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Whole Numbers Worksheet
Whole Numbers Worksheet act as foundational devices in the realm of mathematics, offering an organized yet flexible system for learners to discover and grasp numerical principles. These worksheets
provide an organized strategy to comprehending numbers, nurturing a solid foundation whereupon mathematical proficiency prospers. From the easiest checking exercises to the complexities of advanced
estimations, Whole Numbers Worksheet accommodate learners of diverse ages and skill levels.
Introducing the Essence of Whole Numbers Worksheet
Whole Numbers Worksheet
Whole Numbers Worksheet -
Grade 5 math worksheets on completing whole numbers with mixed numbers Students must find the missing addend which is a fraction to complete the whole number Free fractions worksheets from K5
Learning s online reading and math program
Free math worksheets for basic operations This worksheet generator allows you to make worksheets for addition subtraction division and multiplication of whole numbers and integers including both
horizontal and vertical forms long division etc and simple equations with variables You can make worksheets for
At their core, Whole Numbers Worksheet are vehicles for theoretical understanding. They encapsulate a myriad of mathematical concepts, guiding students via the labyrinth of numbers with a collection
of interesting and deliberate exercises. These worksheets transcend the boundaries of conventional rote learning, urging energetic interaction and fostering an intuitive understanding of numerical
Supporting Number Sense and Reasoning
Free Multiplying Fractions With Whole Numbers Worksheets
Free Multiplying Fractions With Whole Numbers Worksheets
Reading Writing Whole Numbers Date Period Write each as a numeral 1 seven million 7 000 000 2 five million forty thousand 5 040 000 3 four million thirty 4 000 030 4 four hundred million four Create
your own worksheets like this one with Infinite Pre Algebra Free trial available at KutaSoftware Title Reading Writing
These Grade 6 free printable whole numbers worksheets can be used to understand the whole numbers revise class lessons or improve your kids math skills We use whole numbers to count things like 1 2 3
and so on They are also called natural numbers or counting numbers
The heart of Whole Numbers Worksheet depends on growing number sense-- a deep comprehension of numbers' significances and interconnections. They encourage expedition, welcoming learners to dissect
math procedures, decode patterns, and unlock the mysteries of series. With provocative challenges and rational problems, these worksheets end up being gateways to honing reasoning skills, nurturing
the analytical minds of budding mathematicians.
From Theory to Real-World Application
Grade 3 Whole Numbers Worksheets www grade1to6
Grade 3 Whole Numbers Worksheets www grade1to6
Number Words 1 to 20 by QKidz Matching number names and numerals 30 40 by Teacher Angela Numbers by Ali Wheeler ourmixsj85 Addition and Subtraction of Fractions by mikmik22 Adding Two Digit Numbers
Without Regrouping
Whole Number Even and Odd Worksheets With Algebra on its way it is imperative that students in elementary education have the abilitiy to successfully complete even and odd worksheets Turtle Diary is
able to present this particular logic to students using comparative questions and word problems that help organize number logic for each child
Whole Numbers Worksheet act as conduits connecting theoretical abstractions with the palpable truths of daily life. By infusing useful circumstances right into mathematical workouts, learners witness
the significance of numbers in their surroundings. From budgeting and measurement conversions to understanding statistical information, these worksheets equip trainees to wield their mathematical
expertise past the boundaries of the classroom.
Diverse Tools and Techniques
Adaptability is inherent in Whole Numbers Worksheet, employing an arsenal of instructional devices to cater to varied discovering designs. Aesthetic aids such as number lines, manipulatives, and
electronic resources serve as companions in envisioning abstract concepts. This diverse method makes certain inclusivity, accommodating students with different preferences, strengths, and cognitive
Inclusivity and Cultural Relevance
In a significantly diverse globe, Whole Numbers Worksheet accept inclusivity. They transcend cultural limits, integrating examples and troubles that reverberate with learners from varied backgrounds.
By incorporating culturally pertinent contexts, these worksheets promote an environment where every student really feels stood for and valued, improving their connection with mathematical principles.
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Whole Numbers Worksheet chart a training course towards mathematical fluency. They impart perseverance, critical thinking, and analytic skills, crucial attributes not only in maths yet in numerous
facets of life. These worksheets empower learners to navigate the elaborate terrain of numbers, nurturing a profound gratitude for the elegance and logic inherent in mathematics.
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In an era marked by technological innovation, Whole Numbers Worksheet flawlessly adjust to electronic systems. Interactive interfaces and electronic resources increase traditional discovering,
providing immersive experiences that go beyond spatial and temporal limits. This amalgamation of typical approaches with technical technologies advertises a promising age in education, promoting a
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Verdict: Embracing the Magic of Numbers
Whole Numbers Worksheet illustrate the magic inherent in maths-- a captivating journey of exploration, exploration, and proficiency. They go beyond traditional pedagogy, working as stimulants for
stiring up the flames of curiosity and inquiry. Via Whole Numbers Worksheet, learners start an odyssey, opening the enigmatic globe of numbers-- one problem, one option, at a time.
Rounding Whole Numbers Worksheet
Comparing And Ordering Whole Numbers Worksheets
Check more of Whole Numbers Worksheet below
Worksheets Rounding Whole Numbers Mreichert Kids Worksheets
Whole Number Arithmetic Worksheets Worksheet Resume Examples
Multiplying Fractions And Whole Numbers Worksheets
Ordering Fractions Decimals And Whole Numbers Worksheet Have Fun Teaching
Worksheets For CBSE Class 6 Maths Whole Numbers With Answers Class 6 Maths 6 Class Free Math
Dividing Whole Numbers Worksheet Grade1to6
Free Printable Math Worksheets For Basic Operations Whole Numbers
Free math worksheets for basic operations This worksheet generator allows you to make worksheets for addition subtraction division and multiplication of whole numbers and integers including both
horizontal and vertical forms long division etc and simple equations with variables You can make worksheets for
Operations With Whole Numbers Skills Worksheet
OPERATIONS WITH WHOLE NUMBERS SKILLS QUESTIONS Q1 Write the following in words a 245 b 3894 c 9602 Q2 Write the following in extended form a 3643 b 805 c 212 435 Q3 Compare the values of the
nominated digits in each of the following a 9180 823 8s b 42 375 120 2s Q4 Write the following as numbers
Free math worksheets for basic operations This worksheet generator allows you to make worksheets for addition subtraction division and multiplication of whole numbers and integers including both
horizontal and vertical forms long division etc and simple equations with variables You can make worksheets for
OPERATIONS WITH WHOLE NUMBERS SKILLS QUESTIONS Q1 Write the following in words a 245 b 3894 c 9602 Q2 Write the following in extended form a 3643 b 805 c 212 435 Q3 Compare the values of the
nominated digits in each of the following a 9180 823 8s b 42 375 120 2s Q4 Write the following as numbers
Ordering Fractions Decimals And Whole Numbers Worksheet Have Fun Teaching
Whole Number Arithmetic Worksheets Worksheet Resume Examples
Worksheets For CBSE Class 6 Maths Whole Numbers With Answers Class 6 Maths 6 Class Free Math
Dividing Whole Numbers Worksheet Grade1to6
Whole Numbers Class 6 Worksheet Fun And Engaging Exercise
Add And Subtract Whole Numbers Worksheet
Add And Subtract Whole Numbers Worksheet
Divide Whole Numbers By Decimals Worksheets For Kids Online SplashLearn | {"url":"https://szukarka.net/whole-numbers-worksheet","timestamp":"2024-11-08T01:36:40Z","content_type":"text/html","content_length":"26397","record_id":"<urn:uuid:511743ee-1b3b-4f87-94ef-bfce06501359>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00297.warc.gz"} |
How do you convert g to mg? + Example
How do you convert g to mg?
