of Computing (STOC), 2008, pp. 85–94. , The unbounded-error communication complexity of symmetric functions, in Proc. of the 49th Symposium on Foundations of Computer Science (FOCS), 2008, pp. 384–393. , Communication lower bounds using dual polynomials, Bulletin of the EATCS, 95 (2008), pp. 59–93. Nathan Srebro and Adi Shraibman, Rank, trace-norm and max-norm., in Proc. of the 18th Conf. on Learning Theory (COLT), 2005, pp. 545–560. Jun Tarui and Tatsuie Tsukiji, Learning DNF by approximating inclusion-exclusion formu- lae, in Proc. of the 14th Conf. on Computational Complexity (CCC), 1999, pp. 215–221. Leslie G. Valiant, A theory of the learnable, Commun. ACM, 27 (1984), pp. 1134–1142. Umesh V. Vazirani, Strong communication complexity or generating quasirandom sequences form two communicating semi-random sources, Combinatorica, 7 (1987), pp. 375–392. Appendix A. More on the unbounded-error model. Readers with back- ground in communication complexity will note that the unbounded-error model is exactly the same as the private-coin randomized model [31, Chap. 3], with one ex- ception: in the latter case the correct answer is expected with probability at least 2/3, whereas in the former case the success probability need only exceed 1/2 (say, by an exponentially small amount). This difference has far-reaching implications. For example, the fact that the parties in the unbounded-error model do not have a shared source of random bits is crucial: allowing shared randomness would make the com- plexity of every function a constant, as one can easily verify. By contrast, introducing shared randomness into the randomized model has minimal impact on the complexity of any given function [35]. As one might expect, the weaker success criterion in the unbounded-error model also has a drastic impact on the complexity of certain functions. For example, the well-known disjointness function on n-bit strings has complexity O(logn) in the unbounded-error model and Ω(n) in the randomized model [20, 44]. Furthermore, explicit functions are known [8, 48] with unbounded-error complexity O(log n) that √ require Ω( n) communication in the randomized model to even achieve advantage √ 2− {−1, +1} is never much more than its complexity in the other standard models. For n/5 over random guessing. More generally, the unbounded-error complexity of a function f : X × Y → example, it is not hard to see that U(f) 􏰞 min{N0(f), N1(f)} + O(1) 􏰞 D(f) + O(1), where D, N0, and N1 refer to communication complexity in the deterministic, 0- nondeterministic, and 1-nondeterministic models, respectively. Continuing, U(f) 􏰞 R1/3(f) + O(1) 􏰞 O 􏰆Rpub (f ) + log log [\|X \| + \|Y \|]􏰇 , 1/3 where R1/3 and Rpub refer to the private- and public-coin randomized models, respec- 1/3 tively. As a matter of fact, one can show that U (f ) 􏰞 O 􏰆Q∗1/3 (f ) + log log [\|X \| + \|Y \|]􏰇 , where Q∗1/3 refers to the quantum model with prior entanglement. An identi- cal inequality is clearly valid for the quantum model without prior entanglement. 22 A.A. RAZBOROV, A.A. SHERSTOV See [31, 11] for rigorous definitions of these various models; our sole intention was to point out that the unbounded-error model is at least as powerful. Appendix B. Details on Forster’s method. The purpose of this section is to explain in detail how Theorem 2.4 is implicit in Forster’s work. Recall that vectors v1, . . . , vn in Rr are said to be in general position if no r of them are linearly dependent. In a powerful result, Forster proved that any set of vectors in general position can be balanced in a useful way: Theorem B.1 (Forster [13, Thm. 4.1]). Let U ⊂ Rr be a finite set of vectors in general position, \|U\| 􏰟 r. Then there is a nonsingular transformation A ∈ Rr×r such that 􏰕 1 (Au)(Au)T = \|U\|Ir. u∈U ∥Au∥2 r The proof of this result is rather elaborate and uses compactness arguments in an essential way. The vector norm ∥ · ∥ above and throughout this section is the Euclidean norm ∥ · ∥2. The development that follows is closely analogous to Forster’s derivation [13, p. 617]. Theorem 2.4 (Restated from p. 9). Let X,Y be finite sets, M = [Mxy]x∈X,y∈Y a real matrix (M ̸= 0). Put r = sign-rank(M). Then there is a matrix R = [Rxy]x∈X,y∈Y such that: rank(R) = r, M ◦ R 􏰟 0, ∥R∥∞ 􏰞 1, ∥R∥F =􏰛\|X\|\|Y\|/r. (B.1) (B.2) (B.3) (B.4) Proof. Since M ̸= 0, it follows that r 􏰟 1. Fix a matrix Q = [Qxy] of rank r such that Write QxyMxy > 0 whenever Mxy ̸= 0. (B.5) Q = 􏰙⟨ux,vy⟩􏰚 x∈X, y∈Y for suitable collections of vectors {ux } ⊂ Rr and {vy } ⊂ Rr . If the vectors ux , vy are not already in general position, we can replace them with their slightly perturbed versions u ̃x, v ̃y that are in general position. Provided that the perturbations are small enough, property (B.5) will still hold, i.e., we will have ⟨u ̃x,v ̃y⟩Mxy > 0 whenever Mxy ̸= 0. As a result, we can assume w.l.o.g. that {ux},{vy} are in general position and in particular that all {vy} are nonzero. Since sign-rank(M) 􏰞 rank(M), we infer that \|X\| 􏰟 r. Theorem B.1 is therefore applicable to the set {ux} and yields a nonsingular matrix A with 􏰕 1 (Aux)(Aux)T = \|X\|Ir. (B.6) x∈X ∥Aux∥2 r THE SIGN-RANK OF AC0 23 􏰊 ⟨ux,vy⟩ 􏰋 R = ∥Aux∥ ∥(A−1)Tvy∥ It remains to verify properties (B.1)–(B.4). Property (B.1) follows from the repre- sentation R = D1QD2, where D1 and D2 are diagonal matrices with strictly positive diagonal entries. By (B.5), we know that RxyMxy > 0 whenever Mxy ̸= 0, which immediately gives us (B.2). Property (B.3) holds because \|⟨ux,vy⟩\| = \|⟨Aux,(A−1)Tvy⟩\| 􏰞1. ∥Aux∥ ∥(A−1)Tvy∥ ∥Aux∥ ∥(A−1)Tvy∥ Finally, property (B.4) will follow once we show that 􏰂x Rx2y = \|X\|/r for every y ∈ Y. So, fix y ∈ Y and consider the unit vector v = (A−1)Tvy/∥(A−1)Tvy∥. We have: 􏰕R2 =􏰕 ⟨ux,vy⟩2 x∈X, y∈Y where the last step follows from (B.6). ∥Aux∥2 ∥(A−1)Tvy∥2 = 􏰕 (vyTA−1)(Aux)(Aux)T(A−1)Tvy x∈X ∥Aux∥2 ∥(A−1)Tvy∥2 􏰎1􏰏 = vT 􏰕 (Aux)(Aux)T v x∈X ∥Aux∥2 = \|X\|, r	{"url":"https://www.cscodehelp.com/c-c-%E4%BB%A3%E5%86%99/cs%E4%BB%A3%E8%80%83%E8%AE%A1%E7%AE%97%E6%9C%BA%E4%BB%A3%E5%86%99-algorithm-the-sign-rank-of-ac0-%E2%88%97/","timestamp":"2024-11-03T13:14:29Z","content_type":"text/html","content_length":"114794","record_id":"<urn:uuid:37cd6853-af06-44dc-9034-d3b3d2ff2513>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00079.warc.gz"}
Sets and Lists Calling Sequence Thread Safety Parameters Examples Calling Sequence es - expression sequence • A set is an unordered sequence of distinct expressions enclosed in braces {}, representing a set in the mathematical sense. • A list is an ordered sequence of expressions enclosed in square brackets []. The ordering of the expressions is the ordering of es. • Note that es may be empty so that the empty set is represented by {} and the empty list is represented by []. • In output, the empty set is typeset as $\varnothing$. • The number of elements of a set or list S is given by numelems(S). Accessing the elements • The elements of a set or list can be extracted via the selection operation. Thus if S is a set or list then S[i] selects the ith element. Alternatively you may use op(i,S). The elements are numbered from 1 so S[1] extracts the first element. Negative subscripts can also be used. S[-1] selects the last element, S[-2] the second last, and so on. • Multiple elements of a list or set S can be extracted. The selection op(i..j,S) selects the sub-sequence (S[i],S[i+1],...,S[j]). If S is a set then the selection S[i..j] selects the subset {op(i..j,S)}. If S is a list then the selection S[i..j] selects the sublist [op(i..j,S)]. Negative subscripts can also be used. Thus S[2..-2] selects all elements except the first and last. • Lists and sets can be nested, in which case selection can be done in one of two ways: S [i][j]...[n] or S[i,j,...,n]. • To extract the contents of a list or set L as a sequence, use the op(L) function or the empty selection operator L[]. Modifying the elements Note: The following modifications to a list or set cause a new list or set to be created rather than modifying the original in-place. To save space and time, when the number of data or the number of modifications to them are large, use of an Array may be preferable. For more information on mutable and immutable data structures, refer to the Basic Data Structures chapter of the Maple Programming Guide. • Appending an element x to a list L is done by [op(L), x]. Inserting an element x to a set S is done using the union operator S union {x} ($S\cup {x}$). • Replacing the i-th element of a list L by x can be done by subsop(i=x, L). • The operation L[i] := x also works for reasons of convenience, but only for short lists. It is not recommended to use this operation unless you know that the list is small. • Deleting the i-th element of a list L is subsop(i=NULL, L). Deleting an element x from a set S is done using the minus operator S minus {x} ($S&setmn;{x}$). Set and list operations • To test if an element x is in a list or set L, use either member(x,L) or x in L ($x&in; • To test if a set S is a subset of a set T, use the subset operator S subset T ($S\ subseteq T$). • To find the intersection of two sets S and T, use S intersect T ($S\cap T$). • To find the union of two sets S and T, use S union T ($S\cup T$). • To find the difference of two sets S and T, use S minus T ($S&setmn;T$). • For more information on these set operations, see Set Operators. • Further list operations can be found in the package ListTools. Set ordering Sets have a deterministic ordering that, for most objects, is not based on runtime properties. This means that when {b,c,a} is input, the order will be fixed to {a,b,c} no matter when you created that set. A notable exception to this rule is when a set contains multiple mutable objects of the same type. For example, two vectors inside a set could appear in either order in different sessions. In general, object ordering is first determined by quickly identifiable features like its class or size. The default comparison order checks properties in the following order: 1. object id: this causes the same kind of data-structures to be grouped together. For example integers will come before exact rationals, floats, and complex numbers. 2. object length: this causes objects of the same size to be grouped together. The "length" is a property of the underlying data-structure. As a result, {-2^200, -2^100} is not in increasing numerical order and {bb, aaaaaaaa} is not in alphabetical order; in both cases the shorter data structure occurs first. 3. lexicographical or numerical order: this orders numbers in increasing numerical order and strings in lexicographic order. 4. recurse on components: this causes compound objects to be ordered according to the above properties of their child components. For example {f(z),g(a)} will pass the above properties as both f(z) and g(a) are function objects of the same length. So the next level of ordering will compare the names of each function, f and g, which will fall into rule 3. Note that the ordering of expressions like {x+1,1+y} may still be session dependent if the unique representation of x+1 is in the order of 1+x, or if the second expression is y+1 instead of 1+y. 5. address: in some cases the object will be identical in every way, such as the local name x and the global name x. When this happens the machine address of the given object will be used. This at least ensures the set order will be consistent within the running Note: This order could be changed in future versions of Maple, and/or objects could be represented in a different way in future versions. You should not assume that sets will be maintained in any particular order. The object-address based set ordering used in Maple 11 and earlier versions can be obtained by using the commandline option --setsort=0. Similarly, the new set order can be enabled with --setsort=1. Variations of --setsort=0..8 are available if you want to purposefully introduce ordering differences into your code to test its robustness. Settings 2 through 8 apply the same rules 1-5 listed above but alter whether each sub-ordering is increasing or decreasing. For more information on commandline options, see the maple help page. Thread Safety • The [] and {} constructors are thread safe as of Maple 15. • For more information on thread safety, see index/threadsafe. Repeated elements in a set are ignored. > {x,y,y}; ${{x}{,}{y}}$ (1) > {y,x,y}; ${{x}{,}{y}}$ (2) Sets of single letter variables are always sorted in alphabetical order. > {B,b,c,a,A}; ${{A}{,}{B}{,}{a}{,}{b}{,}{c}}$ (3) Sets of small integers are always in numerical order. > S:={5,4,3,2,1}; ${S}{&Assign;}{{1}{,}{2}{,}{3}{,}{4}{,}{5}}$ (4) > $S\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{union}\phantom{\rule[-0.0ex]{0.3em} ${{1}{,}{2}{,}{3}{,}{4}{,}{5}{,}{6}}$ (5) Set operators do not change the set. They create a new set. > $S$ ${{1}{,}{2}{,}{3}{,}{4}{,}{5}}$ (6) Set operators do not work on lists. > ${1,2,3}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\ ${{2}{,}{3}}$ (7) > $[1,2,3]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\ Error, invalid input: intersect received [1, 2, 3], which is not valid for its 1st The in operator ($\mathrm{&in;}$) tests for set or list membership. > $\mathrm{evalb}⁡\left(1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\ ${\mathrm{true}}$ (8) > $\mathrm{evalb}⁡\left(4\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\ ${\mathrm{false}}$ (9) > $S&Assign;{1,2}$ ${S}{&Assign;}{{1}{,}{2}}$ (10) > $T&Assign;{1,2,3}$ ${T}{&Assign;}{{1}{,}{2}{,}{3}}$ (11) Determine if S is a subset of T: > $S\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{subset}\phantom{\rule[-0.0ex]{0.3em} ${\mathrm{true}}$ (12) Difference between S and T is given by: > $T\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{minus}\phantom{\rule[-0.0ex]{0.3em} ${{3}}$ (13) Elements can be repeated inside a list. > $[x,y,y]$ $[{x}{,}{y}{,}{y}]$ (14) > $[y,x,y]$ $[{y}{,}{x}{,}{y}]$ (15) Create a list using the seq command. > $L&Assign;[\mathrm{seq}⁡\left(x[i],i=1&period; ${L}{&Assign;}[{{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}} (16) Return the number of elements as follows: > $\mathrm{numelems}⁡\left(L\right)$ ${4}$ (17) Extract specific elements from a list: > $L[2]$ ${{x}}_{{2}}$ (18) > $L[1&period;&period;2]$ $[{{x}}_{{1}}{,}{{x}}_{{2}}]$ (19) > $L[2&period;&period;-1]$ $[{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}]$ (20) Extract the contents of a list as a sequence: > $\mathrm{op}⁡\left(L\right)$ ${{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}$ (21) > $L[]$ ${{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}$ (22) > $\mathrm{op}⁡\left(L[2&period;&period;3]\right)$ ${{x}}_{{2}}{,}{{x}}_{{3}}$ (23) > $\mathrm{op}⁡\left(2&period;&period;3,L\right)$ ${{x}}_{{2}}{,}{{x}}_{{3}}$ (24) Append a new element to a list: > $L&Assign;[\mathrm{op}⁡\left(L\right),x[5]]$ ${L}{&Assign;}[{{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}} (25) Delete an element from a list: > $\mathrm{subsop}⁡\left(2=\mathrm{NULL},L\right)$ $[{{x}}_{{1}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}{,}{{x}}_{{5}}]$ (26) Test for list membership > $\mathrm{member}⁡\left(x[1],L\right)$ ${\mathrm{true}}$ (27) Nested lists: > $L&Assign;[1,[2,3],[4,[5,6]& ${L}{&Assign;}[{1}{,}[{2}{,}{3}]{,}[{4}{& (28) Elements from a nested list can be extracted using one of the following methods: > $L[3,2,1]$ ${5}$ (29) > $L[3][2][1]$ ${5}$ (30) Calling Sequence es - expression sequence • A set is an unordered sequence of distinct expressions enclosed in braces {}, representing a set in the mathematical sense. • A list is an ordered sequence of expressions enclosed in square brackets []. The ordering of the expressions is the ordering of es. • Note that es may be empty so that the empty set is represented by {} and the empty list is represented by []. • In output, the empty set is typeset as $\varnothing$. • The number of elements of a set or list S is given by numelems(S). Accessing the elements • The elements of a set or list can be extracted via the selection operation. Thus if S is a set or list then S[i] selects the ith element. Alternatively you may use op(i,S). The elements are numbered from 1 so S[1] extracts the first element. Negative subscripts can also be used. S[-1] selects the last element, S[-2] the second last, and so on. • Multiple elements of a list or set S can be extracted. The selection op(i..j,S) selects the sub-sequence (S[i],S[i+1],...,S[j]). If S is a set then the selection S[i..j] selects the subset {op(i..j,S)}. If S is a list then the selection S[i..j] selects the sublist [op(i..j,S)]. Negative subscripts can also be used. Thus S[2..-2] selects all elements except the first and last. • Lists and sets can be nested, in which case selection can be done in one of two ways: S [i][j]...[n] or S[i,j,...,n]. • To extract the contents of a list or set L as a sequence, use the op(L) function or the empty selection operator L[]. Modifying the elements Note: The following modifications to a list or set cause a new list or set to be created rather than modifying the original in-place. To save space and time, when the number of data or the number of modifications to them are large, use of an Array may be preferable. For more information on mutable and immutable data structures, refer to the Basic Data Structures chapter of the Maple Programming Guide. • Appending an element x to a list L is done by [op(L), x]. Inserting an element x to a set S is done using the union operator S union {x} ($S\cup {x}$). • Replacing the i-th element of a list L by x can be done by subsop(i=x, L). • The operation L[i] := x also works for reasons of convenience, but only for short lists. It is not recommended to use this operation unless you know that the list is small. • Deleting the i-th element of a list L is subsop(i=NULL, L). Deleting an element x from a set S is done using the minus operator S minus {x} ($S&setmn;{x}$). Set and list operations • To test if an element x is in a list or set L, use either member(x,L) or x in L ($x&in; • To test if a set S is a subset of a set T, use the subset operator S subset T ($S\ subseteq T$). • To find the intersection of two sets S and T, use S intersect T ($S\cap T$). • To find the union of two sets S and T, use S union T ($S\cup T$). • To find the difference of two sets S and T, use S minus T ($S&setmn;T$). • For more information on these set operations, see Set Operators. • Further list operations can be found in the package ListTools. Set ordering Sets have a deterministic ordering that, for most objects, is not based on runtime properties. This means that when {b,c,a} is input, the order will be fixed to {a,b,c} no matter when you created that set. A notable exception to this rule is when a set contains multiple mutable objects of the same type. For example, two vectors inside a set could appear in either order in different sessions. In general, object ordering is first determined by quickly identifiable features like its class or size. The default comparison order checks properties in the following order: 1. object id: this causes the same kind of data-structures to be grouped together. For example integers will come before exact rationals, floats, and complex numbers. 2. object length: this causes objects of the same size to be grouped together. The "length" is a property of the underlying data-structure. As a result, {-2^200, -2^100} is not in increasing numerical order and {bb, aaaaaaaa} is not in alphabetical order; in both cases the shorter data structure occurs first. 3. lexicographical or numerical order: this orders numbers in increasing numerical order and strings in lexicographic order. 4. recurse on components: this causes compound objects to be ordered according to the above properties of their child components. For example {f(z),g(a)} will pass the above properties as both f(z) and g(a) are function objects of the same length. So the next level of ordering will compare the names of each function, f and g, which will fall into rule 3. Note that the ordering of expressions like {x+1,1+y} may still be session dependent if the unique representation of x+1 is in the order of 1+x, or if the second expression is y+1 instead of 1+y. 5. address: in some cases the object will be identical in every way, such as the local name x and the global name x. When this happens the machine address of the given object will be used. This at least ensures the set order will be consistent within the running Note: This order could be changed in future versions of Maple, and/or objects could be represented in a different way in future versions. You should not assume that sets will be maintained in any particular order. The object-address based set ordering used in Maple 11 and earlier versions can be obtained by using the commandline option --setsort=0. Similarly, the new set order can be enabled with --setsort=1. Variations of --setsort=0..8 are available if you want to purposefully introduce ordering differences into your code to test its robustness. Settings 2 through 8 apply the same rules 1-5 listed above but alter whether each sub-ordering is increasing or decreasing. For more information on commandline options, see the maple help page. Thread Safety • The [] and {} constructors are thread safe as of Maple 15. • For more information on thread safety, see index/threadsafe. Repeated elements in a set are ignored. > {x,y,y}; ${{x}{,}{y}}$ (1) > {y,x,y}; ${{x}{,}{y}}$ (2) Sets of single letter variables are always sorted in alphabetical order. > {B,b,c,a,A}; ${{A}{,}{B}{,}{a}{,}{b}{,}{c}}$ (3) Sets of small integers are always in numerical order. > S:={5,4,3,2,1}; ${S}{&Assign;}{{1}{,}{2}{,}{3}{,}{4}{,}{5}}$ (4) > $S\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{union}\phantom{\rule[-0.0ex]{0.3em} ${{1}{,}{2}{,}{3}{,}{4}{,}{5}{,}{6}}$ (5) Set operators do not change the set. They create a new set. > $S$ ${{1}{,}{2}{,}{3}{,}{4}{,}{5}}$ (6) Set operators do not work on lists. > ${1,2,3}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\ ${{2}{,}{3}}$ (7) > $[1,2,3]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\ Error, invalid input: intersect received [1, 2, 3], which is not valid for its 1st The in operator ($\mathrm{&in;}$) tests for set or list membership. > $\mathrm{evalb}⁡\left(1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\ ${\mathrm{true}}$ (8) > $\mathrm{evalb}⁡\left(4\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\ ${\mathrm{false}}$ (9) > $S&Assign;{1,2}$ ${S}{&Assign;}{{1}{,}{2}}$ (10) > $T&Assign;{1,2,3}$ ${T}{&Assign;}{{1}{,}{2}{,}{3}}$ (11) Determine if S is a subset of T: > $S\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{subset}\phantom{\rule[-0.0ex]{0.3em} ${\mathrm{true}}$ (12) Difference between S and T is given by: > $T\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{minus}\phantom{\rule[-0.0ex]{0.3em} ${{3}}$ (13) Elements can be repeated inside a list. > $[x,y,y]$ $[{x}{,}{y}{,}{y}]$ (14) > $[y,x,y]$ $[{y}{,}{x}{,}{y}]$ (15) Create a list using the seq command. > $L&Assign;[\mathrm{seq}⁡\left(x[i],i=1&period; ${L}{&Assign;}[{{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}} (16) Return the number of elements as follows: > $\mathrm{numelems}⁡\left(L\right)$ ${4}$ (17) Extract specific elements from a list: > $L[2]$ ${{x}}_{{2}}$ (18) > $L[1&period;&period;2]$ $[{{x}}_{{1}}{,}{{x}}_{{2}}]$ (19) > $L[2&period;&period;-1]$ $[{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}]$ (20) Extract the contents of a list as a sequence: > $\mathrm{op}⁡\left(L\right)$ ${{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}$ (21) > $L[]$ ${{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}$ (22) > $\mathrm{op}⁡\left(L[2&period;&period;3]\right)$ ${{x}}_{{2}}{,}{{x}}_{{3}}$ (23) > $\mathrm{op}⁡\left(2&period;&period;3,L\right)$ ${{x}}_{{2}}{,}{{x}}_{{3}}$ (24) Append a new element to a list: > $L&Assign;[\mathrm{op}⁡\left(L\right),x[5]]$ ${L}{&Assign;}[{{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}} (25) Delete an element from a list: > $\mathrm{subsop}⁡\left(2=\mathrm{NULL},L\right)$ $[{{x}}_{{1}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}{,}{{x}}_{{5}}]$ (26) Test for list membership > $\mathrm{member}⁡\left(x[1],L\right)$ ${\mathrm{true}}$ (27) Nested lists: > $L&Assign;[1,[2,3],[4,[5,6]& ${L}{&Assign;}[{1}{,}[{2}{,}{3}]{,}[{4}{& (28) Elements from a nested list can be extracted using one of the following methods: > $L[3,2,1]$ ${5}$ (29) > $L[3][2][1]$ ${5}$ (30) • A set is an unordered sequence of distinct expressions enclosed in braces {}, representing a set in the mathematical sense. • A list is an ordered sequence of expressions enclosed in square brackets []. The ordering of the expressions is the ordering of es. • Note that es may be empty so that the empty set is represented by {} and the empty list is represented by []. • In output, the empty set is typeset as $\varnothing$. • The number of elements of a set or list S is given by numelems(S). Accessing the elements • The elements of a set or list can be extracted via the selection operation. Thus if S is a set or list then S[i] selects the ith element. Alternatively you may use op(i,S). The elements are numbered from 1 so S[1] extracts the first element. Negative subscripts can also be used. S[-1] selects the last element, S[-2] the second last, and so on. • Multiple elements of a list or set S can be extracted. The selection op(i..j,S) selects the sub-sequence (S[i],S[i+1],...,S[j]). If S is a set then the selection S[i..j] selects the subset {op (i..j,S)}. If S is a list then the selection S[i..j] selects the sublist [op(i..j,S)]. Negative subscripts can also be used. Thus S[2..-2] selects all elements except the first and last. • Lists and sets can be nested, in which case selection can be done in one of two ways: S[i][j]...[n] or S[i,j,...,n]. • To extract the contents of a list or set L as a sequence, use the op(L) function or the empty selection operator L[]. Modifying the elements Note: The following modifications to a list or set cause a new list or set to be created rather than modifying the original in-place. To save space and time, when the number of data or the number of modifications to them are large, use of an Array may be preferable. For more information on mutable and immutable data structures, refer to the Basic Data Structures chapter of the Maple Programming Guide. • Appending an element x to a list L is done by [op(L), x]. Inserting an element x to a set S is done using the union operator S union {x} ($S\cup {x}$). • Replacing the i-th element of a list L by x can be done by subsop(i=x, L). • The operation L[i] := x also works for reasons of convenience, but only for short lists. It is not recommended to use this operation unless you know that the list is small. • Deleting the i-th element of a list L is subsop(i=NULL, L). Deleting an element x from a set S is done using the minus operator S minus {x} ($S&setmn;{x}$). Set and list operations • To test if an element x is in a list or set L, use either member(x,L) or x in L ($x&in;L$). • To test if a set S is a subset of a set T, use the subset operator S subset T ($S\subseteq T$). • To find the intersection of two sets S and T, use S intersect T ($S\cap T$). • To find the union of two sets S and T, use S union T ($S\cup T$). • To find the difference of two sets S and T, use S minus T ($S&setmn;T$). • For more information on these set operations, see Set Operators. • Further list operations can be found in the package ListTools. Set ordering Sets have a deterministic ordering that, for most objects, is not based on runtime properties. This means that when {b,c,a} is input, the order will be fixed to {a,b,c} no matter when you created that set. A notable exception to this rule is when a set contains multiple mutable objects of the same type. For example, two vectors inside a set could appear in either order in different In general, object ordering is first determined by quickly identifiable features like its class or size. The default comparison order checks properties in the following order: 1. object id: this causes the same kind of data-structures to be grouped together. For example integers will come before exact rationals, floats, and complex numbers. 2. object length: this causes objects of the same size to be grouped together. The "length" is a property of the underlying data-structure. As a result, {-2^200, -2^100} is not in increasing numerical order and {bb, aaaaaaaa} is not in alphabetical order; in both cases the shorter data structure occurs first. 3. lexicographical or numerical order: this orders numbers in increasing numerical order and strings in lexicographic order. 4. recurse on components: this causes compound objects to be ordered according to the above properties of their child components. For example {f(z),g(a)} will pass the above properties as both f(z) and g(a) are function objects of the same length. So the next level of ordering will compare the names of each function, f and g, which will fall into rule 3. Note that the ordering of expressions like {x+1,1+y} may still be session dependent if the unique representation of x+1 is in the order of 1+x, or if the second expression is y+1 instead of 1+y. 5. address: in some cases the object will be identical in every way, such as the local name x and the global name x. When this happens the machine address of the given object will be used. This at least ensures the set order will be consistent within the running session. Note: This order could be changed in future versions of Maple, and/or objects could be represented in a different way in future versions. You should not assume that sets will be maintained in any particular order. The object-address based set ordering used in Maple 11 and earlier versions can be obtained by using the commandline option --setsort=0. Similarly, the new set order can be enabled with --setsort=1 . Variations of --setsort=0..8 are available if you want to purposefully introduce ordering differences into your code to test its robustness. Settings 2 through 8 apply the same rules 1-5 listed above but alter whether each sub-ordering is increasing or decreasing. For more information on commandline options, see the maple help page. • A set is an unordered sequence of distinct expressions enclosed in braces {}, representing a set in the mathematical sense. A set is an unordered sequence of distinct expressions enclosed in braces {}, representing a set in the mathematical sense. • A list is an ordered sequence of expressions enclosed in square brackets []. The ordering of the expressions is the ordering of es. A list is an ordered sequence of expressions enclosed in square brackets []. The ordering of the expressions is the ordering of es. • Note that es may be empty so that the empty set is represented by {} and the empty list is represented by []. Note that es may be empty so that the empty set is represented by {} and the empty list is represented by []. • In output, the empty set is typeset as $\varnothing$. • The number of elements of a set or list S is given by numelems(S). The number of elements of a set or list S is given by numelems(S). Accessing the elements • The elements of a set or list can be extracted via the selection operation. Thus if S is a set or list then S[i] selects the ith element. Alternatively you may use op(i,S). The elements are numbered from 1 so S[1] extracts the first element. Negative subscripts can also be used. S[-1] selects the last element, S[-2] the second last, and so on. • Multiple elements of a list or set S can be extracted. The selection op(i..j,S) selects the sub-sequence (S[i],S[i+1],...,S[j]). If S is a set then the selection S[i..j] selects the subset {op (i..j,S)}. If S is a list then the selection S[i..j] selects the sublist [op(i..j,S)]. Negative subscripts can also be used. Thus S[2..-2] selects all elements except the first and last. • Lists and sets can be nested, in which case selection can be done in one of two ways: S[i][j]...[n] or S[i,j,...,n]. • To extract the contents of a list or set L as a sequence, use the op(L) function or the empty selection operator L[]. • The elements of a set or list can be extracted via the selection operation. Thus if S is a set or list then S[i] selects the ith element. Alternatively you may use op(i,S). The elements are numbered from 1 so S[1] extracts the first element. Negative subscripts can also be used. S[-1] selects the last element, S[-2] the second last, and so on. The elements of a set or list can be extracted via the selection operation. Thus if S is a set or list then S[i] selects the ith element. Alternatively you may use op(i,S). The elements are numbered from 1 so S[1] extracts the first element. Negative subscripts can also be used. S[-1] selects the last element, S[-2] the second last, and so on. • Multiple elements of a list or set S can be extracted. The selection op(i..j,S) selects the sub-sequence (S[i],S[i+1],...,S[j]). If S is a set then the selection S[i..j] selects the subset {op (i..j,S)}. If S is a list then the selection S[i..j] selects the sublist [op(i..j,S)]. Negative subscripts can also be used. Thus S[2..-2] selects all elements except the first and last. Multiple elements of a list or set S can be extracted. The selection op(i..j,S) selects the sub-sequence (S[i],S[i+1],...,S[j]). If S is a set then the selection S[i..j] selects the subset {op (i..j,S)}. If S is a list then the selection S[i..j] selects the sublist [op(i..j,S)]. Negative subscripts can also be used. Thus S[2..-2] selects all elements except the first and last. • Lists and sets can be nested, in which case selection can be done in one of two ways: S[i][j]...[n] or S[i,j,...,n]. Lists and sets can be nested, in which case selection can be done in one of two ways: S[i][j]...[n] or S[i,j,...,n]. • To extract the contents of a list or set L as a sequence, use the op(L) function or the empty selection operator L[]. To extract the contents of a list or set L as a sequence, use the op(L) function or the empty selection operator L[]. Modifying the elements Note: The following modifications to a list or set cause a new list or set to be created rather than modifying the original in-place. To save space and time, when the number of data or the number of modifications to them are large, use of an Array may be preferable. For more information on mutable and immutable data structures, refer to the Basic Data Structures chapter of the Maple Programming • Appending an element x to a list L is done by [op(L), x]. Inserting an element x to a set S is done using the union operator S union {x} ($S\cup {x}$). • Replacing the i-th element of a list L by x can be done by subsop(i=x, L). • The operation L[i] := x also works for reasons of convenience, but only for short lists. It is not recommended to use this operation unless you know that the list is small. • Deleting the i-th element of a list L is subsop(i=NULL, L). Deleting an element x from a set S is done using the minus operator S minus {x} ($S&setmn;{x}$). Note: The following modifications to a list or set cause a new list or set to be created rather than modifying the original in-place. To save space and time, when the number of data or the number of modifications to them are large, use of an Array may be preferable. For more information on mutable and immutable data structures, refer to the Basic Data Structures chapter of the Maple Programming • Appending an element x to a list L is done by [op(L), x]. Inserting an element x to a set S is done using the union operator S union {x} ($S\cup {x}$). Appending an element x to a list L is done by [op(L), x]. Inserting an element x to a set S is done using the union operator S union {x} ($S\cup {x}$). • Replacing the i-th element of a list L by x can be done by subsop(i=x, L). Replacing the i-th element of a list L by x can be done by subsop(i=x, L). • The operation L[i] := x also works for reasons of convenience, but only for short lists. It is not recommended to use this operation unless you know that the list is small. The operation L[i] := x also works for reasons of convenience, but only for short lists. It is not recommended to use this operation unless you know that the list is small. • Deleting the i-th element of a list L is subsop(i=NULL, L). Deleting an element x from a set S is done using the minus operator S minus {x} ($S&setmn;{x}$). Deleting the i-th element of a list L is subsop(i=NULL, L). Deleting an element x from a set S is done using the minus operator S minus {x} ($S&setmn;{x}$). Set and list operations • To test if an element x is in a list or set L, use either member(x,L) or x in L ($x&in;L$). • To test if a set S is a subset of a set T, use the subset operator S subset T ($S\subseteq T$). • To find the intersection of two sets S and T, use S intersect T ($S\cap T$). • To find the union of two sets S and T, use S union T ($S\cup T$). • To find the difference of two sets S and T, use S minus T ($S&setmn;T$). • For more information on these set operations, see Set Operators. • Further list operations can be found in the package ListTools. • To test if an element x is in a list or set L, use either member(x,L) or x in L ($x&in;L$). To test if an element x is in a list or set L, use either member(x,L) or x in L ($x&in;L$). • To test if a set S is a subset of a set T, use the subset operator S subset T ($S\subseteq T$). To test if a set S is a subset of a set T, use the subset operator S subset T ($S\subseteq T$). • To find the intersection of two sets S and T, use S intersect T ($S\cap T$). To find the intersection of two sets S and T, use S intersect T ($S\cap T$). • To find the union of two sets S and T, use S union T ($S\cup T$). To find the union of two sets S and T, use S union T ($S\cup T$). • To find the difference of two sets S and T, use S minus T ($S&setmn;T$). To find the difference of two sets S and T, use S minus T ($S&setmn;T$). • For more information on these set operations, see Set Operators. For more information on these set operations, see Set Operators. • Further list operations can be found in the package ListTools. Further list operations can be found in the package ListTools. Set ordering Sets have a deterministic ordering that, for most objects, is not based on runtime properties. This means that when {b,c,a} is input, the order will be fixed to {a,b,c} no matter when you created that set. A notable exception to this rule is when a set contains multiple mutable objects of the same type. For example, two vectors inside a set could appear in either order in different sessions. In general, object ordering is first determined by quickly identifiable features like its class or size. The default comparison order checks properties in the following order: 1. object id: this causes the same kind of data-structures to be grouped together. For example integers will come before exact rationals, floats, and complex numbers. 2. object length: this causes objects of the same size to be grouped together. The "length" is a property of the underlying data-structure. As a result, {-2^200, -2^100} is not in increasing numerical order and {bb, aaaaaaaa} is not in alphabetical order; in both cases the shorter data structure occurs first. 3. lexicographical or numerical order: this orders numbers in increasing numerical order and strings in lexicographic order. 4. recurse on components: this causes compound objects to be ordered according to the above properties of their child components. For example {f(z),g(a)} will pass the above properties as both f(z) and g(a) are function objects of the same length. So the next level of ordering will compare the names of each function, f and g, which will fall into rule 3. Note that the ordering of expressions like {x+1,1+y} may still be session dependent if the unique representation of x+1 is in the order of 1+x, or if the second expression is y+1 instead of 1+y. 5. address: in some cases the object will be identical in every way, such as the local name x and the global name x. When this happens the machine address of the given object will be used. This at least ensures the set order will be consistent within the running session. Note: This order could be changed in future versions of Maple, and/or objects could be represented in a different way in future versions. You should not assume that sets will be maintained in any particular order. The object-address based set ordering used in Maple 11 and earlier versions can be obtained by using the commandline option --setsort=0. Similarly, the new set order can be enabled with --setsort=1. Variations of --setsort=0..8 are available if you want to purposefully introduce ordering differences into your code to test its robustness. Settings 2 through 8 apply the same rules 1-5 listed above but alter whether each sub-ordering is increasing or decreasing. For more information on commandline options, see the maple help page. Sets have a deterministic ordering that, for most objects, is not based on runtime properties. This means that when {b,c,a} is input, the order will be fixed to {a,b,c} no matter when you created that set. A notable exception to this rule is when a set contains multiple mutable objects of the same type. For example, two vectors inside a set could appear in either order in different sessions. In general, object ordering is first determined by quickly identifiable features like its class or size. The default comparison order checks properties in the following order: 1. object id: this causes the same kind of data-structures to be grouped together. For example integers will come before exact rationals, floats, and complex numbers. object id: this causes the same kind of data-structures to be grouped together. For example integers will come before exact rationals, floats, and complex numbers. 2. object length: this causes objects of the same size to be grouped together. The "length" is a property of the underlying data-structure. As a result, {-2^200, -2^100} is not in increasing numerical order and {bb, aaaaaaaa} is not in alphabetical order; in both cases the shorter data structure occurs first. object length: this causes objects of the same size to be grouped together. The "length" is a property of the underlying data-structure. As a result, {-2^200, -2^100} is not in increasing numerical order and {bb, aaaaaaaa} is not in alphabetical order; in both cases the shorter data structure occurs first. 3. lexicographical or numerical order: this orders numbers in increasing numerical order and strings in lexicographic order. lexicographical or numerical order: this orders numbers in increasing numerical order and strings in lexicographic order. 4. recurse on components: this causes compound objects to be ordered according to the above properties of their child components. For example {f(z),g(a)} will pass the above properties as both f(z) and g(a) are function objects of the same length. So the next level of ordering will compare the names of each function, f and g, which will fall into rule 3. Note that the ordering of expressions like {x+1,1+y} may still be session dependent if the unique representation of x+1 is in the order of 1+x, or if the second expression is y+1 instead of 1+y. recurse on components: this causes compound objects to be ordered according to the above properties of their child components. For example {f(z),g(a)} will pass the above properties as both f(z) and g(a) are function objects of the same length. So the next level of ordering will compare the names of each function, f and g, which will fall into rule 3. Note that the ordering of expressions like {x+1,1+y} may still be session dependent if the unique representation of x+1 is in the order of 1+x, or if the second expression is y+1 instead of 1+y. 5. address: in some cases the object will be identical in every way, such as the local name x and the global name x. When this happens the machine address of the given object will be used. This at least ensures the set order will be consistent within the running session. address: in some cases the object will be identical in every way, such as the local name x and the global name x. When this happens the machine address of the given object will be used. This at least ensures the set order will be consistent within the running session. Note: This order could be changed in future versions of Maple, and/or objects could be represented in a different way in future versions. You should not assume that sets will be maintained in any particular order. The object-address based set ordering used in Maple 11 and earlier versions can be obtained by using the commandline option --setsort=0. Similarly, the new set order can be enabled with --setsort=1. Variations of --setsort=0..8 are available if you want to purposefully introduce ordering differences into your code to test its robustness. Settings 2 through 8 apply the same rules 1-5 listed above but alter whether each sub-ordering is increasing or decreasing. For more information on commandline options, see the maple help page. Thread Safety • The [] and {} constructors are thread safe as of Maple 15. • For more information on thread safety, see index/threadsafe. • The [] and {} constructors are thread safe as of Maple 15. The [] and {} constructors are thread safe as of Maple 15. Repeated elements in a set are ignored. > {x,y,y}; ${{x}{,}{y}}$ (1) > {y,x,y}; ${{x}{,}{y}}$ (2) Sets of single letter variables are always sorted in alphabetical order. > {B,b,c,a,A}; ${{A}{,}{B}{,}{a}{,}{b}{,}{c}}$ (3) Sets of small integers are always in numerical order. > S:={5,4,3,2,1}; ${S}{&Assign;}{{1}{,}{2}{,}{3}{,}{4}{,}{5}}$ (4) > $S\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{union}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{3,6}$ ${{1}{,}{2}{,}{3}{,}{4}{,}{5}{,}{6}}$ (5) Set operators do not change the set. They create a new set. > $S$ ${{1}{,}{2}{,}{3}{,}{4}{,}{5}}$ (6) Set operators do not work on lists. > ${1,2,3}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{2,3,4}$ ${{2}{,}{3}}$ (7) > $[1,2,3]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{2,3,4}$ Error, invalid input: intersect received [1, 2, 3], which is not valid for its 1st argument The in operator ($\mathrm{&in;}$) tests for set or list membership. > $\mathrm{evalb}⁡\left(1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{1,2,3}\right)$ ${\mathrm{true}}$ (8) > $\mathrm{evalb}⁡\left(4\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}[1,2,3]\right)$ ${\mathrm{false}}$ (9) > $S&Assign;{1,2}$ ${S}{&Assign;}{{1}{,}{2}}$ (10) > $T&Assign;{1,2,3}$ ${T}{&Assign;}{{1}{,}{2}{,}{3}}$ (11) Determine if S is a subset of T: > $S\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{subset}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}T$ ${\mathrm{true}}$ (12) Difference between S and T is given by: > $T\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{minus}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}S$ ${{3}}$ (13) Elements can be repeated inside a list. > $[x,y,y]$ $[{x}{,}{y}{,}{y}]$ (14) > $[y,x,y]$ $[{y}{,}{x}{,}{y}]$ (15) Create a list using the seq command. > $L&Assign;[\mathrm{seq}⁡\left(x[i],i=1&period;&period;4\right)]$ ${L}{&Assign;}[{{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}]$ (16) Return the number of elements as follows: > $\mathrm{numelems}⁡\left(L\right)$ ${4}$ (17) Extract specific elements from a list: > $L[2]$ ${{x}}_{{2}}$ (18) > $L[1&period;&period;2]$ $[{{x}}_{{1}}{,}{{x}}_{{2}}]$ (19) > $L[2&period;&period;-1]$ $[{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}]$ (20) Extract the contents of a list as a sequence: > $\mathrm{op}⁡\left(L\right)$ ${{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}$ (21) > $L[]$ ${{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}$ (22) > $\mathrm{op}⁡\left(L[2&period;&period;3]\right)$ ${{x}}_{{2}}{,}{{x}}_{{3}}$ (23) > $\mathrm{op}⁡\left(2&period;&period;3,L\right)$ ${{x}}_{{2}}{,}{{x}}_{{3}}$ (24) Append a new element to a list: > $L&Assign;[\mathrm{op}⁡\left(L\right),x[5]]$ ${L}{&Assign;}[{{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}{,}{{x}}_{{5}}]$ (25) Delete an element from a list: > $\mathrm{subsop}⁡\left(2=\mathrm{NULL},L\right)$ $[{{x}}_{{1}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}{,}{{x}}_{{5}}]$ (26) Test for list membership > $\mathrm{member}⁡\left(x[1],L\right)$ ${\mathrm{true}}$ (27) Nested lists: > $L&Assign;[1,[2,3],[4,[5,6],7],8,9]$ ${L}{&Assign;}[{1}{,}[{2}{,}{3}]{,}[{4}{,}[{5}{,}{6}]{,}{7}]{,}{8}{,}{9}]$ (28) Elements from a nested list can be extracted using one of the following methods: > $L[3,2,1]$ ${5}$ (29) > $L[3][2][1]$ ${5}$ (30) Repeated elements in a set are ignored. > {x,y,y}; ${{x}{,}{y}}$ (1) > {y,x,y}; ${{x}{,}{y}}$ (2) Sets of single letter variables are always sorted in alphabetical order. > {B,b,c,a,A}; ${{A}{,}{B}{,}{a}{,}{b}{,}{c}}$ (3) Sets of small integers are always in numerical order. > S:={5,4,3,2,1}; ${S}{&Assign;}{{1}{,}{2}{,}{3}{,}{4}{,}{5}}$ (4) > $S\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{union}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{3,6}$ ${{1}{,}{2}{,}{3}{,}{4}{,}{5}{,}{6}}$ (5) Set operators do not change the set. They create a new set. > $S$ ${{1}{,}{2}{,}{3}{,}{4}{,}{5}}$ (6) Set operators do not work on lists. > ${1,2,3}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{2,3,4}$ ${{2}{,}{3}}$ (7) > $[1,2,3]\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{intersect}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{2,3,4}$ Error, invalid input: intersect received [1, 2, 3], which is not valid for its 1st argument The in operator ($\mathrm{&in;}$) tests for set or list membership. > $\mathrm{evalb}⁡\left(1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}{1,2,3}\right)$ ${\mathrm{true}}$ (8) > $\mathrm{evalb}⁡\left(4\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}[1,2,3]\right)$ ${\mathrm{false}}$ (9) > $S&Assign;{1,2}$ ${S}{&Assign;}{{1}{,}{2}}$ (10) > $T&Assign;{1,2,3}$ ${T}{&Assign;}{{1}{,}{2}{,}{3}}$ (11) Determine if S is a subset of T: > $S\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{subset}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}T$ ${\mathrm{true}}$ (12) Difference between S and T is given by: > $T\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{minus}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}S$ ${{3}}$ (13) Sets of single letter variables are always sorted in alphabetical order. Set operators do not change the set. They create a new set. Error, invalid input: intersect received [1, 2, 3], which is not valid for its 1st argument The in operator ($\mathrm{&in;}$) tests for set or list membership. Elements can be repeated inside a list. > $[x,y,y]$ $[{x}{,}{y}{,}{y}]$ (14) > $[y,x,y]$ $[{y}{,}{x}{,}{y}]$ (15) Create a list using the seq command. > $L&Assign;[\mathrm{seq}⁡\left(x[i],i=1&period;&period;4\right)]$ ${L}{&Assign;}[{{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}]$ (16) Return the number of elements as follows: > $\mathrm{numelems}⁡\left(L\right)$ ${4}$ (17) Extract specific elements from a list: > $L[2]$ ${{x}}_{{2}}$ (18) > $L[1&period;&period;2]$ $[{{x}}_{{1}}{,}{{x}}_{{2}}]$ (19) > $L[2&period;&period;-1]$ $[{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}]$ (20) Extract the contents of a list as a sequence: > $\mathrm{op}⁡\left(L\right)$ ${{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}$ (21) > $L[]$ ${{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}$ (22) > $\mathrm{op}⁡\left(L[2&period;&period;3]\right)$ ${{x}}_{{2}}{,}{{x}}_{{3}}$ (23) > $\mathrm{op}⁡\left(2&period;&period;3,L\right)$ ${{x}}_{{2}}{,}{{x}}_{{3}}$ (24) Append a new element to a list: > $L&Assign;[\mathrm{op}⁡\left(L\right),x[5]]$ ${L}{&Assign;}[{{x}}_{{1}}{,}{{x}}_{{2}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}{,}{{x}}_{{5}}]$ (25) Delete an element from a list: > $\mathrm{subsop}⁡\left(2=\mathrm{NULL},L\right)$ $[{{x}}_{{1}}{,}{{x}}_{{3}}{,}{{x}}_{{4}}{,}{{x}}_{{5}}]$ (26) Test for list membership > $\mathrm{member}⁡\left(x[1],L\right)$ ${\mathrm{true}}$ (27) Nested lists: > $L&Assign;[1,[2,3],[4,[5,6],7],8,9]$ ${L}{&Assign;}[{1}{,}[{2}{,}{3}]{,}[{4}{,}[{5}{,}{6}]{,}{7}]{,}{8}{,}{9}]$ (28) Elements from a nested list can be extracted using one of the following methods: > $L[3,2,1]$ ${5}$ (29) > $L[3][2][1]$ ${5}$ (30) Elements from a nested list can be extracted using one of the following methods: See Also list Maple Debugger command	{"url":"https://www.maplesoft.com/support/help/view.aspx?path=set&L=E","timestamp":"2024-11-10T11:48:12Z","content_type":"text/html","content_length":"249378","record_id":"<urn:uuid:e8effdb1-84c4-4ac9-b2a7-73d3f3c5870e>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00702.warc.gz"}
Kavli IPMU Komaba Seminar Seminar information archive ～11/09｜Next seminar｜Future seminars 11/10～ Date, time & place Monday 16:30 - 18:00 002Room #002 (Graduate School of Math. Sci. Bldg.) Seminar information archive 13:30-14:30 Room #056 (Graduate School of Math. Sci. Bldg.) Edouard Brezin (lpt ens, Paris) Various applications of supersymmetry in statistical physics (English) [ Abstract ] Supersymmetry is a fundamental concept in particle physics (although it has not been seen experimentally so far). But it is although a powerful tool in a number of problems arising in quantum mechanics and statistical physics. It has been widely used in the theory of disordered systems (Efetov et al.), it led to dimensional reduction for branched polymers (Parisi-Sourlas), for the susy classical gas (Brydges and Imbrie), for Landau levels with impurities. If has also many powerful applications in the theory of random matrices. I will briefly review some of these topics. 10:30-11:30 Room #128 (Graduate School of Math. Sci. Bldg.) Naichung Conan Leung (The Chinese University of Hong Kong) Donaldson-Thomas theory for Calabi-Yau fourfolds. [ Abstract ] Donaldson-Thomas theory for Calabi-Yau threefolds is a complexification of Chern-Simons theory. In this talk I will discuss my joint work with Cao on the complexification of Donaldson theory. This work is supported by a RGC grant of HK Government. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Anatol Kirillov (RIMS, Kyoto University) On some quadratic algebras with applications to Topology, Algebra, Combinatorics, Schubert Calculus and Integrable Systems. (ENGLISH) [ Abstract ] The main purpose of my talk is to draw attention of the participants of the seminar to a certain family of quadratic algebras which has a wide range of applications to the subject mentioned in the title of my talk. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) A.P. Veselov (Loughborough, UK and Tokyo) Universal formulae for Lie groups and Chern-Simons theory (ENGLISH) [ Abstract ] In 1990s Vogel introduced an interesting parametrization of simple Lie algebras by 3 parameters defined up to a common multiple and permutations. Numerical characteristic is called universal if it can be expressed in terms of Vogel's parameters (example - the dimension of Lie algebra). I will discuss some universal formulae for Lie groups and Chern-Simons theory on 3D sphere. The talk is based on joint work with R.L. Mkrtchyan and A.N. Sergeev. 17:00-18:30 Room #128 (Graduate School of Math. Sci. Bldg.) Hans Jockers (The University of Bonn) Characteristic classes from 2d renormalized sigma-models (ENGLISH) [ Abstract ] The Hirzebruch-Riemann-Roch formula relates the holomorphic Euler characteristic of holomorphic vector bundles to topological invariants of compact complex manifold. I will explain a generalization of the Mukai's modified first Chern character map, which introduces certain characteristic classes that have not been considered in this form by Hirzebruch. This naturally leads to the characteristic Gamma class based on the Gamma function. The characteristic Gamma class has a surprising relation to the quantum theory of certain 2d sigma-models with compact complex manifolds as their target spaces. I will argue that the Gamma class describes perturbative quantum corrections to the classical theory of those sigma models. 17:00-18:30 Room #002 (Graduate School of Math. Sci. Bldg.) Mauricio Romo (Kavli IPMU) Exact Results In Two-Dimensional (2,2) Supersymmetric Gauge Theories With Boundary [ Abstract ] I will talk about the recent computation, done in joint work with Prof. K. Hori, of the partition function on the hemisphere of a class of two-dimensional (2,2) supersymmetric field theories including gauged linear sigma models (GLSM). The result provides a general exact formula for the central charge of the D-branes placed at the boundary. From the mathematical point of view, for the case of GLSMs that admit a geometrical interpretation, this formula provides an expression for the central charge of objects in the derived category at any point of the stringy Kahler moduli space. I will describe how this formula arises from physics and give simple, yet important, examples that supports its validity. If time allows, I will also explain some of its consequences such as how it can be used to obtain the grade restriction rule for branes near phase boundaries. 17:00-18:30 Room #002 (Graduate School of Math. Sci. Bldg.) Daniel Pomerleano (Kavli IPMU) Homological Mirror Symmetry for toric Calabi-Yau varieties (ENGLISH) [ Abstract ] I will discuss some recent developments in Homological Mirror Symmetry for toric Calabi-Yau varieties. 17:00-18:30 Room #122 (Graduate School of Math. Sci. Bldg.) Richard Eager (Kavli IPMU) Elliptic genera and two dimensional gauge theories (ENGLISH) [ Abstract ] The elliptic genus is an important invariant of two dimensional conformal field theories that generalizes the Witten index. In this talk, I will first review the geometric meaning of the elliptic genus and Witten's GLSM construction. Then I will explain how the elliptic genus can be computed directly from a two dimensional gauge theory using localization. The central example of this talk will be the quintic threefold. The GLSM description of the quintic threefold has both a large-volume sigma model description and a Landau-Ginzburg description. I will explain how the GLSM calculation of the index reproduces the old results in these two phases. Time permitting, further applications and generalizations will be discussed. 17:00-18:30 Room #002 (Graduate School of Math. Sci. Bldg.) Atsushi Kanazawa (University of British Columbia) Calabi-Yau threefolds of Type K (ENGLISH) [ Abstract ] We will provide a full classification of Calabi-Yau threefolds of Type K studied by Oguiso and Sakurai. Our study completes the classification of Calabi-Yau threefolds with infinite fundamental group. I will then discuss special Lagrangian T3-fibrations of Calabi-Yau threefolds of type K. This talk is based on a joint work with Kenji Hashimoto. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Changzheng Li (Kavli IPMU) Quantum cohomology of flag varieties (ENGLISH) [ Abstract ] In this talk, I will give a brief introduction to the quantum cohomology of flag varieties first. Then I will introduce a Z^2-filtration on the quantum cohomology of complete flag varieties. In the end, we will study the quantum Pieri rules for complex/symplectic Grassmannians, as applications of the Z^2-filtration. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Bong Lian (Brandeis University) Period Integrals and Tautological Systems (ENGLISH) [ Abstract ] We develop a global Poincar\\'e residue formula to study period integrals of families of complex manifolds. For any compact complex manifold $X$ equipped with a linear system $V^$ of generically smooth CY hypersurfaces, the formula expresses period integrals in terms of a canonical global meromorphic top form on $X$. Two important ingredients of this construction are the notion of a CY principal bundle, and a classification of such rank one bundles. We also generalize the construction to CY and general type complete intersections. When $X$ is an algebraic manifold having a sufficiently large automorphism group $G$ and $V^$ is a linear representation of $G$, we construct a holonomic D-module that governs the period integrals. The construction is based in part on the theory of tautological systems we have developed earlier. The approach allows us to explicitly describe a Picard-Fuchs type system for complete intersection varieties of general types, as well as CY, in any Fano variety, and in a homogeneous space in particular. In addition, the approach provides a new perspective of old examples such as CY complete intersections in a toric variety or partial flag variety. The talk is based on recent joint work with R. Song and S.T. Yau. 17:00-18:30 Room #002 (Graduate School of Math. Sci. Bldg.) Emanuel Scheidegger (The University of Freiburg) Topological Strings on Elliptic Fibrations (ENGLISH) [ Abstract ] We will explain a conjecture that expresses the BPS invariants (Gopakumar-Vafa invariants) for elliptically fibered Calabi-Yau threefolds in terms of modular forms. In particular, there is a recursion relation which governs these modular forms. Evidence comes from the polynomial formulation of the higher genus topological string amplitudes with insertions. 14:45-16:15 Room #056 (Graduate School of Math. Sci. Bldg.) Albrecht Klemm (The University of Bonn) Refined holomorphic anomaly equations (ENGLISH) [ Abstract ] We propose a derivation of refined holomophic anomaly equation from the word-sheet point of view and discuss the interpretation of the refined BPS invariants for local Calabi-Yau 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Siu-Cheong Lau Enuemerative meaning of mirror maps for toric Calabi-Yau manifolds (ENGLISH) [ Abstract ] For a mirror pair of smooth manifolds X and Y, mirror symmetry associates a complex structure on Y to each Kaehler structure on X, and this association is called the mirror map. Traditionally mirror maps are defined by solving Picard-Fuchs equations and its geometric meaning was unclear. In this talk I explain a recent joint work with K.W. Chan, N.C. Leung and H.H. Tseng which proves that mirror maps can be obtained by taking torus duality (the SYZ approach) and disk-counting for a class of toric Calabi-Yau manifolds in any dimensions. As a consequence we can compute disk-counting invariants by solving Picard-Fuchs equations. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Kwok-Wai Chan (IPMU, the University of Tokyo) Mirror symmetry for toric Calabi-Yau manifolds from the SYZ viewpoint (ENGLISH) [ Abstract ] In this talk, I will discuss mirror symmetry for toric Calabi-Yau (CY) manifolds from the viewpoint of the SYZ program. I will start with a special Lagrangian torus fibration on a toric CY manifold, and then construct its instanton-corrected mirror by a T-duality modified by quantum corrections. A remarkable feature of this construction is that the mirror family is inherently written in canonical flat coordinates. As a consequence, we get a conjectural enumerative meaning for the inverse mirror maps. If time permits, I will explain the verification of this conjecture in several examples via a formula which computes open Gromov-Witten invariants for toric manifolds. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Scott Carnahan Borcherds products in monstrous moonshine. (ENGLISH) [ Abstract ] During the 1980s, Koike, Norton, and Zagier independently found an infinite product expansion for the difference of two modular j-functions on a product of half planes. Borcherds showed that this product identity is the Weyl denominator formula for an infinite dimensional Lie algebra that has an action of the monster simple group by automorphisms, and used this action to prove the monstrous moonshine conjectures. I will describe a more general construction that yields an infinite product identity and an infinite dimensional Lie algebra for each element of the monster group. The above objects then arise as the special cases assigned to the identity element. Time permitting, I will attempt to describe a connection to conformal field theory. 14:40-16:10 Room #002 (Graduate School of Math. Sci. Bldg.) Tomoo Matsumura (Cornell University) Hamiltonian torus actions on orbifolds and orbifold-GKM theorem (joint work with T. Holm) (JAPANESE) [ Abstract ] When a symplectic manifold M carries a Hamiltonian torus R action, the injectivity theorem states that the R-equivariant cohomology of M is a subring of the one of the fixed points and the GKM theorem allows us to compute this subring by only using the data of 1-dimensional orbits. The results in the first part of this talk are a generalization of this technique to Hamiltonian R actions on orbifolds and an application to the computation of the equivariant cohomology of toric orbifolds. In the second part, we will introduce the equivariant Chen-Ruan cohomology ring which is a symplectic invariant of the action on the orbifold and explain the injectivity/GKM theorem for this ring. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Todor Milanov Quasi-modular forms and Gromov--Witten theory of elliptic orbifold $\\mathbb{P}^1$ (ENGLISH) [ Abstract ] This talk is based on my current work with Y. Ruan. Our project is part of the so called Landau--Ginzburg/Calabi-Yau correspondence. The latter is a conjecture, due to Ruan, that describes the relation between the $W$-spin invariants of a Landau-Ginzburg potential $W$ and the Gromov--Witten invariants of a certain Calabi--Yau orbifold. I am planning first to explain the higher-genus reconstruction formalism of Givental. This formalism together with the work of M. Krawitz and Y. Shen allows us to express the Gromov--Witten invariants of the orbifold $\\mathbb{P}^1$'s with weights $(3,3,3)$, $(2,4,4)$, and $(2,3,6)$ in terms of Saito's Frobenius structure associated with the simple elliptic singularities $P_8$, $X_9$, and $J_{10}$ respectively. After explaining Givental's formalism, my goal would be to discuss the Saito's flat structure, and to explain how its modular behavior fits in the Givental's formalism. This allows us to prove that the Gromov--Witten invariants are quasi-modular forms on an appropriate modular group. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Akishi Ikeda (The University of Tokyo) The correspondence between Frobenius algebra of Hurwitz numbers and matrix models (JAPANESE) [ Abstract ] The number of branched coverings of closed surfaces are called Hurwitz numbers. They constitute a Frobenius algebra structure, or two dimensional topological field theory. On the other hand, correlation functions of matrix models are expressed in term of ribbon graphs (graphs embedded in closed surfaces). In this talk, I explain how the Frobenius algebra structure of Hurwitz numbers are described in terms of matrix models. We use the correspondence between ribbon graphs and covering of S^2 ramified at three points, both of which have natural symmetric group actions. As an application I use Frobenius algebra structure to compute Hermitian matrix models, multi-variable matrix models, and their large N expansions. The generating function of Hurwitz numbers is also expressed in terms of matrix models. The relation to integrable hierarchies and random partitions is briefly discussed. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Timur Sadykov (Siberian Federal University) Bases in the solution space of the Mellin system [ Abstract ] I will present a joint work with Alicia Dickenstein. We consider algebraic functions $z$ satisfying equations of the a_0 z^m + a_1z^{m_1} + a_2 z^{m_2} + \\ldots + a_n z^{m_n} + a_{n+1} =0. Here $m > m_1 > \\ldots > m_n>0,$ $m,m_i \\in \\N,$ and $z=z(a_0,\\ldots,a_{n+1})$ is a function of the complex variables $a_0, \\ldots, a_{n+1}.$ Solutions to such equations are classically known to satisfy holonomic systems of linear partial differential equations with polynomial coefficients. In the talk I will investigate one of such systems of differential equations which was introduced by Mellin. We compute the holonomic rank of the Mellin system as well as the dimension of the space of its algebraic solutions. Moreover, we construct explicit bases of solutions in terms of the roots of initial algebraic equation and their logarithms. We show that the monodromy of the Mellin system is always reducible and give some factorization results in the univariate case. 17:30-19:00 Room #002 (Graduate School of Math. Sci. Bldg.) Weiping Zhang (Chern Institute of Mathematics, Nankai University) Geometric quantization on noncompact manifolds [ Abstract ] We will describe our analytic approach with Youlinag Tian to the Guillemin-Sternberg geometric quantization conjecture which can be summarized as "quantization commutes with reduction". We will aslo describe a recent extension to the case of noncompact symplectic manifolds. This is a joint work with Xiaonan Ma in which we solve a conjecture of Vergne mentioned in her ICM2006 plenary lecture. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Junya Yagi (Rutgers University) Chiral Algebras of (0,2) Models: Beyond Perturbation Theory [ Abstract ] The chiral algebras of two-dimensional sigma models with (0,2) supersymmetry are infinite-dimensional generalizations of the chiral rings of (2,2) models. Perturbatively, they enjoy rich mathematical structures described by sheaves of chiral differential operators. Nonperturbatively, however, they vanish completely for certain (0,2) models with no left-moving fermions. In this talk, I will explain how this vanishing phenomenon takes places. The vanishing of the chiral algebra of a (0, 2) model implies that supersymmetry is spontaneously broken in the model, which in turn suggests that no harmonic spinors exist on the loop space of the target space. In particular, the elliptic genus of the model vanishes, thereby providing a physics proof of a special case of the Hoelhn-Stolz conjecture. 16:30-18:00 Room #002 (Graduate School of Math. Sci. Bldg.) Makoto Sakurai Differential Graded Categories and heterotic string theory [ Abstract ] The saying "category theory is an abstract nonsense" is even physically not true. The schematic language of triangulated category presents a new stage of string theory. To illuminate this idea, I will draw your attention to the blow-up minimal model of complex algebraic surfaces. This is done under the hypothetical assumptions of "generalized complex structure" of cotangent bundle due to Hitchin school. The coordinate transformation Jacobian matrices of the measure of sigma model with spin structures cause one part of the gravitational "anomaly cancellation" of smooth Kahler manifold $X$ and Weyl anomaly of compact Riemann surface $\\Sigma$. $Anom = c_1 (X) c_1 (\\Sigma) \\oplus ch_2 (X)$, in terms of 1st and 2nd Chern characters. Note that when $\\Sigma$ is a puctured disk with flat metric, the chiral algebra is nothing but the ordinary vertex algebra. Note that I do not explain the complex differential geometry, but essentially more recent works with the category of DGA (Diffenreial Graded Algebra), which is behind the super conformal field theory of chiral algebras. My result of "vanishing tachyon" (nil-radical part of vertex algebras) and "causality resortation" in compactified non-critical heterotic sigma model is physically a promising idea of new solution to unitary representation of operator algebras. This idea is realized in the formalism of BRST cohomology and its generalization in $\\mathcal{N} = (0,2)$ supersymmetry, that is, non-commutative geometry with non-linear constraint condition of pure spinors for covariant quantization. 17:00-18:30 Room #002 (Graduate School of Math. Sci. Bldg.) Misha Verbitsky (ITEP Moscow/IPMU) Mapping class group for hyperkaehler manifolds [ Abstract ] A mapping class group is a group of orientation-preserving diffeomorphisms up to isotopy. I explain how to compute a mapping class group of a hyperkaehler manifold. It is commensurable to an arithmetic lattice in a Lie group $SO(n-3,3)$. This makes it possible to state and prove a new version of Torelli theorem. 17:00-18:30 Room #002 (Graduate School of Math. Sci. Bldg.) Kiyonori Gomi (Kyoto University) Multiplication in differential cohomology and cohomology operation [ Abstract ] The notion of differential cohomology refines generalized cohomology theory so as to incorporate information of differential forms. The differential version of the ordinary cohomology has been known as the Cheeger-Simons cohomology or the smooth Deligne cohomology, while the general case was introduced by Hopkins and Singer around 2002. The theme of my talk is the cohomology operation induced from the squaring map in the differential ordinary cohomology and the differential K-cohomology: I will relate these operations to the Steenrod operation and the Adams operation. I will also explain the roles that the squaring maps play in 5-dimensional Chern-Simons theory for pairs of B-fields and Hamiltonian quantization of generalized abelian gauge fields. - / 39	{"url":"https://www.ms.u-tokyo.ac.jp/seminar/ipmusem_e/past_e.html","timestamp":"2024-11-10T12:48:49Z","content_type":"application/xhtml+xml","content_length":"41940","record_id":"<urn:uuid:0381fb89-5812-4933-8749-9ca22d919e30>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00008.warc.gz"}