1 Answer
Both $m g$ and $g$ represents some mass. Therefore it is easy to convert between both of them.
The $g$ stands from gram in both units. The $m$ stand for milli, so we get milligram. This means that this is $\frac{1}{1000}$ gram.
1 gram consists of 1000 mg. So if we want to convert gram to mg, we multiply the amount of gram with 1000.
So for example, 5 grams gives 5000 mg.
In the image below, a stair is drawn with the units. We can see $g$ in the middle and $m g$ at the right end. Now every step we take we have to do the action that is written with it.
So if you go to the right down, we have to multiply the number with 10. If you go up, you have to divide it by 10.
To get from $g$ to $m g$ we have to do 3 times the $\times 10$, thus:
$10 \times 10 \times 10 = 1000$
So in one step you can multiply the $g$ with 1000 to get the $m g$.
For more information check the metric system unit conversion!
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LuciadLightspeed coordinate reference systems
A coordinate reference system (CRS) specifies the coordinate system that is used to define locations on the Earth or on a flat surface representing the Earth. Based on this definition, there are two
types of CRSes:
• Geodetic reference systems (ILcdGeodeticReference) that represent geographic locations on an ellipsoidal (or a spherical) surface using longitude and latitude coordinates. Geodetic reference
systems are based on an ellipsoid that approximates the shape of the earth. A commonly used geodetic reference system is the World Geodetic System 1984 (WGS 1984).
• Cartesian or grid reference systems (ILcdGridReference) that represent geographic locations on a flat surface using x and y coordinates. A Cartesian reference system is usually based on a
geodetic reference system and requires a map projection to represent the curved surface of the earth on a flat surface as shown in Figure 1, “Projecting the Earth’s surface to a flat surface&
Figure 1. Projecting the Earth’s surface to a flat surface
Each CRS is associated with a geodetic datum (ILcdGeodeticDatum). A geodetic datum typically defines an ellipsoid and its position in relation to the Earth. The ellipsoid’s position is specified by
linking the ellipsoid to a number of physical points on the surface of the Earth. Alternatively, the ellipsoid’s position can be specified by means of a translation and a rotation relative to a known
geodetic datum, for example the one used by the WGS 1984.
Refer to the API reference for more details on ILcdGeodeticReference and ILcdGridReference and their implementations. Model, world, and view coordinates introduces the model, view, and world
coordinate systems and the relationships between them.
Model, world, and view coordinates
The position of the model data is defined at three different levels in LuciadLightspeed using:
• Model coordinates defined within a coordinate system determined by the model reference. The model reference (ILcdModelReference), or CRS associated with an ILcdModel, usually is an
ILcdGeodeticReference or ILcdGridReference.
• World coordinates defined within a coordinate system determined by the world reference. The world reference (ILcdXYZWorldReference), or CRS associated with an ILspView, usually is an
ILcdGridReference for 2D or an ILcdGeocentricReference for 3D. For an ILcdGXYView, the associated world reference (ILcdXYWorldReference) usually is an ILcdGridReference.
• View coordinates defined by the 2D screen coordinates (or pixel coordinates).
Both the world and the view coordinate systems are associated with an ILspView or an ILcdGXYView. The world level is an intermediate level between the model and the view, which is used internally in
the ILspView and ILcdGXYView, mainly for performance reasons. One of the benefits of LuciadLightspeed is that the model references of your data can differ from each other and from the world
reference. LuciadLightspeed automatically converts the model data to the world reference. All model data is displayed in the same reference system in the view and you do not need to transform any
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Stata For Categorical Data Analysis - UMass PDF Free Download
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data AnalysisStata version 14IllustrationCategorical Data AnalysisSpring 2016I. Single 2x2 Table . . . .1.2.3.4.Tests of
Association using tabi w direct entry of counts . .Tests of Association using tabulate . .(Cohort Design) Using the command cs .(Case-Control Design) Using the command cc .23456II. Stratified
Analysis of K 2x2 Tables . . .71. How to Enter Tabular Data . .2. Descriptives – Numerical .3. Descriptives – Graphical .3a. Bar Graph of % Event, Over Strata 3b. Odds Ratio (95% CI), Over Strata .4.
Mantel Haenszel Test of Null: Homogeneity of Odds Ratio .5. Mantel Haenszel Test of Null: Odds RatioCOMMON 1 9121414151718III. 2xC Table Analysis of Trend . . .191. Descriptives – Numerical .2.
Descriptives – Graphical .2a. Mean % Event (95% CI), Over Dose .2b. Odds Event (95% CI), Over Dose 3. Chi Square Test of General Association using tabulate and tabchi.4. Test of Trend 2xC Table using
tabodds .202323242526IV. RxC Table Analysis of Trend using nptrend . . 27V. Chi Square Goodness of Fit Test using chitesti . . . . .28 1. Teaching\stata\stata version 14\stata version 14 – SPRING
2016\Stata for Categorical Data Analysis.docxPage 1 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data AnalysisI - Single 2x2 TableIntroduction to ExamplesExample 1Example 1 is used in Section 1.1 There is not an actual data set.