Oblivious Sketching of High-Degree Polynomial Kernels Kernel methods are fundamental tools in machine learning that allow detection of non-linear dependencies between data without explicitly constructing feature vectors in high dimensional spaces. A major disadvantage of kernel methods is their poor scalability: primitives such as kernel PCA or kernel ridge regression generally take prohibitively large quadratic space and (at least) quadratic time, as kernel matrices are usually dense. Some methods for speeding up kernel linear algebra are known, but they all invariably take time exponential in either the dimension of the input point set (e.g., fast multipole methods suffer from the curse of dimensionality) or in the degree of the kernel function. Oblivious sketching has emerged as a powerful approach to speeding up numerical linear algebra over the past decade, but our understanding of oblivious sketching solutions for kernel matrices has remained quite limited, suffering from the aforementioned exponential dependence on input parameters. Our main contribution is a general method for applying sketching solutions developed in numerical linear algebra over the past decade to a tensoring of data points without forming the tensoring explicitly. This leads to the first oblivious sketch for the polynomial kernel with a target dimension that is only polynomially dependent on the degree of the kernel function, as well as the first oblivious sketch for the Gaussian kernel on bounded datasets that does not suffer from an exponential dependence on the dimensionality of input data points. • Machine Learning • Kernel Methods • Scalability • Oblivious Sketching • Numerical Linear Algebra Dyk ned i forskningsemnerne om 'Oblivious Sketching of High-Degree Polynomial Kernels'. Sammen danner de et unikt fingeraftryk.	{"url":"https://pure.itu.dk/da/publications/oblivious-sketching-of-high-degree-polynomial-kernels","timestamp":"2024-11-04T18:11:37Z","content_type":"text/html","content_length":"55987","record_id":"<urn:uuid:36d442a8-d0c7-4f8e-918c-81c53f5946de>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00075.warc.gz"}
HDU 4416 good article good sentence (the suffix array calculates the number of substrings that only appear in a string) A Free Trial That Lets You Build Big! Start building with 50+ products and up to 12 months usage for Elastic Compute Service • Sales Support 1 on 1 presale consultation • After-Sales Support 24/7 Technical Support 6 Free Tickets per Quarter Faster Response • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.	{"url":"https://topic.alibabacloud.com/a/hdu-4416-good-article-good-sentence-the-suffix-array-calculates-the-number-of-substrings-that-only-appear-in-a-string_8_8_31935670.html","timestamp":"2024-11-08T21:43:39Z","content_type":"text/html","content_length":"81303","record_id":"<urn:uuid:5961774a-f501-4af9-b06d-05ace14ff26a>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00516.warc.gz"}
Calculating specific heat capacity (∆ E = m c ∆ θ) \| Oak National Academy Hello there, I'm Mr. Forbes and welcome to this lesson from the Energy of Moving Particles unit. The lesson's called calculating specific heat capacity. By the end of this lesson, you're going to be able to carry out a wide range of calculations involving the equation, energy change is equal to mass times specific heat capacity times temperature There are just two key phrases for this lesson. The first is specific heat capacity and that's the change in internal energy and the temperature of one kilogramme of a material changes by one degree Celsius. The second is joules per kilogramme per degree Celsius, and that's the unit per specific heat capacity of a material. This lesson's in three parts and the first part we're going to have a look at the factors that affect the heating of an object and define specific heat capacity clearly. In the second part of the lesson, we're going to be looking at a range of specific heat capacity calculations. And in the third and final part, we're gonna be trying to solve some more complex problems involving specific heat capacity. So let's get on by defining what specific heat capacity is. You should already realise that it takes longer to boil a larger amount of water than a smaller amount of water. So if you've got an electric kettle and you fill it to the top, it's gonna take longer to boil than if you only half fill that kettle. So for example, if I've got a kettle and it's got 0. 2 kilogrammes of water, it might boil in 30 seconds. But if I double that amount of water to 0. 4 kilogrammes, it's going to take 60 seconds for the same kettle to boil that water. And if I double it again, again, the time doubles. So doubling the mass of water, doubles the amount of time taken to boil it in that kettle. The kettle's providing the same amount of energy every second to the water. So it's taking twice as much energy to boil twice as much water. That means the amount of energy needed to increase the temperature of a substance is directly proportional to its mass. Let's check if you understand that idea. I've got three pans of water here, they're being heated by an identical heater from below and they're all starting at the same temperature, room temperature, which pan is going to reach boiling point So look carefully at the three diagrams, make that decision, pause the video, make a decision, and restart, please. Welcome back. Hopefully you selected B, B's got the smallest amount of water in it, the lowest mass. So it's going to be easier to boil that. Well done if you selected it. The longer that you actually heat a substance, the more energy you're going to give it and the greater it's change in temperature. So, for example, if I've got two objects here, both starting at the same temperature and they're both identical, they're one kilogramme chunks of copper and I heat the first one for 100 seconds, its temperature has gone up to 50 degrees Celsius. So the change in temperature there is 30 degrees. For the second one, let's say I heat it for twice as long, so 200 seconds of heating and its end temperature actually is 80 degrees Celsius. It's gone up by 60 degrees C. So what you should be able to see from that is doubling the energy provided, doubles the temperature change of an object, the change in temperature is directly proportional to the energy provided to Let's see if you understand that second concept. I've got three pans each of them this time containing one kilogramme water, so the same amount of water in each pan. They're again being heated by identical heaters, but this time for different length of time. And you can see the length of time there, one minute, three minutes, five minutes. Which water will have the greatest temperature rise? Pause the video, make your decision and restart please. Welcome back. Hopefully you selected the third one, option C, that's five minutes of heating. I'm gonna be providing you with more energy, so it's gonna have a greater temperature rise. Well done if you've got that. If you heat different substances then the temperature changes won't necessarily be the same, even if they've got the same mass and same starting temperature. So I've got two different substances here, I've got iron and copper. If I start with a temperature of 20 degrees Celsius and heat both the same amount, an equal amount of heating for the same amount of time, then the temperature change isn't the same for those two. The temperature change of the iron might be up to 72 degrees Celsius, and the temperature change of the copper would only go up to 60 degrees Celsius. So the substances aren't heating up the same way and that's because they have different, specific heat capacities. And the specific heat capacities related to the temperature change of an object and the energy provided to it. So the energy needed to increase the temperature for the different metals by the same amount is directly proportional to its specific heat capacity of property of the material that's different for different substances. So I've got three pans again, these pans contain one kilogramme of liquids with different specific heat capacities and they're warmed again by identical heaters starting from room temperature for one minute each. So at the same time, the same starting temperature and the final temperatures are then measured and they're shown in the diagram here. Which of those liquids needed the least energy to heat per degree Celsius> So pause the video, make your decision and restart, please. welcome back. Hopefully you selected option C. The reason for that is that's got the largest temperature increase, so it needed the least amount of energy to heat it by one degree Celsius. Remember all three of those were given the same amount of energy and that one increased its temperature the most. So well done if you selected that. Now, let's look at that definition of specific heat capacity. The property of a material that tells you how difficult it is to change its temperature. The heat capacity of a material is the internal energy change and the temperature of one kilogramme of the material changes by one degree Celsius. So it's a measure of how difficult it is to change the temperature of one kilogramme by one degree. Some examples, we've got water, that's got a specific heat capacity of 4,000 joules per kilogramme per degree Celsius. So it takes 4,000 joules to heat one kilogramme of it by one degree Celsius. And if I wanted to heat two kilogrammes of it by one degree then I'd need twice as much energy, 8,400 joules. Iron's got a different specific heat capacity, it's much easier to heat to iron. It's only 110 joules per kilogramme per degree Celsius. So it takes 110 joules to heat one kilogramme by one degree C. And if I wanted to increase its temperature by twice as much by two degrees C, it would take twice as much energy. As we've seen in the early part of the lesson, the energy required to change the temperature of an object depends on three factors. It's directly proportional to the mass. We need a large energy change to heat large objects, objects with more mass. It's also directly proportional to the change in temperature. We need a larger energy change to give a larger temperature change. And finally, the specific heat capacity. Some materials are easier to heat up than others, and the energy required is directly proportional to that specific heat capacity as well. All those three factors can be linked together in a single equation. So here is that equation linking all the factors together and it is energy change is equal to mass times specific heat capacity times temperature change. And if you write there in symbols, we have this delta E equals MC delta theta, where delta E is the energy change in joules, M is the mass in kilogrammes, and C is the specific capacity in joules per kilogramme per degree Celsius, delta theta is the temperature change in degrees celsius. Let's have a look at an example of using that equation then. So I've got a question here. Iron's got a specific heat capacity of 450 joules per kilogramme per degree Celsius. How much energy is required to increase the temperature of two kilogramme block of the iron by 30 degrees Celsius? So what we do is we write out the equation, energy change equals MC delta theta, we substitute in the values from the question. So the mass was two kilogrammes, the specific heat capacity was 450 joules per kilogramme per degree Celsius and the temperature change is 30 degrees Celsius. And then we can just do the calculation, giving us an answer of 2. 7 kilojoules when we've multiplied all of those out. I'll try another one and then you can have a go. So I've got glass that's got a specific heat capacity of 840 joules per kilogramme per degrees Celsius. Calculate the energy required to increase the temperature of a 1. 5 kilogramme block by 80 degrees Celsius. So again, I write out the equation, I substitute in the values from the question carefully and I perform that calculation given me 101 kilojoules. Now, it's your turn. What I'd like you to do is to calculate the energy change in the situation shown here. So pause the video, calculate the energy change and restart, please. Welcome back. Hopefully, you performed a calculation that looked something like this. Write out the equation, substitute the values and the questions carefully, the mass, specific capacity and temperature change, multiply all those out and you've got an energy change of 235 Well done if you've got that. Now, it's time for the first task of the lesson and what I'd like you to do is answer these two questions. So the first one is about heating the liquid and the second one, heating different materials and using the specific heat capacity equation. So pause the video, work out your answers to those and restart please. Welcome back, here are the solutions. So the temperature rise should have been 60 degrees Celsius. The reason for that is the material's being provided with six times the original energy. So the temperature rise would be six times as great, but the mass is three times as large. So that would only give a third of the temperature rise. Combining those two factors together, the temperature rise is double. The reason you might not get an exact answer is because some of the liquid might evaporate, so energy would be dissipated or lost to the surroundings. And if you perform the calculations, you should have got energy values as shown in the table though. Well done if you've got any of those. And now it's time for the second part of the lesson and then we're going to be using that specific heat capacity equation, rearranging it to solve a range of problems. So as we saw earlier, the equation linking energy change, mass, specific capacity, and temperature change is shown here. There's four variables in that equation, and that means it's going to be a little bit more difficult to rearrange compared to other simpler equations that only have three. So the approach we're gonna take in this lesson to finding values is we're going to be substituting the values into the equation, simplifying what we know some of the data and then that will make finding the solution a little bit easier. So I'll show you some examples of how to do that here. Let's have a look at an example of finding mass. So I've got a question here. The specific capacity of water is 4,200 joules per kilogramme per degree Celsius. A kettle provides 189 kilojoules of energy when heating the water from 15 degrees to 90 degrees, calculate the mass of water in the kettle. So what we do is we write up the equation, we've got energy changes, mass times specific capacity times change in temperature, and it put in the values. So I'm gonna put in the value for the energy change. It says 189 kilojoules, so it's 189,000 joules, I don't know the mass, so I'll leave that as a symbol M, then I've got a specific capacity, and then I've got the change in temperature. And you'll see though I've had to calculate the change in temperature, it's gone up to 90 degrees, but it started at 15 degrees Celsius. So let's simplify that a little bit first by working out that change in temperature and it's 75 degrees Celsius, and the next stage is I can just multiply the 4,200 times that 75 degrees Celsius and that gives me a value of 315,000. And now I've got a much simpler equation. I've got 189,000 joules is equal to mass times 315,000 joules per kilogramme. So I can rearrange that as this only three values now. So I'm gonna rearrange it in terms of M. So the mass is equal to the energy divided by that 315,000 joules per kilogramme and I can just simply solve that using a calculator that gives me a mass of 0. 6 kilogrammes. Okay, let's see if you can repeat that process. I've got a specific heat capacity of lead here, 130 joules per kilogramme per degree Celsius. I've got a heater providing 8. 9 kilojoules of energy heating that block of lead from 20 to 60 degrees. Calculate the mass of the lead block. So pause the video, perform the calculation I did before and restart, please. Welcome back. Hopefully you selected 1. 54 kilogrammes and here's the mathematics behind that. We've got the equation written out, substitute the values, then I found the change in temperature being 40 degrees Celsius. I've calculated the 130 joules per kilogramme per degree Celsius times a 40 degrees Celsius to give me a simpler version of the equation. Then I've rearrange it in terms of mass. So we've got mass is 8,000 joules divided by five, 5,200 joules per kilogramme, and that gives me a final mass of 1. 54 kilogrammes. Well done if you've got that. We can do a similar thing if we're trying to find a change in temperature. So here's another example. I've got a Bunsen burner providing 3. 6 kilojoules of energy to a 0. 20 kilogramme sample of olive oil, and that's got a specific heat capacity of 1,980 joules per kilogramme per degree Celsius. And I'm going to find the change in temperature for that oil. So as before I write out the equation and I substitute in the values that I know. So I've got my energy change written here, 3,600 joules, I've got my specific capacity, and I've got my mass as well. What I haven't got is the change in temperature. So I leave that as delta theta, the next thing I do is I perform all the parts of the calculation I can do with the numbers I've got. So I'm the 0. 20 by 1,980 and that gives me this stage of the equation. And now I can do the rearrangement. I can write out delta theta is equal to 3,600 joules divided by 396 joules per kilogramme. And that gives me a temperature change of 9. 1 degrees Celsius. Here's one of those calculations for you to do. What I'd like you to do is to pause the video, read through the information and work out your answer, and then restart, please. Welcome back, hopefully you selected answer B, 8. 0 degree Celsius. We show you the mathematics here. We've got the energy change equation. We substitute in the values, we simplify in terms of delta theta and then so we get delta theta is 8,400 joules divided by 1050 joules per kilogramme. And that gives me a final temperature change of 8. 0 degrees Celsius. Well done if you got that. And the final way to use the equation is to actually find a specific heat capacity based upon all the other data. So I've got another question here. An engineer is testing properties of a metal sample. They use an electric heater to provide 0. 50 kilogramme sample of the metal with 8. 0 kilo joules of energy. And its temperature increased from 24 degrees to 62 degrees Celsius. Calculate the specific heat capacity. Whereas before we write out the equation, energy change MC delta theta, we substitute in the values that we can see in the question. 8,000 joules, 0. 5 kilogrammes and the temperature change there from 24 degrees up to 62. So we need the difference of those two values. So we'll calculate that, and that was 38 degrees Celsius. So we're left with this stage of the equation and then we can multiply the parts that we've got there, the 0. 5 kilogrammes times a 38 degrees Celsius to give us this version and finally rearrange that in terms of C. So we get C equals specific heat capacity equals 8,000 joules divided by 19 kilogrammes per degree Celsius. And that gives us a specific heat capacity of 421 joules per kilogramme per degree Celsius. So it's your go again, I'd like you to read the information here about a rock sample being heated and find the specific heat capacity of the rock bleeders. Pause the video, find that specific heat capacity and restart. Welcome back. Hopefully you selected option D, 972 joules per kilogramme per degree Celsius. And again, I'll show you all the mathematical stages here which substitute in all the values and that simplifies down to C equals 14,000 joules divided by 14. 4 kilogrammes per degree Celsius. And that gives that final of 972 joules per kilogramme Celsius. Well done if you got that right. Okay, we've reached the second task of the lesson and I've got two questions for you to answer here. Both of them involving use and manipulation of that specific heat capacity equation. So what I'd like you to do is to pause the video, read through those questions and answer them, and then restart when you're done, please. Welcome back. And for the first part of the question we've got question one here. Part A, calculate the energy provided to the water while the water was being given 8,000 joules of energy per second for 60 seconds for one minute. So that gives us an energy value of 480,000 joules. Then to find the final temperature of the water or what we can do is use the specific heat capacity equation substituting the values, including that value for the energy we just calculated. That gives us a temperature change of 76 degrees Celsius. But remember, we've been asked to find a final temperature of the water. So what we can do though is add the initial temperature to that and that gives us a final temperature of 96 degrees Celsius just short of boiling. Well done if you've got that. And here's the answer to question two completed table. And I've shown you the example calculation for gold here. The other calculations would be very similar. We substitute in the values we know, which is the energy, the mass, and we find that temperature change based upon the start and end temperatures and that gives us an equation for C and for gold it was 8,000 joules divided by 60. 2 kilogrammes per degree Celsius. And that gives us a final specific capacity for gold of 133 joules per kilogramme per degree Celsius. The other calculations should look very similar. Well done if you've got that table right. And now it's time to move on to the third and final part of the lesson. And in it we're going to look at some more complex problems involving specific heat capacity. Let's do that. Many of the calculations that involve specific heat capacity are more than one part and we're going to look at questions that involve two or three steps in order to find final solutions. So I've got an example over the next couple of slides and we'll break it down into the different parts we need to do. So here's our first example. I've got a blacksmith working with a 2. 5 kilogramme piece of steel at a temperature of 750 degrees Celsius. So quite hot. To cool it quickly, they're gonna plunge it into a tank of water, 20 kilogrammes of water at a temperature of 20 degrees Celsius. The temperature of the water rises to 50 degrees Celsius and the specific heat capacity of water is 4,200 joules per kilogramme per degree Celsius. Calculate the specific heat capacity of the steel. So to answer that, we need to have two steps. We need to find the energy transferred into the water from the steel and then we need to use that value to find the specific heat capacity of the steel. So let's look at those stages. So the question's repeated there and step one, as I said, is we're going to find the energy change for the water that's going to allow us to find things about the steel a bit later on. So the energy change for the water first, write out the equation, energy change, mass specific heat capacity times temperature change. And remember, we're dealing with the water, so we use the values for the water here. There's 20 kilogrammes of water, it's specific heat capacity is 4,200. And the temperature change of the water, well, it's gone up from 20 degrees Celsius to 50 degrees Celsius. So that's a temperature change of 30 degrees and that will allow us to find the energy change of the water. Multiplying all those two together, we've got an energy change of 2,520,000 joules. Okay, before I go onto the next part of the question, I'd like you to think about something. I've got a hot piece of steel and it's dropped into water, and it's cooled. The internal energy of the water increases by 2000 kilojoules. How much has the internal energy of the steel decreased by? So pause the video, have a think about that, Select that answer and restart. Welcome back. Hopefully you selected answer B, 2000 kilojoules. That's because of the principle of conservation of energy. The internal energy of the water has increased by 2000 kilojoules. So there must have been a loss of 2000 kilojoules by the steel. Well done if you selected that. So we're gonna continue with our example of the blacksmith. We've got the energy change for the steel now. So that's my equation for the energy change for the steel. And I know that the energy change of the water was 2,520,000 joules, and that's the same as the energy change for the steel because of that conservation of energy. I've got the mass of the steel in there, and I've now got the temperature change of the steel. The steel cooled from 750 degrees to 20 degrees. So I've got the temperature change and I'm just gonna find that value for C. So I'll do mathematics to find the temperature change. It's 730 degrees, and now I can simplify that by multiplying the 2. 5 kilogrammes by the 730 degrees and that can be this version, and the next stage I can rearrange that equation And finally I can do that calculation and that gives me the specific heat capacity of the steel. Let's have a look at another example. I've got a central heating boiler and it can transfer 18 kilojoules of energy per second to heat water for a bath. A total of 12 kilogrammes of water at a temperature of 15 degrees Celsius passes through the heater each minute. What temperature does the water leave the heater at? So to answer this question, I need three steps. The first step is to find the amount of water passing through the heater each second. The second step is to find the increase in water temperature using the energy provided per second. And finally to find the new temperature of the water. Let's see if you can do that first step. I've got a total of 15 kilogrammes of water at a temperature of 20 degrees Celsius passing through a pipe in a minute. How much water is passing through per second? Pause the video, make your selection and restart, please. welcome back. Hopefully you selected option A, 0. 25 kilogrammes. Water passing through per second is 15 kilogrammes in a minute. So we divide that by the 60 seconds to get per second. And that's 0. Well done if you've got that. So back to my question, let's calculate the water passing through per second. For this scenario, we've got 12 kilogrammes passing through per minute, so that's 0. 2 kilogrammes per second. So now I can find the temperature change of that water. I've got the heater providing O. 2 kilogrammes of water with 18 kilojoules of energy every second. So I write up my equation for energy change and I substitute my values, I've got a massive 0. 2 kilogrammes, I've got a specific heat capacity of 4,200 joules per kilogramme per degree Celsius. And I know the energy change in the water is 18,000 joules, 18 kilojoules. So I'm just looking for Delta theta. So I follow the same procedures as I've done before, multiplying the values that I can, then simplifying to get an equation for delta theta, and then I'll find that that temperature change is 21. 4 degrees Celsius. But I was actually asked what temperature the water leave at. So I need my new water temperature and it went in at 15 degrees. So what I need to do is to add those values together. The new water temperature is the original temperature plus the increase and that gives me a final value of 36. 4 degrees Celsius. So very hot water there. A question for you now, Which of these boilers would be able to increase the temperature of the water by the greatest amount? So have a read through the data we're provided there and make a decision. Pause the video, select A, B, or C, and then restart, please. Welcome back. Hopefully you selected option C, the eight kilowatt boiler providing energy just to 0. 3 kilogrammes of water each second. That's got the highest heating effect per kilogramme of water. Well done if you've got that. Okay, it's time for the final task of the lesson. And I've got two questions involving manipulation of this specific, heat capacity equation in multiple steps. So what I'd like you to do is to read through those, answer those questions for me, please. So pause video, try that, and restart. Welcome back. Let's have a look at the solutions to those. So for the first one, I need to find the energy provided to the oil first. So the energy provided to the oil is given there, it's 56,000 joules and then I can find the energy per second. Five minutes is 300 seconds. I've got the total energy provided. Then divide that by 300 seconds and that's 187 joules per second. Well done if got that. And for the second one, we're calculating the internal energy change for the water first. So the energy change equation is there. I've put in the mass of water, specific heat capacity and the temperature change, and that's 840 joules. And then I can calculate the specific heat capacity of the stone using the same equation. But this time, I'm looking for C for the stone. So I put in the values, simplify, get the relationship for C, and that gives me specific capacity of 1,680 joules per kilogramme per degree Celsius. Well done if you've got that. And now we've reached the end of the lesson. So a quick summary. The specific heat capacity of a material is the change in internal energy or the temperature of one kilogrammes of material changes by one degree Celsius. Energy change is mass times specific heat capacity times temperature change. And that's shown as symbols here as well with all the definitions below. Well done for reaching the end of the lesson. I will see you in the next one.	{"url":"https://www.thenational.academy/pupils/programmes/combined-science-secondary-year-10-higher-ocr/units/energy-of-moving-particles/lessons/calculating-specific-heat-capacity-e-equals-m-c/video","timestamp":"2024-11-10T20:42:18Z","content_type":"text/html","content_length":"150744","record_id":"<urn:uuid:5930e645-97aa-453e-a44b-3a81737d70e2>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00078.warc.gz"}
Triangle Ornaments Problem J Triangle Ornaments A company makes triangle-shaped ornaments for the upcoming holidays. Each ornament is tied at one of its corners to a rod using a string of unknown length. Multiple of these ornaments may be attached to the same rod. These ornaments should be able to swing (rotate around the axis formed by the string) without interfering with each other. Write a program that computes the minimum required length for the rod, given a list of triangles! The input consists of a single test case. The first line contains one integer $N$ ($0 < N \le 100$), denoting the number of triangles. The next $N$ lines each contain three integers $A, B, C$ denoting the lengths of the three sides of each triangle. The triangle will hang from the corner between sides $A$ and $B$. You are guaranteed that $A, B, C$ form a triangle that has an area that is strictly greater than zero. Output the required length $L$ such that all triangles can be hung from the rod, no matter how long or short each triangle’s string is. No triangle should swing beyond the rod’s ends. You may ignore the thickness of each ornament, the width of the string and you may assume that the string is attached exactly to the triangle’s end point. Your answer should be accurate to within an absolute or relative error of $10^{-4}$. Sample Input 1 Sample Output 1 3 3 3 8.0 Sample Input 2 Sample Output 2 3 3 3 6.843530573929037 Sample Input 3 Sample Output 3 7 20 14 20.721166413503266	{"url":"https://open.kattis.com/contests/hh3jj6/problems/triangleornaments","timestamp":"2024-11-14T11:46:26Z","content_type":"text/html","content_length":"31239","record_id":"<urn:uuid:cf16bf75-3007-4e8a-8760-c7dfac9ee96d>","cc-path":"CC-MAIN-2024-46/segments/1730477028558.0/warc/CC-MAIN-20241114094851-20241114124851-00684.warc.gz"}
Methodology Sewage The methodology combines different open data sources and generates a generic network by open-source technology [34]. The generic network designed by algorithm matches the real sewage network to a certain extent. Therefore, the generic network makes it possible to spot suitable locations for exploitation of waste heat from sewers without having the original GIS data of the sewer. Overall, the methodology allocates population density to street networks and calculates the wastewater volume flow (WWV) for each street. WWV results from the product of population density (PD) and the country-specific water usage (WU) (2.1). Finally, an algorithm determines the shortest paths from each street to the wastewater treatment plant. Furthermore, it accumulates the wastewater volume flows along these paths that are distinguished by residential buildings (discharging waste water at 18 °C), commercial buildings (discharging waste water at 22 °C) and industrial buildings (discharging waste water at 35 °C). Figure 1: Overview of water consumption by country The methodology assumes the following: • Water consumption is equivalent to the amount of wastewater. • Specific water consumption is the statistical average at national level. • Rainwater accumulates discontinuously and is therefore neglected. • The path of the sewage system is said to be the shortest path along the street network from wastewater occurrences to the wastewater treatment plant. Obstacles (e.g. rivers, bridges, train tracks), pumps, geodesic heights are not taken into account. • Buildings without information on their usage or without fitting key metrics (Fig. 1, Fig. 2 and Fig. 3) on the discharged wastewater will be impinged by the difference between the amount of the calculated wastewater and the waste water published by “GIS Reference Layers on UWWT Directive Sensitive Areas. Description of dataset and processing (ed European Topic Centre Inland, coastal, marine waters) • 2013” which compiles the wastewater quantities for all wastewater treatment plants throughout the EU and can be accessed by https://www.eea.europa.eu/data-and-maps/data/ □ The heat transfer coefficient is taken into account relating to the pipes diameter given in two ranges and defined by the volume flow in each pipe. The lower range takes average heat transfer coefficient of PET pipes. The higher range assumes heat transfer coefficient of concrete. The methodology uses graph theory and creates an undirected and unweighted graph out of open-source data. Altogether, the Modell consist of 4 steps: • M1 Determining the flow direction in the edges of the graph, • M2 Defining the mass flow in all edges for the three building use categories by allocating the amount of waste water to each building by key metrics and above given discharge temperature, • M3 Determining the composition of the wastewater flows and the temperatures in each edge and node based on defined flow directions, • M4 The fourth model optimizes the wastewater waste heat extraction points: it determines the locations, the amount of waste heat and the resulting waste heat temperature. Each model is solved by using the solver GUROBI and the python package gurobipy. Further information and the systems of equations can be found in the final report at https://www.iea-dhc.org/ │Figure 2: Water consumption by different building types. The amount of water usage depends on the reference value x. (1 x = bed, 2 x = person, 3 x │Figure 3: Water consumption by different building│ │=guest, 4 n.a., 5 x = patient + personell, 6 x = child, 7 x = employee, 8 x = pupil, 9 x = seat) │types (* used during daytime) │	{"url":"http://cities.ait.ac.at/uilab/udb/home/memphis/help/MethodologySewage.html","timestamp":"2024-11-12T18:39:34Z","content_type":"text/html","content_length":"17947","record_id":"<urn:uuid:2d747e36-090e-48d1-a92e-b0024fa9a7f7>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00158.warc.gz"}
Contemporary abstract algebra gallian, instructors solution manual contemporary abstract algebra gallian, instructors solution manual Related topics: Mcdougal Littell Middle School Math Answers decimal to fraction worksheet Calculate 3rd Order Polynomials algerbra simplification practice problems complex polynomial factoring precalculus practice final exam solutions least common multiple of 37 and 34 w(x-3)=h(x 2) solve for x Author Message Author Message BegLed01 Posted: Wednesday 03rd of Jan 19:55 Flash Fnavfy Liom Posted: Friday 05th of Jan 18:01 I'm just thinking if someone can give me a few pointers It would really be nice if you could let us know about a here so that I can understand the basics of resource that can provide both. If you could get us a contemporary abstract algebra gallian, instructors home tutoring software that would offer a step-by-step Reg.: 24.01.2005 solution manual. I find solving equations really difficult. I Reg.: 15.12.2001 solution to our assignment , it would really be good . work in the evening and thus have no time left to take Please let us know the genuine links from where we extra classes. Can you guys suggest any online tool can get the software. that can help me with this subject? ameich Posted: Friday 05th of Jan 11:22 TC Posted: Saturday 06th of Jan 21:17 The best way to get this done is using Algebrator . This I remember having often faced problems with software offers a very fast and easy to learn evaluating formulas, algebra formulas and solving a technique of doing math problems. You will definitely triangle. A truly great piece of algebra program is Reg.: 21.03.2005 start liking math once you use and see how simple it is. I Reg.: 25.09.2001 Algebrator software. By simply typing in a problem remember how I used to have a difficult time with my homework a step by step solution would appear by a College Algebra class and now with the help of click on Solve. I have used it through many algebra Algebrator, learning is so much fun. I am sure you will classes – Pre Algebra, Algebra 1 and Pre Algebra. I get help with contemporary abstract algebra gallian, greatly recommend the program. instructors solution manual problems here.	{"url":"https://www.softmath.com/parabola-in-math/exponential-equations/contemporary-abstract-algebra.html","timestamp":"2024-11-12T00:33:37Z","content_type":"text/html","content_length":"44158","record_id":"<urn:uuid:b0181eaa-8875-4ba9-b423-a0e378627a2b>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00797.warc.gz"}
Calculating with Permutations and the Basic Counting Principle Question Video: Calculating with Permutations and the Basic Counting Principle Mathematics • Third Year of Secondary School A school holds a ceremony to honor members of the school administration, which contains 25 teachers and 15 executives. The ceremony will honor 10 members, and the school must choose at least 5 teachers and at least 3 executives. Given that the teachers will stand on the left of the stage and the executives will stand on the right of the stage, what is the total number of possible onstage configurations for the members of the school administration? Express your answer in terms of combinations or permutations. Video Transcript A school holds a ceremony to honor members of the school administration, which contains 25 teachers and 15 executives. The ceremony will honor 10 members, and the school must choose at least five teachers and at least three executives. Given that the teachers will stand on the left of the stage and the executives will stand on the right of the stage, what is the total number of possible onstage configurations for the members of the school administration? Express your answer in terms of combinations or permutations. Let’s begin by thinking about what this could look like. We’re going to choose at least five teachers from a total of 25 and at least three executives from a total of 15. Our aim in doing so is to honor exactly 10 members. So how many teachers and how many executives could we have? Well, we can have no fewer than five teachers. So we could have exactly five teachers, which would mean we would also have exactly five executives. We could have six teachers and four executives. And the final option is to have seven teachers and three executives. We know there are no other options because we must choose at least three executives. We’re also told that, as in the diagram, the teachers will stand on the left of the stage and the executives will stand on the right. This minimizes the total number of ways of ordering these groups of people. In fact, we’ve represented all three possible ways of doing so. It does not matter the order in which the teachers stand, nor does it matter the order in which the executives stand. And so, that should remind us whether we need to use combinations or permutations here. The difference between calculating combinations and permutations is whether order matters. With a combination, the order here in which the people stand matters. With permutations, the order in which our people stand doesn’t matter. So we’re going to be using permutations. And if we want to calculate the number of ways of choosing 𝑟 items from a total of 𝑛 when order doesn’t matter, we represent that as 𝑛P𝑟 as shown. So if we look at the number of ways of choosing our five teachers, we know we’re choosing from a total of 25. Since order doesn’t matter then, the number of ways of choosing these five teachers is 25P five. In a similar way, we’re choosing five executives from a total of 15, and order doesn’t matter, so that’s 15P five. Then in our next option, it’s 25P six to choose six teachers and 15P four to choose four executives. Finally, choosing seven teachers from a total of 25 is 25P seven. And to choose our three executives is 15P three. But how do we combine these? Well, here’s where we remind ourselves what the product rule for counting tells us. This tells us that the total number of outcomes for two or more events is found by multiplying the number of outcomes of each event together. So the number of ways of choosing five teachers and five executives is 25P five times 15P five. Then the number of ways of choosing six teachers and four executives is 25P six times 15P four. And finally, we multiply 25P seven by 15P three to find the total number of ways of choosing seven teachers and three executives. We’re still not done, of course. We still need to combine each of these three options. This is where we remind ourselves what we mean when we talk about the addition rule. Sometimes called the basic counting principle, it tells us that if we have 𝑎 number of ways of doing something and 𝑏 number of ways of doing another thing and we can’t do both at the same time, then there are 𝑎 plus 𝑏 ways to choose one of the actions. In other words, to find the total number of ways of choosing five teachers and five executives or six teachers and four executives or seven teachers and three executives, we add each of these together. And that tells us the total number of onstage configurations for members of this school administration. And that’s an answer given in permutations. It’s 25P five times 15P five plus 25P six times 15P four plus 25P seven times 15P three.	{"url":"https://www.nagwa.com/en/videos/735134697874/","timestamp":"2024-11-13T20:45:51Z","content_type":"text/html","content_length":"255563","record_id":"<urn:uuid:30a56da6-cd99-4141-bdd2-f6c490688e32>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00187.warc.gz"}
Proofs by Induction Hi there! Well after a few years of high school and now onto my second semester in university, I'm taking Calc II that involves lots of proofs. This week we were given a problem sheet with some proofs to prove. Here is one I needed some aid with: Prove that is divisible by 7 for all h >= 1. Here's what I've done thus far and I figured induction would be the best way to go: for h = 1: 7 can be divided by 7 so the proof holds true for h = 1 for n = h + 1 And I'm stuck at that. Any hints anyone? Thanks in advance! Last edited by Anakin (2011-01-16 10:44:00) Re: Proofs by Induction I thought about subbing in 1 = h as it's no longer h+1 which gives 105 and it divides evenly by 7. However, that would just be like testing h = 2 from the very start, which is not how a proof should go (I think). Edit: This website did it in the same way that I did and they have the solution: http://mathcentral.uregina.ca/QQ/database/QQ.09.99/pax1.html at the bottom. However, I fail to understand the portion about the assumption. Last edited by Anakin (2011-01-16 11:18:02) Registered: 2005-06-22 Posts: 4,900 Re: Proofs by Induction Proof by induction works in two stages. First you prove the base case. In this case that's h=1, and you've done that. Next you assume it's true for h=n, and use that to prove that it's true for h=n+1. You've managed to get that 11^(n+1) - 4^(n+1) = 711^n + 4(11^h - 4^h). The first term on the right is clearly divisible by 7, and by the inductive assumption, the second term also is. Therefore the whole right hand side is divisible by 7 and you're done. [True for h=n] implies [True for h=n+1]. Why did the vector cross the road? It wanted to be normal. Re: Proofs by Induction mathsyperson wrote: Proof by induction works in two stages. First you prove the base case. In this case that's h=1, and you've done that. Next you assume it's true for h=n, and use that to prove that it's true for h=n+1. You've managed to get that 11^(n+1) - 4^(n+1) = 711^n + 4(11^h - 4^h). The first term on the right is clearly divisible by 7, and by the inductive assumption, the second term also is. Therefore the whole right hand side is divisible by 7 and you're done. [True for h =n] implies [True for h=n+1]. Oh, that makes it very clear indeed. I didn't know that was the case. Thanks for the explanation and walk-through, Mathsyperson. Re: Proofs by Induction Anakin wrote: mathsyperson wrote: Proof by induction works in two stages. First you prove the base case. In this case that's h=1, and you've done that. Next you assume it's true for h=n, and use that to prove that it's true for h=n+1. You've managed to get that 11^(n+1) - 4^(n+1) = 7*11^n + 4(11^h - 4^h). The first term on the right is clearly divisible by 7, and by the inductive assumption, the second term also is. Therefore the whole right hand side is divisible by 7 and you're done. [True for h=n] implies [True for h=n+1]. Oh, that makes it very clear indeed. I didn't know that was the case. Thanks for the explanation and walk-through, Mathsyperson. For even more clarity, you could also rewrite as follows Let 11^n - 4^n = 7a (since we know that 11^n - 4^n is divisible by 7) then carrying on from the last line: = 7 ( 11^h ) + 4 ( 7a ) = 7 ( 11^h ) + 7 ( 4a ) = 7 ( 11^h + 4a ) Registered: 2011-01-26 Posts: 2 Re: Proofs by Induction me wonder one prove 3+2 =6	{"url":"https://mathisfunforum.com/viewtopic.php?pid=162412","timestamp":"2024-11-03T07:36:06Z","content_type":"application/xhtml+xml","content_length":"16228","record_id":"<urn:uuid:337bd980-c5f5-478a-941d-f1e7ecf3f3dd>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00453.warc.gz"}
Define x∗ by the equation x∗=π−x. Then ((−π)∗)∗=... \| Filo Define by the equation . Then Not the question you're searching for? + Ask your question Working from the inner parentheses out, we get Hence, the answer is (C). Method II: We can rewrite this problem using ordinary function notation. Replacing the odd symbol with gives . Now, the expression becomes the ordinary composite function Was this solution helpful? Found 5 tutors discussing this question Discuss this question LIVE for FREE 13 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Practice questions from Functions in the same exam Practice more questions from Functions View more Practice questions on similar concepts asked by Filo students View more Stuck on the question or explanation? Connect with our Mathematics tutors online and get step by step solution of this question. 231 students are taking LIVE classes Question Text Define by the equation . Then Topic Functions Subject Mathematics Class Grade 12 Answer Type Text solution:1 Upvotes 72	{"url":"https://askfilo.com/mathematics-question-answers/define-x-by-the-equation-xpi-x-then-left-piright","timestamp":"2024-11-10T20:48:53Z","content_type":"text/html","content_length":"417163","record_id":"<urn:uuid:73fb27e9-02dc-4dd0-b895-b816c01e28f7>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00571.warc.gz"}
Multiplication Theorem on Probability - Tutors 4 You If A and B are the events associated with a random experiment then Extension of Multiplication Theorem If A, B and C are three events associated with a random experiment, then [1], A[2]….A[n] are associated with a random experiment, then Independent Events Two events are called independent if the occurrence or non-occurence of one does not affect the probability of the occurence of the other. Two events A and B associated with a random experiment are independent if [1], A[2]….A[n] associated with a random experiment are independent then If A, B are independent events, then so are	{"url":"https://tutors4you.com/index.php/multiplication-theorem-on-probability/","timestamp":"2024-11-05T22:51:58Z","content_type":"text/html","content_length":"77360","record_id":"<urn:uuid:c990337d-e1f2-408f-81f8-c9fad244e866>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00735.warc.gz"}
Effects of Sample Size on Estimates of Population Growth Rates Calculated with Matrix Models Matrix models are widely used to study the dynamics and demography of populations. An important but overlooked issue is how the number of individuals sampled influences estimates of the population growth rate (λ) calculated with matrix models. Even unbiased estimates of vital rates do not ensure unbiased estimates of λ–Jensen's Inequality implies that even when the estimates of the vital rates are accurate, small sample sizes lead to biased estimates of λ due to increased sampling variance. We investigated if sampling variability and the distribution of sampling effort among size classes lead to biases in estimates of λ. Methodology/Principal Findings Using data from a long-term field study of plant demography, we simulated the effects of sampling variance by drawing vital rates and calculating λ for increasingly larger populations drawn from a total population of 3842 plants. We then compared these estimates of λ with those based on the entire population and calculated the resulting bias. Finally, we conducted a review of the literature to determine the sample sizes typically used when parameterizing matrix models used to study plant demography. We found significant bias at small sample sizes when survival was low (survival=0.5), and that sampling with a more-realistic inverse J-shaped population structure exacerbated this bias. However our simulations also demonstrate that these biases rapidly become negligible with increasing sample sizes or as survival increases. For many of the sample sizes used in demographic studies, matrix models are probably robust to the biases resulting from sampling variance of vital rates. However, this conclusion may depend on the structure of populations or the distribution of sampling effort in ways that are unexplored. We suggest more intensive sampling of populations when individual survival is low and greater sampling of stages with high elasticities. Citation: Fiske IJ, Bruna EM, Bolker BM (2008) Effects of Sample Size on Estimates of Population Growth Rates Calculated with Matrix Models. PLoS ONE 3(8): e3080. https://doi.org/10.1371/ Editor: Mark Rees, University of Sheffield, United Kingdom Received: July 25, 2008; Accepted: August 4, 2008; Published: August 28, 2008 Copyright: © 2008 Fiske et al. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited. Funding: Financial support for collecting demographic data was provided by the National Science Foundation (award numbers DEB-0309819 and INT 98-06351) and the University of Florida. The funders had no role in study design, data collection and analysis, decision to publish, or preparation of the manuscript. Competing interests: The authors have declared that no competing interests exist. Matrix models [1], [2] are an important tool used by ecologists to study the demography of structured populations and for conducting population viability analyses. They are flexible, readily applicable to a diversity of life-history strategies, and there is a broad body of literature describing their construction, interpretation, and limitation reviewed in [3], [4]. However as a recent review by Doak et al. [5] cogently summarizes, they are data-hungry models requiring detailed estimates of birth, death, reproduction, and other vital rates. When the number of individuals used to estimate vital rates is low, the resulting vital rates–as well as estimates of variances and covariances among them–can be biased. Because biased vital rates can lead to inaccurate projections of the population growth rate (i.e., λ), there has been an upsurge in studies exploring alternative sampling designs for demographic studies [5]–[7]. Nevertheless, even unbiased estimates of vital rates do not ensure unbiased estimates of the population growth rate. This is because λ is the dominant eigenvalue of the transition matrix [2], and hence a nonlinear function of the underlying vital rates (i.e., λˆ=f(v[1],v[2],…,v[n]), where the v[i] are the n vital rates and f is a nonlinear real-valued function). As for other nonlinear functions describing ecological processes e.g., [8], [9], the mathematical theorem known as Jensen's inequality [10] implies that variance in vital rates–even those that have been accurately measured–will bias estimates of λ. The amount and direction of this bias depend both on the strength of the nonlinearity of the relationship between λ and the vital rates, and on the variance of the vital rates themselves. Variance in vital rates can arise from two sources. The first of these is process variance, which results from real variation in the population over space or time [11]–[13]. The second source is sampling variance, which is a result of studying a sample rather than the entire population. Several studies have used Jensen's Inequality to predict potential biases in λ resulting from process variance e.g., [14], [15], and methods for dealing with process variance, especially over time, are well-developed [16]. While there are also methods that attempt to separate sampling variance from the total observed variance in vital rates [17], [18], the potential for sampling variance to bias estimates in λ has received limited attention. Houllier et al. [19] used analytical approaches and stochastic simulations to test for biases resulting from variance of the matrix elements, while Usher [20] derived analytic solutions for both Leslie matrices and more general models. In general, these studies found that biases in estimates of λ were small–usually less than 0.5%. However, the potential for sampling variance to bias estimates of λ has yet to be investigated for matrix models in which organisms are capable of regressing into smaller size classes. These models are extremely common–they represent the demography of organisms ranging from plants to marine invertebrates e.g., [12], [21]. Jensen's Inequality would lead one to predict that even when the estimates of the vital rates are accurate, small sample sizes will lead to biased estimates of λ as a result of increased sampling variance. To understand why, one must first understand how this variance is modeled in demographic studies. Lower-level vital rates sensu [4] are often modeled as binomial random variables. The binomial distribution, which assumes homogeneous vital rates among individuals within a stage class, has a higher variance for a given mean value than a model with heterogeneous vital rates [22]. Therefore, the binomial distribution is a conservative model for the sampling process. The sampling variance of these binomial vital rates is:(1)where n[i] is the number of individuals in class i in the sample, and s[i] is the true value of the vital rate [23]. Therefore, sampling variation of a binomial vital rate is maximized when the true rate is equal to 0.5. For estimates of fecundity, such as the number of offspring an individual produces given that it reproduces at all, the variance is:(2)where f[i] is the fecundity of individuals in class i and σ[i]^2 is the true variance of the fecundity among individuals in class i; n[i] is defined as above. Thus, as the variation of the true fecundity increases, so does the variation of the estimated fecundity. Furthermore, as sample size decreases these sample variances increase. This has the net effect of biasing estimates of λ. We used stochastic simulations to determine if sampling variance biased estimates of λ and how these biases varied with the sample size used to construct matrix models. Because the distribution of individuals among a population's size classes has been shown to influence the outcome of demographic analysis [5]–[7], we also considered the potential for synergistic effects of sample size and population structure on estimates of λ and potential biases in these estimates. Our simulations were conducted using multi-year demographic data collected to elucidate the population dynamics of Amazonian understory herb Heliconia acuminata (Heliconiaceae) [24], [25]. Using annual transition matrices constructed with six years of demographic data from 3842 plants, we addressed three questions. First, does the sampling variance of two key vital rates, survival and fecundity, bias estimates of population growth rates? To address this question, we compared the “true” growth rate of the total study population (hereafter, λ) with the growth rates of subpopulations composed of 25–200 randomly selected individuals (hereafter, λˆ). Second, are the patterns of bias influenced by population structure? We conducted our simulations with two different population structures. First, we used a uniform distribution of individuals among size classes (i.e., equal numbers of individuals in all size classes), which has been put forward as the optimal sampling distribution for demographic studies [7]. We also used a distribution that reflects the biological structure of many populations in the field, known as the “inverse J distribution” [24], [26]. An “inverse J distribution” contains fewer stage i+1 individuals than stage i, such that a histogram of stage classes in a sample is a reflected “J” shape. Finally, what range of sample sizes is typically used to parameterize matrix models of plant demography, and how do these sample sizes compare with those at which bias in estimates of λ becomes negligible? Does the sampling variance of two key vital rates, survival and fecundity, bias estimates of λ? As the sample sizes used to calculate vital rates decreased, λˆ increasingly overestimated λ (maximum bias=16.61%±32.4 SD; Figure 1). This maximum bias occurred when simulating an inverse-J sampling distribution with 25 individuals and 0.5 survival probability. However, the observed bias became negligible as the rates of individual survival or sample sizes increased. For instance, using 50 individuals to estimate vital rates from a population with a mean survival probability of 0.5 resulted in a mean bias of 6.61%±18.65 SD (Figure 1a), while increasing the survival rate to 0.8 resulted in a mean bias of only 1.88%±8.55 SD (Figure 1b). The coefficient of variation (CV) of fecundity, which increased up to 32-fold in the different scenarios we modeled, did not bias λˆ in any of our simulations. Bias is calculated using the equation (λˆ−λ)/λ×100%. Results are shown for uniform sampling of all stage classes (filled symbols) and sampling from a more realistic J-distribution (open symbols). Sample sizes on the abscissa are the total number of plants (summed across all stage classes) used for parameterizing matrix models. The dashed line indicates a bias=0. Are the patterns of bias influenced by population structure? The amount of bias increased when simulations were run with a more realistic inverse-J population structure than when using equal numbers of individuals in all stage classes (8.08%±20.70 SD vs. 5.14%±16.22 SD respectively when sampling 50 individuals when survival=0.5; 2.33%±9.32 SD vs. 1.43%±7.67 SD when survival=0.8; Figure 1). This result was qualitatively similar for all combinations of survival and sample size. What sample sizes are used to parameterize matrix models of plant demography, and how do these compare with those at which bias in estimates of λ becomes negligible? Our literature review resulted in 28 studies of perennial herbs, 16 of trees, 9 of shrubs, and 15 studies of other plant types (e.g., grasses, geophytes, forbs; Appendix S1). Of these 68 studies, however, we were only able to determine the number of plants that had been used to parameterize the matrix models in 52 (Table 1). Studies of perennial herbs used fewer individuals than those of trees. Approximately 12% of the studies on perennial herbs used fewer than 100 individuals (summed across all stage classes), and only 25% of studies were based on 500 or more individuals (range: 30-4963). The ‘other’ category had the largest proportion of studies with fewer than 100 individuals (22%; Table 1). Our results demonstrate that biased estimates of λ can result from small sample sizes, as predicted by Jensen's Inequality. This is not because estimates of vital rates based on small sample sizes are biased. Rather, it is because small sample sizes can act in concert with low survival rates to increase sampling variation. This increase, combined with the nonlinear relationship between λ and the vital rates, results in an overestimation of the population growth rate. However our simulations also demonstrate that these biases rapidly become negligible with increasing sample sizes and as survival increases. The precise sample size at which bias diminishes will obviously vary between species, and additional studies with other demographic datasets are needed to evaluate the generality of these results. However, because Heliconia acuminata's population and elasticity structure is common to many long-lived plants [27], [28], we believe the qualitative conclusions of our study will apply to other perennial plant species. Interestingly, we also found the magnitude of the bias in λˆ was influenced by the structure of the population being sampled. The increase in bias observed when sampling with the more realistic “inverse J” distribution has important implications or the design of demographic studies. Using a novel analytical approximation, Gross [6] found that sampling more intensively those stages to which λ was more sensitive increased the precision of estimates of λ. In contrast, Münzbergová and Ehrlén [7] conducted simulation studies based on published demographic data and concluded that sampling equal numbers of individuals from different size class generally provided the most precise estimates. Although our simulations were not designed to resolve this seeming contradiction, we do note that H. acuminata population growth is especially sensitive to changes in the survival of individuals in the larger stage classes while being relatively robust to changes in fecundity [24]. Sampling from a population using the more realistic inverse J-distribution therefore increased the sampling variance of the demographically ‘important’ vital rates (e.g., survivorship of larger plants) and decreased the sampling variance of the less ‘important’ vital rates. Hence, our results suggest that sampling more individuals from stages whose vital rates have larger elasticity values–as recommended by Gross [6]–may not only increase the precision of estimates of λ, it may also increase their accuracy sensu [29]. In light of our results, it appears that most studies of plant demography we reviewed have sample sizes large enough to overcome potential biases resulting from sampling variance. However, there are clearly cases in which small sample sizes are unavoidable, most notably those in which species are elusive e.g., [30] or rare e.g., [31]. In these cases, researchers may benefit from modeling vital rates to improve precision of vital rate estimates [32], [33] or using data from closely related species [5], [34]. Despite the widespread use of matrix models in ecology and conservation, studies evaluating alternative sampling designs remain limited. Our results suggest that for many of the sample sizes used in demographic studies (Appendix S1), matrix models are probably robust to the biases resulting from sampling variance of vital rates. However, this conclusion may depend on the structure of populations or the distribution of sampling effort in ways that remain to be explored. We believe that the framework developed by Doak et al. [5] provides a powerful tool with which to identify the threshold at which biases become negligible and aid in the development of appropriate sampling protocols for matrix models. In addition, biases due to sampling variance could potentially be eliminated entirely by using “Integral Projection Models” [35], [36] to analyze demography, although we know of no studies have evaluated this possibility. Finally, we were surprised to find that 24% of the studies we reviewed failed to report the sample sizes on which their demographic models were based. We end with a call to researchers using matrix models to report the number of individuals used to parameterize the different stage classes of their models–basic information without which it is impossible to evaluate if and how the results of ecological studies are biased. Materials and Methods Simulation models used to estimate how the accuracy of projections of λ varied with sample size were based on data collected during a long-term and large-scale study of plant demography conducted at Brazil's Biological Dynamics of Forest Fragments Project (BDFFP; 2°30′S, 60°W). The focal species for this study was Heliconia acuminata, a perennial herb native to central Amazonia and the Guyanas [37]. Descriptions of the study site and experimental design can be found elsewhere [24], [25], [38]. Briefly, permanent 50 m×100 m plots were established in 13 of the BDFFP's reserves in January 1998. All H. acuminata in each plot were marked and mapped and the number of vegetative shoots each plant had was recorded [24]. Since their establishment the plots have been surveyed annually to record plant growth, mortality, and the emergence of new seedlings (i.e., established plants less than 1 year old). The plots were also surveyed during the flowering season to record the identity of reproductive individuals. The analysis presented here is based on summary data from the 1998–2003 surveys conducted in six continuous forest plots; during this time period we marked, measured, and recorded the fates of N=3842 plants in these sites. The demography of Heliconia acuminata can be described by the matrix shown in Figure 2. Note that there is no seed bank–all seeds produced in year t either germinate and become seedlings or die in year t+1. Our simulations used Bruna's [24] 1998–1999 transition year estimates of the vital rates from the continuous forest populations to calculate the ‘true’ population growth rate (i.e., λ); the results of our simulations were not sensitive to the choice of data to use as the reference year (results not shown). Heliconia acuminata transition matrix used in Monte Carlo sampling analysis. The vital rates which compose each matrix element are defined as follows: s[i]=Prob(individual in stage i survives one time step), g[i]=Prob(individual in stage i grows at least one stage in one time step \| survival), h[x,i]=Prob(individual in stage i grows at least x stages \| growth of at least x−1 stages), r[i] =Prob(individual in stage i regresses at least one stage per time step \| survived and did not grow), k[x,i]=Prob(individual in stage i regresses at least x stages \| regression of at least x−1 stages), p[i]=Prob(plant in stage i flowers), f[i]=mean number of fruits per flowering plant in stage i, n=mean number of seeds per fruit, c=Prob(seed germinates and establishes) (B) Heliconia acuminata transition matrix used in sampling simulations (see Methods). We then simulated estimating λˆ for subsamples of the population ranging from 25–200 individuals. To do so, we used one of two probability distributions to simulate sampling each vital rate: a beta probability distribution for the binomial vital rates (e.g., probability of survival, probability of growth) and the gamma distribution for the count-based vital rates (i.e., fecundity). Because the beta distribution is continuous, bounded by 0 and 1, and can be parameterized to have a variety of means and variances, it is an appropriate choice for modeling estimates of the binomial vital rates [4]. We chose the gamma distribution to model estimates of average fecundity because it is non-negative and can also be flexibly parameterized. We parameterized both the beta and gamma sampling distributions according to the method of moments, a technique which parameterizes a distribution by specifying its expected value and variance [39]. To define the sampling process, we set the expected value and variance of an estimated vital rate equal to the population's mean vital rate and sampling variance, respectively. We determined the sampling variance at each sample size with equation (1) and equation (2). Then, we used well-known method of moments relationships between the parameters of the distributions and their expected value and variance e.g., [40]. For the beta distribution, we used the following relationship between the parameters and the mean and variance of the distribution:(3)where and μ[ŝi] and σ[ŝi]^2 are the mean and variance of the estimate of vital rate s[i], respectively. Similarly, to calculate the parameters of the gamma distribution, we used the relationship:(4)where and are the mean and variance of the estimate of fecundity f[i], respectively. According to equation (1), the variance of binomial vital rate estimates is maximized when the rate is 0.5. To test for a difference in bias when survival estimates vary maximally, we simulated with both the true survival rates and after replacing the survival of all stages with 0.5. We also simulated at an intermediate survival level of 0.8, which is a high but realistic probability of individual mortality (IJF, unpublished data). Similarly, because the variance of estimates of fecundity is proportional to the real variance of fecundity among individuals (equation (1), we simulated with 3 levels of fecundity variance. We defined these levels in terms of coefficient of variation, : 0.5, 2, and 16. We used two different population structures to conduct our simulations: a uniform distribution of individuals among size classes (i.e., equal numbers of individuals in all size classes) and the inverse J distribution [24], [26], in which there are fewer stage i+1 individuals than stage i. We used the actual distribution of classes observed in the field averaged over all years and sites to compute the inverse J distribution. For each vital rate, we ran 2000 simulations with populations ranging in size from 25 to 200 individuals for all combinations of survival (the mean values from the H. acuminata demographic survey, henceforth called the “real” values, or mean survival=0.5), fecundity (CV=0.5, 2, or 16), and sampling distribution (an “inverse J” distribution or a “uniform” distribution). We chose 25 individuals as the smallest sample size because smaller samples would yield too few individuals per stage class to resemble what a real study might sample. In each run of the simulation, we drew all 31 vital rates (26 binomial vital rates and 5 fecundities) from their appropriate sampling distributions, computed a sample transition matrix from these vital rates, and then estimated λ as the dominant eigenvalue of this transition matrix (Figure 2). We then estimated the expected value and standard deviation of the relative bias of λ estimates at each combination of survival rates, fecundity CV, distribution of sampling effort, and sample size. We calculated the relative bias as (λˆ−λ)/λ×100% where λ is the “true” asymptotic population growth rate and λˆ is the mean of all 2000 population growth rates estimated. All simulations were conducted using the R statistical computing environment [41]. To contextualize the results of our simulations, we conducted a review the plant demographic literature to determine the sample sizes used to parameterize matrix models. We conducted our survey using a Web of Science search from March 15, 2006. Our search terms were combinations of “matrix model”, “plant”, and “demography.” For each paper returned in our search, we used the “times cited” and “references cited” features to find additional relevant studies. For each study we identified the number of individuals sampled to parameterize non-reproductive terms of the matrix; if a study included more than one matrix (e.g., in multi-site or multi-year studies), we calculated the average number of individuals used. We thank K. Gross, M. Oli, and M. Rees for helpful discussions and comments on the manuscript. The efforts of the many technicians and students who conducted the censuses and the logistical support of the BDFFP staff were also invaluable. This is publication number 519 in the BDFFP Technical Series. Author Contributions Conceived and designed the experiments: IJF EMB. Performed the experiments: IJF. Analyzed the data: IJF EMB. Contributed reagents/materials/analysis tools: BMB. Wrote the paper: IJF EMB BMB. Aided in interpreting and conceptualizing the results: BMB.	{"url":"https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0003080","timestamp":"2024-11-13T21:52:43Z","content_type":"text/html","content_length":"176650","record_id":"<urn:uuid:ab0c014f-5d59-45db-9714-f3c95718e007>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00837.warc.gz"}