Instead, you enter counts as part of the command you issue.Source: Fisher LD and Van Belle G. Biostatistics: A Methodology for the Health Sciences.New York: Wiley, 1993. Chapter 6 problem 5, page
232. Smith, Delgado and Rutledge (1976) report dataon ovarian carcinoma. Individuals had different numbers of courses of chemotherapy. The 5-year survival datafor those with 1-4 and 10 or more
courses of chemotherapy are shown below.Courses1-4 10Five Year StatusDeadAlive21228Do these data provide statistically significant evidence of an association of five year survival with number
ofcourses of chemotherapy?Example 2 (Example 2 is used in Section 1.2.The data set single2x2.dta contains the following 2x2 table of counts.Exposure (Smoking)YesNoDisease (Lung Cancer)YesNo9312471178
1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docx404989Page 2 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis1a. Tests of Association using the immediate command tabi and direct entry of countsGood to know. Sometimes, you want
to be able to do a quick analysis of count data in a table and you want to, simply,type in the cell counts (instead of taking the time to create a Stata data set). Stata has “immediate” commands that
letyou do just that!Tips:(1) For small to moderate sample sizes, use the option exact to obtain a Fisher Exact Test(2) If the cell sizes are too small, Stata will not allow the option chisq to obtain
a Pearson Chi Square Test;this is alright, since this test is not valid when the cell sizes are too small(3) Stata, however, will allow you to perform a likelihood ratio chi square test. Use the
option lrchi. * Fisher Exact Test (For small to moderate cell sizes). * tabi row1col1 row1col2\row2col1 row2col2, exact.tabi 21 2 \2 8, exact.* tabi row1col1 row1col2\row2col1 row2col2, exact. tabi
21 2\2 8, exact colrow 12 Total----------- ---------------------- ---------1 212 232 28 10----------- ---------------------- ---------Total 2310 33Fisher's exact 1-sided Fisher's exact 0.0000.000. *
Likelihood Ratio (LR) Chi Square Test.* tabi row1col1 row1col2\row2col1 row2col2, lrchi. tabi 21 2 \2 8, lrchi.* tabi row1col1 row1col2\row2col1 row2col2, lrchi. tabi 21 2\2 8,lrchi colrow 12
Total----------- ---------------------- ---------1 212 232 28 10----------- ---------------------- ---------Total 2310 33likelihood-ratio chi2(1) 16.8868Pr 0.000 1. Teaching\stata\stata version 14\
stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 3 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis1b. Tests of Association using the command tabulateNote – The command tabulate is NOT an immediate command. It
requires a stata data set in working memoryPreliminary - Input the stata data set single2x2.dta.Note – This data set is accessible through the internet. Alternatively, you can download it from the
course website.a) In Stata, input directly from the internet using the command useuse e2x2.dta”, clearb) From the course website, right click to download. Afterwards, in Stata, use FILE OPENSee,
ations.html(1) For small to moderate sample sizes, use the option exact to obtain a Fisher Exact Test(2) If the cell sizes are too small, Stata will not allow the option chisq to obtain a Pearson Chi
Square Test;this is alright, since this test is not valid when the cell sizes are too small(3) Stata, however, will allow you to perform a likelihood ratio chi square test. Use the option lrchi. *
Fisher Exact Test. * tabulate rowvariable columnvariable, exact. ** tabulate rowvariable columnvariable, exact. tabulate smoking lungca, exact lungcasmoking 01 Total----------- ----------------------
---------Non-smoker 472 49Smoker 319 40----------- ---------------------- ---------Total 7811 89Fisher's exact 1-sided Fisher's exact 0.0110.010. * Likelihood Ratio (LR) Chi Square Test. * tabulate
rowvariable columnvariable, lrchi. * tabulate rowvariable columnvariable, lrchi. tabulate smoking lungca, lrchi lungcasmoking 01 Total----------- ---------------------- ---------Non-smoker 472
49Smoker 319 40----------- ---------------------- ---------Total 7811 89likelihood-ratio chi2(1) 7.2120Pr 0.007 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical
Data Analysis.docxPage 4 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis1c. (Cohort Design) Using the command csTips:(1) Stata also provides an immediate version of this command for use with
direct entry of cell frequencies. The command is csi(2) For a cohort study design, stata will report the estimated relative risk (risk ratio), the RR. Use option or to obtain the odds ratio.(3) To
obtain a Fisher Exact test, use the option exact(4) Be sure to type help cs to view all the other options possible with this command. * cs diseasevariable exposurevariable, exact. * cs
diseasevariable exposurevariable, exact. cs lungca smoking, exact smoking ExposedUnexposed Total----------------- ------------------------ -----------Cases 92 11Noncases 3147 78-----------------
------------------------ -----------Total 4049 89 Risk .225.0408163 .1235955 Point estimate [95% Conf. Interval] ------------------------ -----------------------Risk difference .1841837
.0434157.3249517Risk ratio 5.5125 1.26221224.07491Attr. frac. ex. .8185941 .2077403.958463Attr. frac. pop .6697588 sided Fisher's exact P 0.01002-sided Fisher's exact P 0.0108 1. Teaching\stata\stata
version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 5 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis1d. (Case-control Design) Using the command ccTips:(1) Stata also provides an immediate version of this command for
use with direct entry of cell frequencies. The command is cci(2) For a case-control study design, stata will report the estimated odds ratio, the OR. Use option or to obtain the odds ratio.(3) To
obtain a Fisher Exact test, use the option exact(4) Be sure to type help cc to view all the other options possible with this command. * cc diseasevariable exposurevariable, exact. * cc
diseasevariable exposurevariable, exact. cc lungca smoking, exactProportion ExposedUnexposed TotalExposed----------------- ------------------------ -----------------------Cases 92 110.8182Controls
3147 780.3974----------------- ------------------------ -----------------------Total 4049 890.4494 Point estimate [95% Conf. Interval] ------------------------ -----------------------Odds ratio
6.822581 1.26362867.65054 (exact)Attr. frac. ex. .8534279 .2086276.9852182 (exact)Attr. frac. pop .6982592 sided Fisher's exact P 0.01002-sided Fisher's exact P 0.0108 1. Teaching\stata\stata version
14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 6 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data AnalysisII – Stratified Analysis of K TablesIntroduction to ExampleIn this illustration, you enter the data in tabular form.
Then, you use the command expand to create the full data set.Note – This is a subset of the data used in the Unit 4 (Categorical Data Analysis) practice problems.Source: Fisher LD and Van Belle G.
Biostatistics: A Methodology for the Health Sciences.New York: Wiley, 1993. Chapter 6 problem 14, page 235.Rosenberg et al (1980) performed a retrospective study of the association of coffee drinking
(exposure)and the occurrence of myocardial infarction (MI) (outcome) among n 494. Information on smokingwas also available. The analysis investigated possible modification of the coffee-MI
relationshipwith smoking status (stratification). The sample size is n 494.Is the association of coffee with myocardial infarction different, depending on smoking status (“effect modification”)?If
the association is the same, regardless of smoking status, is there an association of coffee consumption withmyocardial infarction at all? This is an example of a stratified analysis of an
exposure-disease relationship.A stratified analysis of K 2x2 tables is used to assess:(1) evidence of modification of an exposure-disease relationship by changes in the value of a third(stratifying)
variable; or(2) in the absence of modification, a Mantel-Haenszel analysis of an exposure disease relationshipcontrolling for confounding.The data are in tabular form (next page). 1. Teaching\stata\
stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 7 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data AnalysisNote. You might see tables that are “flipped” - The layout of tables here is the following. For rows, row 1 denotesthe
exposure of interest while row 2 denotes the lack of exposure. For columns, column 1 denotes the outcome ofinterest while column 2 denotes controls. In contrast, Stata defines rows and columns
according to their values, withrow 1 being the lower value and column 1 being the lower value. Thus 0/1 variables, when cross-tabbed, aredisplayed “flipped” in Stata. Bummer. To get around this and
to obtain the right display, you could use 1/2 variables,with the value 1 for exposed (in the case of the row variable) and 1 for the outcome (in the case of the columnvariable).Stratum 1: FORMER
SMOKER (smoking 1)Cups Coffee per day 5 (coffee 1) 5 (coffee 2)MI (mi 1)7 (tally 7)20Control (mi 2)18112Stratum 2: 1-14 CIGARETTES/DAY (smoking 2)Cups Coffee per day 5 (coffee 1) 5 (coffee 2)MI (mi
1)733Control (mi 2)2411Stratum 3: 35-44 CIGARETTES/DAY (smoking 3)Cups Coffee per day 5 (coffee 1) 5 (coffee 2)MI (mi 1)2755Control (mi 2)2458Stratum 4: 45 CIGARETTES/DAY (smoking 4)Cups Coffee per
day 5 (coffee 1) 5 (coffee 2)MI (mi 1)3034 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxControl (mi 2)1717Page 8 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis2a. How to Enter Tabular DataTabular data is convenient for data entry. The result, however, is data in collapsed
form. For example, consider thefirst table on the previous page. It shows that there are 7 individuals who are former smokers (stratum 1) who drank 5 cups of coffee per day (coffee 1) and who
experienced an MI (mi 1). We will enter this record just once and keep trackof the frequency of this observation equal to 7 using a variable called tally.Tip – It is possible to analyze tabular data
in Stata. Each profile of variable values is “weighted” by their frequency ofoccurrence (in our case by the variable tally). However, there are some analyses that we might want to do thatcannot be
performed using tabular data. For this reason, after entering the tabular data, the expand function is usedto create a full data set.Coding ManualTips –(1) Always create a coding manual before
creating a data set(2) Use “lower case” only variable names.Example Variable Variable Labelsmoking Smoking StatusFormatsmokingfcoffeeCups Coffee Per DaycoffeefmiMyocardial InfarctionmiftallyFrequency
weight-. *. ***** Illustration:. set more off.Format code definitions1 Former smoker2 1-14 cigs/day3 35-44 cigs/day4 45 cigs day1 5 cups/day2 Less1 MI2 Non-MI0, 1, 2, .Entering Tabular Data******
Initialize variable names and set initial value to missing.generate smoking .generate coffee .generate mi .generate tally . 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for
Categorical Data Analysis.docxPage 9 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis. *. *****From the top menu, click on the data editor icon. *. *****Enter your data.You should see the following.Note
– Enter your table by table and cell by cell, beginning with the first table (smoking 1for the “Former smokers”) as follows. For each table, enter the data row by row, beginningwith the first row.
When you are done, your completed spreadsheet should be populated exactlyas shown below. Close the data editor window. Don’t worry! Your data is not lost. 1. Teaching\stata\stata version 14\stata
version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 10 of 29
Stata Version 14 – Spring 2016. *. ellabelIllustration for Unit 4 - Categorical Data AnalysisClose the data editor window to return to the command window.Assign variable names. Create value labels.
Assign value labels.variable smoking "Stratum of Smoking"variable coffee "Cups of Coffee/Day"variable mi "MI - Myocardial Infarction"define smokingf 1 "Former Smoker" 2 "1-4 cigs/day" 3 "35-44 cigs/
day" 4 "45 cigs/day"define mif 1 "MI" 2 "Non-MI"define coffeef 1 “5 cups” 2 “Less”values smoking smokingfvalues coffee coffeefvalues mi mif. *. ***** Save tabular data in stata data set called
coffeemi tabular.dta. save "/Users/cbigelow/Desktop/coffeemi tabular.dta"file /Users/cbigelow/Desktop/coffeemi tabular.dta saved. *. ***** Create expanded data set. expand tally(478 observations
created)Check.Save as coffeemi full.dta. drop tally. save "/Users/cbigelow/Desktop/coffeemi full.dta"file /Users/cbigelow/Desktop/coffeemi full.dta saved 1. Teaching\stata\stata version 14\stata
version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 11 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis2b. Descriptives - NumericalStata commands for obtaining numerical descriptions of data have been introduced
previously. The following aresuggestions to use in a stratified analysis of multiple 2x2 tables. * Overall crosstab to look at the data, with some summary statistics. * tab2 rowvariable
columnvariable, row columnCommand is tab2. tab2 coffee mi, row column- tabulation of coffee by mi MI - MyocardialCups of InfarctionCoffee/Day MINon-MI Total----------- ----------------------
---------5 cups 7183 154 46.1053.90 100.00 33.3329.54 31.17----------- ---------------------- ---------Less 142198 340 41.7658.24 100.00 66.6770.46 68.83----------- ----------------------
---------Total 213281 494 43.1256.88 100.00 100.00100.00 100.00. * Crosstab, separately over strata of 3rd variable. Command is tab2 with by. * Sort data first. Command is sort. * by sortvariable:
tab2 rowvariable columnvariable, row column. sort smoking. by smoking: tab2 coffee mi, row column- smoking Former Smoker- tabulation of coffee by mi MI - MyocardialCups of InfarctionCoffee/Day
MINon-MI Total----------- ---------------------- ---------5 cups 718 25 28.0072.00 100.00 25.9313.85 15.92----------- ---------------------- ---------Less 20112 132 15.1584.85 100.00 74.0786.15
84.08----------- ---------------------- ---------Total 27130 157 17.2082.80 100.00 100.00100.00 100.00 100.00100.00 100.00 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for