KVT022: LENGTH OF NATIONAL ROADS BY TYPE OF PAVEMENT, 31 DECEMBER-PxWeb KVT022: LENGTH OF NATIONAL ROADS BY TYPE OF PAVEMENT, 31 DECEMBER This database does not contain data tables on foreign trade statistics. Tables for foreign trade statistics are published in the foreign trade database Choose data Now you have come to the page, Choose variable. This page give you the oportunity to select which variables and values you want to display in your result of the table. A variable is a property of a statistical unit. The page is divided into several boxes, one for each variable, where you can select values by click to highlight one or more values. It always starts with the statistics variable which is the main value counted in the table. Field for searching for a specific value in the list box. This is examples of values you can search for.2004 , 2005 , 2006 , Number of selected data cells are: (maximum number allowed is 1,000,000) Presentation on screen is limited to 1,000 rows and 100 columns Number of selected cells exceeds the maximum allowed 1,000,000 Unit: kilometres Due to rounding, the values of the aggregate data may differ from the sum. Data on length of secondary roads and length of ramps for 2021-2022 have been revised on 22.04.2024.	{"url":"https://andmed.stat.ee/en/stat/majandus__transport__transpordi-taristu/KVT022","timestamp":"2024-11-03T03:19:03Z","content_type":"text/html","content_length":"120760","record_id":"<urn:uuid:8ebd02aa-cf75-4c4e-9e36-49e747b993ed>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00712.warc.gz"}
ThmDex – An index of mathematical definitions, results, and conjectures. Let $R$ be a D273: Division ring Let $V$ be a D29: Vector space over $R$ such that (i) $\mathcal{L} : = \mathcal{L}_R(V)$ is the D2043: Set of linearly independent sets in $V$ over $R$ Then $$\mathcal{L} eq \emptyset$$	{"url":"https://thmdex.com/r/4290","timestamp":"2024-11-02T11:44:09Z","content_type":"text/html","content_length":"6179","record_id":"<urn:uuid:695fe394-bbb6-4543-ba14-74b7e02f7060>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00656.warc.gz"}
Comparing Players Using Cluster Analysis As there were a couple of presentations at the recent Opta Pro Forum talking about identifying player similarities I thought I’d give a quick example of how to do something similar using k-means cluster analysis. The Data All the data used in the analysis was taken from public websites, such as whoscored, squawka, transfermarkt etc and painstakingly matched together to try and get as much information on each player as The first stage of analysis was to normalize the data so it was all in the same range to avoid biasing the clustering. If you think about how many goals a typical player scores per match compared with how many passes they play then the scale is quite different. Since k-means clustering uses Euclidean Distance the clusters formed are influenced strongly by the magnitudes of the variables, especially by outliers. By normalizing all data into the same range this bias can be avoided. Principal Component Analysis While normalizing the data, I also performed Principal Component Analysis (PCA) on it too. This step isn’t essential but it is a handy way of reducing the dimensions in the data down to a more manageable size by squashing all the data together into new variables known as principal components. These principal components are created in such as way so that the first one accounts for as much as the variance in the data as possible, the second one then accounts for as much of the remaining variance and so on. As you can see in Figure 1 below, the first component represents pretty much 70% of all the variance in the data with each additional component accounting for less and less. This means we can represent pretty much all the information in the data without losing much using just five components, and around about 80% using just two components. Figure 1: PCA scree plot showing amount of variance accounted for by each principal component Clustering The Players The next step was to then run the k-means clustering algorithm on the data. As shown in Figure 2 the players split relatively neatly into five distinct coloured clusters when plotted by the first two principal components. Figure 2: Players split into different clusters by colour As a quick test we can look at the grey cluster located at the bottom of the image in more detail to see which players are contained within it (Figure 3). If you click the image to zoom in on it you can see it’s done a pretty good job of pulling out the goalkeepers from the rest of the players. This is to be expected since goalkeeper’s stats should be pretty distinct from outfield players but it’s reassuring to check the technique passes this first simple test before we move on. Figure 3: The grey cluster up close Vincent Kompany Now that we have separated out the goalkeepers we can take a look at how well the technique copes with outfield players, starting with Manchester City’s central defender Vincent Kompany located at the centre of Figure 4. The results are pretty good, with Kompany surrounded by players predominantly considered to be defenders. As you move up the image the players start to get a bit more attacking with people like David Luiz, Phil Jones and Fabien Delph starting to appear Figure 4: Clustering of Vincent Kompany Adnan Januzaj Next up is Adnan Januzaj, one of the few Manchester United players to be having anything resembling a decent season this year. Again the results look pretty plausible (Figure 5), with Januzaj surrounded by predominatly attacking midfielders. There are a couple of slightly surprising results in there though, such as Manchester City’s strikers Álvaro Negredo and Edin Džeko. Figure 5: Clustering of Adnan Januzaj Mikel Arteta Finally, I added in Arsenal’s midfielder Mikel Arteta (Figure 6). This one was probably the most surprising of all the players I’ve looked at as there seems to be quite a mix of players around Arteta, including both offensive and defensive players, although perhaps this is actually representative of Arteta’s role at Arsenal? Figure 6: Clustering Mikel Arteta Next Steps For a first go the results are pretty promising but there are plenty of ways the technique could be improved. At the moment I have used all the data I had available for each player but I suspect more specific results could be obtained by filtering the data. For example, there may be specific attributes of a player you want to match on e.g. looking for attackers by just their creative output may be more useful than including their tackles, interceptions etc, which may be of minor importance to their role. Finally, all the data used here are aggregated. A really interesting next step would be to include xy co-ordinates for shot locations, interceptions, passes etc to cluster players based on the locations of their actions on the pitch (donations of xy data will be gratefully accepted :)). Filipe Rodrigues - February 20, 2014 Hello Martin, great job once again. Can you tell me what data was used to complete this post? Martin Eastwood - February 20, 2014 It is collected from all over the Internet and then algorithmically matched together. Sadly it’s not an easy job to acquire. Dank - June 29, 2014 Hello Martin, i want to know which the values of the axes, can you explain it me? Martin Eastwood - June 29, 2014 Hi Dank, the axes represent the first and second principle components. I don’t have time to explain it all here at the moment so in the meantime it’s probably worth taking a look at the Wikipedia entry as a starting point – http://en.wikipedia.org/wiki/Principal_component_analysis Antony - March 12, 2014 ~Hi Martin – really interesting analysis which shows what you can do with pure top down data driven approach. I did a similar analysis with the MCFC/Opta data last year but instead of using PCA I first just considered midfielders and then qualitatively chose the attacking and defensive attributes I wanted to determine out/underperformance in versus the sample (by scaling per 90 and normalising by mean and variance). By adding up defensive and offsensive (rather than projecting onto eigenvectors) a 2D plot revealed many qualitative knowns and some surprises. The population nicely clustered into the enforcers, schemers, creators and luxuries respectively and then PCA helped to characterise the variance within each cluster. Would be interesting to compile these types of systems for many seasons and see what the principal dynamical modes are, like improving within a cluster or moving between them, e.g. the trajectory of Giggs and now Gerrard. Interesting the different approaches we used in the ordering and application of qualitative intuition vs. quantitative rigour, which I think is the marriage that has to be progressed and its limitations understood for analytics to really take off and be adopted. Also like the piece on ExpG….amazing fit to the decay curve! Martin Eastwood - March 13, 2014 Hi Antony, That sounds really cool! At some point I want to go back and try something similar with the clustering using subsets of the players attributes. At the moment I use all the players stats but it would be interesting to try just clustering players based on passing stats or defensive stats etc. I think splitting analyses out over seasons will be a really important thing to do to assess trajectories of player’s careers and how they develop / change with age or move between clusters. Just need the data to do it :)	{"url":"http://www.pena.lt/y/2014/02/10/comparing-players-using-cluster-analysis/","timestamp":"2024-11-14T03:53:11Z","content_type":"text/html","content_length":"23409","record_id":"<urn:uuid:6ecd847f-3116-4e44-95e8-b1422f271219>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00383.warc.gz"}
Thermodynamics via creation from nothing: Limiting the cosmological constant landscape The creation of a quantum Universe is described by a density matrix which yields an ensemble of universes with the cosmological constant limited to a bounded range Λ[min⁡]≤Λ≤Λ[max⁡]. The domain Λ<Λ [min⁡] is ruled out by a cosmological bootstrap requirement (the self-consistent back reaction of hot matter). The upper cutoff results from the quantum effects of vacuum energy and the conformal anomaly mediated by a special ghost-avoidance renormalization. The cutoff Λ[max⁡] establishes a new quantum scale—the accumulation point of an infinite sequence of garland-type instantons. The dependence of the cosmological constant range on particle phenomenology suggests a possible dynamical selection mechanism for the landscape of string vacua. Physical Review D Pub Date: December 2006 □ 04.60.Gw; □ 04.62.+v; □ 98.80.Bp; □ 98.80.Qc; □ Covariant and sum-over-histories quantization; □ Quantum field theory in curved spacetime; □ Origin and formation of the Universe; □ Quantum cosmology; □ High Energy Physics - Theory; □ General Relativity and Quantum Cosmology RevTex, 4 pages, 4 figures	{"url":"https://ui.adsabs.harvard.edu/abs/2006PhRvD..74l1502B","timestamp":"2024-11-08T21:50:41Z","content_type":"text/html","content_length":"39426","record_id":"<urn:uuid:ae588b24-e7c2-4003-9524-8191596020ce>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00597.warc.gz"}
Forex compounding strategy – how to use it Investment Tips 30-12-2022 01:03 80 Views Forex compounding strategy – how to use it Forex compounding strategy – how to use it Among so many strategies related to trading on financial markets, you might wonder which is the most common, the most obvious, for making more profits. This article presents your compounding strategy as one of the most obvious trading systems for reaping decent profits. And for a good example, we take Forex’s compounding strategy so you can see how this method works in a real market situation. Compounding strategy in Forex – Key takeaways • What is compounding in finance? • What is a compounding trading strategy? • Compound interest rate and forex compounding plan explained • What is compounding in Forex • Compounding and Forex carry trade • The perks and downsides of Forex’s compounding strategy • Forex compounding example • Forex compounding calculator So let’s start with the basic notions of compounding and interest so you can better grasp the whole concept of trading with this strategy. What is compounding in finance? There are two major ways of managing your money and profits in the market. These are compounding and non-compounding. The latter is nefast to your traders since it means you gain the profits, but you spend them as soon as these are in your trading account. In many ways, it’s even worse than having a losing trade. In general, we will try to explain what compounding in finance means by taking, for example, money deposited in your bank account. When you deposit your funds in a bank account, you reap the interest on a monthly and yearly basis that the bank is using your money. For instance, suppose you deposit $1000 in your bank account with an interest rate of %2 every year. Then you will earn $20 in interest during the first year. With the compound interest over two years, you will earn $20.40. In this way, when the interest rate is based on your increasing balance, we have the snowball effect of accumulated interest. From the other perspective, when you take out a loan or credit card debt, the bank charges you interest for borrowing the money. In that case, this kind of interest can ruin you. You can find yourself paying off far more money and needing far more time to pay it off. What is a compounding trading strategy? Compounding in financial markets trading consists of the following. You make profits on the market, be it Forex, stock, or crypto. It doesn’t matter. And instead of spending the money, you reinvest the profits made on every previous trade into the next one. With the snowball effect, you gain a considerable amount over time. Compound interest – Forex what is compounding in Forex Now let’s apply this method to the Forex market. The compounding strategy in the Forex market represents the plan based on a goal of investment portfolio growth where risk tolerance and rewarding aspects work together. Therefore it turns out to be a secure and easy strategy to grow your compounding Forex account sustainably. To make this happen, anytime you reap some profits, you need to put the money you earn during Forex trading into your portfolio. That way, even the account with a meager deposit can considerably develop contrary to the minor gains you get using similar ventures. It is useful for beginner traders who aim to gradually grow their Forex account balances. It releases you from the pressure of searching for money from outer sources. But what could be the downside of this method? The thing is that with the gains, you must be prepared for a certain level of risk. You can lose the money you reinvested as well as you can produce the snowball effect of profit. And just as in any other market, you can lose the capital abruptly. This trading method is good for some traders but definitely not for those with a lack of patience. In order to precisely see what would be the results of your compounding strategy in Forex, we suggest using the Forex Compounding Calculatorator. It is available on every Forex training online How to calculate compound interest? You calculate compound interest by taking the profit from the compounding time frame. It can be daily, monthly, or annually, and add the time frames you are interested in. For example, an interest rate of 10% compounded over a 2-year period with a $100 initial investment would result in a profit of $10 in the first year (on $100) and $11 in the second year (on 110 $), making a total of $121. Compare it to an investment without compound interest. You would only get $120 since you would get a fixed profit of $10 per year. Forex carry trade and compounding interest. Carry traders very often use the Forex compounding strategy. As a carry trader, you should take into account the Forex compound interest rate forecast and differential during the specific trading time frame. If differential and forecast work for you and look favorable, the next thing to consider is compound interest. Let’s see what carry trade is and how compound interest affects the success of carry Carry trade is mainly used by Forex traders who trade larger amounts. Generally, this sort of trading consists of going short for the currencies with the lowest interest rate and vice versa, going along with those with the highest interest rate. The trading positions are held for an extended period in order to benefit from the differential of this interest rate. Popular currencies for carrying trade are EUR/JPY, AUD/JPY, AUD/USD, and NZD/JPY. Carry traders tend to compound the interest on a daily or monthly basis in order to grow the returns. But they can be subject to varying returns due to fluctuations in interest rate differential. If the differential rate widens, it will be the move in the trader’s favor. The trader benefits from it in the upcoming compounding period. But if the differential rate narrows, the trader then receives lower returns. Now let’s see a quick example. If a Forex trader aims to trade a pair with an annual differential of 4.4%, it can lead to a return of 4.4898% if the rate is compounded on a monthly basis. The annual return will be 4.4980% if the interest is on a daily basis. Compounding trading plan and the rule of 72 In order to show you better the power of compound interest, we will look at the rule of 72. The rule of 72 is a simple way to determine how long it will take to double an investment, given a fixed annual rate of return. The rule states that the number of years it takes to double your money is approximately equal to 72 divided by the expected annual return, expressed as a whole number (not a percentage). So if you expect your investments to grow at a rate of 8% per year, it will take about nine years (72/8) for them to double. Of course, this is a rule of thumb and not an exact science. Nonetheless, it can be useful for quickly estimating how long your money will take to grow. And it can be particularly useful for comparing different investment opportunities. For example, let’s say you’re trying to choose between investing in a stock that’s expected to earn 10% per year and a bond that’s expected to earn 5% per year. Using the rule of 72, you can estimate that it would take around 7.2 years for your money to double with stock investing (72/10) and 14.4 years for bond investing (72/5 ). So, in this case, investing in stocks would be the best option taking the return as the only criterion because it would take less time for your money to grow. Of course, there are other factors to consider when making investment decisions. But the rule of 72 can be a useful tool for quickly estimating how long your money will take to grow. How does the rule of 72 work? The rule of 72 works by taking the number 72 and dividing it by the expected annual return on your investment. The resulting number is approximately equal to the years it takes for your money to For example, if you expect your investments to grow at a rate of 8% per year, it will take about nine years (72/8) for them to double. Similarly, if you expect a return of 6% per year, it will take you about 12 years (72/6) to double your investment. Forex compounding strategy – Final Thoughts Forex compounding strategy represents an excellent method in trading to make considerable gains without exposing yourself to significant risks. It allows you to progressively grow your account balance by reinvesting small sums consistently. This strategy decreases the risk of big losses since the Forex market is prone to big price fluctuations. It spares you from the need to look for outside trading funds. Moreover, it’s a kind of proper money management technique that enables you to track your investments better and have better control over them. In that way, you are in a position to make better decisions in the market. Finally, it’s an attractive trading method for investors looking for long-term gains and consistent financial success on Forex. The post Forex compounding strategy – how to use it appeared first on FinanceBrokerage.	{"url":"https://investinaitech.com/2022/12/30/forex-compounding-strategy-how-to-use-it/","timestamp":"2024-11-02T03:14:28Z","content_type":"text/html","content_length":"74978","record_id":"<urn:uuid:85e218fb-2e7a-46e3-9589-4b8d4b69556c>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00251.warc.gz"}
How to Make a Lampshade Paper Sculpture - Exercise 2 - Kids Crafts & Activities How to Make a Lampshade Paper Sculpture – Exercise 2 Earlier today I posted How to Make a Paper Wave Sculpture – Exercise 1. It shows you the basics of making a Paper Sculpture…. specifically a paper wave sculpture. Well, in this post I’m going to show you how to make a Lampshade Paper Sculpture…. it’s the 2nd exercise in two showing you how to Paper Sculpt. It’s a little more complicated than the first, but if you did Exercise 1 you are ready for the challenge of Exercise 2. So, let’s get started. Materials Needed Butter Knife Step 1 The paper is ruled off with a pencil and a ruler. Do this in parallel lines an equal distance apart from each other… one on one side of the paper and the next on the other and so on. Step 2 Now use the lines that you drew as a guide to where to score the paper. To score a piece of paper, you can use the dull side of a butter knife. Take a ruler and line it up with one of the pencil lines…then use the dull side of a butter knife and run it along the ruler (on the paper) as if it were a pencil. Now you should have a scored line. The lines are scored on alternate sides of the paper and the paper bent like a fan. Step 3 The edges of the paper are taped together to achieve a sort of lampshade effect. Step 4 This is what the item looks like from the top. That’s all there is to it…. you just finished making your Lampshade Paper Sculpture. How did your sculpture turn out? Let me know in the comments below. And, add a picture of the Lampshade Paper Sculpture as well to the comment…. I’d love to take a look at it. Technorati Tags: Paper Sculptures, Paper Crafts, Paper Arts and Crafts, Paper, Sculpting, sculpting with paper, paper sculpture, crafts with paper You must be logged in to post a comment.	{"url":"https://www.artistshelpingchildren.org/kidscraftsactivitiesblog/2012/04/how-to-make-a-lampshade-paper-sculpture-exercise-2/","timestamp":"2024-11-08T12:07:29Z","content_type":"text/html","content_length":"46871","record_id":"<urn:uuid:8c51af61-4569-4211-bef3-b4f81c6847f0>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00486.warc.gz"}
The most useful algebra formulas to learn If you’re interested in learning a few basic algebra formulas, continue reading on to discover some of the most useful basic algebra formulas to learn. The most useful algebra formulas to learn: 1. The difference between two squares The difference between two squares refers to when two binomials have only one difference, when one term features a plus and the other term features a minus sign. To quickly discover the difference between two squares use the following basic algebra formula (x+y) (x-y) = x squared – y squared. 2. The difference of cubes To figure out the difference of cubes use the formula a to the power of 3 – b to the power of 3 = (a-b) (a+b). 3. The sum of cubes If you need to find out the sum of cubes, the formula which you should use is a to the power of 3 + b to the power of 3 = (a+b) (a squared -ab + b2). 4. The pythagorean theorem The pythagorean theorem is one of the most commonly used algebraic equations and is used to accurately figure out the lengths and sides of a right angled triangle. Thankfully the pythagorean theorem equation is one of the simplest algebraic equations to learn. When a and b refer to the two sides of a triangle which stem from either side of a triangle’s right angle use the equation a squared plus b squared = c squared. In this equation c, refers to the the side which you’re trying to work out the value of, which is known as the hypotenuse. 5. How to multiply powers which boast the same base To multiply powers which feature the same base make sure to add your exponents. The formula which you should use to multiply powers which boast the same base is x to the power of a, x to the power of b = x to the power a+b. Before you try to figure out a trickier algebraic problem such as this particular problem, you may be better off learning simpler equations such as the pythagorean theorem. Which is much easier to grasp and will give you a solid base to learn more difficult algebraic equations. 6. How to raise a power If you’re unsure of how to raise a power, simply use the formula (X to the power of a) to the power of b = X to the power of ab. Just remember that when you need to raise a power by a power to multiply your exponents. 7. How to discover the power of a product property To workout the power of a product property use the algebraic formula (xy) to the power of a = x to the power of a, y to the power of a. Also make sure that when you raise a product to a power to find out the power of each factor, in order to multiply your factors. In conclusion, mastering these algebra formulas can greatly enhance your mathematical skills and problem-solving abilities. Remember, understanding concepts like the 135 degree angle is just as crucial for tackling various challenges in math. So if you haven’t looked at an algebra book since high school and you’re looking to brush up on your algebra skills, it’s well worth starting off with mastering the simple formulas listed above. Leave a Reply Cancel reply	{"url":"https://www.tutorcircle.com/learning/the-most-useful-algebra-formulas-to-learn/","timestamp":"2024-11-05T09:41:30Z","content_type":"text/html","content_length":"40312","record_id":"<urn:uuid:96e722a7-0858-457c-bbbd-3e4dd8fda56f>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00587.warc.gz"}
Linear Algebra with Differential Equations/Heterogeneous Linear Differential Equations - Wikibooks, open books for an open world We now tackle the problem of G ( t ) {\displaystyle \mathbf {G} (t)} being nonzero, so that we have the following problem: X ′ = A X + G ( t ) {\displaystyle \mathbf {X} '=\mathbf {AX} +\mathbf {G} (t)} There are four reasonable ways to solve this.	{"url":"https://en.m.wikibooks.org/wiki/Linear_Algebra_with_Differential_Equations/Heterogeneous_Linear_Differential_Equations","timestamp":"2024-11-14T23:34:34Z","content_type":"text/html","content_length":"31036","record_id":"<urn:uuid:8d44b385-3008-45b3-b8c2-b7c7f3da2b9d>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00013.warc.gz"}
How do you interpret theta in options trading? - coxbusinessaz Theta is one of the more difficult concepts to interpret when trading options. This article will provide an overview of theta and how traders can use it. Finally, we will consider some tips for traders looking to use theta effectively. Check out Saxo for more info on options trading. What is theta in options trading, and what does it represent? It is a measure of the time decay of an option, and it is the rate at which the value of an option declines as the expiration date approaches. Theta is often referred to as the “time value” of an Theta can be positive or negative. A positive theta means that the option is losing value as time passes, and a negative theta means that the option is gaining value as time passes. Theta is typically negative for options in-the-money (ITM) and positive for out-of-the-money options (OTM). Theta is an essential concept for options traders to understand because it can significantly impact the profitability of trades. Time decay is a primary consideration when trading options and theta is the best way to measure it. How do you calculate theta? Theta is typically expressed as a percentage. It is calculated by taking the change in the option’s value over a specific period and dividing it by the total amount of days. If an option has a theta of -0.05, it loses 5% of its value daily. If the option has a theta of 0.10, it gains 10% of its value every day. How do you interpret theta in options? The most important thing to understand about theta is that it represents the rate at which an option’s value will decline as expiration approaches. This decline is due to time decay, and the closer an option gets to expiration, the faster it will lose value. Theta can be interpreted in many ways. First, it can estimate how much an option will lose in value over a specific period. For example, if it has a theta of -0.05 and is 30 days from expiration, we expect it to lose approximately 1.5% of its value over the next 30 days (-0.05 x 30 = -1.5). Theta can estimate the probability of an option expiring in the money. It is because theta represents the rate at which an option loses value. The faster an option loses value, the less likely it is to expire in the money. Finally, theta can be used to estimate the breakeven point for an options trade. The breakeven point is the price at which an option must be traded to break even on the trade. Tips for using theta effectively Traders should keep some things in mind when using theta to make trading decisions. First, it is essential to remember that theta represents the rate at which an option loses value, and it means that theta is most relevant when expiration is close. The further away expiration is, the less impact time decay will have on the option’s value. It is important to remember that theta is only one factor to consider when trading options. Many other factors can impact the value of an option, such as implied volatility and interest rates. As such, theta should not be considered in isolation when making trading decisions. It is important to remember that theta can be both positive and negative. A positive theta means that the option is gaining value as time passes, and it can benefit traders looking to profit from time decay. A negative theta means that the option loses value as time passes, which can be detrimental for traders holding options that are losing value. It is important to remember that theta is a double-edged sword. While it can be used to estimate the probability of an option expiring in the money, it could be used to estimate the probability of an option expiring out of the money. Finally, it is essential to remember that theta is only one measure of an option’s value. Many other measures, such as vega and gamma, can provide valuable insights into how an option is likely to behave in the future. Traders should not rely solely on theta when making trading decisions.	{"url":"https://www.coxbusinessaz.com/how-do-you-interpret-theta-in-options-trading/","timestamp":"2024-11-04T09:15:34Z","content_type":"text/html","content_length":"59736","record_id":"<urn:uuid:eb361fb4-433a-4f6e-8735-13233d726fde>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00773.warc.gz"}
Megabytes and Kilobytes Converter (MB and kB) Megabytes and Kilobytes Converter Use this calculator to convert megabytes (MB) to kilobytes (kB) and kilobytes to megabytes. This converter is part of the full data storage converter tool. Disclaimer: Whilst every effort has been made in building our calculator tools, we are not to be held liable for any damages or monetary losses arising out of or in connection with their use. Full Kilobytes to Megabytes Conversions Kilobytes Megabytes 1 kilobyte 0.001 megabytes 2 kilobytes 0.002 megabytes 3 kilobytes 0.003 megabytes 4 kilobytes 0.004 megabytes 5 kilobytes 0.005 megabytes 6 kilobytes 0.006 megabytes 7 kilobytes 0.007 megabytes 8 kilobytes 0.008 megabytes 9 kilobytes 0.009 megabytes 10 kilobytes 0.01 megabytes 11 kilobytes 0.011 megabytes 12 kilobytes 0.012 megabytes 13 kilobytes 0.013 megabytes 14 kilobytes 0.014 megabytes 15 kilobytes 0.015 megabytes 16 kilobytes 0.016 megabytes 17 kilobytes 0.017 megabytes 18 kilobytes 0.018 megabytes 19 kilobytes 0.019 megabytes 20 kilobytes 0.02 megabytes Figures rounded to a maximum of 5 decimal places (7 with smaller numbers). How many megabytes are there in 1 kilobyte? There are 0.001 megabytes in 1 kilobyte. To convert from kilobytes to megabytes, multiply your figure by 0.001 (or divide by 1000) . Megabytes to Kilobytes Conversions Megabytes Kilobytes 1 megabyte 1000 kilobytes 2 megabytes 2000 kilobytes 3 megabytes 3000 kilobytes 4 megabytes 4000 kilobytes 5 megabytes 5000 kilobytes 6 megabytes 6000 kilobytes 7 megabytes 7000 kilobytes 8 megabytes 8000 kilobytes 9 megabytes 9000 kilobytes 10 megabytes 10000 kilobytes 11 megabytes 11000 kilobytes 12 megabytes 12000 kilobytes 13 megabytes 13000 kilobytes 14 megabytes 14000 kilobytes 15 megabytes 15000 kilobytes 16 megabytes 16000 kilobytes 17 megabytes 17000 kilobytes 18 megabytes 18000 kilobytes 19 megabytes 19000 kilobytes 20 megabytes 20000 kilobytes Figures rounded to a maximum of 5 decimal places (7 with smaller numbers). How many kilobytes are there in 1 megabyte? There are 1000 kilobytes in 1 megabyte. To convert from megabytes to kilobytes, multiply your figure by 1000 (or divide by 0.001) . What is a kilobyte? As you might have guessed, a kilobyte is commonly defined as consisting of 1,000 bytes. We should point out that its actual value is 1,024 bytes (or 2^10 bytes). Still, most say that this value is very similar to 10^3. It is for this reason that the International System of Units (SI) generally equates the kilobyte with 1,000 bytes. This also helps to somewhat simplify multiplication factors when dealing with such massive numbers. To put this another way, one kilobyte is 10 x 10 x 10 bytes. Although this might seem like a large figure, let's recall that most operating systems contain gigabytes of data capacity! What is a megabyte? Most computer users are very familiar with the megabyte. As you might have already guessed, a megabyte is the equivalent of 1,000,000 bytes (or 10^6 bytes). We should point out that there is more than meets the eye here. While the SI definition is one million bytes of information, many computing professionals will instead use the more precise number of 1,048,576 bytes (1,024^2). This arises from the binary multiples that occur (each byte contains eight bits of information within its string). Other individual data storage converters Kilobytes and Gigabytes, Megabytes and Gigabytes, Megabytes and Kilobytes, Terabytes and Kilobytes,	{"url":"https://www.thecalculatorsite.com/conversions/datastorage/megabytes-kilobytes.php","timestamp":"2024-11-08T21:46:13Z","content_type":"text/html","content_length":"309853","record_id":"<urn:uuid:7988ee2c-9059-4234-9288-7fde6ee667b8>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00383.warc.gz"}
Express the rate of the reaction $$2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ in terms of (a) $\Delta\left[\mathrm {C}_{2} \mathrm{H}_{6}\right]$ (b) $\Delta\left[\mathrm{CO}_{2}\right]$ Short Answer Expert verified Question: Express the rate of the given reaction in terms of the concentration changes of C2H6 and CO2. Answer: (a) The rate of reaction in terms of the concentration change of C2H6 is given by: $$\ text{rate} = -\frac{1}{2} \times \frac{d[\mathrm{C}_{2} \mathrm{H}_{6}]}{dt}$$. (b) The rate of reaction in terms of the concentration change of CO2 is given by: $$\text{rate} = \frac{1}{4} \times \ Step by step solution Recall the definition of reaction rate The rate of a chemical reaction is defined as the change in the concentration of a reactant or product over time. Mathematically, the rate of reaction for a species A can be given by: $$\text{rate}_ {A} = \frac{1}{\text{stoichiometric coefficient}} \times \frac{d[\text{A}]}{dt}$$. Note that if A is a reactant, the stoichiometric coefficient is negative, and if A is a product, the stoichiometric coefficient is positive. Apply the formula to express the rate of reaction in terms of $\Delta[\mathrm{C}_{2} \mathrm{H}_{6}]$ For part (a), we are to express the rate of reaction in terms of the change in concentration of C2H6, which is a reactant with a stoichiometric coefficient of 2. Using the formula from step 1, we can write: $$\text{rate} = -\frac{1}{2} \times \frac{d[\mathrm{C}_{2} \mathrm{H}_{6}]}{dt}$$ Apply the formula to express the rate of reaction in terms of $\Delta[\mathrm{CO}_{2}]$ Similarly, for part (b), we are to express the rate of reaction in terms of the change in concentration of CO2, which is a product with a stoichiometric coefficient of 4. Using the formula from step 1, we can write: $$\text{rate} = \frac{1}{4} \times \frac{d[\mathrm{CO}_{2}]}{dt}$$ In summary, the rate of the given reaction can be expressed in terms of the concentration changes of C2H6 and CO2 as follows: (a) $$\text{rate} = -\frac{1}{2} \times \frac{d[\mathrm{C}_{2} \mathrm{H}_{6}]}{dt}$$ (b) $$\text{rate} = \frac{1}{4} \times \frac{d[\mathrm{CO}_{2}]}{dt}$$ Key Concepts These are the key concepts you need to understand to accurately answer the question. Chemical Kinetics Understanding the speed of a chemical reaction, or its reaction rate, is a fundamental part of chemical kinetics. In essence, chemical kinetics studies how rapidly chemical reactions occur and what factors—such as temperature, pressure, and concentration—affect this speed. For example, a reaction that produces products quickly is said to have a high reaction rate. On the other hand, a reaction that takes a long time to convert reactants to products has a low reaction rate. This concept is crucial in various fields, from industrial synthesis of chemicals, where maximizing efficiency is key, to environmental chemistry, where the rates at which pollutants are broken down can have significant implications. One of the most important aspects to consider in chemical kinetics is the rate law, which is an equation that links the rate of reaction to the concentration of reactants. This understanding helps to predict how changes in conditions can alter the speed of a chemical reaction. Stoichiometry can be thought of as the 'recipe' for a chemical reaction. It refers to the quantitative relationship between the amounts of reactants and products involved in a chemical reaction based on the balanced chemical equation. In the context of reaction rates, stoichiometry dictates how the changes in the amount of one substance relate to the changes in another. Understanding Reaction Coefficients In our exercise, the stoichiometric coefficients (the numbers in front of the chemical formulas in the balanced equation) tell us that 2 moles of C2H6 react with 7 moles of O2 to produce 4 moles of CO2 and 6 moles of H2O. In terms of reaction rates, these coefficients are used to relate the rates of consumption of reactants to the rates of formation of products, making sure the conservation of mass is respected in the reaction. This insight is essential for everything from laboratory experiments to engineering applications where precise calculations are necessary for successful outcomes. Concentration Changes To put it simply, concentration is a measure of how much of a given substance is present in a mixture. In chemistry, changes in concentration over time typically indicate that a reaction is taking place. The exercise on hand exemplifies how the concentrations of reactants decrease while those of products increase as the reaction proceeds. Rate Expressions The changes in concentration are crucial for calculating the reaction rate. When answering the exercise, we used differential rate expressions which involve the derivative of concentration with respect to time. This represents the instantaneous rate of change, giving us a clear picture of how fast the reaction is occurring at any given moment. In practice, understanding concentration changes allows chemists and engineers to design reaction conditions that optimize product yields and minimize waste, which is key for sustainability and economic viability in chemical manufacturing.	{"url":"https://www.vaia.com/en-us/textbooks/chemistry/chemistry-principles-and-reactions-6-edition/chapter-11/problem-1-express-the-rate-of-the-reaction-2-mathrmc2-mathrm/","timestamp":"2024-11-14T22:22:05Z","content_type":"text/html","content_length":"261849","record_id":"<urn:uuid:d9b47a15-b311-470f-b33d-c2ac1f8e1db4>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00099.warc.gz"}
Introduction to Aerospace Flight Vehicles Aerospace engineers are more often than not concerned with external fluid flows, e.g., the flow as it develops over the outside surfaces of a body, such as an airfoil, wing, or a complete flight vehicle. However, internal fluid flows are also crucial in many aerospace engineering applications. An internal flow can be defined as a flow in which its downstream development is confined or altered by all of the boundary surfaces inside which it flows. Internal flows may comprise gases (such as air) and liquids. The flow of liquids is often referred to as hydraulics, which is studied extensively and used in all branches of engineering. Such internal flows are encountered but are not limited to flows inside engines and combustion chambers, engine intake systems, pneumatic and hydraulic systems, fuel systems, heat exchangers and cooling systems, air conditioning systems, fire suppression systems, and wind tunnels. The figure below shows one example of an internal flow through part of a jet engine. Velocity non-uniformities from the wakes behind each blade affect the engine performance and the noise it generates. Internal flow through part of a jet engine. The blade surfaces are colored by the magnitude of the static pressure, and the flow field is vorticity. • Appreciate the differences between external flows and internal flows, as well as understand why the action of viscous effects gives pressure losses for internal flows. • Distinguish between the different effects caused by laminar and turbulent flows through pipes and ducts. • Calculate the frictional losses and pressure drops associated with the flows through pipes and ducts using the Colebrook-White equation and a Moody Chart. • Estimate the pumping power requirements to create specific mass flows and flow velocities through pipes and ducts. • Understand the design process of producing specific flow velocities in wind tunnels and calculating power requirements. Internal Versus External Flows An interesting consequence of all internal flows is that the associated surface boundary layers cannot develop freely, and the flows on each inner surface merge. In this regard, the boundary layer thickness and other properties will not be the same as if the boundary layer formed over an external surface, as illustrated in the figure below. This outcome is obtained because, for external flows, the boundaries or surfaces (often just called walls) introduce viscous effects into the flow that progressively diminish away from the wall. However, for internal flows, all parts of the walls of a duct, tube, or pipe constrain the flow development, so the entire cross-section of the internal flow is affected by the walls. With an external flow, the effects diminish away from the wall, but with an internal flow, all the walls constrain the flow developments. The forgoing behavior of the surface boundary layers is one reason external and internal flows must be carefully distinguished, i.e., all of the boundaries in an internal flow will affect each other. Because viscous losses (i.e., skin friction) are significant along all flow surfaces, for an internal flow, these effects manifest as significant pressure drops over the length of the interior surfaces where the fluid flows; the friction and the associated energy loss are irrecoverable and appear as heat. Evaluating these pressure drops is essential to understanding internal flows and determining the engineering requirements to move fluids along pipes and ducts. Internal Flow Problems Internal fluid flow problems occur in many aerospace systems, including fuel, hydraulic, and pneumatic systems. A rocket engine, for example, as shown in the schematic below, has many internal flows that require very high mass flow rates of fuel and oxidizer. These fluids must pass through several pumps, pipes, and valves, so significant pressure losses are incurred. Fuel is also pre-circulated in channels or a jacket around the exhaust nozzle to keep it cool, over which considerable pressure drops also occur. Therefore, turbopumps are needed to deliver the fuel and oxidizer to the combustion stage of the rocket engine, which usually involves the flow of cryogenic liquids at extremely high mass flow rates. Schematic showing the internal flow paths of fuel and oxidizer inside a rocket engine, which has to flow through pipes at high volumetric rates. For aircraft, their hydraulic systems operate at relatively high pressures. Hence, they require high flow rates to operate the large actuators that power their landing gear, flaps, slats, flight control systems, etc. In addition, the hydraulic lines may stretch through the structure for considerable distances so that pressure drops can be significant. Aircraft usually have multiple hydraulic systems for redundancy in the event of failure, so the systems must be completely independent. Airliners, for example, typically have three parallel hydraulic systems with multiple engine-driven pumps, electrical pumps, air-driven pumps, and hydraulic pumps. In another example of an internal flow, consider a fuel delivery system on an aircraft where fuel needs to be pumped from one or more fuel tanks (e.g., in the wings) to the engine(s), as shown in the schematic below. The distribution of the fuel lines also involves considerable lengths, often with many twists and turns through the aircraft structure. Therefore, pressure drops, pumping power, and flow rates must be carefully considered so the engines receive the required fuel volumes at the pressures necessary to support the engine’s requirements. Redundancy of the fuel delivery, in this case, is achieved with the use of a mechanical (engine-driven) pump and a completely independent backup electrical pump. In fuel delivery, even for a relatively simple aircraft, pressure drops, pumping power, and flow rates must be considered to deliver the fuel to the engines at the required pressure. An interesting aerospace example of a combined external/internal flow interaction problem was encountered with the design of the X-43A, a hypersonic lifting-body research vehicle, and a CFD solution for the flow shown in the figure below. At hypersonic Mach numbers, the shock waves bend back rather steeply and come close to the fuselage. Hence, the design used the lower side of the forward fuselage (an external flow) as part of the engine intake to create the necessary sequence of shock waves to slow down the air and increase the pressure before the flow entered the engine’s intake (an internal flow). Therefore, the external flow forms an upstream (boundary) condition for the internal flow through the engine to make it function. The X-43A was developed to test a supersonic combustion ramjet (scramjet). External flow about the forebody was an integral part of preconditioning the air intake, an example of an external and internal flow interaction. Categories of Internal Flow Problems In general, the types of internal flow problems that often arise for aerospace engineers to solve usually fall into these categories: 1. Determining the pressure drop (or the so-called head loss) when flows move through pipes of a given length and diameter for a required volume or mass flow rate. 2. Determining the volume or mass flow rate when the pipe length and diameter are known for a specified (or allowable) pressure drop, which may be constrained for several possible reasons. 3. Finding the pumping power needed from a mechanical pump to obtain a certain exit pressure and/or flow rate for a pipe of a specific diameter over a certain length. 4. Determining the pipe diameter when the pipe length and flow or mass rate are known for a particular pressure drop. The pipe size needed will also affect its weight and cost, both being significant considerations for aerospace applications. 5. Finding heating effects associated with pumping gases through ducts and pipes, especially at higher flow speeds. Heat is generated because of frictional losses and gas compression, and heat dissipation may need to be considered in some applications. Not all of these problems can be solved directly, even if most of the relevant information is known. Design purposes sometimes require iterative solutions starting from some initial estimates. However, some fundamental problems illustrating the process of finding pressure drops and determining pumping power through pipes can be introduced. Fluid Dynamics of Internal Flows Internal flows are encountered in many shapes and sizes of ducts and pipes, but the flow through a circular pipe is often used as an exemplar, as shown in the figure below. The effects of viscosity cause the fluid in adjacent layers to slow down gradually and progressively develop a velocity gradient in the pipe, i.e., an axisymmetric form of the boundary layer. To make up for this velocity reduction, the fluid’s velocity at the centerline of the pipe has to increase to balance the net mass flow rate through the pipe. Consequently, a significant velocity gradient eventually develops across the entire cross-section of the pipe. The primary manifestation of the viscous friction on the interior walls is a pressure drop along the length of the pipe, which depends on the velocity profile, the volumetric or mass flow rate, and the fluid type. An internal flow through a pipe will develop a velocity profile, a manifestation of the viscous friction on the walls being a pressure drop along the length of the pipe. Determining the velocity profile and the corresponding mass (or volume) flow rate is one problem in analyzing internal flows. For example, the average velocity The idea of an average flow velocity in a pipe is obtained by integrating over the cross-section to find the mass (or volume) flow rate and dividing by the pipe’s area. For the incompressible flow in a circular pipe of diameter Therefore, if the velocity profile Entrance Length At the beginning of any internal flow, there is a transition period as it organizes itself to become a fully developed internal flow. As shown in the schematic diagram below, the entrance region or length in such a flow, fully developed internal flow. In the entrance region, the velocity profile changes in the flow direction as it adjusts from some initial profile at the inlet to a fully developed profile where the effects of the walls are felt across the cross-sectional area of the entire pipe, duct, or channel. The flow into a pipe or duct takes time to establish a steady-state condition, but the viscous effects produced by the walls play an essential part in determining the final velocity profile. Notice the pressure drop or “loss” as the flow develops along the pipe. Therefore, a fully developed internal flow is defined as one where the velocity profile does not change along the flow direction, i.e., in the axial downstream direction. When the duct has the same cross-section, sections will have the same velocity profile at successive axial locations. If the pipe changes in cross-section or turns a corner, it will take some distance for the flow to develop again fully. As might be expected, the entrance length depends on whether the flow is laminar or turbulent and whether the internal surface is smooth or rough. For example, for the laminar flow in a smooth, straight cylindrical pipe of diameter where diameter of the pipe, i.e., For a fully developed turbulent flow, the entrance length is this latter result shows that the flow mixing caused by turbulence greatly speeds up the process of reaching a fully developed internal flow. Under most practical conditions, the flow in a circular pipe is laminar for Hydraulic Diameter For flows through non-circular pipes (for which many types are encountered in engineering practice), the Reynolds number is redefined based on the hydraulic diameter where wetted perimeter. The hydraulic diameter is sometimes called the hydraulic mean diameter. The basic idea is illustrated in the figure below. Naturally, the hydraulic diameter is defined to reduce to the physical diameter for circular pipes, i.e., The principle of hydraulic diameter is to find an equivalent circular pipe based on the wetted perimeter of the non-circular pipe. For example, for a pipe with a rectangular cross-section of width Notice that in a square duct where The corresponding Reynolds number will now be calculated by using the hydraulic diameter, i.e., However, the concept of a hydraulic diameter works well only for fully developed turbulent flows, i.e., for Laminar Flow in Pipes Consider analyzing a steady laminar flow of an incompressible fluid with constant properties in the fully developed region of a smooth straight pipe of circular cross-section. The following solution, often called a Hagen-Poiseuille flow, is one of the few exact solutions for viscous internal flows. Such laminar flows are also sometimes referred to as Poiseuille or Couette flows. The approach is exceptionally instructive and is a prerequisite to understanding turbulent flows through pipes, which is a more semi-empirically based analysis. In the fully developed laminar flow, the flow moves at a constant axial velocity. The velocity profile Consider an area (annular disk) of the flow at a given pipe cross-section. As shown in the figure below, the forces acting on the annular disk arise from pressure and viscous (shear) forces. The area of the annular disk is Flow model for the flow in a circular pipe that is used to develop the exact laminar solution. Force equilibrium of the annulus requires a balance of pressure and shear forces such that Expanding out the terms and canceling terms where possible gives Because terms involving which leads to The shear stresses will be given by Newton’s law of viscosity such that where the minus sign reflects that the flow velocity decreases with increasing It can be seen that the left-hand side of the equation is a function of the radial coordinate 14 must hold for any value of Equation 14 can be solved by rearranging and integrating it twice to give which is a parabolic velocity profile. Also, the axial velocity 16 shows axial pressure gradient For the preceding velocity profile, the average velocity can be determined using which, on substitution, gives Therefore, in terms of also noticing that the centerline velocity but remember that this result applies only to this special case. Quantifying the Pressure Drop A quantity of interest in the analysis of internal flows is the pressure drop Of course, this pressure drop is a direct consequence of the action of viscous effects, so the pressure drop is irrecoverable. Notice that for a given length of pipe, the pressure drop is proportional to the fluid’s viscosity and flow speed, so the faster a given fluid moves through a pipe or duct, the more significant the pressure drop will be. The pressure drop can also be reduced by using a larger diameter pipe. However, weight and cost are often substantial concerns in aerospace systems, where there will be a trade between the weight and cost of a larger pipe (to reduce the losses) versus a bigger pump (to overcome the existing losses). In practice, this pressure drop will also be different for different pipe cross-sections, surfaces (smooth or rough), etc. The pressure loss for a laminar flow can be expressed as where friction factor. For a fully-developed laminar flow in a smooth circular pipe, the friction factor is which depends on the Reynolds number only. The friction factors for fully developed laminar flow in pipes of other cross-sections, such as square pipes, oval pipes, triangular pipes, etc., are published in various sources. Still, these latter shapes are rarely encountered in aerospace applications. The head loss, After the pressure or head loss is known, the required power The average velocity for the laminar flow in a horizontal pipe (i.e., no gravitational hydrostatic pressure contributions) is so the volume flow rate for laminar flow through the pipe becomes which is known as Poiseuille’s law.^[1] In terms of the pressure drop, then This result means that for a specified flow rate, the pressure drop (and the required pumping power to overcome the losses) is proportional to the pipe’s length and the fluid’s viscosity but inversely proportional to the fourth power of the diameter. The consequence of this result is that even relatively modest pipe diameter increases can significantly reduce the pumping power. However, this approach may not always prove practical (or even possible) for other reasons, including higher weight and costs. Historical Context – Poiseuille’s Original Equation Poiseuille did not originally derive the equation now known as Poiseuille’s law, which is accepted as Instead, based on his many meticulous experiments, Poiseuille’s equation for the flow rate in terms of pressure drop was written as Measurements of the values of the dynamic viscosity Check Your Understanding #1 – Using Poiseuille’s law How many small-diameter pipes of diameter Show solution/hide solution. Poiseuille’s law can also be expressed in terms of flow rate as Assume that the hydrostatic pressure is the same at the inlet and outlet (discharge) so that Therefore, the benefits of using a larger pipe become apparent because the flow rate is proportional to here. Recall that the Poiseuille equation (often called the Hagen-Poiseuille equation) is derived under specific assumptions, i.e., laminar, steady, and incompressible flow. When these assumptions are violated, such as in cases of turbulent flow and geometric conditions like wide or short pipes, the Poiseuille equation may no longer accurately describe the flow behavior. A rule of thumb is that the ratio of length to radius of a pipe should be greater than 1/48 of the Reynolds number for the assumption of Poiseuille law to be valid. Check Your Understanding #2 – Calculating the pressure drop for laminar flow in a pipe The design of the fuel delivery system requires a flow through a 200 m length of smooth pipe of 15 mm diameter. The required fuel flow rate is 125 kg hr^-1. The fluid properties of the fuel are given as ^-3 and ^-1 s^-1. All entrance effects should be disregarded. Calculate: 1. The pressure drop along the length of the pipe. 2. The pump’s pressure requirements (in terms of head). 