Categorical Data Analysis.docxPage 12 of 29
Stata Version 14 – Spring 2016----some output omittedIllustration for Unit 4 - Categorical Data Analysis----- smoking 45 cigs/day- tabulation of coffee by mi MI - MyocardialCups of InfarctionCoffee/
Day MINon-MI Total----------- ---------------------- ---------5 cups 3017 47 63.8336.17 100.00 46.8850.00 47.96----------- ---------------------- ---------Less 3417 51 66.6733.33 100.00 53.1250.00
52.04----------- ---------------------- ---------Total 6434 98 65.3134.69 100.00 100.00100.00 100.00.****Tip! Here is a much more compact display of descriptives over strataCommand is tabulate with
options summarize and means.tabulate stratumvariable rowvariable, summarize(columnvariable) meansWARNING – In order to obtain mean Percent, the variable must be 0/1generate mi01 mireplace mi01 0 if
mi 2generate coffee01 coffeereplace coffee01 0 if coffee 2label variable coffee01 “Cups of Coffee/Day”label define coffeef2 0 “Less” 1 “5 cups”label values coffee01 coffeef2tabulate smoking coffee01,
summarize(mi01) meansMeans of mi01Stratum of Cups of Coffee/DaySmoking Less5 cups Total----------- ---------------------- ---------Former Sm .15151515.28 .171974521-4 cigs/ .75 .22580645
.5333333335-44 cig .48672566 .52941176 .545 cigs/ .66666667 .63829787 .65306122----------- ---------------------- ---------Total .41764706 .46103896 .43117409KeyIt works!! Because the variable mi01
that we created is coded as 0 NON MI and 1 MI, the value of thesample mean of mi01 will be equal to the % who experience MI. Thus, we see the following:(1) Overall, 43% experienced an MI(2) Among
former smokers whose coffee consumption is “LESS”, 15% experienced an MIEtc. 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 13 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis2c. Descriptives - GraphicalStata also offers several graphical options. Two are shown here. One is a bar graph, which
is often used but notalways a great choice. The second is a plot of the odds ratios, together with their 95% confidence limits.2c. (a) Bar GraphImportant! – The following graph requires that your
column variable (outcome) be coded 0/1.****Bar Graph to display % experiencing outcome, over exposure, separately for eachvalue of the 3rd variable (the stratification variable).Command is graph bar
with options over and overgraph bar columnvariable, over(rowvariable) over(stratificationvariable). ** Bare bones. graph bar mi01, over(coffee) over(smoking). ** Graph again, this time with lots of
aesthetics. graph bar mi01, over(coffee, gap(10)) over(smoking, gap(80)) outergap(50) ytitle("ProportionExperiencing MI") title("Coffee and MI") subtitle("by Smoking Status") ylabel(0(.2)1)caption
("mi01 bar.png", size(vsmall)) 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 14 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis2c. (b) Odds Ratios and 95% CIPreliminary! – This is for the brave. It requires a number of steps. * Graph to display
odds ratio and 95% CI, overall and by strata. * Command is graph twoway (scatter orvariable rowvariable) (rcap lower upper rowvariable)Note – There are fancier ways of doing this, but the syntax can
be complicated.and 95% confidence limits and use these to create a little data set for plottingapplication of the stata graph command graph twoway. Note (see next page) thatobservations, one for each
of the 4 strata of smoking, plus a 5th observation forHere, I obtain the ORusing a simplemy data set has 5the overall. *. ***** Obtain stratum specific OR and 95% CI limits. Command is mh with option
by( ). mhodds mi01 coffee01, by(smoking)Maximum likelihood estimate of the odds ratioComparing coffee01 1 vs. coffee01 0by -----------------------------------smoking Odds Ratiochi2(1)P chi2[95% Conf.
Interval]---------- -----------------Former S 2.1777782.420.11970.796945.951151-4 cigs 0.09722219.810.00000.027170.3479135-44 ci 1.1863640.250.61390.610312.3061245 cigs ------------- some output
omitted --.***** Create a new "little" data set containing the information to be plottedcleargenerate or .generate high .generate low .generate smoking . *. ****Click on the DATA EDITOR icon to enter
the data.You should have the following. 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 15 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis. *. **** Graph the odds ratios. graph twoway (scatter or smoking, msymbol(d)) (rcap low high smoking), yline(1,lwidth
(thin)lpattern(dash) lcolor(black)) xlabel(0 "Overall" 1 "Former" 2 "1-4 cigs" 3 "35-44 cigs" 4 "45 cigs", angle(45)) title("Relative Odds Mycardial Infarction") subtitle("Associated with HighCoffee
Consumption") ytitle("Odds Ratio, 95% CI") legend(off) caption("mi or.png",size(vsmall)) 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage
16 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis2d. Mantel-Haenszel Test of Null: Homogeneity of the Odds RatioThe command cc , will produce the results of both the
Mantel-Haenszel test of homogeneity of odds ratios and theMantel-Haenszel test of the common odds ratio 1. Recall - “cc” stands for “case-control”. * Mantel-Haenszel Test of Null: Homogeneity of Odds
Ratio over K strata of 2x2. * Command is cc outcome exposure, by(stratificationvariable). cc mi01 coffee01, by(smoking)Stratum of Smoki OR[95% Conf. Interval]M-H Weight----------------- rmer Smoker
2.177778.67521636.3609922.2929941-4 cigs/day .0972222.0281342.321265910.5635-44 cigs/day 1.186364.58054952.4295078.0487845 cigs/day .8823529.3533522.2044475.897959----------------- ude
1.192771.79764631.781013M-H combined ----------------------------------------Test of homogeneity (M-H)chi2(3) 19.92 Pr chi2 0.0002Test that combined OR 1:Mantel-Haenszel chi2(1) Pr chi2 pretation:The
Mantel Haenszel test of homogeneity of odds ratio is statistically significant(Chi Square with df 3 19.92, p-value .0002). The assumption of the null hypothesisof no association, when applied to the
observed data, has led to an extremely unlikelyevent. The null hypothesis is rejected. Conclude that there is statistically significantevidence that the association of high coffee consumption with
event of myocardialinfarction is different, depending on smoking status.I find no effect modification. Next, I would like to assess Confounding After assessing effect modification using the Mantel
Haenszel test of equality of stratum specific odds ratios anddetermining that there is no significant evidence of variations in the exposure-outcome relationship by level of the 3rdvariable, the
reasonable next question is: Is there confounding? Unfortunately there is no test of the equality ofthe Null: Crude Odds Ratio equality of the Mantel-Haenszel odds ratio.Suggestion: Compare the crude
and adjusted. You might want to compute the relative difference as a measure ofconfounding. We have what we need from the output above. The % difference is nearly 54%
suggestingconfounding.----------------- ude 1.192771.79764631.781013(exact)M-H combined ----------------------------------------- 0.7751256 - 1.192771 Relative difference x 100% 53.9% 0.77521256 1.
Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 17 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis2e. Mantel-Haenszel Test of Null: Odds RatioCOMMON 1The command cc , in addition to producing the Mantel-Haenszel test
of homogeneity of odds ratios, will alsoproduce the Mantel-Haenszel test of the common odds ratio 1. Recall. * Mantel-Haenszel Test of Null: Odds RatioCOMMON 1. * Command is cc outcome exposure, by
(stratificationvariable). cc mi01 coffee01, by(smoking)Stratum of Smoki OR[95% Conf. Interval]M-H Weight----------------- rmer Smoker 2.177778.67521636.3609922.2929941-4 cigs/day
.0972222.0281342.321265910.5635-44 cigs/day 1.186364.58054952.4295078.0487845 cigs/day .8823529.3533522.2044475.897959----------------- ude 1.192771.79764631.781013M-H combined
----------------------------------------Test of homogeneity (M-H)chi2(3) 19.92 Pr chi2 0.0002Test that combined OR 1:Mantel-Haenszel chi2(1) Pr chi2 pretation:In real world practice, because we have
evidence of effect modification of thecoffee-MI relationship, depending on smoking status, we would not actually performthis test.The results shown here indicate that, on average, there is no
association ofhigh coffee consumption with event of myocardial infarction Chi Square on df 1 1.65,p-value .1992). 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for
Categorical Data Analysis.docxPage 18 of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data AnalysisIII – 2xC Table Analysis of TrendIntroduction to ExampleSource: Tuyns AJ, Pequignot G and Jenson OM (1977) Le cancer
de l’oesophage en Ille-et-Villaine en function des niveaux deconsummation d’alcool et de tabac. Bull Cancer 64: 45-60.The following are excerpted data from a case-control study of the relationship
between alcohol consumption at 4increasing levels (“doses”) and case-control status for the disease of esophageal cancer.CasesControlsTotal0-3929386415Alcohol Consumption (g/day)
40-7980-119755128087355138120 452267Total200775975Because the study design is case-control, an appropriate measure of association is the odds ratio measure ofassociation. We are specifically
interested in how the relative odds of esophageal cancer changes with increasingalcohol consumption. Thus, there are at least two research questions:1. Does the odds of esophageal cancer differ by
level of alcohol consumption?(Test of general association)HO: No association between exposure and diseaseHA: Any association between exposure and disease (unspecified)2. If the odds of esophageal
cancer differs by level of alcohol consumption, then does the odds of esophagealcancer increase with increasing level of alcohol consumption? (Test of trend)HO: No association between exposure and
diseaseHA: Monotone increasing (or decreasing) association between exposure and disease (trend)Preliminary - Input the stata data set esophageal cancer.dta.Note – This data set is also accessible
through the internet. Alternatively, you can download it from the course website.(a) In Stata, input directly from the internet using the command useuse ageal cancer.dta”, clear(b) From the course
website, right click to download. Afterwards, in Stata, use FILE OPENSee, ations.html 1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docxPage 19
of 29
Stata Version 14 – Spring 2016Illustration for Unit 4 - Categorical Data Analysis3a. Descriptives - NumericalA quick look at the data set using the command describe reveals that these data are in
tabular form. describeContains data from al cancer.dtaobs:10vars:325 Mar 2011 --storagedisplayvaluevariable nametypeformatlabelvariable lfloat%9.0galcoholfAlcohol g per
--------------------------Sorted by:. * Expand tabular data to produce full data set. * expand countvariableCommand is expand. * expand countvariable. expand tally(224 observations created). tab2
case alcohol [fweight tally]- tabulation of case by alcohol Alcohol g per daycase 0-39g40-7980-119g120 Total----------- -------------------------------------------- ---------control 473195 92case
2995 25----------- ----
1. Teaching\stata\stata version 14\stata version 14 – SPRING 2016\Stata for Categorical Data Analysis.docx Page 5of 29 1c. (Cohort Design) Using the command cs Tips: (1) Stata also provides an
immediate version of this command for use with direct entry | {"url":"https://wixlib.com/document/f469_stata-for-categorical-data-analysis-umass.html","timestamp":"2024-11-03T06:56:57Z","content_type":"text/html","content_length":"107105","record_id":"<urn:uuid:3f10747c-3ba6-494a-aac6-d2bac5c74187>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00530.warc.gz"} |
Diluted Earnings Per Share | Examples | Advantages and Limitations
Updated July 25, 2023
Definition of Diluted Earnings Per Share
The term “diluted EPS” is an acronym for diluted earnings per share. It refers to the distribution of the company’s net earnings to the stockholders based on the assumption that all the
securities (such as convertible preferred shares, employee stock options, warrants, convertible bonds, etc.) that have the option to convert into common shares will be duly executed during the
given period of time.
In other words, the calculation of diluted earnings per share considers the impact of common shares and convertible securities to show what might happen to earnings distribution if all the
convertible securities were turned into common shares overnight. In this way, the underlying theory for diluted EPS is exactly the opposite of basic EPS.
The formula for diluted EPS can be derived by dividing the difference between the net income generated by the company and preferred dividends paid (adjusted for payment made to the dilutive
securities) by the summation of the weighted average number of common shares outstanding and conversion of dilutive securities. The mathematical representation of the formula is as below:
Diluted EPS = (Net Income – Preferred Dividend + Paid out to Dilutive Securities) / (Weighted Average No. of Common Shares Outstanding + Conversion of Dilutive Securities)
Examples of Diluted Earnings Per Share(With Excel Template)
Let’s take an example to understand the calculation of Diluted Earnings Per Share in a better manner.
Example #1
Let us take the example of a company named GHJ Inc. to illustrate the computation of diluted Earnings Per Share. In 2018, the company registered a net income of $20 million, out of which it paid out
$2 million to the owners of the 1 million preference shares in the form of dividends. 60% of the preference shares are convertible into 1:1 common stock. Further, it had 500,000 bonds (par value of
$10, interest rate 4%, convertible to 1 common stock each) besides the weighted average number of common shares outstanding of 5 million. Calculate the diluted EPS of GHJ Inc. for 2018 if the
effective tax rate is 30%.
The formula to calculate Paid out to Dilutive Securities is as below:
Paid out to Dilutive Securities = 60% * Preference Dividend + No. of Convertible Debt * Par Value * Interest Rate * (1 – Tax Rate)
• Paid out to Dilutive Securities = 60% * $2 million + 0.5 million * $10 * 4% * (1 -35%)
• Paid out to Dilutive Securities = $1.34 million
The formula to calculate the Conversion of Dilutive Securities is as below:
Conversion of Dilutive Securities = 60% * No. of Preference Shares + No. of Convertible Debt
• Conversion of Dilutive Securities = 60% * 1 million + 0.5 million
• Conversion of Dilutive Securities = 1.10 million
The formula to calculate Diluted Earnings per Share is as below:
Diluted EPS = (Net Income – Preferred Dividend + Paid out to Dilutive Securities) / (Weighted Average No. of Common Shares Outstanding + Conversion of Dilutive Securities)
• Diluted EPS = ($20.0 million – $2.0 million + $1.34 million) / (5.0 million + 1.1 million)
• Diluted EPS = $3.17 per share
Therefore, GHJ Inc. managed a diluted EPS of $3.17 per share during the year 2018.
Example #2
Let us take the example of Apple Inc. to show how diluted Earnings Per Share is calculated in most cases practically. As per the latest annual report 2018, the company generated a net income of
$59.53 billion. Further, during the year, the company’s weighted average number of common shares outstanding was 4.96 billion, while the impact of conversion of the dilutive securities was 0.04
billion. Determine the diluted EPS of the company for the year 2018.
The formula to calculate Diluted Earnings Per Share is as below:
Diluted EPS = (Net Income – Preferred Dividend + Paid out to Dilutive Securities) / (Weighted Average No. of Common Shares Outstanding + Conversion of Dilutive Securities)
• Diluted EPS = ($59.53 billion – 0 + 0) / (4.96 billion + 0.04 billion)
• Diluted EPS = $11.91 per share
Therefore, Apple Inc.’s diluted EPS comes down to $11.91 per share for the year 2018.