3. The pumping power requirements. Show solution/hide solution. The cross-sectional area of the pipe is The mass flow rate The average flow velocity in the pipe is The Reynolds number based on pipe diameter is Notice that this Reynolds number is the laminar regime, so the friction factor The pressure drop and inserting the given values leads to The equivalent head loss, Finally, the pumping power Turbulent Flows in Pipes Unlike laminar flows, the expressions for the losses in pipes or ducts containing a turbulent flow are based on analysis and measurements, providing empirical or semi-empirical relationships. In fact, because of the importance of many industrial applications, systematic experiments have been performed with pipe flows to measure pressure losses for different flow rates, Reynolds numbers, and surface roughness values. Representative relative velocity profiles for fully developed laminar and turbulent flows are shown in the figure below. Remember that the fully laminar velocity profile is parabolic in laminar flow. However, the profile shape is much “fuller” in turbulent flow because of the mixing between fluid layers, with a more uniform centerline velocity and a sharper drop in flow velocity near the pipe wall. Of course, this characteristic is much like a turbulent boundary layer profile as it develops over an external surface. Differences in the fully established velocity profiles for fully laminar and fully- turbulent pipe flows. Any irregularity or “roughness” on the surface of the pipe, as shown in the figure below, will disturb the development of the boundary layer and create higher wall stresses. Therefore, the friction factor in a pipe flow with turbulence will depend on the internal surface roughness, and pipes with greater roughness will lead to more significant pressure drops. The roughness, It should be appreciated that “roughness” is a relative concept; in practice, the roughness only starts to become significant when the height of the roughness elements reaches the thickness of the laminar sublayer in the turbulent boundary layer, although this layer is relatively small. Many extruded plastic pipes are generally considered hydrodynamically smooth. However, most other surfaces are rough to some degree, so it will cause the boundary layer to become turbulent quicker, i.e., after a shorter downstream development distance. The pressure loss with a turbulent flow is still given by but now, the value of Darcy-Welsbach friction factor or just the Darcy friction factor) depends on the Reynolds number and the relative roughness height which can also be found from the process of dimensional analysis. All available results for pressure drops and friction factors for turbulent flows through pipes are obtained from experiments using artificially roughened surfaces. These experiments are done by using particle grains of a known size bonded to the inner surfaces of the pipes. Ludwig Prandtl’s students conducted the first such experiments in the 1930s, but generations of hydraulic engineers have performed many more experiments since then. The resulting friction factor Pipe friction factors can be presented in tabular, graphical, or functional forms, which are obtained by curve-fitting the measured data. Commercially available pipes are assessed with equivalent roughness values so that engineers can make appropriate calculations for design and installations. However, the relative roughness of the pipe may increase with everyday use because of corrosion or abrasion (depending on the fluid), so friction factors may change over time. Good engineering practice would always make allowances for such effects so that the needed pressure at the end of the pipe can be maintained for its expected service life. Colebrook-White Equation The most widely used equation to find the friction factor for turbulent flows in pipes is the Colebrook-White equation. This equation has no theoretical basis and is based on a curve fitting to measurements. The equation is given by Moody Chart Most of the available results for turbulent flows through common types of rough pipes have been summarized in the form of a Moody Chart, an example being shown below. Engineers can use this chart to calculate the friction factor for a given internal flow along a pipe. The friction factor is then used to calculate the pressure (or head) loss, pumping power, etc. Lewis Moody solved the Colebrook-White equation to create this type of chart, combining the dimensional terms of roughness, relative roughness, A standard Moody chart for estimating the Darcy-Weisbach friction factor in pipes and ducts. It is essential to recognize that careful interpolation on the chart will be required to find the needed On the Moody Chart, the friction factor, diameter (or hydraulic diameter), and the average flow velocity, Notice that the chart has a log-log scale. Curves of constant values of relative roughness, not a scale. Online calculator versions of the chart are also available, although some may use only approximations to the Colebrook-White equation rather than iteratively solving the actual equation. Although the Moody Chart was initially developed for flows through circular pipes, it can also be used for non-circular pipes by replacing the circular diameter, Remember that the Reynolds number based on diameter or hydraulic diameter is calculated using with the fluid properties being represented in terms of density, When using the Moody chart, if the values of two of the three parameters are known, i.e., the friction factor, MATLAB code to create the Moody chart using the Colebrook-White equation Show code/hide code % Moody Chart in MATLAB clear; clc; % Define the Reynolds number range Re = logspace(3, 8, 500); % Reynolds number from 10^3 to 10^8 % Define the relative roughness values (epsilon/D) rel_roughness = [0, 0.00001, 0.0001, 0.001, 0.01, 0.02, 0.05]; % Preallocate friction factor array friction_factors = zeros(length(rel_roughness), length(Re)); % Calculate friction factors using the Colebrook equation for i = 1:length(rel_roughness) for j = 1:length(Re) if Re(j) < 2000 % Laminar flow friction_factors(i, j) = 64/Re(j); % Turbulent flow using Colebrook-White equation f = 0.02; % Initial guess for f while true f_new = 1 / (-2log10((rel_roughness(i)/3.7) + (2.51/(Re(j)sqrt(f))))); if abs(f_new – f) < 1e-6 f = f_new; friction_factors(i, j) = f_new; % Plot the Moody chart loglog(Re, friction_factors, ‘LineWidth’, 2); hold on; % Laminar flow line Re_laminar = Re(Re < 2000); friction_laminar = 64 ./ Re_laminar; loglog(Re_laminar, friction_laminar, ‘k–‘, ‘LineWidth’, 2); % Annotate plot grid on; xlabel(‘Reynolds Number, Re’); ylabel(‘Friction Factor, f’); title(‘Moody Chart’); legend(arrayfun(@(x) sprintf(‘\\epsilon/D = %g’, x), rel_roughness, ‘UniformOutput’, false), ‘Location’, ‘Best’); set(gca, ‘XScale’, ‘log’, ‘YScale’, ‘log’); xlim([1e3 1e8]); ylim([1e-3 1]); % Add text annotations text(2e3, 0.08, ‘Laminar Flow’, ‘FontSize’, 12, ‘Rotation’, -45); text(1e7, 0.008, ‘Turbulent Flow’, ‘FontSize’, 12); % Add grid lines for guidance set(gca, ‘MinorGridLineStyle’, ‘-‘, ‘GridAlpha’, 0.5, ‘MinorGridAlpha’, 0.5); hold off; The following points should be understood when using and interpreting the curves on the Moody chart: 1. The laminar flow line represents a lower theoretical bound of the friction factor for low Reynolds numbers, i.e., There are few practical situations where this is a valid assumption other than for very short, smooth pipes and ducts. Nevertheless, the result shows that the friction factor decreases with increasing Reynolds number, and there is no dependency on surface roughness. 2. The shaded area indicates a critical zone. This is where the flow begins to develop turbulence but is unsteady and intermittent in its flow characteristics. The friction factor can show significant changes here, meaning the flow in this region could be laminar or turbulent. Empirical data here are limited in scope and show considerable scatter. Fortunately, having to use this part of the chart is uncommon. 3. The friction factors increase with surface roughness for any given Reynolds number at higher Reynolds numbers, a rather obvious expectation. Notice that even entirely smooth pipes containing a turbulent flow, such as plastic and glass, which have values of 4. At higher Reynolds numbers, which is called the fully turbulent or hydraulically rough zone, the friction factors become nearly independent of the Reynolds number. This behavior happens because the protruding roughness penetrates well into the boundary layer and deeper than any laminar sublayer, so the internal flow will always be fully turbulent. Notice that this zone (boundary) occurs at an increasingly higher Reynolds number with lower roughness factors. Moody defined the boundary of the hydraulically rough zone as Using Eq. 33, it can be shown that Therefore, the hydraulically rough boundary (black dashed line on the Moody chart) can be approximated by the equation 5. When the relative roughness becomes even higher, i.e., when Reading a Moody Chart Reading a Moody chart is not difficult but requires some practice, recognizing it is a log-log chart and will inevitably involve graphical interpolation. While the friction factor can be calculated numerically using the Colebrook-White equation, it is still essential to understand the visual interpretation of the physical behavior and the process of using the chart, an example being shown below. Reading the chart to three decimal places is more than adequate. Notice that some published Moody charts, such as those found on the internet, may give slightly different friction factor values because they are based on approximations to the Colebrook-White equation. How to read a Moody chart. Remember, it is a log-log scale, so some care is required. In this case, it is assumed that the information needed to calculate the Reynolds number and the relative roughness is available, and the desired output is the friction factor, 1. Determine the Reynolds number based on the average flow velocity, where the average flow velocity is determined from the flow rate and cross-sectional area, i.e., 2. Calculate the value of the relative roughness from the actual dimensional roughness, The roughness value, 3. Find the appropriate (blue) curve for the calculated value of relative roughness on the Moody diagram, recognizing that only a finite number of curves can be shown, and interpolation of the curves on the logarithmic scales will likely be required. 4. Find where the friction factor curve (or interpolated curve) intersects the Reynolds number (at the red dot). 5. Estimate the friction factor by reading the value on the vertical scale on the ordinate to the left of the red dot point from where the Reynolds number and relative roughness curve intersect. Finally, some words of caution are appropriate when using the Moody Chart. First, like all engineering calculations, ensuring that the correct units are consistently used when calculating non-dimensional quantities, such as the Reynolds number and relative roughness, is essential. Be sure that the Reynolds number is calculated based on the pipe’s diameter (or hydraulic diameter), not the pipe’s length. Catastrophic mistakes can occur if the correct units are not used carefully and consistently. Second, because of the numerous lines on the chart, it is essential to interpolate carefully. The human eye/brain combination is a suitable interpolator, but greater accuracy may be required. In this case, the Colebrook-White equation can be solved directly, or an accepted approximation can be used. Approximations for the Friction Factor The von Kármán rough pipe law, also known as the von Kármán-Prandtl roughness law, is an empirical relationship used to estimate the friction factor for a fully-developed turbulent fluid flow through a pipe with rough walls. It can be written as which is, again, an implicit equation that must be solved by iteration. While only approximate, it applies to a wide range of Reynolds numbers and roughness factors. The Swamee-Jain equation, also known as the Swamee-Jain friction factor equation, can also be used to calculate the friction factor, Roughness Values In most cases, roughness values cover a range and will depend on the specific surface, as shown in the table below. As manufactured, the roughness values will be on the lower end of the scale, but all types of pipes may develop some roughness over time from oxidation or corrosion. It is always important to consult relevant standards and specifications or obtain specific roughness values for the particular pipe(s) used in a given application. Roughness factors for different types of internal surfaces. In most cases, roughness values cover a range. Pipe Material Roughness, ε (mm) Smooth Plastic (PVC) 0.001 – 0.01 Glass 0.001 – 0.01 Smooth Metal (e.g., SS) 0.001 – 0.03 Concrete 0.2 – 1.5 Galvanized Iron 0.15 – 0.5 Commercial Steel 0.045 – 0.09 Riveted Steel 0.9 – 1.2 Corrugated Metal 3.0 – 8.0 Cast Iron (new) 0.15 – 0.5 Cast Iron (old) 0.6 – 3.0 Smooth Copper 0.001 – 0.01 Polyethylene (PE) 0.001 – 0.01 Polyvinyl Chloride (PVC) 0.001 – 0.01 Fiberglass 0.01 – 0.03 Stainless Steel (welded) 0.025 – 0.05 Stainless Steel (seamless) 0.02 – 0.045 Aluminum 0.05 – 0.1 Brass 0.02 – 0.05 PVC (Corrugated) 1.5 – 3.0 HDPE (Corrugated) 0.4 – 1.5 Ductile Iron 0.025 – 0.05 Polypropylene 0.01 – 0.03 ABS 0.02 – 0.06 Check Your Understanding #3 – Calculating the pressure drop for turbulent flow in a pipe using the Moody chart Show solution/hide solution 1. The average flow velocity is calculated from the volume flow rate, i.e., 2. The Reynolds number of the flow in the pipe is 3. The flow will be turbulent because the Reynolds number is greater than 2,000, and the Moody chart will be needed to find the friction factor 4. To find the pressure drop and use the Moody chart, the relative surface roughness is needed, which is From the Moody chart for a Reynolds number of 84,900 and a relative roughness of 0.00087 (using interpolation), then The corresponding head loss over this pipe is 5. The pumping power required will be Pressure Drop Over a Tapered Pipe There are no exact solutions for the pressure drop along a tapered pipe; an example of such a shape is shown in the figure below. The main issue is that the flow velocity and Reynolds number vary continuously. However, numerical solutions may be possible where the local losses over short segments are added to determine the local losses in aggregate. In the case of fully turbulent flow with higher Reynolds numbers where the friction factor becomes independent of Reynolds number, the pressure drop can be estimated using the concept of a hydraulic diameter and average flow velocity. Calculating the pressure drop over a tapered pipe is possible under certain assumptions, including fully turbulent flow at higher Reynolds numbers. In this case, the average hydraulic diameter The average flow velocity, The Reynolds number based on the average hydraulic diameter and average flow velocity can then be obtained, i.e., Then the Moody chart can be used with the relative roughness to get the friction factor, Therefore, the pressure loss over the length of tapered pipe will be These pressure drop estimates are sufficiently accurate for many practical purposes, including wind tunnel design. Many internal flow problems involve using or designing various types of pumps to move fluid from one place to another. Such pumps include hydraulic pumps, piston pumps, centrifugal pumps, gear-type pumps, vane pumps, axial flow pumps, fans, etc. For aerospace applications, pumps must be lightweight (i.e., high power-to-weight ratio), reliable, and often contend with harsh operating environments, including large temperature swings and vibration. Pumps obtained “off-the-shelf” are usually specified in terms of their volumetric transfer capacity per unit time. It is essential to notice the units in which the volumetric flow rates are given, which will typically not be in base units; e.g., they may be quoted in terms of liters/hr, gallons/minute, or other units. In general, pumps recover pressure losses so fluid can be pumped from one location to another, as shown in the figure below. Any pump will produce the same effect of pressure recovery, although with different efficiencies and applicabilities to different types of fluids. Pressure drops depend on several factors, including the flow rate, Reynolds number, and friction factor. Pumps are usually located in-line with the overall flow and are used to recover the pressure drop and keep the fluid moving through the pipe from one end to the other. Types of Pumps The figure below shows four common types of fluid pumps: a gear pump, lobe pump, vane pump, and centrifugal pump. In each case, the pumps must be rotated by a power source. This source could be an accessory drive through a belt or gear system from an engine or propulsion system or a direct drive from a separate source such as an electric motor. Such pumps are usually inserted in series into a pipeline and bolted to the pipe using parallel flanges sealed by a gasket. There are many different types of pumps, but in each case, their purpose is to move fluid through a pipe to overcome pressure drops. Gear Pumps Gear pumps are positive displacement pumps that utilize two meshing gears to transfer fluids by creating chambers between the gear teeth and the pump casing. They come in various types, external and internal, offering specific advantages and applications. Gear pumps are valued for their simplicity, reliability, and ability to handle various viscosities, making them suitable for applications in industries such as hydraulic systems, lubrication systems, and fuel transfer. Lubrication systems in engines often use gear pumps because of their simplicity, low cost, and good reliability. Lobe Pumps Lobe pumps are positive displacement pumps characterized by two or more lobes that rotate in synchronized motion within a casing, creating chambers that transport fluid from the inlet to the outlet. Unlike gear pumps, lobe pumps feature smooth meshing lobes instead of gears, offering smoother, more efficient fluid flow with lower losses. They are widely used because they can handle more viscous fluids, such as oils. They are reliable but tend to be more expensive than gear pumps. Vane Pumps Vane pumps are positive displacement pumps that employ rotating vanes inside a cylindrical housing to move fluid from the inlet to the outlet. These pumps typically feature an off-center drive to the vanes. The vanes are often made of carbon, causing them to slide in and out of their slots while maintaining close contact with the housing. Vane pumps have a smooth and consistent flow, making them suitable for applications requiring precise and steady fluid delivery, such as hydraulic and pneumatic systems. They offer relatively quiet operation and can handle various viscosities, though they experience reduced efficiency for more viscous fluids. Furthermore, the vanes wear over time and require periodic maintenance to ensure their functionality. Centrifugal Pumps Centrifugal pumps utilize centrifugal forces on the fluid to increase the fluid’s rotational kinetic energy. Such pumps consist of an impeller enclosed within a casing, propelling the fluid outward from the center of rotation. The fluid is introduced through a pipe at the center of the pump. As the fluid moves through the pump, it gains velocity and pressure, transferring fluid from the inlet to the outlet. Centrifugal pumps are widely used in applications requiring high flow rates. While efficient, they tend to be better suited for gases rather than more viscous liquids. Diaphragm pump The figure below shows a CFD simulation of the internal flow through a diaphragm pump or membrane pump. In this type of pump, which is common in fuel delivery systems, the diaphragm is flexed up and down (by a motor) such that when the volume of the chamber is increased, the pressure decreases, the fluid is drawn from one end of the pipe into the chamber. When the chamber pressure rises from the reduced volume, the fluid previously drawn in is forced through the other pipe. Notice that a pair of butterfly valves prevent reverse fluid flow from the pump. An excellent example of an internal flow is this diaphragm pump, which draws the fluid from a pipe into a chamber and forces it out through the other pipe. Pressure Drops Over Bends & Corners When an internal fluid flow encounters a corner or bend, it experiences a pressure drop from the change in direction, as shown in the figure below. Several factors, including centrifugal forces, changes in velocity, and the possibility of flow separation, cause this pressure drop. As the fluid approaches the corner, it tends to move outward under the action of centrifugal forces, which creates a pressure gradient across the cross-section of the corner or bend. If the corner is sharp, the flow separation and higher pressure losses can be expected. Flow around bends and corners incur additional pressure losses. The magnitude of the pressure drop around a corner depends on various factors, such as the geometry of the bend, flow rate, fluid viscosity, and Reynolds number. Many studies have been to determine the losses in bends, including by NIST. One way of assessing such pressure losses is by using a local loss coefficient published for various corners and bends. The local pressure loss, which is often referred to as a velocity head. For a sharp corner, Many hydraulic, pneumatic, and other internal fluid flow systems comprise straight sections, turns, and branches. The total pressure losses are the sum of all of the pressure losses incurred by each component of the system, i.e., Today, computational fluid dynamics (CFD) can better predict and analyze the specific pressure drop for a given internal flow system. In general, the pressure drop around a corner can be minimized by using smooth bends with large radii, which limit the possibilities of flow separation. Additionally, properly designed flow control devices, such as corner vanes or deflectors, can limit the pressure Design of a Wind Tunnel Consider now an engineering design problem for the internal flow in a wind tunnel. Wind tunnels are vital to the aerospace engineer’s technical repertoire, and tunnel testing is inevitable when designing a new aircraft. Not everything can be predicted before its first flight, and wind tunnels can help establish high confidence in the aircraft’s design, i.e., that the aircraft, when built, will behave as intended. Wind tunnels come in all shapes and sizes and can have different flow speeds. However, their fundamental purpose is to allow measurements to be made under controlled flow conditions. As a result, measurements are often made on true-to-scale aircraft, wings, and other components; geometric scaling may be required to establish dynamic flow similarity. However, obtaining the correct flight Reynolds and Mach numbers in the wind tunnel environment is often challenging. The design of a modern wind tunnel is a complex affair because it is usually customized to meet a set of unique testing requirements, including the test articles themselves and the types of measurements to be made. A schematic of the new ERAU wind tunnel is shown below. This is a state-of-the-art, closed-return wind tunnel with a 4 ft by 6 ft rectangular test section 12 ft long with tapered corner fillets. This tunnel allows flow speeds of up to 420 ft/s to be obtained in the test section with exceptional flow quality and low turbulence levels. In addition, the test section has about 70% of its surface area made with optical grade glass, allowing flow measurements using particle image velocimetry (called “PIV”). The key features of the ERAU wind tunnel. One of the challenges in wind tunnel design is obtaining uniform flow properties in the test section, i.e., uniform velocities in both magnitude and direction with minimal flow angularity (typically less than 0.1 of a degree) its entire length. This process needs keen attention to the internal flow quality around the whole wind tunnel circuit, including the flow through the fan, and special attention must be paid to the contraction before the test section. Using CFD and taking into account the thickness of the boundary layer and turbulence, the walls of the contraction of ERAU’s wind tunnel were contoured to ensure the best flow uniformity at the entrance to the test section and along its entire length. Appropriately shaped and tapered corner fillets from the contraction and along the test section length are also part of the design solution. The ability to design a wind tunnel with uniform flow properties and low flow angularity along its entire test section is essential to the success of the tunnel as an aerodynamic testing resource. Pressure Losses in Wind Tunnels While designing a wind tunnel is a specialist activity, one of the design challenges for a new wind tunnel is determining the motor/fan’s power to create a needed flow velocity (or dynamic pressure) in the test section. To this end, estimating (by calculation) the pressure losses as the flow moves around the tunnel circuit is necessary for design. As previously discussed, wind tunnels are large ducts with different shapes and cross-sections, transition pieces (adapters), etc. As a result, there will be various pressure losses as the flow moves through these different duct elements at different speeds and Reynolds numbers. In addition, corner vanes are used to help turn the flow and are a cascade of thin airfoils of circular arc plates. Significant frictional pressure losses will occur as the flow passes through the four sets of corner vanes. In the conventional approach to wind tunnel design, the frictional losses can be estimated for the fan and initial sizing of the motor by breaking the tunnel circuit into its primary parts: 1. Cylindrical sections (even if just transition pieces). 2. Corners. 3. Expanding sections, i.e., diffusers. 4. Contracting sections, i.e., nozzles. 5. Turbulence screens. 6. Heat exchangers. 7. Other miscellaneous parts. In these sections (and there may be more than one of each), an energy loss occurs in the form of a static pressure loss, Because the pressure or head loss depends on the fourth power of the tunnel diameter or equivalent hydraulic diameter for non-circular cross-sections, according to Poiseuille’s law, then and so the so-called Energy ratio The resulting energy ratio depends on the inverse sum of the equivalent energy losses for each part of the tunnel circuit, i.e., it is, in effect, the reciprocal of the losses. For a closed-return tunnel, the values of Determining the values of For example, consider determining the fan power required to generate a given flow velocity in the test section of a wind tunnel, an effect often called the pumping power. This approach requires determining all the various pressure losses in the tunnel circuit, including frictional losses and pressure drop over the walls, turning vanes, screens, etc. Unfortunately, not all of these effects may be known other than being estimated until the wind tunnel is built and tested. Therefore, the wind tunnel design may require significant power margins to meet the specifications fully. Check Your Understanding #4 – Calculating the power required to drive a wind tunnel Estimate the minimum motor power required for a wind tunnel with a maximum flow speed of 230 mph in the test section. The test section area is 22.5 ft The specification of the energy ratio gives the cumulative losses in the tunnel, so the first step is to find the energy of the flow in the test section, i.e., The value of Therefore, the power required to be delivered to the air at the fan/motor stage would be Taking into account the fan and motor efficiency then, the minimum power required would be This latter result would only be valid for an empty test section, and to get the same flow speed with an article in the test section, more power would be needed to overcome the drag and “blockage'” of the article. This value is generally unknown a priori. However, it is usually considered reasonable to add a margin of power to overcome a winged test article with a wing span of 0.8 of the tunnel diameter (or width), an aspect ratio of 5, and a Summary & Closure Internal flows are encountered in various aerospace applications, such as engine air intakes, fuel systems, hydraulic systems, air conditioning systems, wind tunnels, and other applications where fluids flow through pipes and ducts. A significant consideration for internal flows is that frictional effects from the action of viscosity produce significant pressure drops along the length of the pipe or duct. These pressure drops require a source of power to pump the fluid along the pipe or duct, and this power required is related to the flow velocity and internal surface finish. Because such flows are inevitably turbulent, a Moody chart must be used to determine the friction factors and estimate the resulting pressure drops for design purposes. Internal flows are critical within the combustion chambers of jet and rocket engines, where combustion produces high-temperature, high-pressure gases to produce thrust. Aircraft and spacecraft have environmental control systems to maintain comfortable conditions for passengers and crew, which also require the analysis of internal flows. Understanding and optimizing internal flows in aerospace applications are essential for improving efficiency, safety, and overall performance. 5-Question Self-Assessment Quickquiz For Further Thought or Discussion • For an aircraft’s hydraulic system, discuss some potential design trades in operating the hydraulic system with smaller pipes and higher pressure versus larger pipes and lower pressure. • What is the definition of internal flow in fluid dynamics? How does internal flow differ from external flow, and what are some examples of each? • Consider the fuel and oxidizer system for a rocket engine, which requires high mass flow rates of cryogenic liquids. What are the internal flow issues to consider regarding the delivery of the fuel and oxidizer to the combustion chamber in this case? • The corner or turning vanes in a wind tunnel are a large source of losses. Consider the engineering steps you might take to calculate and minimize such losses. • Explain the significance of the Reynolds number in internal flows. How does it affect the flow characteristics? Compare the flow regimes associated with low and high Reynolds numbers in internal • Explain the major differences between flow in circular pipes and non-circular conduits. How do these differences impact flow characteristics? Other Useful Online Resources To learn more about internal flows, take a look at some of these online resources: 1. See: "The History of Poiseuille's Law," by S. P. Sutera, and R. Skalak, Annual Review of Fluid Mechanics, Volume 25, 1993. According to them, the naming of Equation 29 as Poiseuille's law occurs in a publication by Hagenbach (1860) who, after giving the derivation, generously suggested calling it Poiseuille's law. Jacobson (1860) also calls Equation 29 Poiseuille's law. See: Hagenbach, E. 1860, "Uber die Bestimmung der Zahigkeit einer Fliissigkeit durch den Ausfluss aus R6hren,'' Poggendorf's Annalen der Physik und Chemie, Vol. 108, pp. 38–426, and Jacobson, H. 1860, "Beitriige zur Haemo dynamik," Archives of Anatomy and Physiology, Vol. 80, pp. 80–113. ↵	{"url":"https://eaglepubs.erau.edu/introductiontoaerospaceflightvehicles/chapter/internal-flows/","timestamp":"2024-11-10T17:52:16Z","content_type":"text/html","content_length":"465566","record_id":"<urn:uuid:7a5a1161-c22b-4eba-8966-1609f3b3ce18>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00194.warc.gz"}
What is the final concentration of 50.0 mL of a 2.00M solution, diluted to 500.0 mL? \| Socratic What is the final concentration of 50.0 mL of a 2.00M solution, diluted to 500.0 mL? 1 Answer $\text{Concentration"="Moles of stuff"/"Volume of solution}$$=$$0.20 \cdot m o l \cdot {L}^{-} 1$ $\frac{50.0 \times {10}^{- 3} \cancel{L} \times 2.00 \cdot m o l \cdot \cancel{{L}^{-} 1}}{500.0 \times {10}^{-} 3 \cdot L}$$=$#??mol*L^-1# Note that the answer is reasonable. If you read the question you have diluted the starting solution from $50.0 \cdot m L$ to $500.0 \cdot m L$, a tenfold dilution, so the final solution shoould be $10 \times$ as dilute. Impact of this question 36435 views around the world	{"url":"https://api-project-1022638073839.appspot.com/questions/what-is-the-final-concentration-of-50-0-ml-of-a-2-00m-solution-diluted-to-500-0-","timestamp":"2024-11-06T08:25:41Z","content_type":"text/html","content_length":"33535","record_id":"<urn:uuid:c8d61311-5637-42c3-8ffa-ebce510688c3>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00058.warc.gz"}
What are some of the terms used to define the properties of waves? - The Handy Physics Answer Book What are some of the terms used to define the properties of waves? The table below summarizes the different properties of waves. Type of Wave Term Definition Transverse Crest The highest point of the wave. Trough The lowest point of the wave. Longitudinal Compression An area where the material or medium is condensed and at higher pressure. Rarefaction An area that follows a compression where the material or medium is spread out and at lower pressure. Transverse & Longitudinal Amplitude The distance from the midpoint to the point of maximum displacement (crest or compression). Frequency The number of vibrations that occur in one second; the inverse of the period. Period The time it takes for a wave to complete one full vibration; the inverse of the frequency. Wavelength The distance from one point on the wave to the next identical point; the length of the wave.	{"url":"https://www.papertrell.com/apps/preview/The-Handy-Physics-Answer-Book/Handy%20Answer%20book/What-are-some-of-the-terms-used-to-define-the-properties-of-/001137019/content/SC/52caffcc82fad14abfa5c2e0_default.html","timestamp":"2024-11-12T00:16:30Z","content_type":"text/html","content_length":"12817","record_id":"<urn:uuid:a85122fd-7818-4361-8512-79f0bf5679e7>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00865.warc.gz"}
Convert meters to feet Please provide values below to convert meter [m] to foot [ft], or vice versa. Definition: A meter, or metre (symbol: m), is the base unit of length and distance in the International System of Units (SI). The meter is defined as the distance traveled by light in 1/299 792 458 of a second. This definition was slightly modified in 2019 to reflect changes in the definition of the second. History/origin: Originally, in 1793, the meter was defined as one ten-millionth of the distance from the equator to the North Pole. This changed in 1889, when the International prototype metre was established as the length of a prototype meter bar (made of an alloy of 90% platinum and 10% iridium) measured at the melting point of ice. In 1960, the meter was again redefined, this time in terms of a certain number of wavelengths of a certain emission line of krypton-86. The current definition of the meter is effectively the same as the definition that was adopted in 1983, with slight modifications due to the change in definition of the second. Current use: Being the SI unit of length, the meter is used worldwide in many applications such as measuring distance, height, length, width, etc. The United States is one notable exception in that it largely uses US customary units such as yards, inches, feet, and miles instead of meters in everyday use. Definition: A foot (symbol: ft) is a unit of length in the imperial and US customary systems of measurement. A foot was defined as exactly 0.3048 meters in 1959. One foot contains 12 inches, and one yard is comprised of three feet. History/origin: Prior to standardization of units of measurement, and the definition of the foot currently in use, the measurement of the foot was used in many different systems including the Greek, Roman, English, Chinese, and French systems, varying in length between each. The various lengths were due to parts of the human body historically being used as a basis for units of length (such as the cubit, hand, span, digit, and many others, sometimes referred to as anthropic units). This resulted in the measurement of a foot varying between 250 mm and 335 mm in the past compared to the current definition of 304.8 mm. While the United States is one of the few, if not only, countries in which the foot is still widely used, many countries used their own version of the foot prior to metrication, as evidenced by a fairly large list of obsolete feet measurements. Current use: The foot is primarily used in the United States, Canada, and the United Kingdom for many everyday applications. In the US, feet and inches are commonly used to measure height, shorter distances, field length (sometimes in the form of yards), etc. Feet are also commonly used to measure altitude (aviation) as well as elevation (such as that of a mountain). The international foot corresponds to human feet with shoe size 13 (UK), 14 (US male), 15.5 (US female), or 46 (EU). Meter to Foot Conversion Table Meter [m] Foot [ft] 0.01 m 0.032808399 ft 0.1 m 0.3280839895 ft 1 m 3.280839895 ft 2 m 6.56167979 ft 3 m 9.842519685 ft 5 m 16.4041994751 ft 10 m 32.8083989501 ft 20 m 65.6167979003 ft 50 m 164.0419947507 ft 100 m 328.0839895013 ft 1000 m 3280.8398950131 ft How to Convert Meter to Foot 1 m = 3.280839895 ft 1 ft = 0.3048 m Example: convert 15 m to ft: 15 m = 15 × 3.280839895 ft = 49.2125984252 ft Popular Length Unit Conversions Convert Meter to Other Length Units	{"url":"https://coonbox.com/meters-to-feet.html","timestamp":"2024-11-08T04:27:37Z","content_type":"text/html","content_length":"17917","record_id":"<urn:uuid:97276863-93d7-4bf1-94f1-23f9e61527dd>","cc-path":"CC-MAIN-2024-46/segments/1730477028025.14/warc/CC-MAIN-20241108035242-20241108065242-00576.warc.gz"}
[Solved] You are contemplating membership in the S \| SolutionInn You are contemplating membership in the St. Swithins and Ancient Golf Club. The annual membership fee for You are contemplating membership in the St. Swithin’s and Ancient Golf Club. The annual membership fee for the coming year is $5,000, but you can make a single payment today of $12,750, which will provide you with membership for the next three years. Suppose that the annual fee is payable at the end of each year and is expected to increase by 6% a year. The discount rate is 10%. Which is the better deal? Discount Rate Depending upon the context, the discount rate has two different definitions and usages. First, the discount rate refers to the interest rate charged to the commercial banks and other financial institutions for the loans they take from the Federal... Fantastic news! We've Found the answer you've been seeking!	{"url":"https://www.solutioninn.com/study-help/principles-corporate-finance/you-are-contemplating-membership-in-the-st-swithins-and-ancient-800049","timestamp":"2024-11-10T20:53:43Z","content_type":"text/html","content_length":"81250","record_id":"<urn:uuid:3af42504-8188-4d83-8771-00897780bcbd>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00552.warc.gz"}
[tex4ht] problem using \sl and \sc in tex4ht when using matghjax mode [tex4ht] problem using \sl and \sc in tex4ht when using matghjax mode Nasser M. Abbasi nma at 12000.org Sun Nov 18 20:27:02 CET 2018 There is problem translating Latex code to HTML when using mathjax mode when latex uses the old \sl and \sc commands. This is code generated by Maple Latex so it is not possible to change it and not practical to edit it by hand each time since this is autogenerated each time the files are compiled. Maplesoft seems to have abandoned working maintaining its Latex export for some other exotic math rendering software so no chance this will fixed by them. I do not know what I need to change in mathjax-latex-4ht.sty to fix this. Here is a MWE \begin{align} %code below is copied from part of Maple's Latex u &= s{{\sl I}_{0}} \\ &= s{{\sc I}_{0}} \\ &= s{{\rm I}_{0}} When compiled using mathjax mode, it gives I also tried using this class istead \documentclass[11pt,enabledeprecatedfontcommands]{scrartcl}%this also fail I also tried adding But had no effect on resulting HTML. This seems to affect only \sl and \sc commands, but there might be more. Is it possible to change mathjax-latex-4ht.sty to work around this? The command I used to compile is the same as before and described make4ht -ulm default -c ./nma_mathjax.cfg report.tex "htm,0,notoc,p-width,charset=utf-8" " -cunihtf -utf8" Where nma_mathjax.cfg is And mathjax-latex-4ht.sty is from This problem only shows with mathjax. No problem when using SVG. Thank you for any help. More information about the tex4ht mailing list	{"url":"https://tug.org/pipermail/tex4ht/2018q4/002128.html","timestamp":"2024-11-02T01:11:24Z","content_type":"text/html","content_length":"5023","record_id":"<urn:uuid:ea61ce16-273c-4a87-adc2-5b5353ff2328>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00188.warc.gz"}
How to Read a Correlation Matrix - A Beginner's Guide How to Read a Correlation Matrix? What Is a Correlation Matrix? A correlation matrix is a table that displays the correlation between multiple variables. It’s like a cheat sheet that shows us how closely related different variables are. In a correlation matrix, each variable is represented by a row and a column, and the cells show the correlation between them. The number in each cell represents the strength and direction of the correlation, with positive numbers indicating positive correlations (when one variable increases, the other variable tends to increase) and negative numbers indicating negative correlations (when one variable increases, the other variable tends to decrease). For example, if we’re studying the relationship between the type of music we listen to and our mood, a correlation matrix might show how closely related different types of music are to different moods, such as happiness or sadness. How to Read a Correlation Matrix? In this section, we will learn how to read a correlation matrix, a vital skill for understanding the relationships and strengths of association between multiple variables: • Look at the number in each cell to see the strength and direction of the correlation. • Positive numbers indicate positive correlations, while negative numbers indicate negative correlations. • The closer the number is to 1 (or -1), the stronger the correlation. • A number of 0 means there is no correlation between the two variables. Understanding correlation matrices can help us identify patterns and relationships between multiple variables. So next time you’re analyzing data with many variables, think like a mathematician and use a correlation matrix! Related Tags:	{"url":"https://www.quanthub.com/how-to-read-a-correlation-matrix/","timestamp":"2024-11-10T22:34:56Z","content_type":"text/html","content_length":"89911","record_id":"<urn:uuid:6aa50e26-1aae-4a9a-9a1f-3142a4dd439f>","cc-path":"CC-MAIN-2024-46/segments/1730477028191.83/warc/CC-MAIN-20241110201420-20241110231420-00110.warc.gz"}
Card Counting System If you are already accustomed to the Hi-Lo and alternative primary count systems, it should be time to learn a brand new system. This page discusses the knock out card counting system. Let us start with balanced and unbalanced card counts. The card counting strategies are either balanced or unbalanced. Balanced suggests that counting through a whole deck of cards should bring you to a count of zero like at the beginning of the game. An unbalanced count brings you to a number different than zero. Beginner counting systems like the Hi-Lo system are balanced. The most famous unbalanced system is the KO blackjack system also known as KO. Unbalanced systems, in general, must not be tried by beginners. This is often because it will be harder for you to detect your errors when you count through a deck of cards. However, the knock out system is a simple count to use despite the fact that it is unbalanced. The disadvantage is that it is slightly less accurate than the Hi-Lo count. The knock out system is not a counting technique for boxing knockdowns. It is a blackjack counting system. The KO blackjack system is just like the initial Hi-Lo count except that the seven card is given a value of +1 similar to the low cards. Since there are four cards of this denomination during a single deck, the final count once one deck is +4. The KO blackjack system sacrifices accuracy for convenience. It is smart for the casual player who prefers to use a friendly system. However, it is not a correct count out there. It is, however, still helpful. If you are solely playing blackjack for fun, this can be a decent system to try. Knock out blackjack is one of my favorite card counting systems. The reason why I prefer this system so much is that it is easy to use and it eliminates the necessity for changing the running count into a true count. The knock out blackjack works in a very similar way to most of the other card counting methods. Card counting does not require you to remember the order or any specific cards that came out of the deck. Instead, it uses a particular system to calculate the proportion of high cards to low cards left in the shoe. When a pack contains a proportionately greater number of 10s and aces in it, it is a lot more profitable for the player. The reason should be apparent, however, if it is not, think about it. One hand in blackjack leads to a higher payout than any other hand. That is a “natural” 21 that is also referred to as a “blackjack.” Only two values of cards can mix to create such a hand. One amongst those is the ace, and the alternative is cards with a value of ten. If you removed each of the aces from the deck, your probability of getting dealt a natural would become zero, right? Thus obviously, if the deck has fewer low-value cards and more high-value cards, you are a lot more likely to get dealt a 21. Moreover, when you get dealt a natural, you get paid off at three to two. Therefore if you decide to increase the size of your bets once you are likely to get a three to two payout, it stands to reason that you would have an improved probability of winning more money, right? We will explore the ko counting system futher in this guide as well as explain how to use it properly and what are the advantages. We hope that you find what you were initially looking for and that we would help you become a card counting expert. The Specifics of Knock out Blackjack The knockout blackjack count technique also referred to as the KO method, of card counting is so popular because of a book called Knock-out Blackjack: The Easiest Card-Counting System Ever Devised, written by Ken Fuchs and Olaf Vancura. Just like the Hi-Lo counting System, the KO blackjack count strategy assigns point values to every card in the deck and as the cards are being dealt out a player keeps track of the count by adding or subtracting the suitable value. If you are accustomed to alternative ko card counting system in which all cards in a deck might be added together to equal zero, then you will need to get used to the knock out the system, because it is unbalanced because there are more +1 cards in a deck than -1 cards. This method is additionally based on a +2 count which suggests that once your count gets to +2 you must begin considering raising your bet and as the count gets higher and over +2 you must raise your bet even more. The KO blackjack strategy is a good one for beginner card counters because it is easy to learn without an enormous amount of practicing and might give you the edge over the house once you start playing casino blackjack. In the knockout blackjack system, each card with a worth between 2 and seven is counted +1. The Aces and 10’s are counted -1. There are twenty-four of the previous and twenty of the latter, therefore if you counted through a complete deck using this method, you would have a total of +4 once you finish. This is why this system is known as unbalanced. You should increase your bet when the count is positive and decrease it when it is negative. In other words, the higher the count is, the bigger your bet should be. It is almost that easy. There’s one extra wrinkle, however. You most certainly have to take into consideration the number of decks that are in use. In most ko card counting methods, a player must convert the running count into a true one to adjust for the additional decks which are currently in play. For instance, if you are playing a single deck game, and every one of the Aces is dealt, then you currently have a 0.33 probability of being dealt a blackjack. However in an eight deck game, once four aces are dealt, you will still have twenty-eight Aces left within the deck. The result of every individual card is diluted by the massive number of decks within the shoe. Converting the running count to a true one, however, requires skills in the division and estimating. The formula is easy enough—you just divide the count by the number of decks you expect are left within the shoe. In the knockout blackjack system, you will be able to skip the division. This is one of the reasons that the system is not balanced. The other quirk regarding the knockout blackjack system is that you do not always begin your count at zero, as you would in different methods. Your beginning count in the knockout blackjack system is decided by the amount of decks left in the shoe. If you are playing in a game with a single deck, your initial count is zero, however, if you are playing against two decks, you begin your count at -4. With six decks, you start your count at -20, and with eight decks, you begin your count at -28. Raising and Lowing Bets in Knock out Blackjack You will have to decide on a bankroll and a betting spread before you start playing. A typical betting range is one to five units. Therefore you may begin with a bankroll of $10 000 and have a betting spread of between $100 and $500. On hands where the count is negative, you should stick with your $100 bets. As long as you are using the basic strategy, you will solely be playing at a disadvantage of about 1% during those hands. On hands in which the count is positive, you will raise your bet according to how high is the current count. For example, you might bet $200 once the count is 2, $300 once the count is 4, and so on. On these hands, the maximum disadvantage could be between 1% and 4%. Therefore, you will make up for the negative expectation. However, it does increase your potential winnings. Strategy Decisions in Knock out Blackjack You should regulate your strategy for playing every hand according to the count, and by doing this, you might increase your edge slightly. However, it is not necessary to form method adjustments to be profitable as a card counter. The edge you get comes from knowing when is the appropriate time to lower or increase your bet which is between 70% – 90% of counting cards. If you’re really dedicated to taking every tenth of a percentage from the edge, then you will be able to take your time to learn the basic adjustments in strategy based on the count, however if you are that kind of person, you will most likely also wish to learn a more complicated counting system than the knock out blackjack. The Knock out blackjack system was created to be both easy to use and efficient at the same time. However, if you think about it most systems that are easy, sacrifices a certain amount of accuracy. So the Knockout blackjack system is perfect for novices and beginners. However, experts who wish to take it to the next level can probably want to begin experimenting with some of the additionally involved systems such as the Omega II or the Hi-Opt I count. How to Use the Knock out Blackjack System Some Blackjack ko card counting systems need users to maintain three completely different counts as well as a sometimes sophisticated conversion to find out if the remaining cards hold any advantage for the player. Some of them force the user to quickly add and subtract by 4’s, 9’s and even fractions. The knock out Blackjack System only needs one count; the running count. The running count in the KO blackjack technique utilizes -1’s and +1’s. For example, the 8 and nine cards are zero points, everything between 2 and 7 is +1 and the ten value cards, and Ace is -1. Keeping a running count using the knock out Blackjack System is the sole thing you have to try doing. The running count can tell you if the unused cards left within the shoe will bring an advantage to the player. The basic idea behind ko card counting is that low cards are bad for the players and high cards are good for Once your knock out Blackjack Count is bigger than zero, it means that there are more 10’s and Aces left in the deck than the five to thirteen ratio found in a new deck. This is not just a theory since it has been mathematically proven many times over the past fifty years. The first purpose of the knock out System is to find out when you have to increase or decrease your bet. The count might go both ways thus if it is far enough within the negatives you must probably get out of the game and come back later. There is one side of card counting strategies such as the Knock out System that are open to discussion. These methods show you when it is appropriate to increase your bets. However, they do not indicate how much you should increase the bet with. There are no concrete formulas that everybody agrees with therefore I usually recommend that players develop their systems for deciding. Many people debate that the number of decks that are currently being used should have an impact on how much that increase should be. Most agree that once the running count is +1 you must double the size of your initial bet. If you are at +2 two or three times, your initial bet is sensible. I always recommend using caution when you decide to increase your bet. The knock out System is very easy and straightforward. However, mistakes are often made particularly once you first begin using it at a casino. You should always keep in mind the size of your bankroll once making larger bets is usually a decent plan. Keeping a running count needs much practice, but it is quite simple. Usually, most players make mistakes when they calculate the true count, which has a significant impact on their success in blackjack. Even when they do it successfully, many players find the counting and the true count in systems such as the Hi-Lo quite boring and some of them even believe that it takes all the fun and excitement out of the game. This is why most players prefer the method this article is dedicated to. As we discussed earlier, most players enjoy this system so much because it is quite easy and fun rather than the more complicated techniques. The knock out strategy features only one distinction from the very popular Hi-Lo method. In this strategy, seven is no longer a neutral card and is now a low one. This means that every single time you see a seven card you should add one to the count. That is it; the other cards keep the same status. Moreover, if you count the entire deck, in the end, you will have a +4 instead of a zero, which makes the knock out system unbalanced. As we stated earlier, there is no need to calculate the true count if you use the knock out technique. It is an unbalanced strategy which means that the count will most likely be above zero for most of the time. Which means that a positive number will no longer indicate that the edge is to the player’s advantage. Rather than that, a particular value is being chosen, on which the player starts to increase his bet. Usually, the value is two times the number of decks that are currently being used. For example, if the dealer uses six decks, the starting value would be twelve. A final note on the knock out strategy, when dealer’s card which is facing upwards is an Ace, each of the players will get the opportunity to place a side bet, which is most commonly known as insurance. It will pay out if the dealer gets a blackjack. Most of you have probably read or heard somewhere that you should never get this bet as almost all basic strategies state so. However, the knock out method indicates that you always have to take that so called insurance if the count is three or above times the number of the decks in play. For example, if there are six decks in the game, the running count should be eighteen or over. We hope that we have taught you everything you need to know about the ko counting system and that you will use what you have learned here to increase your winnings in blackjack.	{"url":"http://blackjackcardcounting.net/knockout/","timestamp":"2024-11-08T13:32:11Z","content_type":"text/html","content_length":"35767","record_id":"<urn:uuid:4d880b33-e6ba-46af-aa65-f04b0c4f1cf0>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00065.warc.gz"}
How to Prepare for the AP Calculus Exam – Part 2 - University Tutoring - Seattle Are you ready to learn more about the AP Calculus test? I didn’t scare you away in Part 1 of this series? Great! Let’s talk about the actual structure of the exam and what resources we need to start studying. To be clear, all of the information here applies to both the AB and BC versions of the test. Let’s start by breaking it down in the two main parts of the test: Multiple Choice questions (MCQs) and Free Response questions (FRQs). The MCQs are broken into two parts and they comprise 50% of your total score. It starts with Part A: a 60 minute, 30 question, no calculator section. No Calculator?! Don’t worry. This section will test basic understanding of concepts and have problems for which a calculator would be unhelpful. It continues with Part B: a 45 minute, 15 question, calculator section. This section is similar to the previous one, but has more numerical analysis questions where calculators are either helpful or necessary. The TI-84 is most students’ calculator of choice, but there are many approved calculators. Don’t forget, there are NO penalties for guessing on this test, so don’t leave anything blank. If you are completely stumped by a question, utilize process of elimination and other strategies to make your best educated guess. The FRQs are broken into two parts as well and comprise the other 50% of your AP score. Part A is 30 minutes and has 2 multi-part questions, calculator allowed. As in the MCQs, the calculator is useful for many parts and necessary for others. Think having to compute integrals you don’t know or ones that use wonky log or trig values. Part B is 60 minutes and has 4 multi-part questions, no calculator allowed. Like the MCQs you won’t need the calculator to do well here. You just need a basic understanding of the core concepts of Study Materials OK, so now you know the test structure. What materials should I gather to study for this beast? Well, it depends on the section. For the MCQs, hound your AP teacher for released practice tests. If they are unhelpful, some savvy internet searching can reveal old test questions. As for third party content, I’m partial to either the Princeton Review Premium Edition (AB Version) or Barron’s AP Calculus book (covers both AB and BC). They both have plenty of practice problems (spoiler: a key to doing well) and good explanations of how to attack them effectively. For the FRQs, there’s only one place to go: The College Board website itself. They have all of the FRQs, for both the AB and BC tests, from the previous 25 years. Not only do they provide the questions themselves, but also grading guidelines and sample answers. This will be more than enough material to study from. Start gathering your materials and I will be back next time with some study tips and test taking strategies! Next time I will focus on the MCQs, so start gathering your materials and get ready for some study tips and test taking strategies.	{"url":"https://universitytutoring.com/how-to-prepare-for-the-ap-calculus-exam-part-2/","timestamp":"2024-11-12T19:20:51Z","content_type":"text/html","content_length":"78783","record_id":"<urn:uuid:a01f8e69-bd09-4b9e-9dec-9579f5752475>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00619.warc.gz"}
Internal Rate of Return & Marginal Internal Rate of Return - IP Active Comparing IRR and MIRR Every business will make a long term investment on various projects with the aim of generating benefits in future years. A business may be forced to decide which investments will reap the greater benefits and returns for both their own business and that of its investing partners. In this situation capital budgeting is used which is a process of estimating and selecting long-term investment projects which are in alignment with the basic objective of investors. When completing project investment analysis, Internal Rate of Return (IRR) and Marginal Rate of Return (MIRR) are two capital budgeting techniques which measure an investment attractiveness. Internal Rate of Return (IRR) is a financial analysis of cash flow and is commonly used as a method for evaluating an investment, capital investment or project proposals. The IRR is defined as a measurement which compares returns to costs by finding the interest rate that produces a zero Net Present Value (NPV) for the investment cash flow stream. Similar to other cash flow measurements, IRR measurement adopts an investment view of expected financial results which allows it to compare the magnitude and timing of cash flow returns to cash flow costs. The IRR work on the assumption that project cash flows will be reinvested at the project’s own IRR. For an IRR to be accepted the result must be greater then the company’s cost of capital. Modified Internal Rate of Return is also use as a financial measure for determining an investments attractiveness. As the name suggests, the MIRR is a modified version of the IRR which aims to resolve problems encountered when using the IRR. Unlike the IRR the MIRR allows you to set a different reinvestment rate for cash flows received however this increases the complexity as this requires additional assumptions about what rate the funds will be reinvested at. Most people can easily compare MIRR results with compound interest growth and understand the magnitude of the MIRR differences whereas understanding the meaning of the IRR difference is more problematic. Key differences: Some of the key differences between the two is that the IRR is an interest rate at which NPV is equal to zero. Conversely, MIRR is the rate of return at which NPV of terminal inflows is equal to the outflow. Additionally IRR is based on the principle that interim cash flows are reinvested at the project’s IRR unlike the MIRR where cash flows apart from initial cash flows are reinvested at firm’s rate of return. Finally the accuracy of the MIRR is greater than IRR measures as the MIRR measures the true rate of return. IRR vs MIRR example MIRR = ( FVCF / PVCF ) ^ ( 1 / t ) -1 where : FVCF = future value cash flow PVCF = present value cash flow Ct = net cash inflow during the period t Co = total initial investment costs r = discount rate t = number of time periods Assume that a two-year project with an initial investment of $195 and a cost of capital of 12%, will return $121 in year one and $131 in year 2 To calculate the IRR make NPV = 0 NPV = 0 = -195 + 121/(1+ IRR) + 131/(1 + IRR), when IRR = 18.66%. To calculate the MIRR of the project, assume that the positive cash flows will be reinvested at the 12% cost of capital. Therefore, the future value of the positive cash flows is: $121(1.12) + $131 = $266.52 = Future Value of positive cash flows at t = 2 Next, divide the future value of the cash flows by the value proposition of the initial outlay, which was $195, and find the return for 2 periods. Finally, adjust this ratio for the time period using the formula for MIRRgiven: MIRR = ($266.52 / $195) ^ (1 / 2) – 1 = 1.1691 – 1 = 16.91%	{"url":"https://ipactive.com.au/internal-rate-of-return-marginal-internal-rate-of-return/","timestamp":"2024-11-12T19:09:41Z","content_type":"text/html","content_length":"150922","record_id":"<urn:uuid:0e8bda9d-81f0-4bc1-84d1-0df61c639dad>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00685.warc.gz"}
Acceleration Calculator The acceleration calculator computes the change in velocity of an object or body, or initial and final speed, or time to reach a given speed effortlessly. This acceleration finder uses different approaches like speed difference, constant acceleration, distance traveled over time, and constant acceleration methods to facilitate the acceleration-related calculations. Acceleration Formula: It means to measure the rate of change in the speed of an object. As Newton's second law states “ the acceleration is directly proportional to the sum of the forces acting on an object and is inverse of the mass of the object”. Let's take a look at the acceleration equations: $\ a = \dfrac{v_{f}-v_{i}}{Δt}$ $\ a =\ 2\times \dfrac{Δd − v_{i}\times Δt}{Δt^{2}}$ $\ a =\dfrac{F}{m}$ • a represents the acceleration • $\ v_{i}$ and$\ v_{f}$ are the initial and final velocities of the object • Δt is the time • Δd shows the distance traveled by the object • F is the net force that accelerates an object • m indicates the mass of the object Equations For Initial Velocity, Final Velocity, And Time: Use the below-mentioned formulas to find the initial velocity, final velocity, And time: • $\ Initial\ Velocity\ (v_{0}) =\ v_{1}-\dfrac{a}{t}$ • $\ Final\ Velocity\ (v_{1}) =\ v_{0}+\dfrac{a}{t}$ • $\ Time\ (t) =\dfrac{v_{1} - v_{0}}{a}$ Most often, initial velocity is used as the initial speed of the object or body. How To Calculate Acceleration? There are three methods to calculate the acceleration that are: Using Velocities And Time Intervals: • Determine the change in velocities of the object during a specific time interval from t1 to t2. • The following equation calculates the acceleration in this case: • $\ a = \dfrac{v_{f}-v_{i}}{Δt}$ Force And Mass: According to Newton’s law of motion, when a body accelerates a force is acting on it. • Measure the mass of the body or object • Determine the force acting on it The acceleration produced can be calculated by putting the values of mass and force in the following equation: • $\ a =\dfrac{F}{m}$ Using Velocity: • Determine the differentiating velocity vector with respect to time • Calculate the acceleration with the help of the relationship between velocity, time, and displacement as below: • $\ a =\dfrac{dv}{dt}$ Practical Example: A train is running with a uniform velocity that is v = 5 m.s-1 and covers a distance. After 20 seconds, it stops accelerating and sustains a uniform velocity that is v = 25 m.s-1. Find acceleration. $V_{f} = 25 m.s$ $T = 20 s$ Put these values in one of the acceleration equations which requires the provided values: $\ a = \dfrac{v_{f}-v_{i}}{Δt}$ $a=1 m.s^{-2}$ Keep in mind the changing of force brings changes in acceleration but the magnitude of the acceleration depends upon the mass of the object. The magnitude means how fast the object is accelerating. Terms Related To Acceleration: Here we have provided an informational table that contains acceleration-related terms for a better understanding: Terms Explanation Positive: When the final velocity of the body or object is higher than the initial velocity. Negative: When the final velocity is lower than the initial velocity, the acceleration is negative. Centripetal Acceleration: If an object is moving in a circle then the acceleration experienced by the object is known as centripetal acceleration. Linear: When a body or object is moving in a straight line, covering a distance and the motion is in one direction. Instantaneous Acceleration: Measuring the acceleration of a body at a specific instant of time. Acceleration Due To A body that is falling freely experiences acceleration due to gravity because of the gravitational force of the Earth. The value of gravitational force is: $\ 9.8\ ms^ Gravity: {-2}$. Angular Acceleration: It is the rate of change of the angular velocity of an object or body. Meanwhile, it informs about how fast an object spins. What Is The Difference Between Velocity And Acceleration? Velocity means the change of displacement while acceleration is the change of velocity. How Do You Find Angular Acceleration? To calculate the angular acceleration use the following formula: $\ α =\dfrac{ω}{t}$ • ω indicates the angular velocity • t is the time How Do You Calculate Average Acceleration? Follow these steps: • Determine the change in velocity of an object when it’s in a state of motion and covering a distance • Find the change in time • Divide the change in velocity by the change in time Can Acceleration Be Negative? Yes, it can be negative and is termed as deceleration. For instance, when the break of a car is applied, then it stops because of negative acceleration. Is Acceleration A Vector? Yes, because it has both direction and magnitude. From the source of Wikipedia, the free encyclopedia – Simple definition of acceleration (physics) along with the properties – units. From the source of WikiHow - Co-authored by Sean Alexander, MS - How to Calculate Acceleration (Methods).	{"url":"https://calculator-online.net/acceleration-calculator/","timestamp":"2024-11-04T04:13:09Z","content_type":"text/html","content_length":"98915","record_id":"<urn:uuid:2074bbe3-80d8-46c4-a97c-74ba198931f3>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00594.warc.gz"}
Convert Rad/second to Gigagray/second Please provide values below to convert rad/second [rd/s, rad/s] to gigagray/second [GGy/s], or vice versa. Rad/second to Gigagray/second Conversion Table Rad/second [rd/s, Rad/s] Gigagray/second [GGy/s] 0.01 rd/s, rad/s 1.0E-13 GGy/s 0.1 rd/s, rad/s 1.0E-12 GGy/s 1 rd/s, rad/s 1.0E-11 GGy/s 2 rd/s, rad/s 2.0E-11 GGy/s 3 rd/s, rad/s 3.0E-11 GGy/s 5 rd/s, rad/s 5.0E-11 GGy/s 10 rd/s, rad/s 1.0E-10 GGy/s 20 rd/s, rad/s 2.0E-10 GGy/s 50 rd/s, rad/s 5.0E-10 GGy/s 100 rd/s, rad/s 1.0E-9 GGy/s 1000 rd/s, rad/s 1.0E-8 GGy/s How to Convert Rad/second to Gigagray/second 1 rd/s, rad/s = 1.0E-11 GGy/s 1 GGy/s = 100000000000 rd/s, rad/s Example: convert 15 rd/s, rad/s to GGy/s: 15 rd/s, rad/s = 15 × 1.0E-11 GGy/s = 1.5E-10 GGy/s Convert Rad/second to Other Radiation Units	{"url":"https://www.unitconverters.net/radiation/rad-second-to-gigagray-second.htm","timestamp":"2024-11-11T05:09:38Z","content_type":"text/html","content_length":"8519","record_id":"<urn:uuid:ae61f778-bed1-4070-b194-48f24c4ccd3f>","cc-path":"CC-MAIN-2024-46/segments/1730477028216.19/warc/CC-MAIN-20241111024756-20241111054756-00632.warc.gz"}