Source Link: Apple Inc. Balance Sheet
Example #3
Let us take the example of Walmart Inc. to illustrate the diluted Earnings Per Share calculation. In 2018, the company booked a net income of $10.52 billion, paying out $0.66 billion against
non-controlling interest. Further, the diluted weighted average no. of common shares outstanding was 3.01 billion. Calculate the diluted EPS of Walmart Inc. for the year 2018.
The formula to calculate Diluted Earnings Per Share is as below:
Diluted EPS = (Net Income – Paid to Non-Controlling Interest + Paid out to Dilutive Securities) / Diluted Weighted Average No. of Common Shares Outstanding
• Diluted EPS = ($10.52 billion – $0.66 billion + 0) / 3.01 billion
• Diluted EPS = $3.28 per share
Therefore, Walmart Inc.’s diluted EPS for the year 2018 stood at $3.28 per share.
Source: Walmart Annual Reports (Investor Relations)
Advantages of Diluted Earnings Per Share
Some of the major advantages of diluted earnings per share are:
• It determines the Earnings Per Share while considering the potential impact of converting all dilutive securities into common stock.
• Lower scope for manipulation given that all convertible securities are captured.
• It helps in comparing the EPS among peers with varying capital structures.
Limitations of Diluted Earnings Per Share
Some of the major limitations of diluted earnings per share are:
• It involves a complex set of calculations that includes conversion, bond tax benefits, etc.
• It is still exposed to the vagaries of financial account manipulation.
So, it can be concluded that diluted earnings per share are a more comprehensive indicator of a company’s earnings than the basic EPS. However, it needs to analyze in conjunction with the stock price
and the outstanding number of common shares to draw true financial insights.
Recommended Articles
This is a guide to Diluted Earnings Per Share. Here we discuss how to calculate Diluted EPS along with practical examples. We also provide a downloadable Excel template. You may also look at the
following articles to learn more – | {"url":"https://www.educba.com/diluted-earnings-per-share/","timestamp":"2024-11-09T10:40:57Z","content_type":"text/html","content_length":"330234","record_id":"<urn:uuid:0ddcdb44-9ae8-48cb-870f-0d1712ce0ee8>","cc-path":"CC-MAIN-2024-46/segments/1730477028116.75/warc/CC-MAIN-20241109085148-20241109115148-00823.warc.gz"} |
Python Datetime Vs. NumPy Datetime: 8 Differences You Should Know
Python datetime module helps in handling datetime values. These values are of the type DateTime class format. This DateTime format is nothing but a data type class that has additional accessors and
attributes such as adding time differences, changing timezones, etc.
NumPy datetime objects are used by almost every individual using Pandas. Although Pandas has its class called Timestamps, which under the hood, uses the best of the two: NumPy vectorized interface
and Python datetime ease of use. There are differences in how native Python datetime and NumPy datetime64 handle data. In this article, I will discuss some of these distinctions.
1. Data Representation
Native Python and NumPy both follow different formats to store date-time attributes. The Python datetime format stores data as a group of integers for every bit of information. It means that year,
month, day, minute, hour, seconds, and so on have their integer representation.
Interestingly, the NumPy follows an offset representation using the Unix epoch. The most common epoch is 1st January 1970. It calculates the datetime based on the number of seconds past the epoch.
These values are stored in a signed int64 array.
To have a better understanding, you can visit Epoch Converter, and try different epochs to get different datetime. For example, here I have passed the epoch as the number of seconds in a day to get
the next date post the baseline.
Epoch Converter website
Similarly, you can try the vice versa of converting the human-readable date to an epoch value.
2. Time Delta & Resolution Granularity
Time deltas allow us to move dates forward or backward by the defined value. Python datetime supports time deltas up to limited units starting the week to just microseconds. NumPy supports 13 units
from Year to Attoseconds for time deltas and therefore offers a deeper shift of dates.
Another difference between Python and NumPy datetime is the resolution. In Python, the data resolution may not be the same as the unit but NumPy data resolution is the same as the unit.
Let's understand this a bit deeper. For example, if we check the resolution for a Python datetime object with only date, it defaults the resolution to ‘days’ and if we check resolution for Python
datetime object with time, suppose hour, it defaults the resolution to ‘microseconds’ instead of ‘hour’. It makes the Python datetime objects resolution inconsistent with the units provided. See the
code example below:
In the case of NumPy datetime, the resolution is the same as the units provided. The resolution for NumPy objects can be seen using the ‘dtype’ attribute. This also shows that NumPy embeds the units
as resolutions.
3. Range
The range of native Python datetime has been set, hardcoded, between 1 to 9999 Years. It is done so because the Python integer can support a very large number, which is not achievable as of now. In
the case of NumPy datetime, the range is flexible and a bit dynamic. The range is dependent on the int64 range along with the unit's range. Smaller units generally have a smaller range and so does
the datetime range.
Have a look at Python datetime min-max dates:
In the case of NumPy, the range decreases with the usage of smaller units. In the example below, you can see the max NumPy datetime with ‘day’ as a unit. When a lower unit, microseconds is chosen,
the max date is decreased as well as time elements are introduced.
4. Converting in Other Units
We usually convert/streamline our datetime units to a standard unit so that the values are compatible for further calculations. In Python datetime, we create a new object for converting into other
units. On the other hand, in NumPy datetime, we can simply use the “astype” method for converting between any unit. Here, the ease of usage aligns towards NumPy datetime.
See an example below where we converted the NumPy datetime default units into months and microseconds.
5. Shifting Using Arithmetic operators
This difference sets apart the NumPy datetime from Python datetime. NumPy supports operator overloading of addition and subtraction for integers. This means along with time delta support for datetime
objects, you can add/sub numbers directly from datetime objects considering the unit defined. On the other hand, Python datetimes only support time delta add/sub. See the examples below:
In the case of NumPy, operator overloading is accepted. Also, as NumPy supports vectorization, you can simply add/sub integers from a NumPy array of datetime objects, making it super easy.
6. Missing Values
We all are aware of the fact that NumPy and Pandas are known to handle missing values with ease. We have the “NaN” which represents a “not a number” placeholder for any type of missing data. This
placeholder is special as it propagates with calculations and does not throw exceptions.
Similarly, in the case of datetime values, we have a special “NaT’, not a time placeholder that can be present in the NumPy datetime arrays. It smartly converts “None” or empty strings into the “NaT’
On the other hand, Python datetime serves missing values as “None” as the placeholder. The problem with this placeholder is that it does not propagate to further calculations and throws exceptions.