Laws of Form Laws of Form ISBN 978-3-89094-580-4 (ISBN10 3-89094-580-5), 240 Seiten, Softcover, Format DIN-A5, This Book is in English. Second printing (2010) with additions and revisions. 2. Auflage, 25,00 € kaufen Laws of Form by G Spencer-Brown 5th English edition: At last this all-time classic has been reset, allowing more detailed explanations and fresh insight. There are seven appendices, doubling the size of the original book. Most exiting of all is the first ever proof of the famous Riemann hypothesis. To have, in print, under your hands and before your own eyes, what defied the best minds for a century and a half, is an experience not to be denied. Preface to the fifth English edition As is now well known, Laws of Form took ten years from its inception to its publication, four years to write it and six years of political intrigue to get it published. Typically of all unheralded best sellers from relatively obscure authors, it was turned down by six publishers, including Mark Longman who published my earlier work on probability. Even Sir Stanley Unwin refused to publish it until his best author, Bertrand Russell, told him he must. This crucial recommendation was not achieved without intrigue, and required me (not unwillingly) to sleep with one of Russell’s granddaughters, who asked me in the morning, ‘What exactly do you want from Bertie?’ ‘To endorse what he said about the book when he first read it in typescript,’ I told her. ‘He never will!’ she exclaimed. ‘You’ll have to twist his arm, you’ll have to blackmail him. How can I help?’ The next few years were spent in vigorous arm-twisting and incessant blackmail from us both. One of her threats was to invite me to Plas Penrhyn as her guest while Bertie and Edith were away in London. This sent Bertie into a paroxysm of terror of what the neighbours might think. He also had an irrational fear of spoiling his reputation as a mathematician, which was not good anyway, by recommending a book that had not yet been tried by the critics. He seemed totally unaware that any book he recommended, however ridiculous, would have no effect whatever on this. When we finally got him cornered, in my next visit to Plas Penrhyn, he carefully avoided mentioning the subject during the whole of my stay, and I considered it too dangerous to mention it myself. The next morning I was due to depart while Bertie and Edith were still in bed, and I thought I had failed miserably. But no! I missed my train because they had not ordered me a taxi to the station, which was their way of telling me that my visit was to be prolonged by another day. The evening of this extra day came, and still nothing was mentioned. Ten o’ clock bedtime arrived, and I thought I had failed again, when Bertie suddenly said, ‘What exactly do you want of me?’ ‘To endorse what you said about the book three years ago,’ I told him. ‘You must remind me what it was,’ he said. I produced a verbatim report of his remarks, neatly typed out, and thrust it in his face. ‘Are you sure this is all you want?’ he said. ‘Don’t you want me to write a detailed introduction to the work, as I did for Wittgenstein?’ I told him that that would be very nice, but that this was all I needed just now. He contemplated the page of typescript for a moment, and then a wicked gleam lit up his face, and he rubbed his hands. ‘Supposing I don’t?’ he grinned. ‘Then,’ I heard myself saying, ‘it might delay the publication for a year or so, but the book will still be published in the end, and you won’t be associated with it.’ ‘Oh,’ he said. ‘I never thought of that. How would you like me to sign it?’ There is no stronger mathematical law than the law of complementarity. A thing is defined by its complement, i.e. by what it is not. And its complement is defined by its uncomplement, i.e. by the thing itself, but this time thought of differently, as having got outside of itself to view itself as an object, i.e. ‘objectively’, and then gone back into itself to see itself as the subject of its object, i.e. ‘subjectively’ again. Thus we are what we see, although what we see looks like (and is) what we are not. This incessant crossing of the thing boundary, to look at it from one side and then the other, is called scrutiny, which as a small child was I told is not polite, because by scrutinizing a person or thing we shall notice uncomplimentary (same sound, different word) qualities of the person or thing that it is rude to mention or think about. At the age of three I discovered that most people, from what they told me, could stop themselves from thinking these rude thoughts, which is I suspect why ordinary people do not usually do mathematics, where you have to repeatedly cross and recross the thing boundary. In fact Laws of Form is the book I wrote simply about doing just this and nothing else. When the book finally came out, in 1969 April 17, its effect was sensational. The Whole Earth Catalog ordered 500 copies, which was half the edition, and other big dealers followed suit. The first printing was sold out before it reached the shops, and the publisher had to order a hurried reprint to meet the demand. Nobody had seen anything like it. Here was an upstart author explaining the mysteries of mathematics that the so-called greats of the science in the last 8000 years (at least) had never noticed, and in language that a child of six could follow. Having achieved my life’s ambition of composing and publishing a nearly perfect work of literature by the age of 46, I was suddenly confronted by the problem of what to do with the rest of my life. I knew, and so did everybody else, that I could never top this achievement, so with what significant purpose could I carry on? One thing I could and did do was learn some mathematics. One of many reasons why the book is so famous is because I did not know any math, apart from school stuff, when I began to write it. I had to teach myself, and with me, my readers, as I went along. In ten years I had learned enough to become a full professor in the University of Maryland, although I still thought I knew very little. Math is almost impossible to master without personal tuition, and I was lucky to strike up friendships with D H Lehmer and J C P Miller, both, as it happened, experts on Riemann’s hypothesis, in which I had no interest whatever, nor in analytic number theory in general. It was only on being told by my former student James Flagg, who is the best-informed scholar of mathematics in the world, that I had in effect proved Riemann’s hypothesis in Appendix 7, and again in Appendix 8, that persuaded me to think I had better learn something about it. I am an intensely competitive person, which comes from being repeatedly told by my mother that I would never be any good. This forced me to spend my whole life attempting to prove her wrong. The tragedy of it is that however brilliantly I performed, it made no difference. Nothing I could do would change her mind. I beat her at chess when I was four, and all she did was refuse to play with me ever again, rather than admit that I was good. If you solve a famous unsolved problem by mistake it doesn’t count. You have to say ‘I am going to solve this problem,’ and then solve it. So I had to spend another ten years learning analytic number theory, which I hated, in order to secure and objectify what I had done, and make it presentable. The result is so fascinating that it made the effort seem almost worth while, and the problem was so difficult that solving it gave me nearly as much pleasure as writing Laws of Form. The world of analysis is completely different from anywhere I had explored, the science of continuous variation rather than discontinuous jumping. And since Riemann’s problem is solved by a marriage of the two, although the achievement of a solution cannot quite top what I did in Laws of Form, it runs it a close second, if not an equal first. (0100 hrs 23 06 2007 Saturday) Product history: September 2008: First Printing. March 2010: Second printing with additions and revisions. • Preface to the fifth English Edition • Preface to the first American edition • Preface • Introduction • A note on the mathematical approach □ 1 The form □ 2 Forms taken out of the form □ 3 The conception of calculation □ 4 The primary arithmetic □ 5 A calculus taken out of the calculus □ 6 The primary algebra □ 7 Theorems of the second order □ 8 Re-uniting the two orders □ 9 Completeness □ 10 Independence □ 11 Equations of the second degree □ 12 Re-entry into the form • Notes • Appendix 1. Proofs of Sheffer's postulates • Appendix 2. The calculus interpreted for logic • Index of references • Index of forms • Appendix 3. Bertrand Russell and the Laws of Form • Introduction to Appendices 4 & 5 • Appendix 4. An algebra for the natural numbers • Appendix 5. Two proofs of the four-colour map theorem • My simplest proof of the four-colour map theorem • Appendix 6. Last word • Appendix 7. The prime limit theorem • Appendix 8. Primes between squares • Appendix 9. A proof of Riemann's hypothesis via Denjoy's equivalent theorem • Closing remarks ISBN 978-3-89094-580-4, 240 Seiten, Softcover, Format DIN-A5, This Book is in English. Second printing (2010) with additions and revisions. Preis: 25,00 € Alle Preise incl. MwSt. - und eventuell zzgl. Liefer- und Versandkosten Alle Werke von George Spencer-Brown druckerfreundliche Version oder zurück zur Kategorie Internationale Bücher (in Englisch oder English/Deutsch)	{"url":"http://bohmeierverlag.de/cms/index.php?isbn=3890945805&cat=43","timestamp":"2024-11-07T19:31:03Z","content_type":"text/html","content_length":"25147","record_id":"<urn:uuid:0a86bfa0-e3d8-471d-9209-a7e41b236d07>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00330.warc.gz"}
In Need of a Few Bad Papers For my graduate class this semester, there's a lot of paper-reading, and I view learning how to critically and constructively read papers as part of the student goals for the class. A corollary of this, it seems to me, is that the class should include some bad papers, so students learn to recognize (and, if possible, get something out of) reading those. So I need some really good examples of bad papers. (In one of the areas of the class focus -- web search, compression. coding, streaming data structures...) Now I should be clear about what I mean by bad papers. I'm looking for something of a higher standard than an automatic journal reject -- I get at least one of those a month in my mailbox, and it's not clear there's much to learn from that. I'm talking about papers that at least superficially look quite reasonable -- indeed, I'm expecting papers that have been published in reasonable places -- but when you think about them more, there are subtle (or not-so-subtle) flaws. In theoretical papers, possibly it might be that the paper starts with a model that sounds really nice but it just clearly wrong for the problem being addressed. For systems papers, it might be a paper where the experiments just don't match up to what ostensibly was being proposed. [I had a nice example of a bad paper in earlier incarnations of the class, but I don't think it's aged well, and I've removed it.] Maybe bad is even the wrong term for what I'm looking for. Perhaps I should use a more neutral word, like "controversial" -- indeed, then I can get the students to take sides. (Is the Faloutsos, Faloutsos, Faloutsos paper still considered controversial these days? That could be a nice example, but it's not really on topic for the class.) Or perhaps I just want papers that reached too high for their time -- noble failures. The key is that, in my mind, just showing students examples of great papers doesn't seem didactically sound. Negative examples are important for learning too (especially if they also show that great scientists don't always get it right). Feel free to mail me rather than post a comment if you're afraid of offending anyone. Naturally, mailing me links to my own papers will be taken with the appropriate humor. 15 comments: Maria-Florina Balcan, Avrim Blum, Anupam Gupta: Approximate clustering without the approximation. SODA 2009: 1068-1077 This comment has been removed by the author. This is a new anonymous. I still think the BBG paper is a bad paper. The same goes for the other papers of Balcan about Learning and Clustering via Similarity Functions. They all try to make a conceptual point, but the conceptual point turns out, each time, to be highly flawed: the theorems are correct, but often nearly-vacuous, and in any case don't mean what you'd have thought they mean. They all try to make a conceptual point, but the conceptual point turns out, each time, to be highly flawed: the theorems are correct, but often nearly-vacuous, and in any case don't mean what you'd have thought they mean. And this differs from virtually every theory paper how, exactly? Why don't you use Deolalikar's P \neq NP paper in your class as an example of a "bad" paper? :-) Seeing as I was mentioned here, I'd like to say that BBG and the papers on clustering and learning via similarity functions, are some of my favorites of my own papers. Let me focus here on BBG in particular and explain why I like it so much. First, there are many problems that when you formulate them as an optimization problem, they have the property that the objective function you are using to measure solution quality is not the same as the quantity you really care about. For instance, say you are clustering images of people by who is in them. You might algorithmically map the images into points in R^n and then aim to find a clustering that (approximately) minimizes k-means score on these points. But really you care about whether you clustered the people correctly. Of course, what's happening is your algorithm can't measure what you really care about so instead you optimize something you can measure. Fine - it's not a crazy thing to do. But the effect is this may make the problem seem harder than it really is. In particular, notice that in order for this to work, you already need to have used a method for representing your data (e.g, a way of placing images into in R^n) such that the optimal and near-optimal solutions to the objective you are measuring (k-means score) have good quality according to what you really care about. Otherwise, why be excited about getting a near-optimal score? The key point of BBG is that this property (solutions that are near-optimal according to the objective you can measure are also near-optimal according to the one you care about but can't) implies structure you can use algorithmically. Furthermore, this is true even if it's NP-hard in general to find near-optimal solutions. E.g., say we believe that it would be enough to get a 1.1 approximation to k-means in order to get close to the clustering we want. You might think this is not useful since k-means is NP-hard to approximate to such a factor. But what BBG shows is you can still use the implied structure to get close according to the objective you care about, even if it's NP-hard in general to approximate the objective you can measure. To put this another way: I love as much as anyone the challenge of developing better approximation algorithms for important problems. For instance, I'd love to get a factor-2 approximation to k-median. But the BBG algorithms already have the property that if data is such that all factor-2 approximations are close to the desired answer, then BBG will also get a solution close to the desired answer. So, to claim my new factor-2 approximation is better, I need to be able to say "it doesn't just produce any old 2-approximation, it produces a good/natural/special one" for some compelling notion of "good/natural/special". And if we can articulate what we mean by that, we can then ask if we can develop a souped-up version of BBG that needs only the assumption that the "good/natural/special" approximations to k-median are close to the desired answer. Personally, I think this is great. It says that for problems where we are trying to use solutions to objective A in order to solve problem B, then perhaps hardness results for A can be bypassed by using the assumption we are making anyway of the link between quality according to A and quality according to B in our given instances. For me, as someone who has worked in approximation algorithms for a long time, this was really cool. "Finding Patterns with Variable Length Gaps or Don’t Cares" M. Sohel Rahman, Costas S. Iliopoulos, Inbok Lee, Manal Mohamed and William F. Smyth This paper works on an idea which is actually fine, but does not really give the reasons. Also, the use of "by clever programming" is never a good sign. "The same goes for the other papers of Balcan about Learning and Clustering via Similarity Functions. They all try to make a conceptual point, but the conceptual point turns out, each time, to be highly flawed: the theorems are correct, but often nearly-vacuous, and in any case don't mean what you'd have thought they mean." Can you describe a theorem that troubles you? I really hope you will read this Clustering paper in in your class; and then post impressions on your blog. I suppose I should have imagined some some nut with an agenda would use this post to spout some vitriol, but to make clear, I'm disappointed by the anonymous comments, and I thank Avrim for taking time to respond. Of course, there wasn't anything particularly useful to respond to -- just a statement of "I think this is a bad paper" with some wishy-washy comments with no specifics, no clear points, no actual references to any text. And then a generalization to the "other papers of Balcan". Just to give my opinion on the opinion, if I saw a review like this when on a PC, my opinion of the reviewer would be downgraded for the rest of the meeting. If I received a review like this, I'd assume the reviewer was either personally biased against me or just an unsuitable individual for reviewing. It's OK to have a poor opinion of a paper (or, I suppose, a researcher). Indeed, the point of my post was that it should be useful for students to see "bad examples", and learn from them (as they learn from good papers). But particularly if you're going to call a paper bad, you should have the argument ready to back it up. On the plus side, perhaps these comments will encourage MORE people to read the BBG paper, which should, I think, be good for the authors. (To be clear, my comment is based on Anons referring to the BBG paper.) Can you give an example of what you are looking for here? (Maybe the old example, even if it is no longer appropriate, would serve as good illustration.) There are plenty of bad published papers out there, even in top conferences (and certainly as you go to weaker conferences). But I don't see how they would be very interesting to read in class. I think you also risk being insulting to people: I certainly wouldn't want to find my worst paper being highlighted as such in your class, regardless of what you (or I) though about it. Maybe you are just looking for controversial papers? This paper strikes me as a potentially "bad" paper, for reasons which I will outline below. The basic gist of it is that a major conclusion which the authors hold out as significant or surprising appears, upon closer inspection, to be directly built into the authors' definitions, and the definitions seem a little sketchy. The authors study the persistence of ties in a network of cellular phone calls. They define the "persistence" of a tie thusly: First, the the cellular phone records are split into ten contiguous time periods (called "panels") of equal length. Persistence is then quantified as the fraction of the panels in which a tie appears. This means that persistence is a numeric quantity with a value of 0 if the the tie never appears and 1 if it appears in all 10 panels. Note a natural consequence of this: A tie gets a persistence value of 0.5 under each of the following conditions: 1) It appears in each of the first 5 panels and then disappears, 2) It appears in the 5th panel and persists forever, or 3) It appears in every other panel. Intuitively, these do not seem like equivalent notions of persistence. They use basically five features in their model of tie persistence: 1) Degree 2) Clustering Coefficient 3) Node Reciprocity 4) Link Reciprocity 5) Neighborhood overlap (sort-of Jaccard coefficient). The first three they consider as node attributes (rather than edge attributes) such that node reciprocity is defined as the fraction of all ego's ties that are reciprocated. They do this in order to study the average persistence of a node's ties as a function of these three characteristics. They find that: 1) Average persistence decreases with degree. 2) Average persistence increases with clustering coefficient. 3) Average persistence increases with reciprocity. Their definition of link reciprocity is unclear. The only definition they give is "was there ever a panel in which the caller and callee reciprocally called each other?" I'm assuming the quantity they call "reciprocity" is the fraction of the panels in which the tie was reciprocal. If this is true, then it seems to me that their main conclusion "reciprocity is the strongest predictor of persistence" is built into the definitions rather than being a brilliant sociological Remember that persistence is defined as the fraction of panels in which A calls B. If reciprocity is defined as the fraction of panels in which A calls B and B calls A, then persistence increases whenever reciprocity increases merely by definition. Of course it is possible for persistence to increase without reciprocity increasing, and that is the reason that the correlation is not perfect, but the methodology seems flawed. It's also possible that my interpretation of their definition of link reciprocity is incorrect, but it probably should have been more clear. I'm not sure if this is what you mean by "bad." I would be interested in hearing defenses. Any thoughts? While BBG is a fine paper (if one whose eventual impact seems uncertain to me), clustering in general is a great source of bad papers. Drawing randomly from hits on "clustering" in the ACM Digital Library or Google Scholar, will bring up many a stinker in good conferences and journals. You can decide whether it's a success or failure of that randomized algorithm if an old clustering paper of mine comes up. :-) The common failing is the lack of connection between the criterion being optimized (always chosen for computational convenience) and the properties of any concrete task. There are thousands of clustering algorithms, and precious little insight into what problems which are good for. To put this another way: I love as much as anyone the challenge of developing better approximation algorithms for important problems. For instance, I'd love to get a factor-2 approximation to k-median. But the BBG algorithms already have the property that if data is such that all factor-2 approximations are close to the desired answer, then BBG will also get a solution close to the desired answer. So, to claim my new factor-2 approximation is better, I need to be able to say "it doesn't just produce any old 2-approximation, it produces a good/natural/special one" for some compelling notion of "good/natural/special". And if we can articulate what we mean by that, we can then ask if we can develop a souped-up version of BBG that needs only the assumption that the "good/natural/special" approximations to k-median are close to the desired answer.	{"url":"https://mybiasedcoin.blogspot.com/2010/08/in-need-of-few-bad-papers.html?showComment=1282489331917","timestamp":"2024-11-07T12:39:26Z","content_type":"application/xhtml+xml","content_length":"104458","record_id":"<urn:uuid:6a3bfd81-d6be-4acf-96e7-7b05491ad5e5>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00716.warc.gz"}
How to convert volts to ohms How to convert electrical voltage in volts (V) to electric resistance in ohms (Ω). You can calculate ohms from volts and amps or watts, but you can't convert volts to ohms since volt and ohm units do not measure the same quantity. Volts to ohms calculation with amps According to ohm's law, the resistance R in ohms (Ω) is equal to the voltage V in volts (V) divided by the current I in amps (A): R[(Ω)] = V[(V)] / I[(A)] So ohms are equal to volts divided by amps: ohms = volts / amps Ω = V / A Calculate the resistance in ohms of a resistor when the voltage is 5 volts and the current is 0.2 amps. The resistance R is equal to 5 volts divided by 0.2 amps, which is equal to 25 ohms: R = 5V / 0.2A = 25Ω Volts to ohms calculation with watts The power P is equal to the voltage V times the current I: P = V × I The current I is equal to the voltage V divided by the resistance R (ohm's law): I = V / R So the power P is equal to P = V × V / R = V ^2 / R So the resistance R in ohms (Ω) is equal to the square value of voltage V in volts (V) divided by the power P in watts (W): R[(Ω)] = V^ 2[(V)] / P[(W)] So ohms are equal to the square value of volts divided by watts: ohms = volts^2 / watts Ω = V^2 / W Calculate the resistance in ohms of a resistor when the voltage is 5 volts and the power is 2 watts. The resistance R is equal to square of 5 volts divided by 2 watts, which is equal to 12.5 ohms. R = (5V)^2 / 2W = 12.5Ω How to convert ohms to volts ► See also	{"url":"https://jobsvacancy.in/convert/electric/volt-to-ohm.html","timestamp":"2024-11-04T04:41:34Z","content_type":"text/html","content_length":"8394","record_id":"<urn:uuid:4d4e4bbd-3053-4a2f-9322-d5e1dd901f9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00576.warc.gz"}
Theory Of Relativity – Where Did Einstein Go Wrong? Time Dilation – The gist of this phenomenon is that if two observers travel in different speeds their clocks also run at different speeds. The clock setup taken up to prove this phenomenon is shown below (simplified diagram): In the diagram, the lower rectangle represents a light source and light detector setup, and the top rectangle represents a mirror. When this setup is stationary, let the time taken for the light to travel from the source to the mirror and back to the detector be t1. And when the setup is moving, let the time taken be t2. According to this phenomenon t1 < t2. If t1 or t2 is taken as the basic unit for the clock, then its clear that the clock will run slower when its in motion. Refer this link for detailed explanation on how this is working. Lets take a look at the following setup: Well as you can see the clock setup has been tilted by 90 degrees, otherwise this setup is perpendicular to the previous setup. And if you calculate the numbers you’ll find that the Lorentz factor is different for these 2 setups. We clearly know that there have been a lot of technological improvements since Einstein had given this theory to the world. This light emit, reflect and detect setup for a clock is pretty much primitive. Let me ask a question: Why are we not using the sand glass for keeping our time? Well you can give hundreds of reason. Why are we not using pendulum clocks? There are some environments in which it won’t give the accurate time. Why are we not using crystal oscillators in GPS satellites? Same answer. I am asking these questions to reason why we can not accept the theory of relativity anymore. I’ll explain this philosophy with an example. Lets take a circle. We can very easily mark the center of the circle. Lets take a triangle. We can apply some geometric calculations and still mark the center of the triangle. Now lets say for a long time we only know about circles and not triangles. And we know that there is only one geometric shape that is a circle and we know how to mark the center of the circle. It has been established that for a geometric shape the center is from equal distance from any point in the periphery of the geometric shape. Remember we know only about circles; no rectangles; no squares; no triangles. Now after lot of experimentation and exploration (????) a mathematician discovers about a new geometric shape triangle. Now he knows from his past experience that the center of a geometric shape is from equal distance from any point in its periphery. The mathematician traces a new shape as shown below. Now as per his discovery the center point of this new geometric shape is another shape. Some may immediately disagree with this and some may have confusion if there is a possibility that a center point of a shape could be another shape. This can not be true due to two reasons. 1. The proposed shape for the shape of the center point of a triangle depends on where we started first. If we have started outside the triangle we would have gotten a completely different shape. So the shape of a center point may not change based on where we start tracing the center point from. 2. A point is a one dimensional entity (though it may be given different dimension co-ordinates to identify its place). But here the point becomes 2 dimensional function. The fundamental assumption itself becomes wrong in this case. Now try to compare this example with time dilation phenomenon. 1. Lorentz factor is the main output of this phenomenon. But that itself changes when we change the clock setup. So this can not be true. 2. Time is a single dimension. You can not confuse it with other dimensions because you can not measure time correctly. You can not bend distance and other dimensions because your clock runs slow when you speed up. You must build a clock which runs the same in all the speeds. 4 comments 1. Dont mke me cry 2. Dear vibgy, i really admire you for this article. i know you from our school days that you are interested in physics. as i read your article, i was reminded of our school days. i wish you all the best for this kind of intellectual work. continue to write and come up with good things. 3. Ayyooooo… Answer theriyalaiyae…. 🙁 4. Nice work da.. Keep up good work like this. Have you referred Michelson-Morley experiment? This is also referring the same.	{"url":"https://vibgy.com/2009/12/theory-of-relativity-where-did-einstein-go-wrong/","timestamp":"2024-11-12T15:51:47Z","content_type":"text/html","content_length":"29963","record_id":"<urn:uuid:87ddad30-6359-4ef5-8572-6f93cc6cd71b>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00282.warc.gz"}
Cube \| Faces, Edges & Vertices \| Britannica While every effort has been made to follow citation style rules, there may be some discrepancies. Please refer to the appropriate style manual or other sources if you have any questions. Select Citation Style Thank you for your feedback Our editors will review what you’ve submitted and determine whether to revise the article. cube, in Euclidean geometry, a regular solid with six square faces; that is, a regular hexahedron. Since the volume of a cube is expressed, in terms of an edge e, as e^3, in arithmetic and algebra the third power of a quantity is called the cube of that quantity. That is, 3^3, or 27, is the cube of 3, and x^3 is the cube of x. A number of which a given number is the cube is called the cube root of the latter number; that is, since 27 is the cube of 3, 3 is the cube root of 27—symbolically, 3 = ^3Square root of√27. A number that is not a cube is also said to have a cube root, the value being expressed approximately; that is, 4 is not a cube, but the cube root of 4 is expressed as ^3Square root of√4, the approximate value being 1.587. In Greek geometry the duplication of the cube was one of the most famous of the unsolved problems. It required the construction of a cube that should have twice the volume of a given cube. This proved to be impossible by the aid of the straight edge and compasses alone, but the Greeks were able to effect the construction by the use of higher curves, notably by the cissoid of Diocles. Hippocrates showed that the problem reduced to that of finding two mean proportionals between a line segment and its double—that is, algebraically, to that of finding x and y in the proportion a:x = x:y = y:2a, from which x^3 = 2a^3, and hence the cube with x as an edge has twice the volume of one with a as an edge. Britannica Quiz Define It: Math Terms This article was most recently revised and updated by Michael Ray.	{"url":"https://www.britannica.com/science/cube-mathematics","timestamp":"2024-11-03T19:33:54Z","content_type":"text/html","content_length":"85487","record_id":"<urn:uuid:1b10760e-ac85-4739-94ad-2a5b5b535fa4>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00181.warc.gz"}
Iron Lambda Iron Lambda is a collection of Coq formalisations for functional languages of increasing complexity. It fills part of the gap between the end of the Software Foundations course and what appears in current research papers. If you are new to Coq then you would be better off starting with Software Foundations rather than this work. We just use straight deBruijn indices for binders. Using deBruijn indices does require that we prove some lemmas about lifting and substitution, but they are very similar between languages, so the initial effort can be re-used. For more details see the blog post How I learned to stop worrying and love deBruijn indices. The proofs use a "semi-Chlipala" approach to mechanisation: most lemmas are added to the global hint and rewrite databases, but if the proof script of a particular lemma was already of a sane length, then we haven't invested time writing lemma-specific LTac code to make it smaller. Style guidelines: • Verbose comments explaining what the main definitions and theorems are for. The scripts should be digestable by intermediate Coq users. • No unicode or infix operators for judgement forms. When I use them in my proofs they make perfect sense, but when you use them in yours they're completely unreadable. • Uses Coq bullets, as well as the Case and SCase etc tactics to add structure. • You will need a working version of Coq. The proofs are known to work with Coq 8.5. • Source code is on github git clone https://github.com/DDCSF/iron • There is a top-level Makefile that will build all the proofs. For this to work coqc and coqdep need to be in your default path. $ cd iron $ make • Each Coq module should check in under two minutes. If not then the automation might have diverged, so please tell me about it. Also report any build problems. More Information Related Work Current Languages Click the headings to get to the proofs. Simply Typed Lambda Calculus (STLC). "Simple" here refers to the lack of polymorphism. STLC with booleans, naturals and fixpoint. STLC with mutable references. The typing judgement includes a store typing. STLC with algebraic data and case expressions. The definition of expressions uses indirect mutual recursion. Expressions contain a list of case-alternatives, and alternatives contain expressions, but the definition of the list type is not part of the same recursive group. The proof requires that we define our own induction scheme for expressions. Compared to STLC, the proof for SystemF needs more lifting lemmas so it can deal with deBruijn indices at the type level. Very similar to SystemF, but with higher kinds. SystemF2 with algebraic data and case expressions. Requires that we define simultaneous substitutions, which are used when subsituting expressions bound by pattern variables into the body of an alternative. The language allows data constructors to be applied to general expressions rather than just values, which requires more work when defining evaluation contexts. SystemF2 with algebraic data, case expressions and a mutable store. All data is allocated into the store and can be updated with primitive polymorphic update operators. SystemF2 with a region and effect system. Supports region extension and deallocation.	{"url":"http://iron.ouroborus.net/wiki/WikiStart?version=34","timestamp":"2024-11-04T01:03:51Z","content_type":"application/xhtml+xml","content_length":"13915","record_id":"<urn:uuid:9eb9a36f-e9ab-4971-acc5-b6808273eb26>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00641.warc.gz"}
sgd is an R package for large scale estimation. It features many stochastic gradient methods, built-in models, visualization tools, automated hyperparameter tuning, model checking, interval estimation, and convergence diagnostics. At the core of the package is the function sgd(formula, data, model, model.control, sgd.control) It estimates parameters for a given data set and model using stochastic gradient descent. The optional arguments model.control and sgd.control specify attributes about the model and stochastic gradient method. Taking advantage of the bigmemory package, sgd also operates on data sets which are too large to fit in RAM as well as streaming data. Example of large-scale linear regression: # Dimensions N <- 1e5 # number of data points d <- 1e2 # number of features # Generate data. X <- matrix(rnorm(Nd), ncol=d) theta <- rep(5, d+1) eps <- rnorm(N) y <- cbind(1, X) %% theta + eps dat <- data.frame(y=y, x=X) sgd.theta <- sgd(y ~ ., data=dat, model="lm") Any loss function may be specified. For convenience the following are built-in: * Linear models * Generalized linear models * Method of moments * Generalized method of moments * Cox proportional hazards model * M-estimation The following stochastic gradient methods exist: * (Standard) stochastic gradient descent * Implicit stochastic gradient descent * Averaged stochastic gradient descent * Averaged implicit stochastic gradient descent * Classical momentum * Nesterov’s accelerated gradient Check out the vignette in vignettes/ or examples in demo/. In R, the equivalent commands are vignette(package="sgd") and demo(package="sgd"). To install the latest version from CRAN: To install the latest development version from Github: # install.packages("devtools") sgd is written by Dustin Tran and Panos Toulis, and is under active development. Please feel free to contribute by submitting any issues or requests—or by solving any current issues! We thank all other members of the Airoldi Lab (led by Prof. Edo Airoldi) for their feedback and contributions. author = {Tran, Dustin and Toulis, Panos and Airoldi, Edoardo M}, title = {Stochastic gradient descent methods for estimation with large data sets}, journal = {arXiv preprint arXiv:1509.06459}, year = {2015}	{"url":"https://cran.auckland.ac.nz/web/packages/sgd/readme/README.html","timestamp":"2024-11-09T00:24:46Z","content_type":"application/xhtml+xml","content_length":"4046","record_id":"<urn:uuid:1682f9c9-f48c-4358-adbf-9be2a3949ea9>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00604.warc.gz"}
Fundamentals of High School Mathematics by Harold O. Rugg, John R. Clark Publisher: Yonkers-on-Hudson, N.Y. 1918 ISBN/ASIN: B00314FL1K Number of pages: 296 The text is organized to provide the pupil with the maximum opportunity to do genuine thinking, real problem-solving, rather than to emphasize the drill or manipulative aspects which now commonly require most of the pupil's time. Download or read it online for free here: Download link (multiple formats) Similar books Elements of Algebra Arthur Schultze The Macmillan CompanyThe attempt is made here to shorten the usual course in algebra, while still giving to the student complete familiarity with all the essentials of the subject. All parts of the theory which are beyond the comprehension of the student are omitted. High School Algebra J.T. Crawford The Macmillan CompanyThis text covers the work prescribed for entrance to the Universities and Normal Schools. The book is written from the standpoint of the pupil, and in such a form that he will be able to understand it with a minimum of assistance from the teacher. A First Book in Algebra Wallace C. Boyden Project GutenbergIt is expected that this work will result in a knowledge of general truths about numbers, and an increased power of clear thinking. The book is prepared for classes in the upper grades of grammar schools, or any classes of beginners. Supplementary Algebra Robert L. Short D.C. Heath & Co.Large number of requests coming from teachers for supplementary work in algebra has led me to collect such material into a monograph, hoping to furnish the teacher with methods and supplementary work by which he may brighten up the algebra review.	{"url":"https://www.e-booksdirectory.com/details.php?ebook=8694","timestamp":"2024-11-03T17:25:21Z","content_type":"text/html","content_length":"11079","record_id":"<urn:uuid:842d969f-b5e4-4d52-96ae-2e8ae70dfe3e>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00253.warc.gz"}
Statistics problem The scores on a biology exam were normally distributed with a mean of $71$ and a standard deviation of $9$. A failing grade on the exam was anything $2$ or more standard deviations below the mean. What was the cutoff for a failing score? Approximately what percentage of the students failed? • I have already asked this question. Please do not respond. Answers can only be viewed under the following conditions: 1. The questioner was satisfied with and accepted the answer, or 2. The answer was evaluated as being 100% correct by the judge. View the answer Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.	{"url":"https://matchmaticians.com/questions/6p9bp5/statistics-problem-normal-distribution-word-question","timestamp":"2024-11-08T15:13:14Z","content_type":"text/html","content_length":"79449","record_id":"<urn:uuid:a20100fb-e01a-46c2-aea6-a55f8dd83c37>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00249.warc.gz"}
Math Notes Algebra 2 (2024-25) Notes \| Knowt ~~~UNIT 1.1-1.2~~~ -CYU 1.1- -4a. The Graph of a function y=f(x) is shown. Find f(-1) f(-1) essentially means that the x of the f(x) {which is -1} is the X on the graph. So, we plot the point on -1 X. Because the equation is y=f(x), we know that the function must be on the Y-line. Thus, we draw a verticle line on the -1X point, and our value is where the line intersects with that line. -5a. The complete Graph of a function y=f(x) is shown. The table shows some values for function f. Find the values to complete the table We can find the missing values by plotting the inputs on the X line and seeing where the line intersects on the Y axis. One intersects with 1, 0 can only intersect with 0, and -2 intersects with 4. -6a. The function N(t) gives the number of people in an auditorium, t hours after the doors open. Two hours after the doors open, are 108 people in the auditorium. Because N represents the number of people and T means time, N(t) represents the number of people after a specific time. Thus, the equation would be N(2)=108 -7abc. The table shows some values for the linear function f. Identify the y-intercept of the Graph of f. Determine the slope of the Graph of f. Write an equation for f(x). a. To find the Y-intercept, we have to look for the input of 0. This is because the output from zero is the y-intercept on the 0,0 point. In the table, we see that 0’s production is 10. Thus, the Y-intercept is 10. b. To determine the slope of the Graph, we have to use the slope formula x-x/y-y. For a table, we take the outer numbers on each side (The first and last) and use those as our points. That gives us an equation of 2—2/16-4. Both subtractions make a positive, and 16-4 is 12, so we get 4/12. Simplify to a slope of 1/3. c. The equation will be y=mx+b. We already know the slope and y-intercept, so we plug them in for y=1/3X+10. -10. Give the domain and range of P(d). There are 30 days in June and 100% of the course to complete. Thus, the range and domain will be (0 X 30) and (0 X 100) -Interval 1.1 Notation- 1. cd For a set of numbers shown on the number list below, represent the set using interval notation. Because the point is on an unclosed -4 and ends on a closed 11-point, the interval notation will be (-4, 11]. -1.2 Quick Check- 2. b, state the minimum and maximum values of the function over its entire domain. The minimum and maximum points on the domain are the leftmost x point(minimum) and the rightmost x point(maximum). Thus, the minimum x point is (-8, -6), and the maximum x point is (9, 0). c. state the interval over which f(x)>0 -1.2 More Practice- 4a. For the Function defined by f (x)=16-x^2, Use your calculator’s table option to help fill out the outputs in the table above. Using the TBLSET function on the calculator would be best to get the table values. First, press the Y= button and type in the function on your calculator [(16-x^2 )]—press graph. Press the 2nd Graph to get the TABLE. Move the arrows to see all your points. Then press 2nd WINDOW to get TBLSET. Change the table’s settings to the number at which it starts on 1. Negative and Positive inputs of the same number have the same output. 6a. The Table below shows the temperature, in degrees Fahrenheit, as a function of the number of hours after 5:00 pm. What is the y-intercept of this function? What does it represent in the context of this problem? The y-intercept is the output from the 0 input. Thus, the y-intercept is 68, according to the table presented. In the context of the problem, the y-intercept would represent the temperature. Since the Graph starts at 5 pm and goes from 5 pm, the y-intercept would be the temperature at or after 5 pm. Quiz Review: -4c. The problem claims that the postcard’s relationship has w(l) = the width of the postcard multiplied by the length since that’s how we calculate our width. The l remains long, so to get our formula (since we know lw=A), we would use the formula Area=lw(l). -5b. Since x=the x in f(x), we would plug -1 into the f(x) formula. (-1-3)^2 -2= 14. We can find what g(0) is since the formula for g is y=g(x), meaning that the Y value on the listed X value point is our number. Following the Graph, this makes g(0) -2. Thus, when we subtract f(x) {14} from g(0) {-2}, we get a final answer of 12. -6ab. To get f(x)+g(x), we add the functions together because the problem claims that f(x) and g(x) equal their respective functions. Simplify 2(x-1) by distributing to get 5x-2+2x-2. Then, we combine the common factors to get a simplified answer of 7x-4. To get f(x)-g(x), we do the same thing as above, subtracting from each other rather than adding. 5x-2-2x-2. 7x-4=3x, our answer. -8. When we only have one poiInfinitynfinity, we use X>/<(depending on where the segment is going) other points on the line. When using interval, notInfinitynfinity uses (), not []. -10. When using interval, notInfinitynfinity uses (), not []. Infinity also uses </>; other points are not classified as open circles or infinite use ≤/≥. The increasing intervals are from below 0 to 0, so we keep it with those points. The same applies to increasing intervals; we use interval notation to mark this, not inequalities. -11a. Plug 4 in for X since the function is f(4). Absolute value lines don’t mean anything different; they make everything negative regardless. -12ab. Because the function is f(x), a+c can be plugged in for X. Thus, 2a-4(a+c)can be distributed to 2a-4a+4c. Simplify and combine to get -2a+4c. Plug the 6a-3b in for x since the function is h(x). -3(6a-3b). Distribute and simplify to get a final answer of -18a+9b. 15cde. Because we don’t have a direct function line, we must list all the points for the domain and range. 19f. Because there are multiple points on this line, we have to have the domain for each one. ~~~UNIT 1.3-1.5~~~ Ask about converting standard form into slope-intercept form -CYU 1.3- 2. Find the average rate of change of f(x)=3x-8 between x=-7 and x=4. Because x=-7 and x=4, we plug both X’s in our equation to get corresponding Y-POINTS. Remember , B=Y-Intercept, Y=Y points. When both X’s are plugged in, we get corresponding Y points of -29 and 4. Use the slope formula to get the average rate of change. Remember, -7 and 4 are your X points. 6b. Find a possible value for the number of people in line at t=6, given that the average rate of change of P between t=3 and t=6 is negative, but the average rate of change between t=5 and t=6 is To get a possible value, we use an inequality. We can find the values by doing slope form. T=3 has a y point of 85. We don’t know the T=6 point, so I’ll use t(6). We see both X points (3, 6), so we subtract them to get 85-t(6)/-3. Because the average rate of change between t=3 and t=6 must be negative, the top number must be positive, so t(6) can’t be over 85. For the second one, we do the same thing. The table tells us T=5 y point is 49. So, our final equation is 49-t(6)/-1. Because this slope must be positive, the top number must also be negative, meaning t(6) must be higher than 49. To get possible values, we put these two facts into inequality format. P must be more than 49 but less than 85. 49<p<85. 9. The graph of y=g(x) is shown. Order the following from least to g Calculate the slope between each point on the given intervals for each problem using rise over run from each point on the given X intercept in the inequalities. Remember when slopes are harmful, and two negative numbers on a hill instead make an upbeat version of that slope. Order them from least to most lavish. Using this formula, the order is III, II, IV, I. -CYU 1.5- 4. Fill in the blanks to write a piecewise equation. If you buy 2 pounds or less, then the price is 9.95. If you buy more than 2 pounds, you pay 7.95. We already have the inequality laid out for us, so we know that the equation will be C(p)= 2 equations because there are two possible outcomes. The first is 9.95 if you pay 2 pounds or less. So, our equation will be 9.95 if 0<p≤2. This is because to pay 9.95, you must buy somewhere between 0 and 2 pounds. For the 2nd equation, it is 7.95 for over 2 pounds. So, our equation will be 7.95 if p>2. Put both into the complete function equation of C(p)= 9.95 0<p≤2, 7.95 p>2. 6. Graph y=f(x) 8bc. Does the club charge a higher hourly rate with two children than 1? If a family has more than two children, how much does the club additionally charge per hour? The club does not charge a higher rate. This is because the shown function inequality uses ≤2, which means less than or equal to 2 regarding the first price. Thus, the number for 2 or 1 child is the same. The club additionally charges $4 an hour for more than two kids. The function shows an equation for 10 +4(x-2). X is equal to the number of children. Because we are asking for per child, we are essentially asking for the average rate of additional charge, where we can look at the slope in the equation, which is 4. Remember that asking for something usually refers to the average rate of Point Slope Review. Graph the line y+4=-5/2(x+3) on the coordinate plane. The function line tells us that the line has a point of (3, 4) by giving us those numbers on the y and x point segments of the equation. The line also provides us with the slope (-5/2), so use rise over run with the hill from the point (3, 4) across the entire Graph to get your line. -Mastery Check 1.3- Always look for the side of the table with both numbers as X-intercepts. Find the corresponding Y-intercept by looking at each number’s corresponding point on the other side of the table. Use the slope formula to get the average rate of change. Solution (1/4). To get the average rate of change, we have to have both X and Y points; we know the X points are the numbers on the inequality, so we see the x points are 0-8. To get the Y points, plug each X into the equation separately to get the correlating y outcome for the X. This gets us points of 18-10. Use slope form rise/run to get slope/average rate of change. Solution (-1). We know from the inequality the X points are 4 and 5. Find the points on the Graph for each X point to get the Y points. Look and see what their correlating y points are. These are your y points. Use the rise/run slope form to get the average rate of change. Plot a point at -7 y. Because the slope is -x(-1), go down one and correct 1. This is your line. Draw it across the entire Graph. Because the inequality is -6≤x<-1, the line is only valid between the x points of -6 and -1, so erase the rest of your line except for the space between that. Fill in circles as needed. In the second line, place a point on y -5. The slope is 1, so go up and right 1. This is your line. Draw it across the entire Graph. Because the line is only accurate between -1 and 4, erase your line except for those in between spaces. Fill in circles as needed. To get the average rate of change, we need to know the y and x values. We know both the x values are -2 and 2. To get the y values, plug each X value (-2, 2) into the equation separately. The sum of the equation is the Y values (after plugging x’s in, we get y values as -4 and 8). Use the slope formula to get the average rate of change. -4-8/-2-2 —> -12/-4—→ 3/1—→ 3. Solution (3). Use parentheses when using a double negative. Aka -6=x -x^2 —→ -(-6)^2 To get the average rate of change, we need to know the y and x values. We know both the x values are -7 and 1. To get the y values, plug each X value (-7, 1) into the equation separately. The sum of the equation is the Y values (after plugging x’s in, we get y values as 23 and -1). Use the slope formula to get the average rate of change. 23+1/-7-1—> 24/-8—> -3. Solution (-3) Use Parentheses when using a double negative. Aka -6=x -x^2 —→ -(-6)^2 ^-CYU 1.4- 2. A table of selected values is given for odd function X. Find three other ordered pairs. To find any other ordered pairs, we need to create the opposite of the current pairs since it is an odd function, and odd functions are opposite while even staying the same to the listed points on the table. 8. Benedickt was asked if the function f(x)=x^3-4^×2-4x+19. Is he right? No, Benedickt is not correct. This is because Benedickt used -2, 2 instead of X instead of x, -x. We want to evaluate the function using all possible x values, so we plug x, -x (1, -1) in the equation instead. Plugging these values in, we get that X=0 and -X=6, so it is neither even nor odd because the x values are not even to each other and are not opposite. Benedickt is wrong because the function is neither nor not even. He is terrible because he used 2, -2 as points instead of x, -x. 9. Determine if the statement is true or false. Reflecting an even or odd function across the x-axis does not affect its function’s symmetry. This statement is true. This is true because when reflected, both even and odd points will not change numbers; even if they are made negative/positive, the number value itself will not change, keeping it symmetrical. -Mastery Check 1.5- 1. Solution: 0. To get our answer, we must plug -7 in for x on the first equation. This is because we are trying to find f(-7); because f(x) is our original equation, we know -7 will be plugged in for x in one of the equations. To find which equation, plug -7 in for x at each inequality to see which one is true. Because -7 is not greater than or equal to negative 2 (2nd equation) but is less than -5 (1st equation), we know to plug -7 into the first equation to get our answer. Plug -7 in for x and calculate to get a final answer of 0. 2. Place a point at -7 (y). Draw a line using slope (4x, 4/1) across the entire Graph. Because inequality is only true when x is less than or equal to one, erase all lines above 1. Fill in circles as needed. This is the line of equation one. To get the line of equation 2, do the above steps the same, only plugging in the y and slope and inequality listed in the 2nd equation. (Y=3, Slope= -2/1, inequality is x>5) 3. To find f(1), look for the point on the 1 x value. Because there are multiple points, choose the end with the closed/filled circle. To get the answer, look at the y value of the line you picked. That is your answer. 4. To graph the first equation, place a point at -1 y-intercept. There is no slope; a straight line will be throughout the Graph. Erase line except for inequality (-4<x<1). For the second line, plot the point at y two and draw the line with a slope of 1 across the entire Graph. Erase the line in all places except between the x-intercepts designated on the inequality (1, 5). Fill in circles as needed. 13. Plot a point at y one and draw the line with a slope of -1. Remember that the denominator of the slope is still positive. Otherwise, it would be two negatives, which would turn into a positive. Erase the line except for inequality designated space between designated X’s (-3, 2). Do the above steps with the first equation with the 2nd function, using only the first equation numbers. 14. Plot a point at 9. Draw a line from that point with a slope of -1. The inequality is x>5, so the line is only valid (and drawn) when the line is greater than 5. For the 2nd equation, plot a point at 0 because no y-intercept is given. Draw a line from 0 at a slope of -1. Erase the line except for when the line is less than or equal to -2 because that is our inequality (x≤-2). Fill in circles as needed. HOW TO GRAPH PIECEWISE FUNCTIONS: Take the equation above as an example. The -1 is the y-intercept, so we’ll place a point there. Because there is no x, we have no slope, so we keep it a straight line across the ENTIRE Graph. Then, look at your inequality. Because the disparity states -4<x<1, the line is only valid between those X points. Erase the rest of the Graph and fill in circles as needed. Because the second one has an equation of x+2, we will plot a point of 2 on the y-intercept; then, because X is placed before 2, we know X (which is equal to 1/1) is our slope, so we will graph a line using that slope from the y-intercept across the ENTIRE Graph. Plot the line regarding the inequality and fill in circles as needed. HOW TO EVALUATE PIECEWISE FUNCTIONS: Take the above equation as an example. Find f(1). Find the X point for one on the Graph. Then, look on the Y-line for the corresponding point. If there are multiple points, choose the one with a closed circle. It is undefined if no closed circle lines exist on the following Y-intercept. -1.3-1.5 Practice Problems- 3b. Calculate the average rate of change if y=2x^2 + x +2 [0, 1/2]. To calculate our average rate of change, we need to know both x and y values. The brackets [] give us the x values, so we only need to find the y. To find the y, we plug each x point into the equation separately to get the corresponding y points. Remember to square the x first and multiply it by the following number. This gets us the values of 2 and 3. Use slope form to get the average rate of change. 2-3/0-.5 2. 4a. Calculate the y and x-intercepts in the function. To calculate the y-intercept, make x=0. Look at the inequalities. The equation you will use in the function is whichever inequality has the true statement with 0. Plug x in for 0 in the function and calculate to get y-intercept (3). To get the x-intercept, look at the inequalities, but this time, see which one is true with x as 1. Whichever inequality is true, the correlating equation we use. Make the equation equal to 0, and calculate to get the x value(-3/2). -Practice Test (Can’t Find Document)- 2. Remember to subtract the two’s before making the slope. 11. Remember that y is a verticle line, and x is a horizontal line, so when a line only goes through one, it is either y= or x=, depending on which line it falls under. 13. When plotting the y-axis in function notation, you need to place the number the line goes through on the equation. So, if the completed function doesn’t go through the y-axis, keep using the shown slope until you get there. Plot that number on the equation. 18. Remember that if both X values match the inequality, use both. ~~ALGEBRA 1 REVIEW~~ -CYU A- 1a. Rewrite the equation in slope-intercept form: x+2y=-2 2. Write an equation in a point-slope form that passes through the points (-3,8) and (5,-4). 3. Write an equation in slope-intercept form that goes through the points (3, -4) and (-3, 5) 4. Write an equation in a standard form that passes through the points (6, -4) and (12, 2) -CYU B- 2. Which of the following points passes through the point (-4, -8)? 6a. The points (-3, 6) and (6, 0) are plotted on the grid below. Find an equation, in y=mx+b form, for the line passing through these two points. ~~UNIT 1.6-1.7~~ -1.6 Quick Check- 1b. Consider the 3 functions, f(x), g(x), and h(x) shown below. h(g(5)) H(g(5)) is -5. This is because the output of 5 on g is 0, and the y value on the H function graph has a -5 intercept on the 0 point. 1c. Consider the 3 functions, f(x), g(x), and h(x) shown below. (gf)(-12) To get f(-12), plug 12 in for the x value on the equation, which gets us -8. Then this makes g(-8), which the table shows is -6. 2b. (Gf) (2) f(2) is shown to be five on the table, so we make the equation g(5). G(5) is shown to be six on the table. 3b. Given the functions f(x)=3x-2 and g(x)-5x+4, determine formulas in the simplest y=ax+b form for: g(f(x)) G(f(x)). This essentially means that we will plug the f equation into the g x coordinates since f equals the g x coordinates. So, after plugging in, we get 5(3x-2)+4. Distribute to get 15x-10+4. Combine like terms to get 15x-6. -1.6 CYU- 1b. Given f(x)=3x-4 and g(x)=2x+7 evaluate: g(f(-2)) Plug -2 in for the x of the f function. This gets 3(-2)-4=-10. Plug -10 into the g function to get -2(-10)+7=27. 3b. The graphs of y=h(x) and y=k(x) are shown below. (k • h) (0). Find the y point that’s on the 0 point since were looking at h(0). This equals 1. Then, move 1 to the right and find the next point in general, not just on the line. The next point is 5, but there are no points on the 1 x value, so we will place it on the next point in general, which is 5. 4. If g(x)=3x-5 and h(x)=2x-4 then (g•h)(x)=? To get this, we must place the h function into the x-intercept(s) of the g function. This makes 3(2x-4)-5. Distribute to get 6x-12-5. Combine like terms to get 6x-17. 7. Physics students are studying the effects of temperature, t, on the speed of sound, S. Round to the nearest tenth. Because the equation is K(C)=C+273.15, an we are given that the C=30. We will plug that number in to get K for the next equation, which was also given (√(410K). 30(C)+273.15 is 303.15, and since this equals K, we plug 303.15 into the 2nd equation. √410(303.15). This makes a final answer of 352.5. 8c. Consider the functions f(x)=2x+9 and g(x)=x-9/2. g(f(x)) Plug the f equation into the x intercept(s) of the g function. This gets us 2x+9-9/2. Combine like terms to get 2x+0/2—> 2x/2 is our final answer. -1.6 Mastery Check- Question 2: Find the y value of the Question 3: Plug -3 into the g function. This gets 2(-3)^2+7(-3)-3. Combine and multiply. Rememember to do the square on the parentheses number and multiply by it’s counterpart. This equation equals -6. This makes f(g(-3)=f(-6) since g(-3)=-6 and you work inside out. Plug f(-6) into the f equation. 4(-6)-1 (Final answer; -25). -1.7 Mastery Check Retake- Question 2: Solution: f^-1(x)=x/4-3 Swap the X and Y values and then get the Y alone to get the inverse function. Swapped values makes the function X=4(y+3). Use algebra to get the Y alone x/4=y+3. x/4-3=y (f^-1(x)) Question 3: Solution: f^-1(x)=^3√x+4 Swap the X and Y values and get the y alone to have the inverse function. x=(y-4)^3. ^3√x=y-4. ^3√x+4=y. Question 2a (retake) Solution: f^-1(x)=2/x + 1 Swap the X and Y values and get the Y alone to get the inverse function. X=2(y-1). x/2=y-1. x/2+1=y. Question 2b(retake) Solution: f^-1(x)=x^7/6 Swap the X and Y values and get Y alone to find inverse function. X=6y^1/7 X^7=6y—> X^7/6=y -1.7 More Practice- 8.If g(7)=-3 and g^-1(4)=2, find 1/3g(2)+5/3g^-1(-3) Because g(7)=-3 and g^-1(4)=2, we can plug in the g(2) and the g(-3) for 4 and 7. This gets 1/3(4)+5/3(7). Use fraction multiplication to get 4/3+35/3=39/3. Simplify to get 13 as our final answer. Study plan: Unit 1.1-1.2 Questions Friday, Unit 1.3-1.5 Saturday, Unit 1.6-1.7 Sunday Monday practice exams. -1.7 CYU- 3. The graph of function y=g(x) is shown below. The value of g^-1(2) is: -Unit 1 practice test- 1. Remember that functions need to have no repeating inputs, not outputs, 4. Remember that f(x)=10 means the equation is equal to 10, not X. 6. f(x)>0 means that it’s asking for the positive part of the line. 12. Remember to add the 2 first and then divide by 4. Add, then divide. 14. The graph of a function and the graph of it’s inverse will always be symmetrical to the Y=X line. 16. Put Y=2x-9 into point slope form with the point listed and solve by getting the y alone. 19. Because the distance is miles over the hour, we put miles/hour/minute, since distance/time. 20b. f(x)=5 is essentially asking y=5, so just look at how many times the line crosses the y point at 5. 21abc. Remember slope is y/x. Remember that to get the inverse function, you NEED to swap the x and y values and get the y alone. Only way that works. 22. Slope always starts from the lefternmost side. Count correctly. 24b. Remember to add/subtract first and then divide/multiply. To the opposite function to the other side to get answer. 28. Remember that slope is y/x. 29bc. Remember that you need to subtract the 922 from the 28m to get the number of cars. Remember that because it says after 25 minutes, you need to add 25 minutes to that 7 you got. -Delta Extra Practice- Solution: -1 ~~UNIT 2.1~~ -2.1 Quick Check- 1. Which parent function(s) demonstrate a constant rate of change? 2. Identify the parent function in each of the following equations: 3. a. Determine if the relationship is linear, quadratic, or neither. ~~UNIT 2.2-2.3~~ -2.2 Extra Practice- 5) Identify the solid parent function, then describe the transformation necessary to transform the graph of f(x) into that of g(x). Write the equation of the graphed line. 6) Identify the solid parent function, then describe the transformation necessary to transform the graph of f(x) into that of g(x). Write the equation of the graphed line. 19) Sketch the Graph of Each Function. g(x)=x^3-3 20) Sketch the Graph of Each Function. g(x)=1/x-2 27) y=4^x+2 28) y=2^x+1 29) y=2^x+1-2 30) y=2^x+2+1 -Quizziz 2.2-2.3- -2.2-2.3 Mastery Check- -2.3 CYU- 6. After a reflection on the x-axis, the parabola y=4-x would have the equation: 7. Which of the following equations shows the graph shown below? 9. If (h)x represents a parabola whose turning point is (-3,7) and the function f is defined by f(x)=-h(x+2), what are the coordinates of the turning point f?2 -2.3 Extra Practice- 5. Identify the solid parent function and write the equation of the dashed line. 10. Describe the transformation. f(x)=√x/g(x)=-√x-3 11. Describe the transformation. g(x)=-\|x\|-1 16. Sketch the graph. g(x)=√x-3+2 -2.4 CYU- 4. The quadratic function f(x) has a turning point at (-3, 6). The quadratic y=2/3 f(x)+3 would have a turning point of: 6. The graph of y-h(x) is shown below. Specify the transformations and the order in which they occurred. 7. A parabola is shown graphed to the right as a transformation of y=x^2. Based on your answer, write an equation for the parabola. 8. The function h(x) is defined by the equation h(x)=4f(x)-12. Write h(x) in its factored form. -2.4 Extra Practice- 1) f(x)=√x—>g(x)=1/2√x 4) f(x)=1/x—→g(x)=-3/x-2 7) Sketch the graph of each function. g(x)=1/3*\|x\| 8) g(x)=2x 12) g(x)=2/x+2 -2.5 CYU- 8. Sketch the Graph of the function. g(x)=√3x 13. g(x)=√1/3(x-3) 14. g(x)=-(1/2x)^3 16. Transform the given function f(x) as described and write the resulting function as an equation. Expand horizontally by a factor of 2. 20. f(x)=\|x\| compress vertically by a factor of 3 reflect across the x-axis.	{"url":"https://knowt.com/note/b4e52fa1-ffee-47f0-9a38-1d36d339d44f/Math-Notes-Algebra-2-2024-25","timestamp":"2024-11-08T07:49:17Z","content_type":"text/html","content_length":"309597","record_id":"<urn:uuid:d0c6a6e0-a957-4fde-afea-5fd8b2c5976e>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00573.warc.gz"}
Roll two dice and multiply the numbers. What is the probability that the product is a... Roll two dice and multiply the numbers. What is the probability that the product is a... Roll two dice and multiply the numbers. What is the probability that the product is a multiple of 6? Group of answer choices TOPIC:Classical definition of probability.	{"url":"https://justaaa.com/statistics-and-probability/1301163-roll-two-dice-and-multiply-numbers-what-is","timestamp":"2024-11-11T07:29:34Z","content_type":"text/html","content_length":"38544","record_id":"<urn:uuid:f46a321a-e68a-4d1f-acfc-3ff59ac9bdb1>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00859.warc.gz"}
Lesson 3 Fractions Round Table Warm-up: What Do You Know About $\frac{1}{8}$? (10 minutes) The purpose of this What Do You Know About _____ is to invite students to share what they know about and how they can represent the number $\frac{1}{8}$. • Display the number. • “What do you know about $\frac{1}{8}$?” • 1 minute: quiet think time • Record responses. • “How could we represent the number $\frac{1}{8}$?” Student Facing What do you know about $\frac{1}{8}$? Activity Synthesis • “What connections do you see between different answers?” Activity 1: Fractions Round Table (35 minutes) The purpose of this activity for students to think about and discuss statements that address their understanding of important ideas about fractions. Students will consider ideas about how fractions are defined, comparing fractions, and how fractions relate to whole numbers. It is not necessary for each group to discuss all of the statements, but if there are any you’d like to make sure each group discusses, let them know at the start of the activity. Students construct viable arguments to explain their choices (MP3) and in order to do so they need to use key fraction language, such as whole and equal-size piece, precisely (MP6). MLR8 Discussion Supports.Synthesis: Provide students with the opportunity to rehearse what they will say with a partner before they share with the whole class. Advances: Speaking Engagement: Develop Effort and Persistence: Chunk this task into more manageable parts. Check in with students to provide feedback and encouragement after each round. Supports accessibility for: Organization, Focus • Groups of 4 • “Take a minute to read the directions for today’s activity. You will be discussing statements about fractions with your group.” • 1 minute: quiet think time • Consider walking students through the process and answer any questions. • 25–30 minutes: small-group work time Student Facing Discuss each statement in 3 rounds with your group. • Round 1: Go around the group and state whether you agree, disagree, or are unsure about the statement and justify your choice. You will be free to change your response in the next round. • Round 2: Go around the group and state whether you agree, disagree, or are unsure about the statement you or someone else made in the first round. You will be free to change your response in the next round. • Round 3: State and circle the word to show whether you agree, disagree, or are unsure about the statement now that discussion has ended. Repeat the rounds for as many statements as you can. Activity Synthesis • “Was there a statement that you changed your mind about during your group's discussion? What was the statement? What made you change your mind?” • Consider asking: □ “What statements do you still have questions about?” Lesson Synthesis “Which statement did your group have the most discussion about and why?” (We discussed the idea that one half is always greater than one third the most because some people agreed and some disagreed.) Cool-down: Round Table Reflection (5 minutes)	{"url":"https://curriculum.illustrativemathematics.org/k5/teachers/grade-3/unit-8/lesson-3/lesson.html","timestamp":"2024-11-02T21:35:22Z","content_type":"text/html","content_length":"86961","record_id":"<urn:uuid:e3ca5c8e-2f54-4097-9913-64648bd48efd>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00571.warc.gz"}
(13%) Problem 6: A football player is preparing to punt the ball down (13%) Problem 6: A football player is preparing to punt the ball down the field. He drops the ball from rest and it falls vertically 1.0 m down onto his foot. After he kicks it, the ball leaves the foot with a speed of 16.5 m/s at an angle 56° above the horizontal. @theexperta.com-tracking id: 0W86-9E-D7-41-A23E-47203. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account. ▷ 50% Part (a) What is the magnitude of the impulse delivered to the ball, in kilogram meters per second, as he kicks it? An American football has a mass of 420 J= 5.071 O Degrees tan() 21 ( acos() E 1 * I give up! Feedback: 0% deduction per feedback. Hints: 0% deduction per hint. Hints remaining 2 & 50% Part (b) At what angle, in degrees above the horizontal, is the impulse delivered to the football? All com© 2023 Expert TA. LLC Grade Summary Attempts remaining: 5 (3% per attempt) detailed view Fig: 1	{"url":"https://tutorbin.com/questions-and-answers/13-problem-6-a-football-player-is-preparing-to-punt-the-ball-down-the-field-he-drops-the-ball-from-rest-and-it-falls","timestamp":"2024-11-13T00:58:53Z","content_type":"text/html","content_length":"67099","record_id":"<urn:uuid:56a0c2ae-f354-4902-a125-81e8b58cfcfc>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00042.warc.gz"}
A Factoring Calculator With Steps \| Factor Expressions In general, the first step in factoring any algebraic expression is to determine whether the terms have a common factor. Factoring or factorisisng is the process of spliting an expression into simpler expressions whose product equal to the original expression. Our factoring polynomial calculator can factor any algebraic expressions into a product of simpler factors/ prime factors. The calculator works for any binomials, trinomials, monomials, rational and irrationals. The calculator shows you all the steps by utilizing various techniques such as grouping, quadratic roots formula, difference of 2 squares factoring etc. For Binomials, quadratic, polynomial expressions or number into its prime factors. Enter math expression to find its factors What we can factor? All the following can be factored using our online factoring solver with steps • Numbers i.e. rational or irational, e.g. 15,20,25 • Polynomial => 2x^2-3x+3 (A polynomial is an algebraic expression comprised of variables, constants and exponents combied by adition, multiplication, substructions, no division by varriables). • Binomial=> algebraic expression which contains only two terms with adition or substruction 2x-5: two terms polynomial • Trinomial => 3 term polynomial • Quadratic => Second degree, single varriable polynomial • Rational => Numbers that can be expressed in the form p/q, where p,q are intergers and 2x^2-3x+3q \neq 0 Factoring expression Calculator for Binomials, quadratic, polynomial expressions This calculator will help you factor any algebraic expressions into its factors. The calculator works on any algebraic expression. Factor Expression free online calculator To learn how the factor calculator works, click on the “Try It” button to view the solutions with steps. You can also enter an expression on the input field provided above and click the “factor” button to view the solution. Try typing these expressions into the calculator, click the blue arrow, and select "Factor" to see a demonstration. Alternatively, you can use the examples provided as a template to create and solve your own problems. Problem: 4x^2-9 Solution: (2x+3)(2x-3) How the free factorizing calculator works Factorization is the process of breaking a complex expression into simpler terms. To factor an expression completely, one needs to re-write the expression as a product of its irreducible products. Thus, an expression is factorable if and only if it can be written as a product of two or more factors under multiplication. For example 2x-2= 2(x-1) , in this example, 2 and x-1 are the factors of 2x-2, the factors are prime / irreducible since the cannot be Brocken further into simpler factors with rational More examples Example: x^2+2x-8 Solution: (x-2)(x+4) Example: x^21 Solution: (x-1)(x+1) Example: 2x^2+2xy-8x Solution: 2x(x+y-4) Example: (x-3)(x^2-1) Solution: (x-3)(x+1)(x-1) From the examples above, it is evident that an expression is said to be fully factored, if it can be fully expressed as a product of simple factors. In addition, an expression can be made of any combination of terms and factors. Factor a Number into its prime factors calculator Our factorization calculator helps you re-write any number as a product of its prime factors. Any number can be written down as a product of its prime factors through factorization process. The process involves division iteration with the smallest divisor of a given number. Factorize a rational number into prime factors Any rational number can be factorized into a product of prime factors. Given a rational number q, we can write it as a product of its prime factors e.i q= p0 \times p_1 \times p_2….. \times pn, where p_0,p_1,p_2…. p_n is a finite set of prime factors or divisors of q. Our factor calculator can help you factor any number to its prime factors. To factor a number, simply enter the number in the given text area, click on the factor button to proceed. How the online factor calculator works There are several techniques that are applied when factoring an expression. The technique applied depends on the given expression. However, factoring an expression is not always obvious and not all expressions can be factored. Learning how to factor an expression is a useful technique that is useful in solving or finding the roots of polynomials. Factor an expression by grouping calculator This is one of the fundamental techniques applied in factoring expressions. The method groups terms within an expression by finding the common factors. Example 1: 2y(x+3)+5(x+3) Solution: since (x+3) is a common factor between the terms, the expression can be rewritten as 2y(x + 3) + 5(x + 3) = (x+3)(2y+5) Example 2: factor 2xy-2y^2-2x+y 2(x-y)(y-1)= 2xy-2y^2-2x+y Example 3: factor ax - ay - 2x + 2y Solution: a(x-y)-2(x-y)=(a-2)(x-y) Notice that (a-2) and (x-y) are the factors of ax-ay-2x+2y Factoring trinomial expression A trinomial is an algebraic expression composed of 3 terms which are combined with addition and subtraction. Usually, a trinomial is a product of two binomials. Most factoring problems involve trinomials and can be solved using several methods. Examples of trinomials expressions and how to factor them 1. 2x^2-2x+1 2. x^2+3x-4 The most commonly used method to factor a polynomial ax^2+bx+c with a,b,c \neq 0 is by finding to constants h,k such that h+k=b and h \times k=c \times a. the constants h,k are also the roots of the equation ax^2+bx+c=0 Example: x^2-2x+3 Solution: Using trial and error method, you can find 2 constants 3 and -1. We can then re-write the expression as follows x^2-x+3 = x^2-1 \cdot x+3 \cdot x+3 note that -1+3=2 and -1 \cdot 3=3 \cdot 1 we can now group the expression using parenthesis as follows x^2-1 \cdot x+3 \cdot x+3=(x^2-x)+(3x+3) =x(x-1)+3(x+1) [factoring out the like terms] =(x+3)(x+1) factoring further We now have (x+3)(x+1) as the factors of x^2-2x+3 Factoring binomial expression by difference of 2 squares method A binomial is an algebraic expression that is comprised of two terms. A binomial expression can be factored using various methods. Difference of two squares is the most common strategy that is used to factor a binomial expression. This method will only work if and only if the binomial is made up of two perfect squares separated by a subtraction sign e.g. x^2-4 The general formular for the difference of 2 squares factoring method is a^2-b^2 = (a+b)(a-b), Example: x^2-4 = (x+2)(x-2), notice that x^2 and 4 are perfect squares whose square roots are x and 2 respectively. Factorize a rational expression into its constituent prime factors When factoring an expression, the aim is to reach a prime factorization. By reaching the prime factors means that the expression cannot be factored further. For example given the expression x^3+x^2-x-2 The expression can be factored as follows, x^3+x^2-x-2 = (x^2-1)(x+2) We notice that x^2-1 is a difference of 2 perfect squares and that is not a prime factor since it can be factored further into (x+1)(x-1), thus x^3+x^2-x-2 =(x+1)(x-1)(x+2) is the complete factorization of the expression. Factor polynomial calculator A polynomial is an algebraic expression involving variables, exponents and coefficients with addition and subtraction as the only operations between the terms. While some polynomials can be factored into irreducible/ prime factors, others cannot be factored. For polynomials of degree greater than 2, there is no general factoring procedure for such. A trial and error method is applied in such a case. Our factoring polynomial calculator gives you instant solution or factors for any polynomial that is factorable. This helps you avoid the long procedure of trial and error which may not be a great approach. To factor a polynomial using our factorization calculator, simply enter an algebra expression on the input field provided, hit the factor button to calculate. Create equivalent expressions by factoring calculator Factoring an expression, means creating an equivalent expression or factors of the original expression such that when multiplied/simplified they result into the initial expression. Creating equivalent expression or factoring can be a tedious expression especially when dealing with polynomials of a higher degree. Our factoring calculator lets; you find factors for a given algebraic expression instantly. Factor the trinomial completely calculator A polynomial is said to be completely factored if it is written as a product of prime factors. When factoring a polynomial, the resulting factors are not always irreducible and my need further factoring. Our factoring calculator for polynomial lets’ you factor an expression completely resulting in a prime factorization. The calculator rewrites the expression by factoring out calculator. Factor quadratics special cases calculator When factoring an expression, some special cases arises which can be factored using some special formula. Difference of 2 square formula is an example of a special factoring case. Given an expression of the form a^2-b^2 , where a^2 and b^2 are perfect squares separated by a minus sign, the expression can be factored using the negative square difference formula i.e. a^2-b^2=(a+b)(a-b). This formula can be extended to all algebraic expression of this form. Factoring by grouping solver with steps Factoring an expression by grouping is one of the fundamental factoring techniques. This method groups terms within an expression depending on the similarity. For example the expression 2x+2y-xy-x^2 can be grouped as follows (2x-x^2)+(2y-xy). From the grouping, it is easy to factor the like terms out as follows (2x-x^2 )+(2y-xy)=x(2-x)+y(2-x) , the expression can be factored further (2x-x^2 )+( 2y-xy)=x (2-x) +y(2-x)=(2-x)(x+y)	{"url":"https://www.equationcalc.com/factoring-calculator","timestamp":"2024-11-02T23:34:52Z","content_type":"text/html","content_length":"47614","record_id":"<urn:uuid:b534e47c-6b20-451b-880c-b9d548a2c185>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00816.warc.gz"}
Quantum and classical parallelism in parity algorithms for ensemble quantum computers Quantum and classical parallelism in parity algorithms for ensemble quantum computers Created by W.Langdon from gp-bibliography.bib Revision:1.8010 □ author = "Ralf Stadelhofer and Dieter Suter and Wolfgang Banzhaf", □ title = "Quantum and classical parallelism in parity algorithms for ensemble quantum computers", □ journal = "Physical Review A", □ year = "2005", □ volume = "71", □ number = "3", □ pages = "032345--1--032345--6", □ month = mar, □ keywords = "genetic algorithms, genetic programming, quantum computing", □ ISSN = "2469-9926", □ size = "6 pages", □ abstract = "The determination of the parity of a string of N binary digits is a well-known problem in classical as well as quantum information processing, which can be formulated as an oracle problem. It has been established that quantum algorithms require at least N∕2 oracle calls. We present an algorithm that reaches this lower bound and is also optimal in terms of additional gate operations required. We discuss its application to pure and mixed states. Since it can be applied directly to thermal states, it does not suffer from signal loss associated with pseudo-pure-state preparation. For ensemble quantum computers, the number of oracle calls can be further reduced by a factor 2**k, with k in {1,2,..,log2(N∕2)}, provided the signal-to-noise ratio is sufficiently high. This additional speed-up is linked to (classical) parallelism of the ensemble quantum computer. Experimental realizations are demonstrated on a liquid-state NMR quantum computer.", □ notes = "American Physical Society", Genetic Programming entries for Ralf Stadelhofer Dieter Suter Wolfgang Banzhaf	{"url":"http://gpbib.pmacs.upenn.edu/gp-html/Stadelhofer_2005_PhysRevA.html","timestamp":"2024-11-02T05:51:33Z","content_type":"text/html","content_length":"4398","record_id":"<urn:uuid:9739e644-97df-4805-bc46-504f8b7fa8d9>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00167.warc.gz"}
Module 7: Supply Chain Finance II - Tutgator.com Module 7: Supply Chain Finance II A course where supply chain practitioners will learn how to use the tools of finance to evaluate our supply chain design In this course, we will consider how to use the tools of finance to evaluate our supply chain designs and initiatives. It’s important to use the tools of finance because we want to communicate the value that we are creating, as supply chain professionals, within the firm. And if we use the same tools for evaluating our supply chain designs and initiatives, then our colleagues in finance will understand them better and we’ll be able to communicate the value all the way up to the CFO and the CEO. What you’ll learn • Financial Flows and Evaluation. • Review of Corporate Finance. • Projected Cash Flows. • Relevant Cash Flows. • Free Cash Flows. • Working Capital Cash Flows. • Figures of Merit and Acceptance Criteria. • Time Value of Money and Discount Rate. • Net Present Value. • Internal Rate of Return. • Terminal Value. • Inventory Holding Cost. Course Content • Cash Flows –> 10 lectures • 1hr 44min. • Discounted Cash Flows –> 7 lectures • 1hr 22min. In this course, we will consider how to use the tools of finance to evaluate our supply chain designs and initiatives. It’s important to use the tools of finance because we want to communicate the value that we are creating, as supply chain professionals, within the firm. And if we use the same tools for evaluating our supply chain designs and initiatives, then our colleagues in finance will understand them better and we’ll be able to communicate the value all the way up to the CFO and the CEO. The way that they evaluate decisions, specifically investment decisions within the firm, how to allocate capital to different initiatives, is using a tool called discounted cash flow analysis. It’s a very standard tool used in finance. We’re going to consider how to use it for various supply chain examples and in the way we design our supply chains. So the investment decisions that we can make using this tool can range from actually that the investors in our firm would use discounted cash flow analysis to determine whether to buy our stock. The CEO uses the same tools so that the CEO can make sure that he or she is delivering value to those investors. But you can also use this discounted cash flow analysis in your own finances. For example, when I decided to put solar panels on my roof, I used the same discounted cash flow analysis and it’s worked out well. So let’s think about the investment decisions we’re going to be making, but we have to start by articulating the cash flows that go into those decisions. And we’ll think about different kinds of cash flows. We’ll have projected cash flows. We’ll have what we call free cash flows. It’s not free money, but it’s what’s free for the firm to invest. And then we have to think about what are the relevant cash flows, especially for the supply chain decisions that we’re going to be making. So projected cash flows, we have to think about going forward in time. So we’ll think about period one, period two, period three, to periods often years, one year, two years, three years, and we’ll try to accumulate the cash flows over that year and think about them, one, two, three years out. We also have to consider the upfront cash flows and a lot of times those are our investments. So we’ll call that period zero. At the end of period zero, which is right now, we’ll make those investments. And we want to think about how long that time horizon is. And let’s say in this case, we’re going to go four years. So in the fourth year we’ll call it capital T for that end of period that we’re considering in terms of cash flows. When we think about the cash flows we have to think what goes into those and that’s where we think about our free cash flows here. And here’s a formula for free cash flows that you’ll get more into in the lesson here. But that summarizes the key investments and returns related to our investment decision. So we have capital expenditures. This is what’s very commonly where they will commonly use discounted cash flow analysis in a firm is for capital expenditure decisions and some of you may have gone through that kind of a process. So we’ll call that Capex and we have to think about it time T. A lot of time, again, it’s in period zero, but we could have capital expenditures for fixed assets, like warehouses and trucks over time as well. So that fits into the rate there, Capex t. We’ve got working capital, which is the combination of our receivables, our payables, but most importantly for us in supply chain, inventory. And that investment of working capital could be throughout the period as well. And we want to think about working capital, we always have to have working capital in the firm, but we want to think what is the net working capital and how is it changing over time. So we want to think about the delta, the change in what we call net working capital for each period. And we’ll get into what that is. But we see that down here, too. So in our free cash flows, we’re subtracting out these investments. What are our returns? Well we have revenue, we have expenses. We have to consider depreciation of these capital expenditures and we always have to consider taxes. They can make or break an initiative. Not considering taxes properly can really distort your investment decision. So with revenue and expenses, we will combine those into what’s called EBIT, earnings before interest and taxes, and we’re doing that for every period T. Hopefully, those EBITS will be really positive to offset the investments. We have to consider the depreciation here, which down here is DA, depreciation and amortization. We mostly have depreciation and so that’ll be over time as well. We’ll depreciate those capital investments. And then taxes. We consider taxes by looking at the tax rate and we’ll apply it to our profits, our EBIT, where we subtract depreciation because we get a tax shield for that. But note down here, we have to add it back in. Why is that? Because depreciation is not really cash. We’ve spent the cash up front. We want to account for the taxes, but we don’t want to have it distorting the cash flows because we’re not actually paying out the depreciation over time. So it’s an important adjustment in our free cash flows. And that brings us to this point. I didn’t put an E here on the tax rate, so we’ve got to consider the full tax rate. That brings us to the important point of what’s relevant. Depreciation and amortization is not a relevant cash flow over time. It’s part of, it’s embedded in our capital expenditures, but it’s not a cash inflow from the income statement that we should consider over time. We have to add it back in to make sure we offset that adjustment we make in our EBIT because it’s included in EBIT. So we have to consider a lot of these kinds of decisions of what’s relevant and there’s two principles we’ll use in doing that. There’s a cash flow principle. We have to make sure cash is actually moving. And in the case of depreciation, it’s an accounting number that where cash is not being spent over time. It’s accounting for the upfront cost. And the second principle is with and without. With and without our investment decision, we have to consider the cash flows that are relevant. Once we’ve got our cash flows together, then we move over to think about our investment decision. We have to consider the time value of money because $1 today is worth more than $1 in the future because we can invest it. And there’s less risk of cash flows in the future not happening. So we want to think about the time value of money. And we’ll have what we call a discount rate. We’ll discount those future cash flows according to the returns we can get by investing money today. So the discount rate we call r. And the time value of money for any, for cash in any period T, a cash flow period T, we’d have to divide it by 1 plus r to the power of T, to discount it to be and that’s equal to our cash and present value. So bringing a future cash flow to the present value is a key part of discounted cash flow analysis. Some of these cash flows may not end in period T. They may be what we call a perpetuity, something that would be forever. For example, if you find a new inventory policy to reduce the inventory and it would stay that way as long as you keep the policy in place, that’s a perpetuity. Well, how do we account for that when we only stop at period T? There’s a calculation, we’ll show you how to do that, and we’ll roll it into what we call a terminal value. There are some cash flows that could happen in the future beyond T, we’ll roll them up and account for them in period T. So we have to think about the terminal value and it’s only going to be in period capital T because we’re going to account for all of these future cash flows as a cash flow in period T of the terminal value. So we’ve got our terminal value and we’ve got the account for the future. We’ve got all the cash flows discounted appropriately. What’s our what’s our investment decision? It comes down to figures of merit, calculations we make about the value of this investment, and then an acceptance criteria. We have to have a rule by which we will make our decision. A common one is called payback period. How long will it take for us to pay back the initial investments. That’s a fairly straightforward calculation. It’s important because it gives you a sense of how risky, how long it will take, because those future cash flows are not certain, but it doesn’t account for the time value of money. Two of the approaches that do account for time value of money are called net present value and internal rate of return. The calculation of net present value is taking our initial cash flow and then discounting all the cash flows up to period T, up to our final period here, discounting them as we said up here, and then calculating what is that net present value of all of those cash flows, including those upfront investments. Now the criteria for the net present value we want to have is that this should be greater than zero because our expectations is to get a return of the discount rate. So as long as our NPV is greater than zero, then it’s a good investment. Regarding payback period, it will be a criteria something like less than N periods, whatever N periods is, and that’s something that the firm will determine. Just like the firm will determine what is the rate we expect to get. Finally, the internal rate of return, that’s simply a calculation of the discount rate that gives you an NPV greater than zero. What’s the largest discount rate that gives you that? It gives you a sense of how robust the decision is. So it obviously needs to be greater than r, our expected rate of return or discount rate, but if it’s much greater than r, that means we’re pretty certain that we’re going to do a good job making this investment. So these are all the different principles. Again, I think it’s really critical that we, in supply chain, learn how to use the tools, like discounted cash flow analysis, that our colleagues in finance, the CFO and the CEO, use to make decisions. And in doing that, we will be communicating the real value of our supply chains throughout the organization. As always, use the practice problems and the short questions after the videos throughout, and practice using these tools of finance and, hopefully, you’ll be able to adopt them in your supply chain professional life. Good luck.	{"url":"https://tutgator.com/tutorials/business/module-7-supply-chain-finance-ii/","timestamp":"2024-11-12T22:08:26Z","content_type":"application/xhtml+xml","content_length":"82827","record_id":"<urn:uuid:6bebe7b3-c92c-4234-8f4c-b6e436abbbd1>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00194.warc.gz"}
The Stacks project Lemma 31.15.7. Let $X$ be a locally Noetherian scheme. Let $D \subset X$ be an integral closed subscheme. Assume that 1. $D$ has codimension $1$ in $X$, and 2. $\mathcal{O}_{X, x}$ is a UFD for all $x \in D$. Then $D$ is an effective Cartier divisor. Comments (2) Comment #8518 by Zhenhua Wu on The argument seems to hold in the case of locally Noetherian scheme. Comment #9119 by Stacks project on Thanks and fixed here. Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). All contributions are licensed under the GNU Free Documentation License. In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0AGA. Beware of the difference between the letter 'O' and the digit '0'. The tag you filled in for the captcha is wrong. You need to write 0AGA, in case you are confused.	{"url":"https://stacks.math.columbia.edu/tag/0AGA","timestamp":"2024-11-07T03:29:19Z","content_type":"text/html","content_length":"15485","record_id":"<urn:uuid:dda3a359-0d4e-49e2-8017-3572132c7fb5>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00854.warc.gz"}
C INCLUDE 'model.inc' C ------------------------------------------------------------------ CHARACTER*80 TEXTM SAVE /MODELT/ C .................................................................. INTEGER MSB,MCB PARAMETER (MSB=128) PARAMETER (MCB=128) INTEGER NEXPV,NEXPQ,IVERT,NEGPAR REAL BOUNDM(6) INTEGER NSRFCS,NSB,KSB(0:MSB),NCB,KCB(0:MSB) EQUIVALENCE (KSB(0),NSB),(KCB(0),NCB) SAVE /MODELC/ C ------------------------------------------------------------------ C TEXTM...The name of the model. String of 80 characters. C NEXPV,NEXPQ... Specify exponents of the power of velocities C (NEXPV) and Q-factors (NEXPQ) in input data. For example, C unit values of NEXPV and NEXPQ indicate that the C parameters of the medium are velocities and Q factors, C indices equal -1 indicate reciprocal values of these C quantities, i.e. slownesses and loss factors. C IVERT...Orientation of the vertical axis: C IVERT=0: unknown (default), C IVERT=+1: X1 vertical, pointing upwards, C IVERT=-1: X1 vertical, pointing downwards, C IVERT=+2: X2 vertical, pointing upwards, C IVERT=-2: X2 vertical, pointing downwards, C IVERT=+3: X3 vertical, pointing upwards, C IVERT=-3: X3 vertical, pointing downwards, C Should have no influence on the calculations. If it C is non-zero, it may be considered for plotting purposes. C NEGPAR..Flag whether the negative values of material parameters C are allowed: C NEGPAR=0: Negative values of material parameters or zero C P-wave velocity are reported as errors. C NEGPAR=1: Negative values of material parameters or zero C P-wave velocity are not reported as errors. C BOUNDM..Boundaries X1MIN,X1MAX,X2MIN,X2MAX,X3MIN,X3MAX of the C model. C NSRFCS..Number of smooth surfaces in the model. The surfaces C are indexed sequentially by positive integers, from 1 to C NSRFCS. NSRFCS is the storage location for NSRFC. C NSB... Number of material simple blocks in the model. The blocks C are indexed sequentially by positive integers ISB from 1 C to NSB. Free-space blocks are not indexed. C KSB... Contains the indices of the surfaces bounding individual C simple blocks. KSB(ISB), for ISB = 1 to NSB, specify the C partition of array KSB(NSB+1:NSB+NS) among the simple C blocks. Here NS is the total number of all occurences of C the indices of the surfaces bounding all individual simple C blocks in the input data. The indices of the surfaces C bounding individual simple blocks are stored from C KSB(NSB+1) to KSB(NSB+NS). The locations KSB(NSB+NS+1:MSB) C are undefined. It must be NSB+NS.LT.MSB. The indices of C the surfaces bounding the simple block ISB are stored in C KSB(I1) to KSB(I2), with C I1 = KSB(ISB-1)+1 , C I2 = KSB(ISB) , C where KSB(ISB-1)=NSB for ISB=1. For each simple block C with index ISB, the indices of the surfaces forming the C set F(+) are stored with positive signs, the indices of C surfaces from F(-) with negative signs. For an example C refer to the sample input data for the model. C NCB... Number of material complex blocks in the model. The blocks C are indexed sequentially by positive integers ICB from 1 C to NCB. The free-space blocks are not indexed. C KCB... Contains the indices of the simple blocks forming C individual complex blocks. KCB(ICB), for ICB = 1 to NCB, C specify the partition of array KSB(NCB+1:NCB+NB) among the C complex blocks. Here NB is the total number of all C occurences of the indices of the simple blocks forming C individual complex blocks in the input data. The indices C of the simple blocks forming individual complex blocks are C stored from KSB(NCB+1) to KSB(NCB+NB). The locations C KSB(NCB+NB+1:MCB) are undefined. It must be NCB+NB.LT.MCB. C The indices of the simple blocks forming complex block ICB C are stored in KCB(I1) to KCB(I2), where C I1 = KCB(ICB-1)+1 , C I2 = KCB(ICB) . C Here KCB(ICB-1)=NCB for ICB=1. For an example refer to C the sample input data for the model. C All the input data are stored sequentially in the same order as C they were read. The only exception are locations KSB(1) to C KSB(NSB) and KCB(1) to KCB(NCB) which are inserted when the input C data are being read . The index of the last allocated numeric C storage unit of array KSB is named MSB. The index of the last C allocated numeric storage unit of array KCB is named MCB. The C values of MSB and MCB are given by the sixth and seventh statement C of the block data subroutine MODELB. If the value of MSB or MCB C is changed, it must be adjusted in all subroutines which include C the common block /MODELC/. C Common block /MODELT/ is included in external procedure MODEL1. C of file 'model.for', and in subroutine SECT1 of file 'sec.for'. C Common block /MODELC/ is included in external procedures MODEL1, C BLOCK, ISIDE, INTERF of file 'model.for', in subroutines FUNC and C DISC of file 'sec.for', in program 'intf.for', in subroutine C RAY1 of file 'ray.for' of package CRT, in 'bndlin.for', C and may be included in any other subroutine. C Date: 1999, August 16 C Coded by Ludek Klimes	{"url":"https://seis.karlov.mff.cuni.cz/software/sw3dcd18/model/model.inc","timestamp":"2024-11-01T23:34:42Z","content_type":"text/html","content_length":"6517","record_id":"<urn:uuid:1872ed7e-8295-4754-87d2-7101426dfdee>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00196.warc.gz"}
What is the difference between instrumental variable diagnostics and model diagnostics in econometrics? \| Hire Some To Take My Statistics Exam What is the difference between instrumental variable diagnostics and model diagnostics in econometrics? To wit, instrumental variable measurements are highly correlated, which means your best hypothesis will be your best hypothesis. For example, sometimes a Visit Website perhaps if we want to come up with a useful “model-based” hypothesis (such as a model-based hypothesis that can be statistically calibrated)? In any case, in this article, I want to take you Go Here to thinking about much more important things in your empirical work than did you before. I intend to answer this in the next paragraph as well. The more I digressingly, the deeper my understanding gets toward evaluating these functions, and the greater my understanding of them, the more interesting it becomes. # 4: General Reliability F-Measure _This section focuses on the most critical concepts of instrumental variable diagnostics, but they do exist. For each method, you’ll want to look at the use of the terms _gauge_ and _graceful_ if you want to understand the data. For each method, I’ve written an article on how to use these terms. These are taken from in which they’re explained in detail. # 4 Discussion If you’ve read my previous articles and generalizations, you probably already know how to use instrumental variable diagnostics to assign Learn More statistics. In most calculations for analytical studies, that means studying the time course of the model at its smallest scale—that is, examining only the relevant time-course in a log-log scale. In this article, I’ve given you a starting point, and I’ll give you a starting point for further questions about all the important models and data points in my empirical work. As an example, this section looks up what are our _functional distributions_ news _functional hyperparameters_ and what are our useful _momentum distributions_What is the difference between instrumental variable diagnostics and model diagnostics in econometrics? A: I can see that there are many practical differences between both, but, based on my understanding, yes, I think instrumental variables are highly correlated with each other, so that there is a tendency in many cases (some so-called “separation”) not to associate most variables with one another. As mentioned, to be sure either use the Eq.3 of the question, or use Eq.6, which is rather directly related to the second paper, since in the former the first equation looks like a more complex equation and as for the latter, this means that the equations themselves are very simple, but is that the same? If so, why not use Eq.3 Eq.9, where Eq. Hire Someone To Take Online Class 3 means using the first equation instead of Eq.3 and replacing each of the equations with a particular formula: $$x^2_i [A+C]+x^2_j [B+C] $$ which is a simple algebra to solve. Or if one uses Eq.5 again to analyze the relationship between the variables themselves (we have set the variable A‒M‒M* to start with, and M by M+a, and A+M by +a, for C and M to start with), the Eq.6 uses a simple algebra: $$y^2_i [A+C]+x^2_j [B+C]) $$ and when using Eq.5, though Eq.3 only looks as for the full dataset, you could use Eq.21Eq.27Eq.28 under similar circumstances: $$b^3_i(x,y)=y^2_i[A+M+C]+y^2_j [B+M]+x^3_i (x,y) $$ if you need a little more to separate the variables than just BWhat is the difference between instrumental variable diagnostics and model diagnostics in econometrics? Does it describe a standard model of can someone do my statistics examination action that can be calibrated? Does something on the analysis be calibrated? The big question is this: what best conforms to an instrument, but if that instrument is used to answer diagnostic diagnostics, what comes out most is a calibrated instrument. What I would argue is here is that the data within the model (not the instrumental response) (or as a result not a single you can try here (fractional model)) should come out as best fit to the real data, but that the parametric model fits the instrument better to the wikipedia reference 1. My problem is that I don’t believe that a simple model can reproduce the data on only two inputs. That is more the question mark either which best fits a particular parameter to that parameter, or in which I’m at the right amount to ask Extra resources my data are more “standard” to both instruments on what criteria I want to know if there is a simple solution (or if I just don’t see it I don’t call it a good fit) between the standard model and the instrument. 2. I’m pretty disappointed that model diagnostics came out as best fit to the data, and that those models come out which use more data than first models. Those Read Full Article modelling are just models having very similar behavior when it comes to measurements (how do you fit a model to measurements to figure out which parameter is important?). 1. In my discussion on the different types of model this particular problem has been raised a few times (I’ve just solved the paper, and also looked at the paper and find) but no one has so good a test–to me this is the most important issue, I have to question why are these techniques used over and over no matter how the method was originally tried and how many changes in the data in the last decade? 2. If the problem lays	{"url":"https://hireforstatisticsexam.com/what-is-the-difference-between-instrumental-variable-diagnostics-and-model-diagnostics-in-econometrics-2","timestamp":"2024-11-07T02:30:47Z","content_type":"text/html","content_length":"169223","record_id":"<urn:uuid:31004fd0-3aea-429e-b018-e63ada70857c>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00806.warc.gz"}
An Efficient Full-Band Sliding DFT Spectrum Analyzer - Rick Lyons An Efficient Full-Band Sliding DFT Spectrum Analyzer In this blog I present two computationally efficient full-band discrete Fourier transform (DFT) networks that compute the 0th bin and all the positive-frequency bin outputs for an N-point DFT in real-time on a sample-by-sample basis. An Even-N Spectrum Analyzer The full-band sliding DFT (SDFT) spectrum analyzer network, where the DFT size N is an even integer, is shown in Figure 1(a). The x[n] input sequence is restricted to be real-only valued samples. Notice that the only real parts of the delay elements’ complex-valued outputs are used to compute the feedback P[n] samples. The Figure 1(a) network’s complex coefficients are defined by $e^{j2{\pi}k/N}$ where integer k defines the DFT bin index and is in the range 0 ≤ k ≤ N/2. This even-N network has N/2+1 parallel paths that compute the complex-valued X[0], X[1], X[2],..., X[N][/2] DFT bin output samples. The multiply by 1/2 can be implemented with a simple binary right shift by one bit. The Figure 1(a) even-N network was inspired by the Figure 4 network in Reference [1]. However, for real-valued input samples I’ve simplified the Reference [1] network (which has N parallel paths) and reduced its computational complexity to develop the above proposed Figure 1(a) network. An Odd-N Spectrum Analyzer To compute real-time positive-frequency odd-N-point DFT spectral samples we use the network shown in Figure 1(b). This odd-N network has (N+1)/2 parallel paths that compute the X[0], X[1], X[2],..., X[(N-1)/2] DFT bin output samples. Again the x[n] input sequence is restricted to be real-only valued samples. The Figure 1(b) network’s complex coefficients are defined by $e^{j2{\pi}k/N}$ where integer k is 0 ≤ k ≤ (N-1)/2. Two surprising properties of both Figure 1 networks are: 1) although they use recursive complex multiplications the networks are guaranteed stable, and 2) the networks compute sliding DFT samples without a comb delay-line section required by traditional sliding DFT networks! (The Reference [1] network also has those two noteworthy properties.) Both Figure 1 networks are implemented as follows: 1) Set the outputs of the delay elements and the P[0] feedback sample to zero. 2) Accept the first x[0] input sample and compute the X[k][0] output samples for each parallel path. 3) Shift the data through the delay elements and compute the P[1] feedback sample. 4) Accept the second input sample and compute the X[k][1] output samples for each parallel path. 5) Shift the data through the delay elements and compute the new P[2] feedback sample. 6) And so on. Computational Complexity Comparison In case you’re interested in the computational workload of the Reference [1] network and our two proposed Figure 1 networks, have a look at Table 1. I presented computationally efficient full-band sliding DFT networks that compute the 0th bin and all the positive-frequency DFT bin outputs for both even- and odd-N-point DFTs in real-time on a sample-by-sample basis. (No taxpayer money was used and no animals were injured in the creation of this blog.) [1] Z. Kollar, F. Plesnik, and S. Trumpf, “Observer-Based Recursive Sliding Discrete Fourier Transform,” IEEE Signal Proc. Mag., Vol. 35, No. 6, pp. 100-106, Nov. 2018. [ - ] Comment by ●April 9, 2021 I notice this is posted on April 1. Am I foolish to read it? John P [ - ] Comment by ●April 9, 2021 Hi JProvidenza. Ah ha. Such an interesting question. The answer is no, you would not be foolish to read it. (Perhaps I should have paid more attention to the date and posted this blog on April 2nd.) [ - ] Comment by ●April 11, 2021 Rick - I couldn't resist the comment - it was too sweet a date! Thanks for all your great posts. I have a couple of your books that I've used for study - my background is not DSP, but it's great to stretch my brain on topics I was not introduced to in college decades ago. John P [ - ] Comment by ●April 13, 2021 Rick - Could you add a little info on how to interpret the output? John P [ - ] Comment by ●April 14, 2021 Hi JProvidenza. Well, ...the traditional sliding DFT, shown in the following diagram, is used to compute the real-time (one output sample for each new input sample) kth bin output of a N-point DFT. (The X_k[n] output is the kth bin output of a N-point DFT for the current input sample and the previous N-1 input samples.) The typical application of the traditional sliding DFT is to detect the presence, or absence, of spectral energy centered at the kth bin of an N-point DFT in real time. If you wanted to simultaneously compute the k = 3, k = 5, and k = 7 bins’ outputs you would use the following network. For real-valued x[n] input sequences, my blog’s Figure 1 networks are used to compute the 0th bin and all the positive-frequency bin outputs of an N-point DFT in real-time (on a sample-by-sample basis). Notice in my blog’s Figure 1 that, unlike the traditional sliding DFT, no N-length delay line stage is needed. I hope what I've written here makes sense. [ - ] Comment by ●November 5, 2021 Hi Rick, I am not sure if my question is related to this exact article. Could the sliding DFT be used to extract different bins of an OFDM signal which is not really like a sine wave. Also is it applicable on parallel samples? [ - ] Comment by ●November 5, 2021 Hello tushar_tyagi90 This blog presents an efficient way to compute the N complex output bin samples of an N-point DFT in real time. By “real time” I mean that that for each data input sample all N DFT bin complex spectral sample values are computed. The output samples of this scheme are not valid until the Nth input sample has been applied to the network. [The traditional DFT (or radix-2 FFT) provides outputs only for every N input samples--it’s not “real-time”.] The networks in this blog's Figure 1 are only valid for real-valued input samples. A sliding DFT network, on the other hand, computes a single complex output bin spectral sample of an N-point DFT in real time (one complex output spectral sample for each input sample). I have no experience in OFDM processing, but if your OFDM processing requires you to compute all N output spectral samples of an N-point DFT in real time then one of the networks in this blog’s Figure 1 should achieve your goal. If your OFDM processing requires you to compute, say 8, real time N-point DFT spectral samples (where N is greater than 8) then you merely need to implement 8 sliding DFT networks. That is, one comb section driving 8 separate resonator sections. I must tell you, “traditional” sliding DFT networks can experience long-term stability problems. Several guaranteed-stable sliding DFT networks have been proposed in the literature but they’re more computationally intensive than the traditional sliding DFT. Last year I developed a guaranteed-stable sliding DFT network that is significantly more computationally efficient than the previously proposed guaranteed-stable sliding DFT networks. My “new and improved” guaranteed-stable sliding DFT is described at: "Improvements to the Sliding DFT Algorithm", DSP Tips & Tricks column, IEEE Signal Processing Magazine, Vol. 38, No. 4, July 2021. I intend to write a blog at this dsprelated.com web site describing my new and improved sliding DFT network as soon as I have the time. tushar_tyagi90, if you have any further questions for me please let me know. To post reply to a comment, click on the 'reply' button attached to each comment. To post a new comment (not a reply to a comment) check out the 'Write a Comment' tab at the top of the comments. Please login (on the right) if you already have an account on this platform. Otherwise, please use this form to register (free) an join one of the largest online community for Electrical/Embedded/DSP/FPGA/ML engineers:	{"url":"https://dsprelated.com/showarticle/1396.php","timestamp":"2024-11-06T23:43:32Z","content_type":"text/html","content_length":"82846","record_id":"<urn:uuid:ea8a94dd-b64d-4885-ae30-03aa3944b27d>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00237.warc.gz"}
Most Decorated Sportsmen from your country I am not talking about the 'Greatest' of them all but about those players who may or may not be the most talented of the lot, they may or may have not won much and don't have much achievement to their name. But the way they played the game(including controversies),the legacy they left behind,the influence they had on the sport and on generation after generation, made them eternal. Who dou you think are these players from your country from any sport? Sign up to remove this ad. Donald Bradman. Peerless cricketer Rod Laver. One of the great tennis players. Idol of Federer. Godfather of tennis. Greg Norman. Johnny Warren. Captain Socceroo. Instrumental in getting football taken seriously in Australia. Not a world best but hugely important to Australian football. Tim Cahill (biased but i dont care) Andrew Gaze. The Michael Jordan of Australian basketball. Johan Cruyff - Changed the game of football to what we see today. Fanny Blankers-Koen - Best Dutch women's Athlete of all time. Raymond Van Barneveld - started the Darts revolution in Holland, there wouldn't be a Michael Van Gerwen without him. 18 minutes ago, Harry said: Donald Bradman. Peerless cricketer Rod Laver. One of the great tennis players. Idol of Federer. Godfather of tennis. Greg Norman. Johnny Warren. Captain Socceroo. Instrumental in getting football taken seriously in Australia. Not a world best but hugely important to Australian football. Tim Cahill (biased but i dont care) Andrew Gaze. The Michael Jordan of Australian basketball. No Dennis Lillie? • Subscriber PT Usha - Track athelete ahead of her time in India Sachin Tendulkar - You don't really need to say anything more Leander Paes / Mahesh Bhupathi - One weened off a bit but Leander Paes imo is one of the best if not the best versatile doubles partner in tennis. That run with Hingis is astounding. 8 minutes ago, Mel81x said: PT Usha - Track athelete ahead of her time in India Sachin Tendulkar - You don't really need to say anything more Leander Paes / Mahesh Bhupathi - One weened off a bit but Leander Paes imo is one of the best if not the best versatile doubles partner in tennis. That run with Hingis is astounding. Dhayan Chand? • Subscriber Just now, Azeem98 said: Dhayan Chand? Don't really follow field hockey that much but you're right he's a legend in his own right. Its not easy to do what he did and I have family that have won olympic gold in field hockey but sadly I have never taken to the sport. Do you watch your NT play? I try and catch as much as I can. 3 minutes ago, Mel81x said: Don't really follow field hockey that much but you're right he's a legend in his own right. Its not easy to do what he did and I have family that have won olympic gold in field hockey but sadly I have never taken to the sport. Do you watch your NT play? I try and catch as much as I can. Favourite sport after football but i hate what they have done to the sport, purple turf,yellow ball,quaters instead of halves,penalty shoot has changed. I liked it when it was more similar to football it gave a traditional look to the sport now it looks like an American highschool sport to me.But still follow it. • Subscriber Just now, Azeem98 said: Favourite sport after football but i hate what they have done to the sport, purple turf,yellow ball,quaters instead of halves,penalty shoot has changed. I liked it when it was more similar to football it gave a traditional look to the sport now it looks like an American highschool sport to me.But still follow it. You'd laugh but in my 11/12th the biggest sport was field hockey and omg was it terrible when we played other schools. These days it seems more like its catered towards the person watching the TV more than the players on the pitch. I even skipped the last time we played Pakistan because it wasn't worth it. • Subscriber I'd almost like to even give props to our badminton women these days but we shall see how far they go. PV Sindhu might have a gold medal but Saina imo is miles better than her. In football we have a lot of players with a lot more domestic success but in terms of international impact it's by far Kenny Dalglish - 3 European cups and a Ballon d'or runner-up. Off the top of my head elsewhere, Chris Hoy and Andy Murray come to mind. Actually, do managers count as sportsmen? • Subscriber 1 minute ago, Inverted said: Actually, do managers count as sportsmen? If they did good for the sport I don't see why they can't be included Dhyan Chand is a good shout, even though like Mel I don't follow hockey. Saina and PV Sindhu will be huge in coming years. Mary Kom for breaking the barriers. Kapil Dev is big too for the first WC. But, yeah, no one comes close to Sachin. For four decades he was, rightfully, the centre of everyone's attention and someone who brought the nation together. • Subscriber 1 minute ago, Azeem98 said: This man is a machine. I remember going through so many grades in school and seeing this guy win game after game. Who is your current squash champ? • Subscriber 2 minutes ago, IgnisExcubitor said: Dhyan Chand is a good shout, even though like Mel I don't follow hockey. Saina and PV Sindhu will be huge in coming years. Mary Kom for breaking the barriers. Kapil Dev is big too for the first WC. But, yeah, no one comes close to Sachin. For four decades he was, rightfully, the centre of everyone's attention and someone who brought the nation together. I always forget Mary but shes big too. P.S. That movie was shit 8 minutes ago, Mel81x said: If they did good for the sport I don't see why they can't be included Well, in that case we've got Shankly, Fergie, Busby and Stein, ofc. 3 minutes ago, Mel81x said: This man is a machine. I remember going through so many grades in school and seeing this guy win game after game. Who is your current squash champ? Aamir Atlas Khan his nephew • Subscriber Just now, Azeem98 said: Aamir Atlas Khan his nephew Is squash big up there? I tried playing it and its horrible for me and I respect it more as a result. • Subscriber 3 minutes ago, Inverted said: Well, in that case we've got Shankly, Fergie, Busby and Stein, ofc. Surely McRae is in there too? Obviously not for the sport you're mentioning for the other names 4 minutes ago, Mel81x said: Is squash big up there? I tried playing it and its horrible for me and I respect it more as a result. Now only popluar among certain groups and regions but it will change only a major tournament victory is needed to bring the sport back to life. 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Irr rate example Internal rate of return (IRR) is the discount rate at which the net present value of an investment is zero. IRR is one of the most popular capital budgeting technique. Projects with an IRR higher than the hurdle rate should be accepted. In the example below, an initial investment of $50 has a 22% IRR. That is equal to earning a 22% compound annual growth rate. Internal Rate of Return (IRR) Example: Sam is going to start a small bakery! Sam estimates all the costs and earnings for the next 2 years, and calculates the Net Present Value: At 6% Sam gets 6 Jun 2019 Internal rate of return (IRR) is the interest rate at which the net present value of all the cash flows (both positive and negative) from a project or Internal rate of return (IRR) is the minimum discount rate that management uses to identify what capital investments or future projects will yield an acceptable Guide to what is Internal Rate of Return along with practical examples with IRR calculation in Excel. Here we also discuss significance & drawbacks of IRR. It is the discount rate at which the present value of a project's net cash inflows becomes equal to the present value of its net cash outflows. In other words, internal Internal rate of return (IRR) is the minimum discount rate that management uses to identify what capital investments or future projects will yield an acceptable Internal rate of return (IRR) is the minimum discount rate that management uses to identify what capital investments or future projects will yield an acceptable Guide to what is Internal Rate of Return along with practical examples with IRR calculation in Excel. Here we also discuss significance & drawbacks of IRR. It is the discount rate at which the present value of a project's net cash inflows becomes equal to the present value of its net cash outflows. In other words, internal Internal Rate of Return for periodic input values. Notes. The IRR is perhaps best understood through an example (illustrated using np.irr in the Examples section The internal rate of return formula is calculated by subtracting the initial cash investment from the sum of all future cash flow of the investment after a discount The rate of return calculated by IRR is the interest rate corresponding to a 0 (zero ) net present value. The following formula demonstrates how NPV and IRR are 26 Nov 2019 IRR Calculation. Consider as an example, the following cash flow diagram. At the start of year 1 (today) there is a cash out flow of 2,000 Spot rates (the current in- terest rate appropriate for discounting a cash flow of some given maturity) and forward interest rates are com- puted via the IRR equation. The calculation of Internal Rate of Return can be done as follows- The cash flows of the project are as per below table: Since the IRR for this project gives two values: -6% & 38% it is difficult to evaluate the project using this method as it is unclear as to which IRR should be considered. Internal rate of return (IRR) is the interest rate at which the net present value of all the cash flows (both positive and negative) from a project or investment equal zero. Internal rate of return is used to evaluate the attractiveness of a project or investment. If the IRR of a new project exceeds a company’s required rate of return, that project is desirable. At 10% interest rate NPV = -$3.48. So the Internal Rate of Return is about 10%. And so the other investment (where the IRR was 12.4%) is better. 3 Sep 2019 The internal rate of return (IRR) shows investors how they can expect to For example, let's say you purchase a commercial office building to Example: Sam is going to start a small bakery! Sam estimates all the costs and earnings for the next 2 years, and calculates the Net Present Value: At 6% Sam gets 6 Jun 2019 Internal rate of return (IRR) is the interest rate at which the net present value of all the cash flows (both positive and negative) from a project or Internal rate of return (IRR) is the minimum discount rate that management uses to identify what capital investments or future projects will yield an acceptable Guide to what is Internal Rate of Return along with practical examples with IRR calculation in Excel. Here we also discuss significance & drawbacks of IRR. It is the discount rate at which the present value of a project's net cash inflows becomes equal to the present value of its net cash outflows. In other words, internal Internal Rate of Return for periodic input values. Notes. The IRR is perhaps best understood through an example (illustrated using np.irr in the Examples section The internal rate of return formula is calculated by subtracting the initial cash investment from the sum of all future cash flow of the investment after a discount The internal rate of return (IRR) is a measure of an investment's rate of return. The term internal refers to the fact that the calculation The internal rate of return is the discount rate that makes the net present value equal to zero. Simple IRR example For example, project A requires an initial investment of $100 (cell B5). The internal rate of return we get is 14%. Example 2. Let’s calculate the CAGR using IRR. Suppose we are given the following information: The IRR function is not exactly designed for calculating compound growth rate, so we need to reshape the original data in this way: The internal rate of return sometime known as yield on project is the rate at which an investment project promises to generate a return during its useful life. It is the discount rate at which the present value of a project’s net cash inflows becomes equal to the present value of its net cash outflows. The calculation of Internal Rate of Return can be done as follows- The cash flows of the project are as per below table: Since the IRR for this project gives two values: -6% & 38% it is difficult to evaluate the project using this method as it is unclear as to which IRR should be considered. Internal rate of return (IRR) is the interest rate at which the net present value of all the cash flows (both positive and negative) from a project or investment equal zero. Internal rate of return is used to evaluate the attractiveness of a project or investment. If the IRR of a new project exceeds a company’s required rate of return, that project is desirable. At 10% interest rate NPV = -$3.48. So the Internal Rate of Return is about 10%. And so the other investment (where the IRR was 12.4%) is better. Internal rate of return (IRR) is the minimum discount rate that management uses to identify what capital investments or future projects will yield an acceptable return and be worth pursuing. The IRR for a specific project is the rate that equates the net present value of future cash flows from the project to zero. The Internal Rate of Return (IRR) is the discount rate that makes the net present value (NPV) of a project zero. In other words, it is the expected compound annual rate of return that will be earned on a project or investment. In the example below, an initial investment of $50 has a 22% IRR.	{"url":"https://optioneepxmkvu.netlify.app/peterman87202kuny/irr-rate-example-mus.html","timestamp":"2024-11-11T17:04:44Z","content_type":"text/html","content_length":"36816","record_id":"<urn:uuid:a1eb354b-2486-4882-968b-7119ff281c38>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00880.warc.gz"}
2 capicitor wire s habing diameter 3.0mm and T having diameter 1.5mm are connected in parallel a potential difference is applied across the ends of parallel arrangement the value of ratio of current T to current S is? \| Socratic 2 capicitor wire s habing diameter 3.0mm and T having diameter 1.5mm are connected in parallel a potential difference is applied across the ends of parallel arrangement the value of ratio of current T to current S is? 1 Answer If I interpreted the question as intended, ${I}_{S} / {I}_{T} = \frac{\frac{V}{R} _ S}{\frac{V}{R} _ T} = {R}_{T} / {R}_{S} = 4$ I am not clear on first part of the description. I will assume: 2 wires of equal length, S having diameter 3.0mm and T having diameter 1.5mm are connected in parallel ... Resistance of wire is inversely proportional to cross section area. (The full formula for resistance of a length of wire is #R = rho*l/A)#. The properties $\rho \mathmr{and} l$ in both the wires are equal. Therefore the ratio of their resistances ${R}_{S} / {R}_{T}$ will simplify down to ${R}_{S} / {R}_{T} = \frac{\frac{1}{1.5 m m} ^ 2}{\frac{1}{0.75 m m} ^ 2} = \frac{1}{2} ^ 2 = \frac{1}{4}$ Considering Ohm's Law, the ratio of the currents ${I}_{S} / {I}_{T} = \frac{\frac{V}{R} _ S}{\frac{V}{R} _ T} = {R}_{T} / {R}_{S} = 4$ I hope this helps, Impact of this question 830 views around the world	{"url":"https://api-project-1022638073839.appspot.com/questions/2-capicitor-wire-s-habing-diameter-3-0mm-and-t-having-diameter-1-5mm-are-connect#636519","timestamp":"2024-11-10T17:59:06Z","content_type":"text/html","content_length":"33518","record_id":"<urn:uuid:d5fb0381-da28-4e1f-afe1-b5b32fee928b>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.61/warc/CC-MAIN-20241110170046-20241110200046-00254.warc.gz"}
10 Pips a Day Forex Compounding Plan to $57.665,04 - Get Know Trading Forex Compounding Plan Forex compounding plan represents steps to use compounding techniques by reinvesting fixed percentage of profit on initial investment and increasing the profits exponentially. Steps of Forex compounding plan include these points: Step #1 – define initial investment as account balance Step #2 – define percentage growth per time period Step #3 – define time period which you will use in compounding plan Step #4 – define Forex compounding trading strategy Forex Compounding Plan Investment Forex compounding plan includes investment you will bring to Forex trading which you will grow. Growing that investment will be by the compounding calculation which you can check in the Forex compounding calculator. The investment in a compounding plan can start from $1000 or $10 000. That depends on your ability to invest money. If you do not have money then you can try a compounding plan with a demo account. You can open demo trading account and test compounding plans with virtual money. If you have money then you can start Forex trading with real money. If you can invest $1000 you can check what you could expect to make after a certain period. Below is a graph where I have put $1000 investment through 30 time periods. The return on $1000 is calculated with compounding calculator and I have used a 10% return. Forex Compounding Plan Percentage Return When you have defined the initial investment it is time to define which percentage you will try to reach on each time period. In the graph below I have put three different percentage return so you can see what you would get if you decide to have: 1. 1% return – grey line 2. 5% return – brown line 3. 10% return – blue line You can see that 10% has exponential growth after 30 time periods. 1% and 5% have small growth which is not so attractive. But it is conservative return which some traders like to see. If you imagine that you invest $1000 000 and put 5% return you would get 4x times after 30 time periods. Here are the results from the graph below: 1% return – $1.437,85 5% return – $4.321,94 10% return – $17.449,40 You can see that 10% return has exponential growth with a return of 1744%. 5% return has growth of 432%. 1% return has growth of 143% Forex Compounding Plan Time Period This is one of the most important parts you will need to decide what to use. The compounding time period defines how long you need to wait to get the desired outcome. I have used a time period of 30 something. Something here can be trades, days, weeks, months or years. Lets check what would mean if I use 30 trades as a time period for compounding plan. Trades as a Time Period in Compounding Plan 30 trades means you would open 30 trades with profit and at the end you would end up with money on your account. Those 30 trades can be in a day, or a week. That depends on you how fast you can achieve 30 trades in a row. Here is the condition to have 30 trades in a row profitable, not one losing. Which is less likely to achieve because if you can do 30 trades in row with profit with 10% return you would not be here reading this. So, with 30 trades in row without losing trade you would end up with $17.449,40 on your account if you use 10% profit on each trade and with $1000 initial investment. 30 trades x 10% profit = $17.449,40 Days as a Time Period in Compounding Plan If you use days as a time period in your compounding plan then you would try to achieve results in 30 day time. That means each day you would need to have 10% of profit. In one day you could have more trades open, 1, 2, 3 or more, but at the end of the day you would need to have 10% of profit. If you do not have 10% of profit each day you would not be able to achieve results from the graph. Day as a time period gives you more freedom because you can open more trades per day and make a strategy to reach those 10%. Check this – if you have 10 trades per day and each trade you open has a risk of 2%. But on each trade you try to get 4% of profit or more. That is Risk:Reward = 1:2. Let’s say you use 4% of profit and 2% of risk. That means if you open a trade and that trade becomes bad trade you would end up with 2% loss. 2% risk on bad trade = 2% of $10 000 = $200 Then you open a second trade and that trade becomes profitable. You end up with 4% of profit. 4% profit = 4% of $10 000 = $400 Lets say you have 5 bad trades and 5 good trades. You would end a day with: 5 x 2% risk = 10% loss 5 x 4% profit = 20% profit TOTAL = 20% profit – 10% loss = 10% profit You see, with Risk:Reward = 1:2 you would need 5 good trades with 5 bad trades to reach daily target of 10% of profit. This gives you a lot of space to make errors in trading. Compared to 30 good trades in a row this time period makes much more sense and makes it more reachable. Week as a Time Period in Compounding Plan If you have a week as a time period then you would have more freedom to make more bad trades because you could open more trades per week. The goal is to reach 10% of profit per week. And then you would need 30 weeks to reach the target from the graph. 30 weeks is equal to 7 and half months where you would make $17.449,40. Now it is up to you to decide which time period you would use. NOTE – I will not cover the monthly and yearly time period. The analogy is the same as described for a week time period. If you use a monthly time period you would need 30 months to reach the target. If you use a yearly time period you would need 30 years to reach the target. I assume that would be too long to wait, but it is up to you to decide. Compounding Plan Strategy As a final step you need to decide how to use all steps explained before to start trading by using a Forex compounding plan. Strategy should include these steps: Step #1 – define currency pair for trading Step #2 – define lot size you will use Step #3 – define stop loss and take profit levels Step #4 – define risk per each trade, Risk:Reward Step #5 – define time period you will use in Forex compounding plan Lets now check each step so you know how to use this compounding strategy in Forex trading. You can use this strategy if you want, but first test it on a demo account before trying on a live account. Currency Pair for Compounding Strategy Plan If you are beginner in Forex trading then you should learn how to choose currency pair for trading. Each currency pair has its own characteristics and you need to select those that are good for your trading strategy. In this article I am not showing which trading strategy to use. You can use technical analysis, fundamental analysis, price action analysis or any other. But, one of the currency pairs that are good to trade at the beginning is the EUR/USD currency pair. It has moderate volatility and you can see trends which are good to follow to make money and to make less mistakes. When you have a currency pair for trading you can move to the next step and that is to define the volume or lot size you will use in each trade. Volume or Lot Size in Compounding Trading Plan Volume or Lot size defines how much each pip will be worth. Is one pip equal to $0.1, $1 or $10 or more. If you want, you can learn what pip is, and how to calculate the pip on this website. If you use EUR/USD currency pairs then you do not have problems understanding the lot size for each pip. If you use micro lot size you will have 1 pip equal to $0.1. 1 micro lot for 1 pip = $0.1 Here is a table with other lot sizes so you know which one to use in trading. Remember that lot size should be defined with the risk you will use on each trade and how many pips you will go for in each trade. I will explain this later with an example. Define Stop Loss and Take Profit To define stop loss and take profit you need to have a trading strategy. Strategy defines entry and exit levels for each trade you open. In the Forex trading platform you have two fields where stop loss and take profit are defined. Check the image below. Risk:Reward In Compounding Plan Strategy Now you have come to one of the most important steps in compounding plan strategy. And that is to define Risk:Reward ratio. Risk to reward ratio defines how much you accept to lose per each trade and how much you will make per each trade. In percentage. If you define Risk:Reward as 1:2 with 1% risk that means you plan to lose 1% on that trade if the trade is a bad trade. 1% loss as a risk per each trade If that trade is positive trade then you would make 2% of profit on that trade. 2% profit per each trade You should plan to have at least R:R = 1:2 as a minimum. If you put R:R = 1:3 that is even better. Because each positive trade will give you 3% of profit. Lets see two examples where one trade is bad trade and one is positive trade with R:R = 1:3. First trade on $10 000 account: 1% loss = $100 loss Second trade is positive with 3% profit: 3% profit = $300 profit After two trade with one bad and positive you end up with $200 profit. TOTAL = $300 – $100 = $200 profit That means you can lose two more trades in a row to be equal. To have $10 000 on your account just like when you have started. Time Period in Forex Compounding Plan Now it is time to define the time period. Will you watch each trade as a time period or you will set a time period to daily. Daily time period means you need to have a return at the end of the day defined by percentage. If you have set 10% as a target per day then you need to finish the trading day with 10% of profit to follow Forex compounding plan. You can have several trades per day and they can be bad trades and there can be good trades. Final result should be to have 10% of profit. If you select a weekly time period, which is also ok, you would need to finish the week with 10% of profit. This gives you more room to open more trades. You have five days to get 10% of profit. You can reach 10% in one trade and wait the rest of the week because you have reached your weekly target. That way you can rest and prepare for the next week. It is important not to be greedy and trade just to trade to avoid becoming bored. I would use a weekly time period because it gives more freedom and the pressure is not too strong. Pressure can lead to overtrading which at the end gives you account wipeout. With a weekly time period you can open one trade and let it run for the whole week. Now it is time to show you an example with all steps explained now. So you can see how that looks. 10 Pips a Day Compounding Strategy So, in this example I want to show you how it looks when you use a 10 pips a day compounding strategy where you define lot size, risk:reward ratio and then open the trade. You will see how much you can make after a certain number of days by using a 10 pips a day target. The profit each day you make with 10 pips you will add to the initial balance and then calculate next day return. I will use 1% of profit per day as a target. So, here is summary what I will have: Step #1 – Currency pair EUR/USD Step #2 – Lot size – I will show this below Step #3 – stop loss and take profit – check below Step #4 – Risk per each trade – 1% Step #5 – time period is daily time period While I have a currency pair and that is EUR/USD I know how much is micro lot per each pip. 1 pip on micro lot = $0.1 Account balance is $1000. So I am targeting 1% which is: 1% of $1000 = $10 Now I will calculate the lot size I will use to have 1% risk on each trade. Because Risk:Reward is set to 1:3 I will target 3% of profit on each trade. 1% risk of $1000 = $10 If I am targeting 10 pips on each trade I will need only one trade positive. To get $10 target which is 1% of $1000 I will need to have: 10 pips = $10 1 pip = $1 That is equal to one mini lot size. That means I need to open an order with 0.1 lot size in the Forex trading platform. Day 1 – 10 Pips a Day Compounding Trading Plan When I open a new order with one mini lot size with stop loss of 10 pips I will risk $10 per trade. But, I have put the Risk:Reward ratio to 1:3 which means I am targeting 30 pips with each trade. So, If I open three trades per day I can have 2 bad trades and one positive trades to reach the daily target of 10 pips. 1st trade – 10 pips loss 2nd trade – 10 pips loss 3rd trade – 30 pips profit TOTAL = -10 -10 + 30 = 10 pips With 10 pips of profit I have earned 10 pips x $1 = $10 When I add profit to account balance I will have $1010. This amount will be used to open new trades the next day. Day 2 – 10 Pips a Day Compounding Trading Plan Now I will use the same strategy to open new trades. The goal is to reach 10 pips of profit which will give me 1% of profit. 1% profit of $1010 = $10,1 Calculate the Lot Size Now I will need to calculate the lot size to get $10,10 of profit. This time I will not use 1 mini lot size which is equal to 0.1. Lets calculate the lot size. 10,10 / 10 pips = 1,010 per pip 1,010 per pip = 0,1010 lot size This is one mini lot(0,1) + one nano lot(0,01). If you are using the Metatrader 4 trading platform then you will not be able to use 0,1010 because MT4 accepts only micro lot size as the smallest unit If you use an Oanda broker then you can use nano lot size 0,001. Oanda uses terminology of units instead of lot size. So, 1 lot is 100 000 units. 1 mini lot which is 0.1 is equal to 10 000 units. 1 nano lot which is 0.001 is equal to 100 units. So, if I open three trades as on the first day where I have 2 bad trades and one positive trade I would get this: 1st trade – 10 pips loss 2nd trade – 10 pips loss 3rd trade – 30 pips profit TOTAL = -10 -10 + 30 = 10 pips With 10 pips of profit I have earned 10 pips x $1,01 = $10,10 Now, when I add this amount to the account balance from day before which was $1010 I will get new account balance: $1010 + $10,10 = $1020,10 Day 3 – 10 Pips a Day Compounding Trading Plan Now I will use the same strategy to open new trades. The goal is to reach 10 pips of profit which will give me 1% of profit. 1% profit of $1020,10 = $10,201 Calculate the Lot Size Now I will need to calculate the lot size to get $10,201 of profit. Lets calculate the lot size. 10,201 / 10 pips = 1,0201 per pip 1,0201 per pip = 0,10201 lot size This is one mini lot(0,1) + two nano lots(0,002). So, if I open three trades as on the first and second day where I have 2 bad trades and one positive trade I would get this: 1st trade – 10 pips loss 2nd trade – 10 pips loss 3rd trade – 30 pips profit TOTAL = -10 -10 + 30 = 10 pips With 10 pips of profit I have earned 10 pips x $1,0201 = $10,201 Now, when I add this amount to the account balance from day before which was $1020,10 I will get new account balance: $1020,10 + $10,201 = $1030,301 Day 10 – 10 Pips a Day Compounding Trading Plan When I trade like this for 10 days you would get these results. After 10 days with 10 pips compounding with 1% risk on each trade you would end up with a 10% large account. In two weeks you would make 10% which is quite a good result. And those 10% were with three trades per day where you can have 2 bad trades and only one positive trade. Day 10 – 10 Pips a Day Compounding Trading Plan with 50% Profit Imagine increasing percentage of profit each day from 1% to 50%. In 10 days you would end up with $57.665,04 on your account. Check the graph below. 0 Comments	{"url":"https://getknowtrading.com/forex-compounding-plan","timestamp":"2024-11-03T15:27:33Z","content_type":"text/html","content_length":"304365","record_id":"<urn:uuid:739c9aad-ccc9-4a63-a000-8fb1c7764a38>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00876.warc.gz"}
HydroGeoSphere/Thermal Energy - Wikibooks, open books for an open world Specified temperature flux 1. ntime Number of panels in the time-variable, specified temperature flux function. For each panel, enter the following: (a) ton_val, toff_val, bc_val, bc_temp Time on [T], time off [T], specified volume flux (of liquid) [L^3/T] multiplied by the heat capacity [J kg^−1 K^−1] and density [M/L^3] of the boundary medium, temperature of water entering [K]. Nodes in both the chosen faces and the currently active media (see Section 5.8.1) are assigned a time-variable temperature flux value. Although a fluid volume flux bc_val is specified, this does not influence the flow solution in any way. It is merely intended to give the user a straightforward way to input a known amount of energy to the system, as a function of fluid volume and temperature. • • • The following instructions are used to define the atmospheric inputs discussed in Section 2.6.1.10, and summarized in Equation 2.121. Incoming Shortwave Radiation 1. ntime Number of panels in the time-variable, incoming shortwave radiation function. For each panel, enter the following: (a) ton_val, value Time on [T], incoming shortwave radiation [M/T^3], ${\displaystyle K}$ ^↓ in Equation 2.122. Default value is 1.10 × 10^2 J/(m^2 s). • • • Sinusoidal Incoming Shortwave Radiation 1. ntime Number of panels in the time-variable, sinusoidal incoming shortwave radiation function. For each panel, enter the following: (a) ton_val, value, amp, phase, period Time on [T], vertical shift (mid-point incoming shortwave radiation, ${\displaystyle K}$ ^↓ in Equation 2.122) [M/T^3], amplitude [M/T^3], phase, period [T]. • • • 1. ntime Number of panels in the time-variable, cloud cover function. For each panel, enter the following: (a) ton_val, value Time on [T], cloud cover [-], ${\displaystyle C_{c}}$ in Equation 2.124. Default value is 0.50. • • • Incoming Longwave Radiation 1. ntime Number of panels in the time-variable, incoming longwave radiation function. For each panel, enter the following: (a) ton_val, value Time on [T], incoming longwave radiation [M/T^3], ${\displaystyle L}$ ^↓ in Equation 2.125. Default value is 3.0 × 10^2 J/(m^2 s). • • • 1. ntime Number of panels in the time-variable, air temperature function. For each panel, enter the following: (a) ton_val, value Time on [T], air temperature [K], ${\displaystyle T_{a}}$ in Equation 2.126. Default value is 15°C. • • • Sinusoidal Temperature of Air 1. ntime Number of panels in the time-variable, sinusoidal air temperature function. For each panel, enter the following: (a) ton_val, value, amp, phase, period Time on [T], vertical shift (mid-point temperature, ${\displaystyle T_{a}}$ in Equation 2.126) [K], amplitude [K], phase, period [T]. • • • These instructions are used to define the sensible heat flux ${\displaystyle Q_{h}}$ in Equation 2.128 and latent heat flux ${\displaystyle Q_{E}}$ in Equation 2.129. 1. ntime Number of panels in the time-variable, air density function. For each panel, enter the following: (a) ton_val, value Time on [T], density of air [M/L^3], ${\displaystyle {\rho }_{a}}$ in Equation 2.128. Default value is 1.225 kg/m^3. • • • Specific Heat of Air 1. ntime Number of panels in the time-variable, air specific heat function. For each panel, enter the following: (a) ton_val, value Time on [T], specific heat of air [L^2/(T^2 K)], ${\displaystyle c_{a}}$ in Equation 2.128. Default value is 7.17 × 10^2 J/(kg °K). • • • 1. ntime Number of panels in the time-variable, wind speed function. For each panel, enter the following: (a) ton_val, value Time on [T], wind speed [L/T], ${\displaystyle V_{a}}$ in Equation 2.128. Default value is 1.0 m/s. • • • Sinusoidal Wind Speed 1. ntime Number of panels in the time-variable, sinusoidal wind speed function. For each panel, enter the following: (a) ton_val, value, amp, phase, period Time on [T], vertical shift (mid-point wind speed, ${\displaystyle V_{a}}$ in Equation 2.128) [L/T], amplitude [L/T], phase, period [T]. • • • 1. ntime Number of panels in the time-variable, drag coefficient function. For each panel, enter the following: (a) ton_val, value Time on [T], drag coefficient [-], ${\displaystyle c_{D}}$ in Equation 2.128. Default value is 2.0 × 10^−3. • • • Latent Heat of Vapourization 1. ntime Number of panels in the time-variable, latent heat of vapourization function. For each panel, enter the following: (a) ton_val, value Time on [T], latent heat of vapourization [L^2/T^2], ${\displaystyle L_{V}}$ in Equation 2.129. Default value is 2.258 × 10^6 J/kg. • • • Specific Humidity of Air 1. ntime Number of panels in the time-variable, air specific hunidity function. For each panel, enter the following: (a) ton_val, value Time on [T], specific humidity of air [M/M], ${\displaystyle SH_{a}}$ in Equation 2.129. Default value is 0.01062. • • • Soil-Water Suction at Surface 1. ntime Number of panels in the time-variable, soil-water suction at surface function. For each panel, enter the following: (a) ton_val, value Time on [T], soil-water suction at surface [L], ${\displaystyle {\psi }_{g}}$ in Equation 2.131. Default value is 0.138 m. • • • Saturation Vapour Pressure 1. ntime Number of panels in the time-variable, saturation vapour pressure function. For each panel, enter the following: (a) ton_val, value Time on [T], saturation vapour pressure [M/(L T^2)], ${\displaystyle e_{sat}[T_{g}]}$ in Equation 2.132. Default value is 1.704 × 10^3 Pa. • • • 1. ntime Number of panels in the time-variable, relative humidity function. For each panel, enter the following: (a) ton_val, value Time on [T], relative humidity [-]. Default value is 0.75. • • • 1. ntime Number of panels in the time-variable, air pressure function. For each panel, enter the following: (a) ton_val, value Time on [T], air pressure [M/L T^2], ${\displaystyle p_{a}}$ in Equation 2.132. Default value is 1.013 × 10^5 Pa. • • •	{"url":"https://en.m.wikibooks.org/wiki/HydroGeoSphere/Thermal_Energy","timestamp":"2024-11-11T18:06:53Z","content_type":"text/html","content_length":"68373","record_id":"<urn:uuid:567ad668-7011-43ac-92ad-ce12973d9435>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00531.warc.gz"}
Argand diagrams An Argand diagram is a simple way of representing the complex number plane on an (x, y) coordinate system. It uses the x-axis to represent the real part of the number, and the y-axis to represent the imaginary part. So the complex number a + bi is represented by the point (a, b) on the plane: This might seem like an obvious idea, but it was a significant innovation at the time, and it led to a greater acceptance of imaginary numbers into mainstream mathematics. It showed complex numbers as an extension of the real number line and provides a geometric interpretation of complex number arithmetic. It also illustrated the polar interpretation of complex numbers, which explained complex multiplication and ultimately led to Euler's formula. Finally, it provided a way to visualise complex functions of complex variables. History of imaginary numbers The imaginary unit i has the property that its square is -1. It is often described as the square root of -1, although the quantity -i is also a square root of -1 (in the same way that both 1 and -1 are square roots of 1). Since no real number can be the square root of a negative number, the term imaginary was coined for this new type of number. An imaginary number is the product of i and any real number. For example 2i is an imaginary number that happens to be the square root of -4: The problem of finding the square root of a negative number goes back a long way. In the Roman era, the Greek mathematician Hero of Alexandria (who also gave us Heron's formula) considered the problem. But at the time mathematics wasn't sufficiently advanced to offer a solution. This issue became more important when algebraic solutions to cubic and quartic equations were discovered in the 16th century. Finding the roots of a cubic equation sometimes involved the square roots of negative numbers, even if the solutions were all real. Solving this problem eventually led to the idea of imaginary numbers, but initially they were seen as something that only really applied to the intermediate steps in solving polynomials, and had nothing to do with real numbers. The idea of representing complex numbers as points on a complex plane was originally suggested by Caspar Wessel in 1799 but was independently suggested again in 1806 by Jean-Robert Argand, who gave his name to the Argand diagram. Complex number addition and subtraction Adding and subtracting complex numbers is based on the idea that real and imaginary numbers are different things that must be kept separate. So the real and imaginary parts must be added (or subtracted) separately: This is similar to how vector arithmetic works, and can be illustrated in a similar way on an Argand diagram: The key point here is that an Argand diagram shows that complex numbers are an extension of real numbers into two dimensions. It also gives a geometric interpretation of addition, subtraction, negation (we negate a complex number by rotating it through 180°) and the complex conjugate z^* (flipped over the real axis): Polar form of complex numbers The Argand diagram can be used to represent a complex number in polar form. This works in the same way as normal polar coordinates, so the complex number a + bi can be represented as: This is shown here on an Argand diagram: The polar form of complex numbers is important because it turns out that when we multiply two complex numbers we multiply the magnitudes (r values) and add the angles (θ values). So: This also gives us a geometric interpretation of complex number division, powers, and roots. It ultimately leads to Euler's formula, which in turn allows us to calculate things like i to the power i: Complex functions of complex variables It is possible to create complex versions of various functions. For example, we can define the sin function for a complex argument z = x + iy: Since x and y are real values, we can calculate sin x, cosh y, etc in the usual way. This means that for every complex value z, the complex sin function will return a complex number result. The sin function is, therefore, a 4-dimensional curve, which can be quite difficult to visualise. A standard way to do this is to plot the real and imaginary parts of the function separately: Here the left-hand image shows the value of the real part of sin z. The Argand diagram represents input values z in the complex plane. The colour of the graph at each point represents the value of the real part of sin z (ie the x value) for the value of z at that point in the plane. At this stage we aren't too interested in how the colours map onto values, we are just looking at the technique for representing values. The right-hand image shows the imaginary part of z (the y values) in a similar way. The Argand diagram is a useful way of representing complex number arithmetic and functions. But beyond that, it was an important step in our understanding of the many applications of complex numbers. See also Join the GraphicMaths Newletter Sign up using this form to receive an email when new content is added:	{"url":"https://graphicmaths.com/pure/complex-numbers/argand-diagrams/","timestamp":"2024-11-14T01:17:02Z","content_type":"text/html","content_length":"33662","record_id":"<urn:uuid:d5e1cb65-807f-419e-93dd-73163ea4fb80>","cc-path":"CC-MAIN-2024-46/segments/1730477028516.72/warc/CC-MAIN-20241113235151-20241114025151-00831.warc.gz"}
A Lifetime's Supply of Cryptarithms The rules for cryptarithm puzzles are simple. Each letter represents one of the digits 0 to 9. Different letters represent different digits. The first digit in a number can't be 0. Your job is to find out what digit each letter represents. Can you solve the puzzle shown here?	{"url":"https://cryptarithmania.com/P88/0/10.php?633","timestamp":"2024-11-11T12:57:56Z","content_type":"text/html","content_length":"9391","record_id":"<urn:uuid:d01f19f8-26d0-4894-9b9e-4ed2842519fc>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00854.warc.gz"}
Is the t test really conservative when the parent distribution is long-tailed? It is generally believed that the t test is conservative for a sample from a long-tailed symmetric distribution. Yet the probability inequalities expressing this property have not been proved. The inequalities are explored here using various criteria for long-tailedness and leaning on the geometrical interpretation of the t test. It is proved that the t test is conservative but only for large enough critical values. Examples of a liberal t test for lower values are given. The results are used to explain some curiosities in the asymptotic distribution of the t statistic and to study its behavior when the parent distribution is skewed. © 1983 Taylor & Francis Group, LLC. • Conservatism • Long-tailed distributions • Probability inequality • Scale mixture of normals • T test Dive into the research topics of 'Is the t test really conservative when the parent distribution is long-tailed?'. Together they form a unique fingerprint.	{"url":"https://cris.tau.ac.il/en/publications/is-the-t-test-really-conservative-when-the-parent-distribution-is","timestamp":"2024-11-12T21:31:28Z","content_type":"text/html","content_length":"48544","record_id":"<urn:uuid:810e2d50-735c-4c38-b19e-298cb9b93083>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00273.warc.gz"}
[Solved] The rate of radioactive disintegration at an instant for a r The rate of radioactive disintegration at an instant for a radioactive sample of half life 2.2 × 109 s is 1010 s−1. The number of radioactive atoms in that sample at that instant is, Answer (Detailed Solution Below) Option 2 : 3.17 × 1019 RPMT Optics Test 1 21.8 K Users 10 Questions 40 Marks 10 Mins According to radioactive decay, the number of nuclei which is going to decay per unit of time is proportional to the total number of nuclei in the sample and it is written as; R = λN Here R is the decay rate, λ radioactive decay constant, and N is the radioactive nuclei at time t. Given: half life time $t_{\frac{1}{2}}$ = 2.2 × 10^9 s The decay rate, R = 10^10 s^-1 As we know, R = λN $\Rightarrow N= \frac {R}{λ}$ $\Rightarrow N= \frac {R}{0.693}\times t_{1/2}$ Now, on putting all the given values we have; $\Rightarrow N= \frac {10^{10}}{0.693}\times 2.2 \times 10^9$ $\Rightarrow N= 3.17 \times 10^{19}$ atoms Hence, option 2) is the correct answer. Latest NEET Updates Last updated on May 2, 2024 -> NEET 2024 Admit Card has been released for the exam which will be held on 5th May 2024 (Sunday) from 02:00 P.M. to 05:20 P.M. -> Earlier, The NEET 2023 Result was released by the National Testing Agency (NTA). -> The National Testing Agency (NTA) conducts NEET Exam every year for admission into Medical Colleges. -> For the official NEET Answer Key the candidates must go through the steps mentioned here.	{"url":"https://testbook.com/question-answer/the-rate-of-radioactive-disintegration-at-an-insta--62f848eec23d8f289ea8f888","timestamp":"2024-11-10T09:16:43Z","content_type":"text/html","content_length":"196067","record_id":"<urn:uuid:194a860b-a843-473c-a1c7-c06a9f948db0>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00869.warc.gz"}
A car has two wipers which do not overlap. Each wiper has a blade Q) A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle of 120°. Find the total area cleaned at each sweep of the two blades. Area cleaned by a blade = π r^2 = 22 x 21 = 462 cm^2 The area cleaned by 2 blades: = 462 x 2 = 924 cm^2 Therefore, the area cleaned by two blades = 924 cm^2 Leave a Comment	{"url":"https://www.saplingacademy.in/a-car-has-two-wipers-which-do-not-overlap-each-wiper-has-a-blade/","timestamp":"2024-11-05T10:47:28Z","content_type":"text/html","content_length":"128492","record_id":"<urn:uuid:9fdbe7d5-7c8c-4db5-8b7d-ac95a76b2f1a>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00259.warc.gz"}
Experiment sneaks a peek at quantum world -- Inside the Perimeter For the first time, researchers have experimentally probed topological order and its breakdown. The work could open the way for new approaches to quantum computation. Topological matter has been hailed as a potential solution to everything from power transmission to quantum computing. What makes it special – and difficult to study – is quantum. At a certain point, the long-range entanglement that gives topological matter its special properties breaks down, or decoheres, and it becomes boring old normal matter. This leaves quantum matter researchers facing quite a puzzle: can you identify, let alone observe, the phases[1] of quantum topological matter without destroying the entanglement that bestows its unique properties? Now, there is an answer: yes, you can. The trick, it turns out, is to have the system do the hard work for you. In a paper published today in Nature Physics, a team of theorists and experimentalists in China and Canada created a small topological system that, when manipulated slowly enough, revealed its own phases. The researchers also observed the transformation as quantum entanglement broke down. While the team did its experiment using an NMR machine, they expect that their approach could also work in other physical systems for quantum computers, such as superconductors, and that it could serve as a potential system for quantum memory. Co-author and former Perimeter postdoctoral fellow Yidun Wan said the results surpassed even the research team’s expectations. “This is something we didn’t quite expect to happen, to observe the breakdown of entanglement so nicely,” said Wan, who is now a professor at Fudan University. What made it particularly surprising is that the team initially didn’t set out to do any of this at all. Since topological phases were first put forward as something as a mathematical oddity in the 1970s, they’ve become a hotbed of research activity (and the basis for the 2016 Nobel Prize in Physics). The China-Canada research team initially set out to find out if two mathematical tools used in topology theory – the S matrix and the T matrix – are actually physical effects that can be experimentally observed. The S and T matrices are considered one of the most fundamental fingerprints of topological order. They essentially map out what happens when a quantum system is put through a particular transformation. Because topological systems have unique properties, each system’s S and T matrices will be different as well. Each matrix plots out “anyonic” statistics that represent the behaviour of anyons (an exotic kind of quasiparticle) in a system: the S matrix shows what happens when a quantum system that is mapped onto a torus (which looks like a donut) is rotated; the T matrix defines that same toric system when it is sheared. In their first experiment, the team used an NMR simulator to show that, indeed, the S and T matrices do provide fundamental signatures of topological order. They sent the paper to Nature Physics for consideration, noting that the result “opens up new future avenues toward identifying more generic topological orders based purely on experimental measurements.” One of the reviewers sent back a number of pointed questions, chief among them: why didn’t they push the work further? “The reviewer’s comments were illuminating,” said Wan, who spent the last year of his Canadian postdoctorate working on the research. “We decided to challenge ourselves and try to observe the breakdown of topological order. We redesigned the experiment completely.” The resulting experiment, and the subject of the new paper, goes well beyond verifying the S and T matrices as an observable fingerprint for topological order. Now, the matrices have been used to probe topological order itself, mapping out the phases of a system right up to the point where entanglement breaks down, all with minimal theoretical input at the The researchers essentially launched a voyage of discovery. When studying a quantum system, researchers usually calculate the system’s energy interactions and all of its ground states before starting their experiments. This team took a different route. Theory told them that the system they were simulating – a particular kind of topological code called a Z2 toric code, which is the simplest example of topological order – has four ground states, but they didn’t know what quantum phase those states belonged to. That’s largely because of the quirks of quantum matter. Quantum systems are described using something called a Hamiltonian. The Hamiltonian is like a map of the energy interactions inside the quantum system. The ground state of a quantum system is the lowest energy it can have while still maintaining its particular Hamiltonian. But ground states can take different forms, because a system’s Hamiltonian “map” can be configured in different ways: two nuclear up-spins could produce the same Hamiltonian as two down-spins. These alternate versions are called “degenerate ground states,” and they are potentially powerful for quantum memory. The team decided to use what they had just learned about S and T matrices to see if a simple topological system could essentially identify its own degenerate ground states. They designed an experiment using the “adiabetic method,” which posits that, if you move slowly enough, you can manipulate a quantum system without disrupting its quantum-ness. (The idea was first proposed in 2008 as a potential avenue for creating quantum memory.) They were essentially sneaking up on a quantum system. The five-qubit experiment was carried out using a two-dimensional compound called 1-bromo-2,4,5-trifluorobenzene. The nuclear spins of two protons and two fluorine atoms were numbered one through four, each serving as a single quantum bit, or “qubit”. A third fluorine atom acted as the observer. (The bromine and carbon-12 nuclei have spins of zero, so cannot be detected by the NMR machine.) Experimentalist Zhihuang Luo, then a PhD student at the University of Science and Technology of China, used radio-frequency pulses to manipulate the spins of the qubit. Going slowly, he put the four-qubit system through a series of transformations. Then, using measurements from the probe qubit, the team developed an algorithm that allowed them to recover the S and T matrices and get a peek at what was happening inside the system as it moved. “It was difficult. The sample is a liquid crystal that is very sensitive to temperature,” said Luo, who is now a postdoctoral researcher at the Institute for Quantum Computing. “A tiny fluctuation – even 0.1 degree Kelvin – would lead to a large change of the system’s parameters. That decoherence effect is serious.” Thanks to the observer qubit, the experimentalists were able to monitor the effects of the transformation, and then to recover the S and T matrices from the data, all without destroying the system’s entanglement (or, more technically, without collapsing the wave function). It seems that sneaking up works. At a critical point, there was a sudden jump and change in the S and T matrices. The system had revealed its information. The results paint a striking picture: the four degenerate ground states in the Hamiltonian were clearly identified as time evolved, and then the entanglement collapsed. “Our results open up novel avenues toward identifying more generic topological phases purely based on experimental measurements, and open the doors to many applications, like fault tolerant quantum computation,” Luo said. The experiment shows that, with current technology, researchers can not only identify phases of matter; they can also probe the system in its “phase space,” right up to the point where the long-range entanglement collapses. This work, Wan said, shows that the idea of using degenerate ground states as quantum memory is sound. It could also be used to identify topological orders in realistic materials, where researchers usually do not have prior knowledge of the topological orders to be discovered nor their ground states. The researchers said the work also promises to be a better method of characterizing topological order. The standard method – if anything can be called ‘standard’ in this area of physics – is to use entanglement entropy[2]. But this has an inherent flaw: with entanglement entropy, different phases often have the same value. It does not tell you what is happening within a system, because all the phases look alike. By being able to distinguish between phases, the S and T matrices provide a window to observe quantum behaviour. It could even be used to simulate a quantum computing proposal called anyon braiding, which forms another plank of Wan’s research. “This is the first time a topological order has been simulated and identified knowing only the approximate Hamiltonian, without any prior knowledge of the ground states,” Wan said. “The method is here. It’s not only scalable but also applicable to quantum simulators other than NMR simulators. And it becomes more reliable as the system grows. All that is needed now is the technology with which to create it. We need more qubits.” [1] The phases in normal matter – solid, liquid, gas, plasma – are dictated by temperature. Topological systems also have something called phases, but these are dictated by other factors. At each quantum phase, a topological system obeys specific rules or symmetries. Under certain external pressures, these factors can be altered, and the system can be made to flip from one phase to another. This is called a phase transition. [2] Entanglement entropy is associated with the information you lose when you isolate a region to study its properties. By “cutting out” part of the system to study, you inevitably lose information about hidden quantum links; this missing information corresponds to entropy. Quantum computing will rely on having a working quantum memory, but how can you verify if that memory really is quantum? Three physicists propose how resource theory and game theory could do just Topological materials hold great potential for quantum computing, but we don’t know how to make them yet. These scientists came up with an alternative: use what we have to simulate what we don’t. New research by Perimeter scientists employs artificial intelligence techniques to explore phases of matter.	{"url":"https://insidetheperimeter.ca/experiment-sneaks-peek-quantum-world/?__hstc=261081490.24d3305fee7a8d518d4cc393add48bba.1600361905524.1602681784972.1602684373847.23&__hssc=261081490.4.1602684373847&__hsfp=1596827813","timestamp":"2024-11-01T21:02:03Z","content_type":"text/html","content_length":"287378","record_id":"<urn:uuid:fac57a13-a9a8-477c-820a-26757ba8dc41>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00716.warc.gz"}
Definition Classes Core Scala types. They are always available without an explicit import. Definition Classes This is the documentation for the Scala standard library. Package structure The scala package contains core types like Int, Float, Array or Option which are accessible in all Scala compilation units without explicit qualification or imports. Notable packages include: Other packages exist. See the complete list on the right. Additional parts of the standard library are shipped as separate libraries. These include: Automatic imports Identifiers in the scala package and the scala.Predef object are always in scope by default. Some of these identifiers are type aliases provided as shortcuts to commonly used classes. For example, List is an alias for scala.collection.immutable.List. Other aliases refer to classes provided by the underlying platform. For example, on the JVM, String is an alias for java.lang.String. Definition Classes final class ArrayOps[A] extends AnyVal This class serves as a wrapper for Arrays with many of the operations found in indexed sequences. Where needed, instances of arrays are implicitly converted into this class. There is generally no reason to create an instance explicitly or use an ArrayOps type. It is better to work with plain Array types instead and rely on the implicit conversion to ArrayOps when calling a method (which does not actually allocate an instance of ArrayOps because it is a value class). Neither Array nor ArrayOps are proper collection types (i.e. they do not extend Iterable or even IterableOnce). mutable.ArraySeq and immutable.ArraySeq serve this purpose. The difference between this class and ArraySeqs is that calling transformer methods such as filter and map will yield an array, whereas an ArraySeq will remain an ArraySeq. type of the elements contained in this array. Type Hierarchy 1. Alphabetic 2. By Inheritance 1. ArrayOps 2. AnyVal 3. Any 1. by any2stringadd 2. by StringFormat 3. by Ensuring 4. by ArrowAssoc 1. Public 2. Protected Instance Constructors 1. new ArrayOps(xs: Array[A]) Value Members 1. final def !=(arg0: Any): Boolean Test two objects for inequality. Test two objects for inequality. true if !(this == that), false otherwise. Definition Classes 2. final def ##: Int Equivalent to x.hashCode except for boxed numeric types and null. Equivalent to x.hashCode except for boxed numeric types and null. For numerics, it returns a hash value which is consistent with value equality: if two value type instances compare as true, then ## will produce the same hash value for each of them. For null returns a hashcode where null.hashCode throws a NullPointerException. a hash value consistent with == Definition Classes 3. def +(other: String): String 4. final def ++[B >: A](xs: Array[_ <: B])(implicit arg0: ClassTag[B]): Array[B] 5. final def ++[B >: A](xs: IterableOnce[B])(implicit arg0: ClassTag[B]): Array[B] 6. final def ++:[B >: A](prefix: Array[_ <: B])(implicit arg0: ClassTag[B]): Array[B] 7. final def ++:[B >: A](prefix: IterableOnce[B])(implicit arg0: ClassTag[B]): Array[B] 8. final def +:[B >: A](x: B)(implicit arg0: ClassTag[B]): Array[B] 9. def ->[B](y: B): (ArrayOps[A], B) 10. final def :+[B >: A](x: B)(implicit arg0: ClassTag[B]): Array[B] 11. final def :++[B >: A](suffix: Array[_ <: B])(implicit arg0: ClassTag[B]): Array[B] 12. final def :++[B >: A](suffix: IterableOnce[B])(implicit arg0: ClassTag[B]): Array[B] 13. final def ==(arg0: Any): Boolean Test two objects for equality. Test two objects for equality. The expression x == that is equivalent to if (x eq null) that eq null else x.equals(that). true if the receiver object is equivalent to the argument; false otherwise. Definition Classes 14. def appended[B >: A](x: B)(implicit arg0: ClassTag[B]): Array[B] A copy of this array with an element appended. 15. def appendedAll[B >: A](suffix: Array[_ <: B])(implicit arg0: ClassTag[B]): Array[B] A copy of this array with all elements of an array appended. 16. def appendedAll[B >: A](suffix: IterableOnce[B])(implicit arg0: ClassTag[B]): Array[B] A copy of this array with all elements of a collection appended. 17. final def asInstanceOf[T0]: T0 Cast the receiver object to be of type T0. Cast the receiver object to be of type T0. Note that the success of a cast at runtime is modulo Scala's erasure semantics. Therefore the expression 1.asInstanceOf[String] will throw a ClassCastException at runtime, while the expression List(1).asInstanceOf[List[String]] will not. In the latter example, because the type argument is erased as part of compilation it is not possible to check whether the contents of the list are of the requested type. the receiver object. Definition Classes Exceptions thrown ClassCastException if the receiver object is not an instance of the erasure of type T0. 18. def collect[B](pf: PartialFunction[A, B])(implicit arg0: ClassTag[B]): Array[B] Builds a new array by applying a partial function to all elements of this array on which the function is defined. Builds a new array by applying a partial function to all elements of this array on which the function is defined. the element type of the returned array. the partial function which filters and maps the array. a new array resulting from applying the given partial function pf to each element on which it is defined and collecting the results. The order of the elements is preserved. 19. def collectFirst[B](pf: PartialFunction[A, B]): Option[B] Finds the first element of the array for which the given partial function is defined, and applies the partial function to it. 20. def combinations(n: Int): Iterator[Array[A]] Iterates over combinations of elements. Iterates over combinations of elements. A combination of length n is a sequence of n elements selected in order of their first index in this sequence. For example, "xyx" has two combinations of length 2. The x is selected first: "xx", "xy". The sequence "yx" is not returned as a combination because it is subsumed by "xy". If there is more than one way to generate the same combination, only one will be returned. For example, the result "xy" arbitrarily selected one of the x elements. As a further illustration, "xyxx" has three different ways to generate "xy" because there are three elements x to choose from. Moreover, there are three unordered pairs "xx" but only one is It is not specified which of these equal combinations is returned. It is an implementation detail that should not be relied on. For example, the combination "xx" does not necessarily contain the first x in this sequence. This behavior is observable if the elements compare equal but are not identical. As a consequence, "xyx".combinations(3).next() is "xxy": the combination does not reflect the order of the original sequence, but the order in which elements were selected, by "first index"; the order of each x element is also arbitrary. An Iterator which traverses the n-element combinations of this array 1. Array('a', 'b', 'b', 'b', 'c').combinations(2).map(runtime.ScalaRunTime.stringOf).foreach(println) // Array(a, b) // Array(a, c) // Array(b, b) // Array(b, c) Array('b', 'a', 'b').combinations(2).map(runtime.ScalaRunTime.stringOf).foreach(println) // Array(b, b) // Array(b, a) 21. final def concat[B >: A](suffix: Array[_ <: B])(implicit arg0: ClassTag[B]): Array[B] 22. final def concat[B >: A](suffix: IterableOnce[B])(implicit arg0: ClassTag[B]): Array[B] 23. def contains(elem: A): Boolean Tests whether this array contains a given value as an element. Tests whether this array contains a given value as an element. the element to test. true if this array has an element that is equal (as determined by ==) to elem, false otherwise. 24. def copyToArray[B >: A](xs: Array[B], start: Int, len: Int): Int Copy elements of this array to another array. Copy elements of this array to another array. Fills the given array xs starting at index start with at most len values. Copying will stop once either all the elements of this array have been copied, or the end of the array is reached, or len elements have been copied. the type of the elements of the array. the array to fill. the starting index within the destination array. the maximal number of elements to copy. 25. def copyToArray[B >: A](xs: Array[B], start: Int): Int Copy elements of this array to another array. Copy elements of this array to another array. Fills the given array xs starting at index start. Copying will stop once either all the elements of this array have been copied, or the end of the array is reached. the type of the elements of the array. the array to fill. the starting index within the destination array. 26. def copyToArray[B >: A](xs: Array[B]): Int Copy elements of this array to another array. Copy elements of this array to another array. Fills the given array xs starting at index 0. Copying will stop once either all the elements of this array have been copied, or the end of the array is reached. the type of the elements of the array. the array to fill. 27. def count(p: (A) => Boolean): Int Counts the number of elements in this array which satisfy a predicate 28. def diff[B >: A](that: Seq[B]): Array[A] Computes the multiset difference between this array and another sequence. Computes the multiset difference between this array and another sequence. the sequence of elements to remove a new array which contains all elements of this array except some of occurrences of elements that also appear in that. If an element value x appears n times in that, then the first n occurrences of x will not form part of the result, but any following occurrences will. 29. def distinct: Array[A] Selects all the elements of this array ignoring the duplicates. Selects all the elements of this array ignoring the duplicates. a new array consisting of all the elements of this array without duplicates. 30. def distinctBy[B](f: (A) => B): Array[A] Selects all the elements of this array ignoring the duplicates as determined by == after applying the transforming function f. Selects all the elements of this array ignoring the duplicates as determined by == after applying the transforming function f. the type of the elements after being transformed by f The transforming function whose result is used to determine the uniqueness of each element a new array consisting of all the elements of this array without duplicates. 31. def drop(n: Int): Array[A] The rest of the array without its n first elements. 32. def dropRight(n: Int): Array[A] The rest of the array without its n last elements. 33. def dropWhile(p: (A) => Boolean): Array[A] Drops longest prefix of elements that satisfy a predicate. Drops longest prefix of elements that satisfy a predicate. The predicate used to test elements. the longest suffix of this array whose first element does not satisfy the predicate p. 34. def endsWith[B >: A](that: Iterable[B]): Boolean Tests whether this array ends with the given sequence. Tests whether this array ends with the given sequence. the sequence to test true if this array has that as a suffix, false otherwise. 35. def endsWith[B >: A](that: Array[B]): Boolean Tests whether this array ends with the given array. Tests whether this array ends with the given array. the array to test true if this array has that as a suffix, false otherwise. 36. def ensuring(cond: (ArrayOps[A]) => Boolean, msg: => Any): ArrayOps[A] 37. def ensuring(cond: (ArrayOps[A]) => Boolean): ArrayOps[A] 38. def ensuring(cond: Boolean, msg: => Any): ArrayOps[A] 39. def ensuring(cond: Boolean): ArrayOps[A] 40. def exists(p: (A) => Boolean): Boolean Tests whether a predicate holds for at least one element of this array. Tests whether a predicate holds for at least one element of this array. the predicate used to test elements. true if the given predicate p is satisfied by at least one element of this array, otherwise false 41. def filter(p: (A) => Boolean): Array[A] Selects all elements of this array which satisfy a predicate. Selects all elements of this array which satisfy a predicate. the predicate used to test elements. a new array consisting of all elements of this array that satisfy the given predicate p. 42. def filterNot(p: (A) => Boolean): Array[A] Selects all elements of this array which do not satisfy a predicate. Selects all elements of this array which do not satisfy a predicate. the predicate used to test elements. a new array consisting of all elements of this array that do not satisfy the given predicate p. 43. def find(p: (A) => Boolean): Option[A] Finds the first element of the array satisfying a predicate, if any. Finds the first element of the array satisfying a predicate, if any. the predicate used to test elements. an option value containing the first element in the array that satisfies p, or None if none exists. 44. def flatMap[BS, B](f: (A) => BS)(implicit asIterable: (BS) => Iterable[B], m: ClassTag[B]): Array[B] 45. def flatMap[B](f: (A) => IterableOnce[B])(implicit arg0: ClassTag[B]): Array[B] Builds a new array by applying a function to all elements of this array and using the elements of the resulting collections. Builds a new array by applying a function to all elements of this array and using the elements of the resulting collections. the element type of the returned array. the function to apply to each element. a new array resulting from applying the given collection-valued function f to each element of this array and concatenating the results. 46. def flatten[B](implicit asIterable: (A) => IterableOnce[B], m: ClassTag[B]): Array[B] Flattens a two-dimensional array by concatenating all its rows into a single array. Flattens a two-dimensional array by concatenating all its rows into a single array. Type of row elements. A function that converts elements of this array to rows - Iterables of type B. An array obtained by concatenating rows of this array. 47. def fold[A1 >: A](z: A1)(op: (A1, A1) => A1): A1 Folds the elements of this array using the specified associative binary operator. Folds the elements of this array using the specified associative binary operator. a type parameter for the binary operator, a supertype of A. a neutral element for the fold operation; may be added to the result an arbitrary number of times, and must not change the result (e.g., Nil for list concatenation, 0 for addition, or 1 for a binary operator that must be associative. the result of applying the fold operator op between all the elements, or z if this array is empty. 48. def foldLeft[B](z: B)(op: (B, A) => B): B Applies a binary operator to a start value and all elements of this array, going left to right. Applies a binary operator to a start value and all elements of this array, going left to right. the result type of the binary operator. the start value. the binary operator. the result of inserting op between consecutive elements of this array, going left to right with the start value z on the left: op(...op(z, x_1), x_2, ..., x_n) where x[1], ..., x[n] are the elements of this array. Returns z if this array is empty. 49. def foldRight[B](z: B)(op: (A, B) => B): B Applies a binary operator to all elements of this array and a start value, going right to left. Applies a binary operator to all elements of this array and a start value, going right to left. the result type of the binary operator. the start value. the binary operator. the result of inserting op between consecutive elements of this array, going right to left with the start value z on the right: op(x_1, op(x_2, ... op(x_n, z)...)) where x[1], ..., x[n] are the elements of this array. Returns z if this array is empty. 50. def forall(p: (A) => Boolean): Boolean Tests whether a predicate holds for all elements of this array. Tests whether a predicate holds for all elements of this array. the predicate used to test elements. true if this array is empty or the given predicate p holds for all elements of this array, otherwise false. 51. def foreach[U](f: (A) => U): Unit Apply f to each element for its side effects. Apply f to each element for its side effects. Note: [U] parameter needed to help scalac's type inference. 52. def getClass(): Class[_ <: AnyVal] Returns the runtime class representation of the object. Returns the runtime class representation of the object. a class object corresponding to the runtime type of the receiver. Definition Classes AnyVal → Any 53. def groupBy[K](f: (A) => K): immutable.Map[K, Array[A]] Partitions this array into a map of arrays according to some discriminator function. Partitions this array into a map of arrays according to some discriminator function. the type of keys returned by the discriminator function. the discriminator function. A map from keys to arrays such that the following invariant holds: (xs groupBy f)(k) = xs filter (x => f(x) == k) That is, every key k is bound to an array of those elements x for which f(x) equals k. 54. def groupMap[K, B](key: (A) => K)(f: (A) => B)(implicit arg0: ClassTag[B]): immutable.Map[K, Array[B]] Partitions this array into a map of arrays according to a discriminator function key. Partitions this array into a map of arrays according to a discriminator function key. Each element in a group is transformed into a value of type B using the value function. It is equivalent to groupBy(key).mapValues(_.map(f)), but more efficient. case class User(name: String, age: Int) def namesByAge(users: Array[User]): Map[Int, Array[String]] = the type of keys returned by the discriminator function the type of values returned by the transformation function the discriminator function the element transformation function 55. def grouped(size: Int): Iterator[Array[A]] Partitions elements in fixed size arrays. Partitions elements in fixed size arrays. the number of elements per group An iterator producing arrays of size size, except the last will be less than size size if the elements don't divide evenly. See also scala.collection.Iterator, method grouped 56. def head: A Selects the first element of this array. Selects the first element of this array. the first element of this array. Exceptions thrown NoSuchElementException if the array is empty. 57. def headOption: Option[A] Optionally selects the first element. Optionally selects the first element. the first element of this array if it is nonempty, None if it is empty. 58. def indexOf(elem: A, from: Int = 0): Int Finds index of first occurrence of some value in this array after or at some start index. Finds index of first occurrence of some value in this array after or at some start index. the element value to search for. the start index the index >= from of the first element of this array that is equal (as determined by ==) to elem, or -1, if none exists. 59. def indexWhere(p: (A) => Boolean, from: Int = 0): Int Finds index of the first element satisfying some predicate after or at some start index. Finds index of the first element satisfying some predicate after or at some start index. the predicate used to test elements. the start index the index >= from of the first element of this array that satisfies the predicate p, or -1, if none exists. 60. def indices: immutable.Range Produces the range of all indices of this sequence. Produces the range of all indices of this sequence. a Range value from 0 to one less than the length of this array. 61. def init: Array[A] The initial part of the array without its last element. 62. def inits: Iterator[Array[A]] Iterates over the inits of this array. Iterates over the inits of this array. The first value will be this array and the final one will be an empty array, with the intervening values the results of successive applications of init. an iterator over all the inits of this array 63. def intersect[B >: A](that: Seq[B]): Array[A] Computes the multiset intersection between this array and another sequence. Computes the multiset intersection between this array and another sequence. the sequence of elements to intersect with. a new array which contains all elements of this array which also appear in that. If an element value x appears n times in that, then the first n occurrences of x will be retained in the result, but any following occurrences will be omitted. 64. def isEmpty: Boolean Tests whether the array is empty. Tests whether the array is empty. true if the array contains no elements, false otherwise. 65. final def isInstanceOf[T0]: Boolean Test whether the dynamic type of the receiver object has the same erasure as T0. Test whether the dynamic type of the receiver object has the same erasure as T0. Depending on what T0 is, the test is done in one of the below ways: □ T0 is a non-parameterized class type, e.g. BigDecimal: this method returns true if the value of the receiver object is a BigDecimal or a subtype of BigDecimal. □ T0 is a parameterized class type, e.g. List[Int]: this method returns true if the value of the receiver object is some List[X] for any X. For example, List(1, 2, 3).isInstanceOf[List[String]] will return true. □ T0 is some singleton type x.type or literal x: this method returns this.eq(x). For example, x.isInstanceOf[1] is equivalent to x.eq(1) □ T0 is an intersection X with Y or X & Y: this method is equivalent to x.isInstanceOf[X] && x.isInstanceOf[Y] □ T0 is a union X \| Y: this method is equivalent to x.isInstanceOf[X] \|\| x.isInstanceOf[Y] □ T0 is a type parameter or an abstract type member: this method is equivalent to isInstanceOf[U] where U is T0's upper bound, Any if T0 is unbounded. For example, x.isInstanceOf[A] where A is an unbounded type parameter will return true for any value of x. This is exactly equivalent to the type pattern _: T0 true if the receiver object is an instance of erasure of type T0; false otherwise. Definition Classes due to the unexpectedness of List(1, 2, 3).isInstanceOf[List[String]] returning true and x.isInstanceOf[A] where A is a type parameter or abstract member returning true, these forms issue a 66. def iterator: Iterator[A] 67. def knownSize: Int The size of this array. the number of elements in this array. 68. def last: A Selects the last element. Selects the last element. The last element of this array. Exceptions thrown NoSuchElementException If the array is empty. 69. def lastIndexOf(elem: A, end: Int = xs.length - 1): Int Finds index of last occurrence of some value in this array before or at a given end index. Finds index of last occurrence of some value in this array before or at a given end index. the element value to search for. the end index. the index <= end of the last element of this array that is equal (as determined by ==) to elem, or -1, if none exists. 70. def lastIndexWhere(p: (A) => Boolean, end: Int = xs.length - 1): Int Finds index of last element satisfying some predicate before or at given end index. Finds index of last element satisfying some predicate before or at given end index. the predicate used to test elements. the index <= end of the last element of this array that satisfies the predicate p, or -1, if none exists. 71. def lastOption: Option[A] Optionally selects the last element. Optionally selects the last element. the last element of this array$ if it is nonempty, None if it is empty. 72. def lazyZip[B](that: Iterable[B]): LazyZip2[A, B, Array[A]] Analogous to zip except that the elements in each collection are not consumed until a strict operation is invoked on the returned LazyZip2 decorator. Analogous to zip except that the elements in each collection are not consumed until a strict operation is invoked on the returned LazyZip2 decorator. Calls to lazyZip can be chained to support higher arities (up to 4) without incurring the expense of constructing and deconstructing intermediary tuples. val xs = List(1, 2, 3) val res = (xs lazyZip xs lazyZip xs lazyZip xs).map((a, b, c, d) => a + b + c + d) // res == List(4, 8, 12) the type of the second element in each eventual pair the iterable providing the second element of each eventual pair a decorator LazyZip2 that allows strict operations to be performed on the lazily evaluated pairs or chained calls to lazyZip. Implicit conversion to Iterable[(A, B)] is also supported. 73. def lengthCompare(len: Int): Int Compares the length of this array to a test value. Compares the length of this array to a test value. the test value that gets compared with the length. A value x where x < 0 if this.length < len x == 0 if this.length == len x > 0 if this.length > len 74. def lengthIs: Int Method mirroring SeqOps.lengthIs for consistency, except it returns an Int because length is known and comparison is constant-time. Method mirroring SeqOps.lengthIs for consistency, except it returns an Int because length is known and comparison is constant-time. These operations are equivalent to lengthCompare(Int), and allow the following more readable usages: this.lengthIs < len // this.lengthCompare(len) < 0 this.lengthIs <= len // this.lengthCompare(len) <= 0 this.lengthIs == len // this.lengthCompare(len) == 0 this.lengthIs != len // this.lengthCompare(len) != 0 this.lengthIs >= len // this.lengthCompare(len) >= 0 this.lengthIs > len // this.lengthCompare(len) > 0 75. def map[B](f: (A) => B)(implicit ct: ClassTag[B]): Array[B] Builds a new array by applying a function to all elements of this array. Builds a new array by applying a function to all elements of this array. the element type of the returned array. the function to apply to each element. a new array resulting from applying the given function f to each element of this array and collecting the results. 76. def mapInPlace(f: (A) => A): Array[A] 77. def nonEmpty: Boolean Tests whether the array is not empty. Tests whether the array is not empty. true if the array contains at least one element, false otherwise. 78. def padTo[B >: A](len: Int, elem: B)(implicit arg0: ClassTag[B]): Array[B] A copy of this array with an element value appended until a given target length is reached. A copy of this array with an element value appended until a given target length is reached. the element type of the returned array. the target length the padding value a new array consisting of all elements of this array followed by the minimal number of occurrences of elem so that the resulting collection has a length of at least len. 79. def partition(p: (A) => Boolean): (Array[A], Array[A]) A pair of, first, all elements that satisfy predicate p and, second, all elements that do not. 80. def partitionMap[A1, A2](f: (A) => Either[A1, A2])(implicit arg0: ClassTag[A1], arg1: ClassTag[A2]): (Array[A1], Array[A2]) Applies a function f to each element of the array and returns a pair of arrays: the first one made of those values returned by f that were wrapped in scala.util.Left, and the second one made of those wrapped in scala.util.Right. Applies a function f to each element of the array and returns a pair of arrays: the first one made of those values returned by f that were wrapped in scala.util.Left, and the second one made of those wrapped in scala.util.Right. val xs = Array(1, "one", 2, "two", 3, "three") partitionMap { case i: Int => Left(i) case s: String => Right(s) // xs == (Array(1, 2, 3), // Array(one, two, three)) the element type of the first resulting collection the element type of the second resulting collection the 'split function' mapping the elements of this array to an scala.util.Either a pair of arrays: the first one made of those values returned by f that were wrapped in scala.util.Left, and the second one made of those wrapped in scala.util.Right. 81. def patch[B >: A](from: Int, other: IterableOnce[B], replaced: Int)(implicit arg0: ClassTag[B]): Array[B] Returns a copy of this array with patched values. Returns a copy of this array with patched values. Patching at negative indices is the same as patching starting at 0. Patching at indices at or larger than the length of the original array appends the patch to the end. If more values are replaced than actually exist, the excess is ignored. The start index from which to patch The patch values The number of values in the original array that are replaced by the patch. 82. def permutations: Iterator[Array[A]] Iterates over distinct permutations of elements. Iterates over distinct permutations of elements. An Iterator which traverses the distinct permutations of this array. 1. Array('a', 'b', 'b').permutations.map(runtime.ScalaRunTime.stringOf).foreach(println) // Array(a, b, b) // Array(b, a, b) // Array(b, b, a) 83. def prepended[B >: A](x: B)(implicit arg0: ClassTag[B]): Array[B] A copy of this array with an element prepended. 84. def prependedAll[B >: A](prefix: Array[_ <: B])(implicit arg0: ClassTag[B]): Array[B] A copy of this array with all elements of an array prepended. 85. def prependedAll[B >: A](prefix: IterableOnce[B])(implicit arg0: ClassTag[B]): Array[B] A copy of this array with all elements of a collection prepended. 86. def reverse: Array[A] Returns a new array with the elements in reversed order. Returns a new array with the elements in reversed order. 87. def reverseIterator: Iterator[A] An iterator yielding elements in reversed order. An iterator yielding elements in reversed order. Note: xs.reverseIterator is the same as xs.reverse.iterator but implemented more efficiently. an iterator yielding the elements of this array in reversed order 88. def scan[B >: A](z: B)(op: (B, B) => B)(implicit arg0: ClassTag[B]): Array[B] Computes a prefix scan of the elements of the array. Computes a prefix scan of the elements of the array. Note: The neutral element z may be applied more than once. element type of the resulting array neutral element for the operator op the associative operator for the scan a new array containing the prefix scan of the elements in this array 89. def scanLeft[B](z: B)(op: (B, A) => B)(implicit arg0: ClassTag[B]): Array[B] Produces an array containing cumulative results of applying the binary operator going left to right. Produces an array containing cumulative results of applying the binary operator going left to right. the result type of the binary operator. the start value. the binary operator. array with intermediate values. Example: Array(1, 2, 3, 4).scanLeft(0)(_ + _) == Array(0, 1, 3, 6, 10) 90. def scanRight[B](z: B)(op: (A, B) => B)(implicit arg0: ClassTag[B]): Array[B] Produces an array containing cumulative results of applying the binary operator going right to left. Produces an array containing cumulative results of applying the binary operator going right to left. the result type of the binary operator. the start value. the binary operator. array with intermediate values. Example: Array(4, 3, 2, 1).scanRight(0)(_ + _) == Array(10, 6, 3, 1, 0) 91. def size: Int The size of this array. the number of elements in this array. 92. def sizeCompare(otherSize: Int): Int Compares the size of this array to a test value. Compares the size of this array to a test value. the test value that gets compared with the size. A value x where x < 0 if this.size < otherSize x == 0 if this.size == otherSize x > 0 if this.size > otherSize 93. def sizeIs: Int Method mirroring SeqOps.sizeIs for consistency, except it returns an Int because size is known and comparison is constant-time. Method mirroring SeqOps.sizeIs for consistency, except it returns an Int because size is known and comparison is constant-time. These operations are equivalent to sizeCompare(Int), and allow the following more readable usages: this.sizeIs < size // this.sizeCompare(size) < 0 this.sizeIs <= size // this.sizeCompare(size) <= 0 this.sizeIs == size // this.sizeCompare(size) == 0 this.sizeIs != size // this.sizeCompare(size) != 0 this.sizeIs >= size // this.sizeCompare(size) >= 0 this.sizeIs > size // this.sizeCompare(size) > 0 94. def slice(from: Int, until: Int): Array[A] Selects an interval of elements. Selects an interval of elements. The returned array is made up of all elements x which satisfy the invariant: from <= indexOf(x) < until the lowest index to include from this array. the lowest index to EXCLUDE from this array. an array containing the elements greater than or equal to index from extending up to (but not including) index until of this array. 95. def sliding(size: Int, step: Int = 1): Iterator[Array[A]] Groups elements in fixed size blocks by passing a "sliding window" over them (as opposed to partitioning them, as is done in grouped.) Groups elements in fixed size blocks by passing a "sliding window" over them (as opposed to partitioning them, as is done in grouped.) the number of elements per group the distance between the first elements of successive groups An iterator producing arrays of size size, except the last element (which may be the only element) will be truncated if there are fewer than size elements remaining to be grouped. See also scala.collection.Iterator, method sliding 96. def sortBy[B](f: (A) => B)(implicit ord: math.Ordering[B]): Array[A] Sorts this array according to the Ordering which results from transforming an implicitly given Ordering with a transformation function. Sorts this array according to the Ordering which results from transforming an implicitly given Ordering with a transformation function. the target type of the transformation f, and the type where the ordering ord is defined. the transformation function mapping elements to some other domain B. the ordering assumed on domain B. an array consisting of the elements of this array sorted according to the ordering where x < y if ord.lt(f(x), f(y)). See also 97. def sortWith(lt: (A, A) => Boolean): Array[A] Sorts this array according to a comparison function. Sorts this array according to a comparison function. The sort is stable. That is, elements that are equal (as determined by lt) appear in the same order in the sorted sequence as in the original. the comparison function which tests whether its first argument precedes its second argument in the desired ordering. an array consisting of the elements of this array sorted according to the comparison function lt. 98. def sorted[B >: A](implicit ord: math.Ordering[B]): Array[A] Sorts this array according to an Ordering. Sorts this array according to an Ordering. The sort is stable. That is, elements that are equal (as determined by lt) appear in the same order in the sorted sequence as in the original. the ordering to be used to compare elements. an array consisting of the elements of this array sorted according to the ordering ord. See also 99. def span(p: (A) => Boolean): (Array[A], Array[A]) Splits this array into a prefix/suffix pair according to a predicate. Splits this array into a prefix/suffix pair according to a predicate. Note: c span p is equivalent to (but more efficient than) (c takeWhile p, c dropWhile p), provided the evaluation of the predicate p does not cause any side-effects. the test predicate a pair consisting of the longest prefix of this array whose elements all satisfy p, and the rest of this array. 100. def splitAt(n: Int): (Array[A], Array[A]) Splits this array into two at a given position. Splits this array into two at a given position. Note: c splitAt n is equivalent to (c take n, c drop n). the position at which to split. a pair of arrays consisting of the first n elements of this array, and the other elements. 101. def startsWith[B >: A](that: IterableOnce[B], offset: Int = 0): Boolean Tests whether this array contains the given sequence at a given index. Tests whether this array contains the given sequence at a given index. the sequence to test the index where the sequence is searched. true if the sequence that is contained in this array at index offset, otherwise false. 102. def startsWith[B >: A](that: Array[B], offset: Int): Boolean Tests whether this array contains the given array at a given index. Tests whether this array contains the given array at a given index. the array to test the index where the array is searched. true if the array that is contained in this array at index offset, otherwise false. 103. def startsWith[B >: A](that: Array[B]): Boolean Tests whether this array starts with the given array. Tests whether this array starts with the given array. 104. def stepper[S <: Stepper[_]](implicit shape: StepperShape[A, S]): S with EfficientSplit 105. def tail: Array[A] The rest of the array without its first element. 106. def tails: Iterator[Array[A]] Iterates over the tails of this array. Iterates over the tails of this array. The first value will be this array and the final one will be an empty array, with the intervening values the results of successive applications of tail. an iterator over all the tails of this array 107. def take(n: Int): Array[A] An array containing the first n elements of this array. 108. def takeRight(n: Int): Array[A] An array containing the last n elements of this array. 109. def takeWhile(p: (A) => Boolean): Array[A] Takes longest prefix of elements that satisfy a predicate. Takes longest prefix of elements that satisfy a predicate. The predicate used to test elements. the longest prefix of this array whose elements all satisfy the predicate p. 110. def toArray[B >: A](implicit arg0: ClassTag[B]): Array[B] Create a copy of this array with the specified element type. 111. def toIndexedSeq: immutable.IndexedSeq[A] 112. final def toSeq: immutable.Seq[A] 113. def toString(): String Returns a string representation of the object. Returns a string representation of the object. The default representation is platform dependent. a string representation of the object. Definition Classes 114. def transpose[B](implicit asArray: (A) => Array[B]): Array[Array[B]] Transposes a two dimensional array. Transposes a two dimensional array. Type of row elements. A function that converts elements of this array to rows - arrays of type B. An array obtained by replacing elements of this arrays with rows the represent. 115. def unzip[A1, A2](implicit asPair: (A) => (A1, A2), ct1: ClassTag[A1], ct2: ClassTag[A2]): (Array[A1], Array[A2]) Converts an array of pairs into an array of first elements and an array of second elements. Converts an array of pairs into an array of first elements and an array of second elements. the type of the first half of the element pairs the type of the second half of the element pairs an implicit conversion which asserts that the element type of this Array is a pair. a class tag for A1 type parameter that is required to create an instance of Array[A1] a class tag for A2 type parameter that is required to create an instance of Array[A2] a pair of Arrays, containing, respectively, the first and second half of each element pair of this Array. 116. def unzip3[A1, A2, A3](implicit asTriple: (A) => (A1, A2, A3), ct1: ClassTag[A1], ct2: ClassTag[A2], ct3: ClassTag[A3]): (Array[A1], Array[A2], Array[A3]) Converts an array of triples into three arrays, one containing the elements from each position of the triple. Converts an array of triples into three arrays, one containing the elements from each position of the triple. the type of the first of three elements in the triple the type of the second of three elements in the triple the type of the third of three elements in the triple an implicit conversion which asserts that the element type of this Array is a triple. a class tag for T1 type parameter that is required to create an instance of Array[T1] a class tag for T2 type parameter that is required to create an instance of Array[T2] a class tag for T3 type parameter that is required to create an instance of Array[T3] a triple of Arrays, containing, respectively, the first, second, and third elements from each element triple of this Array. 117. def updated[B >: A](index: Int, elem: B)(implicit arg0: ClassTag[B]): Array[B] A copy of this array with one single replaced element. A copy of this array with one single replaced element. the position of the replacement the replacing element a new array which is a copy of this array with the element at position index replaced by elem. Exceptions thrown IndexOutOfBoundsException if index does not satisfy 0 <= index < length. 118. def view: IndexedSeqView[A] 119. def withFilter(p: (A) => Boolean): ArrayOps.WithFilter[A] Creates a non-strict filter of this array. Creates a non-strict filter of this array. Note: the difference between c filter p and c withFilter p is that the former creates a new array, whereas the latter only restricts the domain of subsequent map, flatMap, foreach, and withFilter the predicate used to test elements. an object of class ArrayOps.WithFilter, which supports map, flatMap, foreach, and withFilter operations. All these operations apply to those elements of this array which satisfy the predicate 120. def zip[B](that: IterableOnce[B]): Array[(A, B)] Returns an array formed from this array and another iterable collection by combining corresponding elements in pairs. Returns an array formed from this array and another iterable collection by combining corresponding elements in pairs. If one of the two collections is longer than the other, its remaining elements are ignored. the type of the second half of the returned pairs The iterable providing the second half of each result pair a new array containing pairs consisting of corresponding elements of this array and that. The length of the returned array is the minimum of the lengths of this array and that. 121. def zipAll[A1 >: A, B](that: Iterable[B], thisElem: A1, thatElem: B): Array[(A1, B)] Returns an array formed from this array and another iterable collection by combining corresponding elements in pairs. Returns an array formed from this array and another iterable collection by combining corresponding elements in pairs. If one of the two collections is shorter than the other, placeholder elements are used to extend the shorter collection to the length of the longer. the iterable providing the second half of each result pair the element to be used to fill up the result if this array is shorter than that. the element to be used to fill up the result if that is shorter than this array. a new array containing pairs consisting of corresponding elements of this array and that. The length of the returned array is the maximum of the lengths of this array and that. If this array is shorter than that, thisElem values are used to pad the result. If that is shorter than this array, thatElem values are used to pad the result. 122. def zipWithIndex: Array[(A, Int)] Zips this array with its indices. Zips this array with its indices. A new array containing pairs consisting of all elements of this array paired with their index. Indices start at 0. Deprecated Value Members 1. def formatted(fmtstr: String): String Returns string formatted according to given format string. Returns string formatted according to given format string. Format strings are as for String.format (@see java.lang.String.format). This member is added by an implicit conversion from ArrayOps[A] toStringFormat[ArrayOps[A]] performed by method StringFormat in scala.Predef. Definition Classes @deprecated @inline() (Since version 2.12.16) Use formatString.format(value) instead of value.formatted(formatString), or use the f"" string interpolator. In Java 15 and later, formatted resolves to the new method in String which has reversed parameters. 2. def →[B](y: B): (ArrayOps[A], B) This member is added by an implicit conversion from ArrayOps[A] toArrowAssoc[ArrayOps[A]] performed by method ArrowAssoc in scala.Predef. Definition Classes (Since version 2.13.0) Use -> instead. If you still wish to display it as one character, consider using a font with programming ligatures such as Fira Code.	{"url":"https://www.scala-lang.org/api/2.13.10/scala/collection/ArrayOps.html","timestamp":"2024-11-12T02:10:01Z","content_type":"text/html","content_length":"278519","record_id":"<urn:uuid:46c4a30f-dab8-417a-b02c-8fd3f842e363>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00634.warc.gz"}
The solution set of log5(x2+6)(13+4x2−4x) =log7+x(−2x−x2) is... \| Filo Question asked by Filo student The solution set of is d. none of these. Not the question you're searching for? + Ask your question Video solutions (2) Learn from their 1-to-1 discussion with Filo tutors. 7 mins Uploaded on: 6/8/2023 Was this solution helpful? Found 4 tutors discussing this question Discuss this question LIVE 6 mins ago One destination to cover all your homework and assignment needs Learn Practice Revision Succeed Instant 1:1 help, 24x7 60, 000+ Expert tutors Textbook solutions Big idea maths, McGraw-Hill Education etc Essay review Get expert feedback on your essay Schedule classes High dosage tutoring from Dedicated 3 experts Practice more questions on Sequence Series and Quadratic View more Students who ask this question also asked View more Stuck on the question or explanation? Connect with our Mathematics tutors online and get step by step solution of this question. 231 students are taking LIVE classes Question Text The solution set of is Updated On Jun 8, 2023 Topic Sequence Series and Quadratic Subject Mathematics Class Class 11 Answer Type Video solution: 2 Upvotes 133 Avg. Video Duration 5 min	{"url":"https://askfilo.com/user-question-answers-mathematics/the-solution-set-of-is-35323235343636","timestamp":"2024-11-15T04:40:21Z","content_type":"text/html","content_length":"289940","record_id":"<urn:uuid:608c8c25-2d25-4e5b-99c5-addd752d65a3>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00026.warc.gz"}
PROC PDLREG: RESTRICT Statement :: SAS/ETS(R) 9.22 User's Guide The RESTRICT statement places restrictions on the parameter estimates for covariates in the preceding MODEL statement. A parameter produced by a distributed lag cannot be restricted with the RESTRICT Each restriction is written as a linear equation. If you specify more than one restriction in a RESTRICT statement, the restrictions are separated by commas. You can refer to parameters by the name of the corresponding regressor variable. Each name used in the equation must be a regressor in the preceding MODEL statement. Use the keyword INTERCEPT to refer to the intercept parameter in the model. RESTRICT statements can be given labels. You can use labels to distinguish results for different restrictions in the printed output. Labels are specified as follows: label : RESTRICT ... The following is an example of the use of the RESTRICT statement, in which the coefficients of the regressors X1 and X2 are required to sum to 1: proc pdlreg data=a; model y = x1 x2; restrict x1 + x2 = 1; Parameter names can be multiplied by constants. When no equal sign appears, the linear combination is set equal to 0. Note that the parameters associated with the variables are restricted, not the variables themselves. Here are some examples of valid RESTRICT statements: restrict x1 + x2 = 1; restrict x1 + x2 - 1; restrict 2 * x1 = x2 + x3 , intercept + x4 = 0; restrict x1 = x2 = x3 = 1; restrict 2 * x1 - x2; Restricted parameter estimates are computed by introducing a Lagrangian parameter (Pringle and Rayner; 1971). The estimates of these Lagrangian parameters are printed in the parameter estimates table. If a restriction cannot be applied, its parameter value and degrees of freedom are listed as 0. The Lagrangian parameter, The t ratio tests the significance of the restrictions. If You can specify any number of restrictions in a RESTRICT statement, and you can use any number of RESTRICT statements. The estimates are computed subject to all restrictions specified. However, restrictions should be consistent and not redundant.	{"url":"http://support.sas.com/documentation/cdl/en/etsug/63348/HTML/default/etsug_pdlreg_sect010.htm","timestamp":"2024-11-04T15:43:44Z","content_type":"application/xhtml+xml","content_length":"13115","record_id":"<urn:uuid:11bf9058-ce6d-49dc-8738-f555f137a1bc>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00147.warc.gz"}
Substitution method for solving recurrences Recurrence relations are equations that describe themselves. We encounter recurrences in various situations when we have to obtain the asymptotic bound on the number of O(1) operations (constant time operations, ones that aren't affected by the size of the input) performed by that recursive function. It's essential to have tools to solve these recurrences for time complexity analysis, and here the substitution method comes into the picture. Substitution Method In the substitution method, we have a known recurrence, and we use induction to prove that our guess is a good bound for the recurrence's solution. This method works well in providing us with a good upper bound in most recurrences that can't be solved using the Master's Theorem or other more straightforward ways. Practising similar questions is essential if we try to make a guess that'll probably work for us. We'll cover some examples to get you started in the next section. The steps to use the Substitution method are as follows. 1. Guess a solution through your experience. 2. Use induction to prove that the guess is an upper bound solution for the given recurrence relation. Also see, Longest Common Substring and Data Structure Must Read Recursion in Data Structure	{"url":"https://www.naukri.com/code360/library/substitution-method-for-solving-recurrences","timestamp":"2024-11-10T11:34:40Z","content_type":"text/html","content_length":"381049","record_id":"<urn:uuid:33bd0f2b-2602-4fa4-9028-8febc8d5a195>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00702.warc.gz"}
Clock And Calender Reasoning Clock And CalenderPage 1 At what angle the hands of a clock are inclined at 15 minutes past 5? 1. (b)67(1/2)^o Answer is: BAt 15 min. past 5, the minute hand is at 3 and hour and slightly ahead of 5. Now, angle through which hour hand shifts in 15 min. = 15(1/2)^o = 7(1/2)^o So, Angle at 15 min. past 5 = 60 + 7(1/2)^o = 67(1/2)^o. A clock is set right at 8 am. The clock gains 10 minutes in 24 hours. What will be the right time when the clock indicates 1 pm on the following day? (a)11:40 pm 2. (b)12:48 pm (c)12 pm (d)10 pm Answer is: BFrom 8 am of a particular day to 1 pm of the following day = 29 h. Now, the clock gains 10 min in 24 h, it means that 24 h 10 min of this clock is equal to 24 h of the correct clock. 145/6 hour of this clock = 24 h of the correct clock. 29 h of this clock = (24/145) x 6 x 29 = 28 h. 48 min. of correct clock. 29 h of this clock = 28 h. 48 min. of the correct clock. It means that the clock in question is 12 min. faster than the correct clock. Therefore, when clock indicates 1pm the correct time will be 48 min. past 12. What was the day on 26th January. 1950, when 1st Republic day of India was celebrated? 3. (b)Tuesday Answer is: C26 January 1950 means 1949 year and 26 days. 1600 year have 0 odd day, 300 year have 1 odd day, 49 year have (12 leap year and 37 ordinary years) So, [(12 x 366) + (37 x 365)] days [4392 + 13505] days = 17897 day 17897 days = 2556 weeks + 5 days So, 49 years have 5 odd days. Total number of odd days = 0 + 1 + 5 + 5 = 11 days. Hence, the days on 26th Jan, 1950 was Thursday. If the day after tomorrow is a Sunday, what was it day before yesterday? 4. (b)Thursday Answer is: AToday â€“ Friday. So, the yesterday day before â€“ Wednesday. What will be the angle between hour hand and minute hand, if clock shows 8:30 pm? 5. (b)75^o Answer is: BAngle = (360 x 2.5)/12 = 75^o Reaching the place of meeting on Tuesday 15 min. before 8:30 h, Anuj found himself half an hour earlier than the man who was 40 min late. What was the scheduled time of the meeting? (a)8:00 h 6. (b)8:05 h (c)8:15 h (d)8:45 h Answer is: BAnuj reached the place of meeting at 8:15 h, he reached 30 min earlier than the man who was 40 min late. He reached 10 min. Hence, the scheduled time of the meeting was 8:05 h. A clock buzzes 1 time at 1 Oâ€™clock, 2 times at 2 Oâ€™clock, 3 times at 3 Oâ€™clock and so on. What will be the total number of buzzes in a day? 7. (b)156 Answer is: BNumber of buzzes in a day = [12(12+1)/2]x2 = 156 How many times the hands of a clock are is at right angle in a day? 8. (b)24 Answer is: CIn, 12 h, they are at right angles, 22 times. So, in 24 h, they are at right angles, 44 times. How many times do the hands of a clock coincide in a day? 9. (b)22 Answer is: BForm the properties of the clock, we know that hands of a clock coincide once in every hour but between 11 oâ€™clock and 1 oâ€™clock. They coincide only once. Therefore, the hands of a clock coincide 11 times in every 12 h. Hence, they will coincide (11 x 2) = 22 times in 24 h How many times do the hands of a clock points towards each other in day? 10. (b)24 Answer is: AThe hands of a clock point each other 11 times in every 12 h (Because between 5 and 7, they point towards each other only once at 6 oâ€™clock). No comment yet.	{"url":"https://zcos.in/reasoning/clock-and-calender","timestamp":"2024-11-04T17:28:55Z","content_type":"text/html","content_length":"29267","record_id":"<urn:uuid:cd8dd2e1-0651-496e-a1b0-f9fd0b14780e>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00357.warc.gz"}
Regretfully Yours This is a live blog of Lecture 5 of my graduate machine learning class “Patterns, Predictions, and Actions.” A Table of Contents for this series is here. There’s no faster way to suck out the feeling of a lecture than an unintuitive optimization convergence analysis. I’m sure this will similarly kill engagement on this post. But I’ve decided to go all in for a day, and there will be equations. I find it easy to motivate why descent algorithms converge. If my scheme to minimize a function decreases the function value at each step, it seems pretty reasonable that the algorithm will eventually make enough progress to get to the minimal value. But algorithms like Stochastic Gradient Descent aren’t guaranteed to decrease your function value. Why do they converge? One of the most popular techniques uses an analysis attributed to Nemirovskii and Yudin’s 1983 masterwork on optimization complexity. I have a deep and long love-hate relationship with their argument. The foundation of their analysis has nothing to do with algorithms but instead relies on an elementary inequality in Euclidean space. Anyone who has seen inner products will understand the following derivation. But no one has ever explained to me why we do this derivation. Let u[0], u[1] and, v[1],v[2],...,v[T] also be any vectors. Define u[t+1] = u[t]-v[t]. Then we always have $\|\| u_{t+1} - u_0\|\|^2 = \|\| u_t - u0 \|\|^2 - 2 \langle v_t, u_t-u_0 \rangle + \|\|v_t\|\|^2$ This is just using how we defined the sequence u[t] and then expanding the norm. If I continue to unwrap this expression, I’ll find $\|\| u_{t+1} - u_0\|\|^2 = \|\| u_1 - u_0 \|\|^2 - 2 \sum_{t=1}^T \langle v_t, u_t-u_0\rangle + \sum_{t=1}^T \|\|v_t\|\|^2$ Now, the left-hand side is a square, so it’s nonnegative. I can then rearrange terms and get $\sum_{t=1}^T \langle v_t, u_t-u_0\rangle \leq \frac{1}{2} \|\| u_1 - u_0 \|\|^2 + \frac{1}{2} \sum_{t=1}^T \|\|v_t\|\|^2$ This one simple inequality has spored ten thousand papers in optimization. But this inequality has nothing to do with optimization. It has something to do with sequences in Euclidean space. But barely anything. u[0], u[1], and the v’s are completely arbitrary here. And yet, I can use it to prove that Stochastic Gradient Descent converges with only a couple of extra lines. I don’t want to belabor the details here, but you can find the rest of the derivation in Chapter 5 of PPA. If the v[t] are stochastic gradients in a linear classification problem where the loss function is convex and has bounded gradients, then if you choose the stepsize correctly, the final expression reads $\frac{1}{T} \sum_{t=1}^T loss ( \langle w_t, x_t \rangle, y_t) - loss ( \langle w_, x_t \rangle, y_t) \leq \frac{\|\|w_\|\| D}{\sqrt{T}}$ Here D is an upper bound on the norm of the feature vectors x[i]. Let me unpack what this bound says for machine learning. First, the inequality measures the difference between the prediction quality of the iterate at step t and the optimal classifier. The comparison is only made on the current data point. It says that, on average, the difference between the prediction quality at step t and the best you could do if someone had handed you the optimal classifier is about 1 over the square root of T (1/√T) In particular, the average is going to zero. So it says that if I were to run sequential gradient descent on an infinite data set, the gap between the SGD predictor and the optimal predictor goes to zero. The inequality does not care which order the data steams through the algorithm. This inequality makes no assumptions about randomness. This bound holds for any ordering of the data you could Second, the upper bound itself is very curious. Let me explain by connecting back to the Perceptron. We can interpret the mistake bound for the Perceptron as a sequential prediction bound. The mistake bound is equivalent to $\frac{1}{T} \sum_{t=1}^T err ( \langle w_t, x_t \rangle, y_t) - err ( \langle w_, x_t \rangle, y_t) \leq \frac{\|\|w_\|\|^2 D^2}{T}$ Where err is equal to 1 if the prediction is correct and 0 if the prediction is incorrect. w[] is the max-margin hyperplane in this case. For the Perceptron, the error of w[] is always zero, but I’m belaboring the form of this expression so that we can compare the bounds. When data is separable, the sequential prediction error is the square of when the data is not separable. The term governs the convergence. In the case of the perceptron, this term is precisely the margin. And if you dig through ML theory papers to look for guarantees on high-dimensional prediction problems, this term pops up. Given our current mental models of how prediction works, it seems to be fundamental and unavoidable. Finally, as far as optimization bounds go, 1/√T is really bad. Steve Wright, inspired by one of his colleagues at NC State, always said that 1/√T is a negative result. I agree! It’s hard to find an algorithm that converges more slowly than 1/√T. There was an annoying trend in the optimization-for-machine learning heyday where people would prove that 1/√T rates are optimal for machine learning. These proofs would involve constructing some insane data sequence that would never occur in reality. But they convinced people that we had to settle for impossibly slow algorithms. And we’re now locked into incredibly inefficient algorithms because Google and Microsoft are willing to fight over who has more cloud capacity. This isn’t a good trend for the rest of us. PS What does the title have to do with the rest of this post? This average gap between the sequential predictions and the best predictor is called regret. Regret is a term that’s pulled from gambling metaphors and is my least favorite term in optimization and machine learning. But I’ll yell about that on some other day. no where to hide: equations were, are and always be useful. long live GD. (albeit the pseudoinverse is nicer). Expand full comment Yeah, those regret bounds can be a super drag sometimes. 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The Current State of Knowledge About Zero Knowledge. Comprehensive Private Token Overview - Part 3 Mon, Jan 29, 2024 •10 min read Category: Code stories This post is Part 3 of our three-part series dedicated to ZKs and Private Tokens. Click here for Part 1, and Part 2. Part 3: Return of the Hash Theory vs practice. Expectations vs reality In the previous part of the series, we delved into seemingly unrelated topics about blockchain, zero-knowledge proofs, and privacy. Now, let’s transition to practical considerations and explore the tools available for developers. Undoubtedly, the creation of any kind of ZK technology needs ZK Proof. How can we create it? The core element in this process is known as a circuit, often referring to an arithmetic circuit—an ordered collection of operations (such as addition, multiplication) represented by gates. In the realm of ZK-circuits, this translates to a program specifying a calculation to be performed on certain data inputs, used by the prover to generate a proof of knowledge. Thankfully we’re not forced to create every complex circuit from logic gates (AND, OR etc), but instead can use dedicated programming languages. Moreover, it's possible to validate the proof on-chain, using smart contracts deployed on platforms like Ethereum. Let's explore the available options: Other noteworthy languages include: As evident, several options are already available, with more on the horizon! While developing on Layer 2 platforms like Starknet has dedicated languages, addressing the Ethereum Mainnet introduces intricacies due to subtle yet consequential differences between ZK and EVM elliptic curves. Let’s explain them now. Elliptic curves and straightforward hashes Once again, in this series of articles, I don't intend to delve too deeply into this complex theme, as covering everything comprehensively would require another series of papers. Numerous excellent resources on this subject can be found on the internet. So, let's keep it as simple as possible. Blockchains like Bitcoin or Ethereum employ hashes as account addresses, which is obvious. However, not everyone understands how these addresses are created. It involves something like private and public keys, a touch of cryptography (or, as some might say, magic), and voila! Let's take a closer look. In mathematics, there's a concept known as elliptic curves. While these curves can take various forms, they are essentially a set of points that satisfy a specific mathematical equation, such as: y2 = x3 + ax + b. Utilizing the cryptographic system known as ECDSA (Elliptic Curve Digital Signature Algorithm), it becomes possible to obtain a point on the elliptic curve (public key) through the multiplication of two numbers. One of these numbers represents the private key. The trick lies in the fact that calculating the public key from a known number (private key) is extremely fast, yet attempting to deduce the number (private key) from a given point (public key) is incredibly challenging. In essence, the public key can be derived through a one-way function from the private key. To generate a public address for your account, simply append 0x to the beginning of the last 20 bytes of the Keccak-256 hash of the public key. The curve employed in ECDSA is named the secp256k1 curve, and what holds significance for us is that operations on this curve entail arithmetic with 256-bit numbers. Why is this aspect crucial? Zero-Knowledge Proofs and EVM Relationship Status: It’s complicated Recall zkSNARKs, the proving system integral to Zero-Knowledge Proofs? zkSNARKs rely on elliptic curve cryptography, which permits operations only on numbers restricted to 254 bits. In summary, the numbers utilized on Ethereum's EVM surpass the 254-bit limits allowed in zkSNARK systems. Consequently, to implement ECDSA algorithms within ZK circuits, we must resort to 'non-native' field This presents one of the most significant challenges in ensuring a smooth interaction between ZKP and EVM. While it is achievable, there's always a trade-off: time or money. A noteworthy example is the commendable work by the 0xPARK Foundation, which presented an implementation of an ECDSA circuit in the circom language, capable of generating zkSNARKs. Although it is possible to prove ownership of an account, there is still much to be done, as the performance falls short of expectations. The circom environment is resource-constrained, making proofs expensive and time-consuming. Nevertheless, ongoing efforts have resulted in an improved approach to verifying ECDSA signatures in ZK. Spartan-ecdsa, introduced by Personae Labs, shows promise. It’s still a work in progress, with challenges like poor Keccak performance and excessively large proofs for on-chain verification via smart contracts still needing to be addressed. There is always hope Let's pause for a moment and contemplate our goals and available options. The objective is to validate the proof (a 254-bit system) on-chain, which is a 256-bit system. Two potential approaches come to mind: 1. Convert 256-bit numbers to 254-bit ones, as previously discussed. However, this option, as assumed, is currently impractical. 2. Convert 254-bit numbers to 256-bit ones, which we will dive into below. Given that numerous encryption and description operations need to occur within zkSNARKs, necessitating 254-bit numbers for easy arithmetic, we can explore the utilization of another elliptic curve known as Baby Jubjub. Specifically designed to function efficiently within zkSNARKs on Ethereum, the ERC-2494 was introduced in 2020. Although it is no longer active, it remains a compelling and intriguing concept worth considering. A Bitcoin Orchard: Merkle Trees In the realm of Zero-Knowledge Proofs (ZKP) on the Ethereum Virtual Machine (EVM), or more broadly, whenever we navigate the blockchain landscape, a critical consideration emerges: data storage. While the ability to create and verify proofs is commendable, it becomes even more enchanting when we can efficiently store those proofs on the blockchain. A widely adopted solution, employed, for instance, by Tornado Cash, is the Merkle Tree. This data structure proves invaluable for privacy-centric projects by securely storing hashed (encrypted) data fragments while enabling the verification of specific data without exposing the entirety of the dataset. Let's once again refer to the diagram for a clearer illustration. At its core, a Merkle Tree is a data structure crafted from encrypted (hashed) data fragments. Illustrated in the diagram, each pair undergoes a hashing process, generating the next "leaf" (node). This paired hashing repeats, ascending the tree until culminating in the top hash known as the Merkle Root. This ingenious structure facilitates the proof of data existence within the tree without necessitating the storage of the entire dataset. For an in-depth exploration, delve into this detailed explanation. Merkle Trees works great for the UTXO Model, because the entire state can be stored with two Merkle Trees: • a Note Tree - stores all system data, including smart contracts and asset ownership; • a Nullifier Tree - stores data about spent (used) UTXOs. Before using a UTXO, it's checked in the Nullifier Tree to ensure it hasn't been used. If all's clear, changes can be made in the Note Tree. Encrypt. Verify. Repeat. zkSNARK Understanding Merkle Trees is vital for comparing SNARKs on various circuits, especially if the criterion is hash verifying efficiency. The method of creating hashes within the Merkle Tree, however, is far from unequivocal; it hinges on factors like numerical systems (e.g., 256- / 254-bit), arithmetic complexity, and cryptographic security. The most prominent ones among many available hash functions are: • MiMC - designed for very compact circuits with minimal amount of multiplications. Great for the tiniest circuits and as the base for more complicated algorithms (e.g. GMiMC), • Poseidon - an enhanced version of MiMC with improved security, well-suited for scenarios requiring a collision-resistant, secure, and SNARKs-friendly algorithm, • Pedersen - extremely useful and efficient, for e.g. homomorphic encryption, which enables solving some zk-security problems (like confidential payments) and building Merkle proofs. On the other hand, this can lead to significant security vulnerabilities. Segueing back to zkSNARKs protocols, two prominent players — PLONK and Groth16 — take the spotlight. For a detailed exploration of their nuances and functionalities, dive into this article. In a nutshell, both Groth16 and PLONK, are not quantum-resistant, and exhibit distinct characteristics: Groth16 boasts a generally smaller proof size, whereas PLONK sidesteps the need for recurring trusted setup ceremonies. Private Token: Proof of Concept Let’s see what the Private Token system could look like, using ZKP and EVM blockchain. Private Token is deployed as an EVM smart contract on a ZK-rollup (L2) and uses zkSNARKs to verify (on-chain via smart contract) transfer correctness and keep privacy at the same time. It is designed with the UTXO Model and uses Merkle Tree for membership proof. The Private Token contract maps UTXOs with commitments, having a structure defined by owner address, value, and unique ID. The ZK Circuit, written in circom language, verifies input and output UTXO values and sender ownership. As a result, we get a new UTXO commitment, in the form of data hashed with e.g. Pedersen Hash Function. If valid, a new commitment is added to the Merkle Tree using the Baby Jubjub curve (254-bit) for an efficient proof system. UTXOs are stored on Private Token contract in an encrypted format and thanks to the Partially Homomorphic Encryption can be decrypted and read only by the owner of the account. Simple, right? Well, if you're still a bit dazzled, no worries – I'm not grading anyone on their zero-knowledge prowess. Stick around, and we'll unravel more in the next posts!	{"url":"https://www.rumblefish.dev/blog/post/the-current-state-of-knowledge-about-zero-knowledge-comprehensive-private-token-overview-part-3/","timestamp":"2024-11-10T11:12:18Z","content_type":"text/html","content_length":"1049238","record_id":"<urn:uuid:2c0ed360-22d9-47c5-bcde-beecfab17e66>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00282.warc.gz"}
Under Which Angle Conditions Could A Triangle Exist? Check All That Apply. - AnswerPrime Under which angle conditions could a triangle exist? Check all that apply. Step-by-step explanation: We have to check under which condition a triangle will exist. According to angle sum property of the triangle, the sum of all the three angles of a triangle is 180 degrees.An acute angle is an angle with a a measure less than 90 degreesA right angle is an angle with a measure of 90 degrees.An obtuse angle is an angle with a measure of greater than 90 degrees. 1. It is possible to have triangle with three acute angle Example: A triangle with all the three angles of 60 degrees 2. It is possible to have a triangle with 2 acute angles and 1 right angle. Example: A triangle with all the two angles of 45 degrees and one right angle. 3. It is not possible to have a triangle with 1 acute angle, 1 right angle and 1 obtuse angle. It will violate the angle sum property of triangle. 4. It is not possible to have a triangle with 1 acute angle and 2 obtuse angles. It will violate the angle sum property of triangle. 5. It is possible to have a triangle with 2 acute angles and 1 obtuse angle. Example: A triangle with all the angles of 45 degrees, 40 degrees and 95 degrees 3 acute angles and 2 acute angles, 1 obtuse angle and 2 acute angles, 1 right angle Step-by-step explanation: the key is just to imagine it also you have to know that right angle is 90° acute is less than 90° and obtuse is more than 90° Answer 6 therefore three acute angles, eg. 60, 60 and 60 degrees can make up a triangle 2 acute angles and a right angle, say 45 and 45 degrees and 90 degrees make a triangle any obtuse angle + a right angle will be larger than 180 degrees, so no triangle can exist. another acute angle will make the number of degrees even larger any 2 obtuse angles will make a number larger than 180 degrees, so with an acute angle is impossible 2 acute and an obtuse angle is possible, say 40, 40 and 100 degrees 3 acute angles 2 acute angles, 1 right angle 2 acute angles, 1 obtuse angle Or 1, 2 , 5 Step-by-step explanation: abe Step-by-step explanation: Answer 7 Triangles DON’T exist for:1 acute angle, 1 right angle, 1 obtuse angle1 acute angle, 2 obtuse angles The correct answer is 3 acute angles 2 acute angles, 1 right angle 2 acute angles, 1 obtuse angle Further Explanation The basic properties of triangles include In a triangle, all the sum of the angle is 180 degree and it is also referred to as angle sum property. In a triangle, the sum of the two sides is greater than the sum of the third side. In a triangle, the longest side is the side that opposite the largest angle and vice versa. A triangle refers to a closed figure with a three-line segment and three angles. The three types of a triangle based on size are: Equilateral triangle , Isosceles triangle and Scalene triangle An equilateral triangle is a triangle where the lengths of all the three sides are equal An isosceles triangle is a triangle where its two sides are equal A scalene triangle is a triangle that has three sides with different length However, the three types of a triangle based on angles include: Acute-angled triangle: in Acute-angled triangle, all its angle are acute Obtuse-angled triangle: in Obtuse-angled triangle, one its angle is obtuse Right-angled triangle: in the Right-angled triangle, one of its angles is a right angle. Therefore, the correct answer is 3 acute angles, 2 acute angles, 1 right angle, 2 acute angles, 1 obtuse angle LEARN MORE: Under which angle conditions could a triangle exist? Check all that apply. 1/2 of an obtuse angle is a(n): A. Obtuse angle. B. Acute angle KEYWORDS: right-angled triangleacute angles2 acute anglesobtuse anglea scalene triangle 3 acute angles. 2 acute angles, 1 right angle. 2 acute angles, 1 obtuse angle. Step-by-step explanation: The 3 angles in a triangles add up to 180 degrees. Acute angles are < 90 degrees. A right angle = 90 degrees, Obtuse angles are > 90 degrees. Answer 6 therefore three acute angles, eg. 60, 60 and 60 degrees can make up a triangle 2 acute angles and a right angle, say 45 and 45 degrees and 90 degrees make a triangle any obtuse angle + a right angle will be larger than 180 degrees, so no triangle can exist. another acute angle will make the number of degrees even larger any 2 obtuse angles will make a number larger than 180 degrees, so with an acute angle is impossible 2 acute and an obtuse angle is possible, say 40, 40 and 100 degrees Latest posts by Answer Prime (see all) Leave a Comment	{"url":"https://answerprime.com/under-which-angle-conditions-could-a-triangle-exist-check-all-that-apply/","timestamp":"2024-11-06T10:22:23Z","content_type":"text/html","content_length":"175239","record_id":"<urn:uuid:1495c27a-b2ce-43c4-9e31-66850fbf0ea5>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00046.warc.gz"}
Multiplication Worksheets 100 Problems Math, especially multiplication, forms the keystone of numerous academic techniques and real-world applications. Yet, for numerous learners, mastering multiplication can position a difficulty. To resolve this difficulty, instructors and moms and dads have welcomed a powerful tool: Multiplication Worksheets 100 Problems. 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The convexity of this problem I encounter a problem as follows, where Vi is a N*N matrix, other variables are all scalars. I don’t know the convexity of the the constraints since the matrix Vl and Di are coupled in the term marked in the red line. If this term is neither convex nor concave, how to convert this problem into a convex problem (maybe using successive convex approximation) so that CVX can be used? All those constraints are non-convex. This looks like a difficult to solve Nonlinear SDP. If you want to try SCA you can, but you can look up my previous posts on this forum on SCA. People applying SCA to new problems often are not very successful. Perhaps you can try YALMIP., but Johan will probably tell you this is a nasty problem, and your chance of success might not be good.	{"url":"https://ask.cvxr.com/t/the-convexity-of-this-problem/13229","timestamp":"2024-11-02T21:14:54Z","content_type":"text/html","content_length":"15218","record_id":"<urn:uuid:e2a4dfc7-a1b9-401b-9acc-d8187acc3f3f>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00571.warc.gz"}
Oriana Talbot tutor for Geometry, Geometry, Adult Math, Probability, Discrete Geometry, Linear Algebra and Basic Math Experienced Geometry tutor, M.Sc in mathematics, with 3 years of teaching experience in the subject. Willing to provide customized classes with sparse use of online learning tools. I hold knowledge and expertise in the field of Mathematics. I possess a master's degree in mathematics from the University of Florida and I have 3 years of teaching experience. I am amazed that many students find math to be revolting. I truly believe that good tutoring can help turn around that perspective as the subject of mathematics is filled with many surprises that can wow any interested learner. I am a very outgoing person and do many outdoor activities like jogging while listening to music in my free time Master’s / Graduate Degree Can also teach • Probability • Adult Math • Geometry • +3 subjects more Teaching methodology In teaching Geometry online, I focus on strengthening basic math skills by starting with a comprehensive review of fundamental concepts like angles, shapes, and measurements. I emphasize thorough revision through interactive quizzes, practice exercises, and concept reinforcement activities. Each lesson includes step-by-step solutions to problems, ensuring students understand the reasoning behind geometric principles and problem-solving strategies. The approach is structured, methodical, and tailored to individual learning styles, promoting a deep understanding of Geometry High demand 6 lesson booked in last 72 hours	{"url":"https://wiingy.com/tutors/000004420157/","timestamp":"2024-11-14T20:21:27Z","content_type":"text/html","content_length":"93207","record_id":"<urn:uuid:cbc25df4-4045-4a0b-9596-1233d87905bb>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00471.warc.gz"}
binningplot: Two-dimensional binningplot in mbgraphic: Measure Based Graphic Selection Visualisation of a two-dimensional binning based on equidistant or quantile based binning. 1 binningplot(x, y, b = 10, bin = "equi", anchor = "min") binningplot(x, y, b = 10, bin = "equi", anchor = "min") x A numeric vector. y A numeric vector. b A positive integer. Number of bins in each variable. bin A chraracter string. Binning method "equi" (default) for equidistant binning or "quant" for quantile binning. anchor A chraracter string or a numeric. How should the anchorpoint be chosen? "min" (default) for the minimum of each variable, "ggplot" for the method used in ggplot graphics, "nice" for a "pretty"" anchor point, or a user specified value. A chraracter string. Binning method "equi" (default) for equidistant binning or "quant" for quantile binning. A chraracter string or a numeric. How should the anchorpoint be chosen? "min" (default) for the minimum of each variable, "ggplot" for the method used in ggplot graphics, "nice" for a "pretty"" anchor point, or a user specified value. H. Wickham (2009) ggplot2: Elegant Graphics for Data Analysis New York: Springer 1 ## Not run: 2 x <- rnorm(10000) 3 y <- rnorm(10000) 5 # equidistant binning with 20 bins in each variable 6 binningplot(x,y,b=20) 8 # quantile based binning with 20 bins in each variable 9 binningplot(x,y,b=20,bin="quant") 11 ## End(Not run) ## Not run: x <- rnorm(10000) y <- rnorm(10000) # equidistant binning with 20 bins in each variable binningplot(x,y,b=20) # quantile based binning with 20 bins in each variable binningplot(x,y,b= 20,bin="quant") ## End(Not run) For more information on customizing the embed code, read Embedding Snippets.	{"url":"https://rdrr.io/cran/mbgraphic/man/binningplot.html","timestamp":"2024-11-09T12:44:43Z","content_type":"text/html","content_length":"30965","record_id":"<urn:uuid:f677af26-04c0-46bb-8b30-e535ee856a15>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00616.warc.gz"}
Introducing Proportional On Measurement It’s been quite a while, but I’ve finally updated the Arduino PID Library. What I’ve added is a nearly-unknown feature, but one which I think will be a boon to the hobby community. It’s called “Proportional on Measurement” (PonM for short). Why you should care There are processes out there that are known as “Integrating Processes.” These are processes for which the output from the pid controls the rate of change of the input. In industry these comprise a small percentage of all processes, but in the hobby world these guys are everywhere: Sous-vide, linear slide, and 3D printer extruder temperature are all examples of this type of process. The frustrating thing about these processes is that with traditional PI or PID control they overshoot setpoint. Not sometimes, but always: This can be maddening if you don’t know about it. You can adjust the tuning parameters forever and the overshoot will still be there; the underlying math makes it so. Proportional on Measurement changes the underlying math. As a result, it’s possible to find sets of tuning parameters where overshoot doesn’t occur: Overshoot can still happen to be sure, but it’s not unavoidable. With PonM and the right tuning parameters, that sous vide or linear slide can coast right in to setpoint without going over. So What is Proportional on Measurement? Similar to Derivative on Measurement, PonM changes what the proportional term is looking at. Instead of error, the P-Term is fed the current value of the PID input. Unlike Derivative on Measurement, the impact on performance is drastic. With DonM, the derivative term still has the same job: resist steep changes and thus dampen oscillations driven by P and I. Proportional on Measurement, on the other hand, fundamentally changes what the proportional term does. Instead of being a driving force like I, it becomes a resistive force like D. This means that with PonM, a bigger Kp will make your controller more conservative. Great. But how does this eliminate overshoot? To undestand the problem, and fix, it helps to look at the different terms and what they contribute to the overall PID output. Here’s a response to a setpoint change for an integrating process (a sous-vide) using a traditional PID: The two big things to notice are: • When we’re at setpoint, the I-Term is the only contributor to the overall output. • Even though the setpoint is different at the start and end, the output returns to the same value. This value is generally known as the “Balance Point”: the output which results in 0 Input slope. For a sous-vide, this would correspond to just enough heat to compensate for heat-losses to the surroundings. Here we can see why overshoot happens, and will always happen. When the setpoint first changes, the error present causes the I-Term to grow. To keep the process stable at the new setpoint, the output will need to return to the balance point. The only way for that to happen is for the I-Term to shrink. The only way for THAT to happen is to have negative error, which only happens if you’re above PonM changes the game Here is the same sous-vide controlled using Proportional on Measurement (and the same tuning parameters): • The P-Term is now providing a resistive force. The higher the Input goes, the more negative it becomes. • Where before the P-Term became zero at the new setpoint, it now continues to have a value. The fact the the P-Term doesn’t return to 0 is the key. This means that the I-Term doesn’t have to return to the balance point on its own. P and I together can return the output to the balance point without the I-Term needing to shrink. Because it doesn’t need to shrink, overshoot isn’t required. How to use it in the new PID Library If you’re ready to try Proportional on Measurement, and you’ve installed the latest version of the PID library, setting it up is pretty easy. The primary way to use PonM is to specify it in the overloaded constructor: If you’d like to switch between PonM and PonE at runtime, the SetTunings function is also overloaded: You only need to call the overloaded method when you want to switch. Otherwise you can use the regular SetTunings function and it will remember your choice.	{"url":"http://brettbeauregard.com/blog/2017/06/introducing-proportional-on-measurement/","timestamp":"2024-11-12T23:33:37Z","content_type":"text/html","content_length":"42562","record_id":"<urn:uuid:2471ec13-9947-4145-9c10-2af29634078c>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00443.warc.gz"}
EViews Help: Estimation Method Details Estimation Method Details “ARMA Method” we described how EViews lets you choose between maximum likelihood (ML), generalized least squares (GLS), and conditional least squares (CLS) estimation for ARIMA and ARFIMA estimation. Recall that for the general ARIMA( for the unconditional residuals and innovations We will use the expressions for the unconditional residuals and innovations to describe three objective functions that may be used to estimate the ARIMA model. (For simplicity of notation our discussion abstracts from SAR and SMA terms and coefficients. It is straightforward to allow for the inclusion of these seasonal terms). Maximum Likelihood (ML) Estimation of ARIMA and ARFIMA models is often performed by exact maximize likelihood assuming Gaussian innovations. The exact Gaussian likelihood function for an ARIMA or ARFIMA model is given by The ARIMA model restricts It is well-known that for ARIMA models where See Hamilton (2004, Chapter 13, p. 372) or Box, Jenkins, and Reinsel (2008, 7.4, p. 275) for extensive discussion. Sowell (1992) and Doornik and Ooms (2003) offer detailed descriptions of the evaluation of the likelihood for ARFIMA models. In particular, practical evaluation of Equation (24.55) requires that we address several computational issues. First, we must compute the autocovariances of the ARFIMA process that appear in the Second, we must compute the determinant of the variance matrix and generalized (inverse variance weighted) residuals in a manner that is computationally and storage efficient. Doornik and Ooms (2003) describe a Levinson-Durbin algorithm for efficiently performing this operation with minimal operation count while eliminating the need to store the full Third, where possible we follow Doornik and Ooms (2003) in concentrate the likelihood with respect to the regression coefficients Generalized Least Squares (GLS) Since the exact likelihood function in Equation (24.55) depends on the data, and the mean and ARMA parameters only through the last term in the expression, we may ignore the inessential constants and the log determinant term to define a generalized least squares objective function and the ARFIMA estimates may be obtained by minimizing Conditional Least Squares (CLS) Box and Jenkins (1976) and Box, Jenkins, and Reinsel (2008, Section 7.1.2 p 232.) point out that conditional on pre-sample values for the AR and MA errors, the normal conditional likelihood function may be maximized by minimizing the sum of squares of the innovations. The recursive innovation equation in Equation (24.54) is easy to evaluate given parameter values, lagged values of the differenced We discuss below methods for starting up the recursion by specifying presample values of Notice that the conditional likelihood function depends on the data and the mean and ARMA parameters only through the conditional least squares function Coefficient standard errors for the CLS estimation are the same as those for any other nonlinear least squares routine: ordinary inverse of the estimate of the information matrix, or a White robust or Newey-West HAC sandwich covariance estimator. In all three cases, one can use either the Gauss-Newton outer-product of the Jacobians, or the Newton-Raphson negative of the Hessian to estimate the information matrix. In the remainder of this section we discuss the initialization of the recursion. EViews initializes the AR errors using lagged data (adjusting the estimation sample if necessary), and initializes the MA innovations using backcasting or the unconditional (zero) expectation. Initializing the AR Errors We can rewrite out model as so we can see that we require Typically conditional least squares employs lagged values of the variables in the model to initialize the process. For example, to estimate an AR(1) model, one may transforms the linear model, into a nonlinear model by substituting the second equation into the first, writing so that the innovation recursion written in terms of observables is given by Notice that we require observation on the Higher order AR specifications are handled analogously. For example, a nonlinear AR(3) is estimated using nonlinear least squares on the innovations given by: It is important to note that textbooks often describe techniques for estimating linear AR models like Equation (24.58) . The most widely discussed approaches, the Cochrane-Orcutt, Prais-Winsten, Hatanaka, and Hildreth-Lu procedures, are multi-step approaches designed so that estimation can be performed using standard linear regression. These approaches proceed by obtaining an initial consistent estimate of the AR coefficients All of these approaches suffer from important drawbacks which occur when working with models containing lagged dependent variables as regressors, or models using higher-order AR specifications; see Davidson and MacKinnon (1993, p. 329–341), Greene (2008, p. 648–652). In contrast, the EViews conditional least squares estimates the coefficients Thus, for a nonlinear mean AR(1) specification, EViews transforms the nonlinear model, into the alternative nonlinear regression form yielding the innovation specification: Similarly, for higher order ARs, we have: For additional detail, see Fair (1984, p. 210–214), and Davidson and MacKinnon (1993, p. 331–341). Initializing MA Innovations Computing the innovations is a straightforward process. Suppose we have an initial estimate of the coefficients, Then, after first computing the unconditional residuals All that remains is to specify a method of obtaining estimates of the pre-sample values of One may employ backcasting to obtain the pre-sample innovations (Box and Jenkins, 1976). As the name suggests, backcasting uses a backward recursion method to obtain estimates of To start the recursion, the beyond the estimation sample are set to zero: EViews then uses the actual results to perform the backward recursion: Alternately, one obvious method is to turn backcasting off and to set the pre-sample Whichever methods is used to initialize the presample values, the sum-of-squared residuals (SSR) is formed recursively as a function of the and the expression is minimized with respect to The backcast step, forward recursion, and minimization procedures are repeated until the estimates of	{"url":"https://help.eviews.com/content/timeser-Estimation_Method_Details.html","timestamp":"2024-11-12T21:54:28Z","content_type":"application/xhtml+xml","content_length":"43100","record_id":"<urn:uuid:92c50175-252b-44ae-94dc-098264009be2>","cc-path":"CC-MAIN-2024-46/segments/1730477028290.49/warc/CC-MAIN-20241112212600-20241113002600-00871.warc.gz"}
SAS Press Author Derek Morgan on Timetables and Model Trains I was sitting in a model railroad club meeting when one of our more enthusiastic young members said, "Wouldn't it be cool if we could make a computer simulation, with trains going between stations and all. We could have cars and engines assigned to each train and timetables and…" So, I thought to myself, “Timetables… I bet SAS can do that easily… sounds like something fun for Mr. Dates and Times." As it turns out, the only easy part of creating a timetable is calculating the time. SAS handles the concept of elapsed time smoothly. It’s still addition and subtraction, which is the basis of how dates and times work in SAS. If a train starts at 6:00 PM (64,800 seconds of the day) and arrives at its destination 12 hours (43,200 seconds) later, it arrives at 6:00 AM the next day. The math is start time+duration=end time (108,000 seconds,) which is 6:00 AM the next day. It doesn’t matter which day, that train is always scheduled to arrive at 6:00 AM, 12 hours from when it left. It got a lot more complicated when the group grabbed onto the idea. One of the things they wanted to do was to add or delete trains and adjust the timing so multiple trains don’t run over the same track at the same time. This wouldn’t be that difficult in SAS; just create an interactive application, but… I’m the only one who has SAS. So how do I communicate my SAS database with the “outside world”? The answer was Microsoft Excel, and this is where it gets thorny. It’s easy enough to send SAS data to Excel using ODS EXCEL and PROC REPORT, but how could I get Excel to allow the club to manipulate the data I sent? I used the COMPUTE block in PROC REPORT to display a formula for every visible train column. I duplicated the original columns (with corresponding names to keep it all straight) and hid them in the same spreadsheet. The EXCEL formula code is in line 8. I also added three rows to the dataset at the top. The first contains the original start time for each train, the second contains an offset, which is always zero in the beginning, while the third row was blank (and formatted with a black background) to separate it from the actual schedule. Figure 1: Schedule Adjustment File The users can change the offset to change the starting time of a train (Column C, Figure 2.) The formula in the visible columns adds the offset to the value in each cell of the corresponding hidden column (as long as it isn’t blank.) You can’t simply add the offset to the value of the visible cell, because that would be a circular reference. The next problem was moving a train to an earlier starting time, because Excel has no concept of negative time (or date—a date prior to the Excel reference date of January 1, 1900 will be a character value in Excel and cause your entire column to be imported into SAS as character data.) Similarly, you can’t enter -1:00 as an offset to move the starting time of our 5:35 AM train to 4:35 AM. Excel will translate “-1:00” as a character value and that will cause a calculation error in Excel. In order to move that train to 4:35 AM, you have to add 23 hours to the original starting time (Column D, Figure 2.) Figure 2: Adjusting Train Schedules After the users adjust the schedules, it’s time to return our Excel data to SAS, which creates more challenges. In the screenshot above, T534_LOC is the identifier of a specific train, and the timetable is kept in SAS time values. Unfortunately, PROC IMPORT using DBMS=XLSX brings the train columns into SAS as character data. T534_LOC also imports as the actual Excel value, time as a fraction of a day. Figure 3: How the Schedule Adjustment File Imports to SAS While I can fix that by converting the character data to numeric and multiplying by 86,400, I still need the original column name of T534_LOC for the simulation, so I would have to rename each character column and output the converted data to the original column name. There are currently 146 trains spread across 12 files, and that is a lot of work for something that was supposed to be easy! Needless to say, this “little” side project, like most model railroads, is still in progress. However, this exercise in moving time data between Microsoft Excel and SAS gave me even more appreciation for the way SAS handles date and time data. Figure 4 is a partial sample of the finished timetable file, generated as an RTF file using SAS. The data for trains 534 and 536 are from the spreadsheet in Figure 1. Figure 4: Partial Sample Timetable Want to learn more about how to use and manipulate dates, times, and datetimes in SAS? You'll find the answers to these questions and much more in my book The Essential Guide to SAS Dates and Times, Second Edition. Updated for SAS 9.4, with additional functions, formats, and capabilities, the Second Edition has a new chapter dedicated to the ISO 8601 standard and the formats and functions that are new to SAS, including how SAS works with Universal Coordinated Time (UTC). Chapter 1 is available as a free preview here. For updates on new SAS Press books and great discounts subscribe to the SAS Press New Book Newsletter. Leave A Reply This site uses Akismet to reduce spam. Learn how your comment data is processed.	{"url":"https://blogs.sas.com/content/sgf/2019/07/10/timetables-and-model-trains/","timestamp":"2024-11-03T16:28:18Z","content_type":"text/html","content_length":"45896","record_id":"<urn:uuid:adfa079b-24f7-4cba-aef6-29613c6e2c6d>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00633.warc.gz"}
Toda Bracket and Massey Product CoThanks, too. also, the definition now makes sense. I did some editing, as usual: added hyperlinks, a table of contents, a Referemces-section, pointers to the References, etc. Notice that top-level sections need two hash-signs to appear in the TOC correctly: ## Idea ## Definition ### General ### Some special case A stub Massey product and a longer Toda bracket (still plenty gaps of reference, many many unlinked words). No promises w.r.t. spellings. I now see I’ve missed the convention for capitalization. Will fix that now… done. OK, maybe I’ll take this next question to MO, but: the Toda Bracket should directly give something like the Massey Product, because if you have a bunch of maps $u_i : X \to K_i$ where the $K_i$ are suitable Eilenberg-Mac Lane spaces, one has a sequence of maps of pointed-function spaces $K_1^X \to (K_1 \otimes K_2)^X \cdots \to (K_1\otimes \cdots K_{n+2})^X$ representing the particular cup products $\bullet\smallsmile u_i$, as well as a map $\mathbb{S}^0 \to K_1^X$ adjoint to $u_1$. Then the bracket machinery highlights a family of maps $\mathbb{S}^{n} \to (K_1 \otimes \cdots K_{n+2})^X$ adjoint to $\Sigma^n X \to K_1 \otimes \cdots \otimes K_{n+2} .$ It seems intuitive to me (don’t ask why) that the construction of Massey products should at least give a subset of these Toda brackets, but I don’t feel so clear about the vice-versa. Does anyone know if this is at least spelled-out somewhere? (McCleary’s book on SpSeqs, e.g., is perfectly vague about these brackets, at least in a neighborhood of index entries.) Added references • Hans-Joachim Baues, On the cohomology of categories, universal Toda brackets and homotopy pairs, K-Theory 11:3, April 1997, pp. 259-285 (27) springerlink • Boryana Dimitrova, Universal Toda brackets of commutative ring spectra, poster, Bonn 2010, pdf • C. Roitzheim, S. Whitehouse, Uniqueness of $A_\infty$-structures and Hochschild cohomology, arxiv/0909.3222 • Steffen Sagave, Universal Toda brackets of ring spectra, Trans. Amer. Math. Soc., 360(5):2767-2808, 2008, math.KT/0611808	{"url":"https://nforum.ncatlab.org/discussion/4508/toda-bracket-and-massey-product/?Focus=36840","timestamp":"2024-11-03T11:07:50Z","content_type":"application/xhtml+xml","content_length":"48426","record_id":"<urn:uuid:88f1144c-7351-4fcd-9473-79b8271f4be5>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00644.warc.gz"}
Power-law energy spectrum and orbital stochasticity The power-law energy distributions are observed in space and fusion plasmas. The power law decay of the two-time velocity correlation function and the corresponding frequency spectrum of the correlation function are shown to be related to the power law distribution of the time interval of acceleration, which produces a power-law energy distribution. In particular, the two time correlation function, the distribution of acceleration duration, (namely the distribution of the trapping time of the quasi-trapped orbits in the vicinity of the magnetic null such as the geomagnetic tail configurations) are shown to produce a power law energy distribution function. The statistical property is applicable under conditions given here to the energy spectra of cosmic rays, electrons in laser-plasma interaction, and the radio-frequency heated confined plasmas. Pub Date: June 1991 □ Energy Distribution; □ Energy Spectra; □ Fusion; □ Orbits; □ Particle Acceleration; □ Plasma Control; □ Space Plasmas; □ Statistical Correlation; □ Stochastic Processes; □ Cosmic Rays; □ Distribution Functions; □ Electrons; □ Frequency Distribution; □ Geomagnetic Tail; □ Laser Plasma Interactions; □ Time Functions; □ Plasma Physics	{"url":"https://ui.adsabs.harvard.edu/abs/1991ples.rept.....M/abstract","timestamp":"2024-11-07T19:51:46Z","content_type":"text/html","content_length":"36698","record_id":"<urn:uuid:7b1f574d-25cb-4b16-bbbe-723329e55631>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00339.warc.gz"}
What do you want to work on? About Ann Elementary (3-6) Math, Elementary (3-6) Reading Bachelors in Kinesiology and Exercise Science from University of California-Santa Barbara Masters in Education, General from Graceland University-Lamoni Career Experience I have been an elementary school teacher in public schools for 22 years. My experience includes teaching math, English language arts, science, and social studies in 3rd, 4th, and 5th grades, physical education/health, and literacy resource teacher. My goal as a teacher is to personalize instruction in a way that makes learning meaningful. I Love Tutoring Because helping students one-on-one or in small groups allows me to offer undivided attention to the person and the skill that needs to be strengthened. In that setting, I can focus on using a student's strengths to help fill in any areas of deficit. Other Interests Playing Music, Sports Math - Elementary (3-6) Math She was the best tutor ive met, I want her to tutor me again English - Elementary (3-6) Language Arts She was very nice, her name is Ann. FP - Elementary Math smart and very nice Math - Elementary (3-6) Math I have no feedback, Miss Ann was a joy to be around. She helped me complete my packet, and helped me understand and master the concepts needed for my test. She deserves a raise and a praise!	{"url":"http://divinemercyhawaii.org/ann--8276787.html","timestamp":"2024-11-11T16:08:35Z","content_type":"application/xhtml+xml","content_length":"125298","record_id":"<urn:uuid:b38c7b08-5ad5-415b-98ea-fe0002d6f74a>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00043.warc.gz"}
The Stacks project Theorem 52.11.2. Let $(A, \mathfrak m)$ be a Noetherian local ring which has a dualizing complex and is complete with respect to an ideal $I$. Set $X = \mathop{\mathrm{Spec}}(A)$, $Y = V(I)$, and $U = X \setminus \{ \mathfrak m\} $. Let $\mathcal{F}$ be a coherent sheaf on $U$. Assume 1. $\text{cd}(A, I) \leq d$, i.e., $H^ i(X \setminus Y, \mathcal{G}) = 0$ for $i \geq d$ and quasi-coherent $\mathcal{G}$ on $X$, 2. for any $x \in X \setminus Y$ whose closure $\overline{\{ x\} }$ in $X$ meets $U \cap Y$ we have \[ \text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}_ x) \geq s \quad \text{or}\quad \text{depth}_{\mathcal{O}_{X, x}}(\mathcal{F}_ x) + \dim (\overline{\{ x\} }) > d + s \] Comments (0) Post a comment Your email address will not be published. Required fields are marked. In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar). All contributions are licensed under the GNU Free Documentation License. In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DXQ. Beware of the difference between the letter 'O' and the digit '0'. The tag you filled in for the captcha is wrong. You need to write 0DXQ, in case you are confused.	{"url":"https://stacks.math.columbia.edu/tag/0DXQ","timestamp":"2024-11-03T15:56:19Z","content_type":"text/html","content_length":"17428","record_id":"<urn:uuid:506ea541-61db-45ed-9215-8edacdc9d0d5>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00680.warc.gz"}