7. Performance Comparisons
The NumPy is known to be a better Python implementation in terms of performance and speed. In this case, too, NumPy datetime is way faster than the Python datetime objects. In terms of applying
operations on a list/array of datetime objects, NumPy leads as its core implementation is C-arrays which provide the much-needed vectorization. Great comparison with concrete numbers can be found
8. Interconversion
Till now, we have seen that the NumPy datetimes have a border range as compared to the Python datetimes. NumPy has support for the input of smaller range datetime objects and Python datetime is no
exception to this.
Therefore, if we want to convert a Python datetime object into a NumPy datetime object, it can be easily done by passing this Python object directly to the NumPy object and adding the “astype” method
to apply the appropriate datetime data type unit. See one example below:
Converting Python datetime into NumPy datetime
Now, when we move from a broader range of units to a narrow range, some of the unit's information would be distorted or transformed into offsets. NumPy handles it very gracefully and does not require
any extra effort. It provides the astype and tolist method to convert a NumPy datetime to Python datetime. If you want to use the astype method, then you can simply pass the object as a parameter in
astype and the resultant would be the Python datetime. Take a look at the below examples:
Converting NumPy datetime into Python datetime
Note: Any unit conversion going below microseconds returns the offset representation.
This article presents the differences between Python datetime and NumPy datetime objects. I think it is important to understand how these two modules handle the datetime values as it helps in
implementing robust code. The combination of the two interfaces, pandas Timestamps is the perfect example of using the best of both worlds: NumPy vectorization and Python datetime ease of use. You
can explore more about these differences and do let me know if I missed any differences!
If you want to read/explore every article of mine, then head over to my master article list which gets updated every time I publish a new article on any platform!
For any doubts, queries, or potential opportunities_, you can reach out to me via my LinkedIn
Previous Article:
This article has been highly inspired and referred from this PyData Talk: | {"url":"https://plainenglish.io/blog/python-numpy-datetime-8-differences-you-should-know-ecb4111eeeca","timestamp":"2024-11-12T12:59:09Z","content_type":"text/html","content_length":"74773","record_id":"<urn:uuid:998f9614-f049-4f8b-ba8a-59d67b550339>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00407.warc.gz"} |
Counting blanks in a Contacts column
I'm trying to count the number of blank cells in a Contacts column in a remote sheet. Here's the formula:
=COUNTIF({SE-Onboarding Tracker - DM}, "")
The problem is ... it's returning a value of 30 when it should be 1. There's only 20 rows and only one of them doesn't have a Contact stored in the DM column. And yes I'm 100% sure there aren't any
blank rows at the end of my sheet. Someone suggested I use this instead:
=COUNTIF({SE-Onboarding Tracker - DM}, "=")
That returns a value of 20 which is the number of rows including the one with the blank. Any suggestions?
Best Answers
• Sorry about that. I completely missed that one.
Your latest does make sense though. There are always at a minimum 50 rows in a sheet, and you have 10 filled in.
One last thing to try on the original... Is there another column that will always have something in it on every row that needs to be counted? If so you can count where the contact column is blank
and that other column is not blank. There still seems to be some issue with your first sheet considering even when we count non-blanks we get the wrong number, but it is at least one more thing
to try that Support will probably also suggest.
=COUNTIFS({Contact Column}, @cell = "", {Other Column}, @cell <> "")
• I was not aware of the 50 row internal limit. Good to know. I now see why you had me try -10. If my sheet had more than 50 rows, that would have worked but I’ll be honest, I’m not entirely sure
why :-P
Anyway, your formula suggestion gives me the output I’m expecting. Thank you Mr. Newcome. You’re the best!
• What happens if you try this:
=COUNTIF({SE-Onboarding Tracker - DM}, @cell = "") - 10
• Hey Paul,
If I add - 10 to the formula the value changes to 20. The correct value is 1. I have no clue why this is not working.
• I re-read the last post, I changed it to:
=COUNTIF({SE-Onboarding Tracker - DM}, @cell = "") - 10
and the value is now 20 (when it should be 1).
• Double check that you selected the correct column for your cross sheet reference.
If that is correct, what happens when you plug in
=COUNTIF({SE-Onboarding Tracker - DM}, @cell <> "")
• If I change the formula to:
=COUNTIF({SE-Onboarding Tracker - DM}, @cell <> "")
The value is 20. The cross sheet reference is correct (see attached). The sheet has 21 rows; only the first row has a blank in the DM column.
• So if we count blank cells and subtract the 10 from the bottom of the sheet then we get 20. But if we count non-blanks then we also get 20. Basically it sounds as if each of the cells in rows
that are actually being used are being counted as both blank and non-blank at the same exact time.
Are you able to provide a screenshot of the formula actually in the sheet similar to the snippet below?
• Revert back to the original formula. I did a mouse over instead so you could see the value it's returning. This seems like a bug IMO. This source sheet only has 21 rows. I've deleted rows at the
bottom of the the source sheet several times just to be 100% sure there's aren't any blank rows that are throwing off the count.
• And are you able to show the referenced column like so? It does sound like you may be stuck with a bug though.
• That screenshot was in the previous post but thank you for trying to help me. I've opened a support case for this. I was able to duplicate using two new sheets and a very simple example (see
• Sorry about that. I completely missed that one.
Your latest does make sense though. There are always at a minimum 50 rows in a sheet, and you have 10 filled in.
One last thing to try on the original... Is there another column that will always have something in it on every row that needs to be counted? If so you can count where the contact column is blank
and that other column is not blank. There still seems to be some issue with your first sheet considering even when we count non-blanks we get the wrong number, but it is at least one more thing
to try that Support will probably also suggest.
=COUNTIFS({Contact Column}, @cell = "", {Other Column}, @cell <> "")
• I was not aware of the 50 row internal limit. Good to know. I now see why you had me try -10. If my sheet had more than 50 rows, that would have worked but I’ll be honest, I’m not entirely sure
why :-P
Anyway, your formula suggestion gives me the output I’m expecting. Thank you Mr. Newcome. You’re the best!
• Happy to help. 👍️
Yes. Your sheet will always have 50 rows or your current number of used rows plus 10. That is why the -10 would have worked if you had 40+ rows. If I had caught that piece in your original post,
I may have gotten to the resolution a little quicker. Sorry about that.
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VE475 Introduction to Cryptography Assignment 3 solved
Ex. 1 — Finite fields
1. Show that X
2 + 1 is irreducible in F3[X].
2. Why does the multiplicative inverse of 1 + 2X mod X
2 + 1 exist in F3[X]?
3. Find the multiplicative inverse of 1 + 2X mod X
2 + 1, in F3[X].
Ex. 2 — AES
The goal of this exercise is to reshape the decryption of AES such that it has the same structure as
1. First we determine the inverse of each layer.
a) Describe InvShiftRows the inverse operation of ShiftRows.
b) What is the inverse of the layer AddRoundKey?
c) Explain why the transformation InvMixColumns is given by the multiplication by the matrix
We call InvSubBytes the lookup table inverse of SubBytes.
2. Describe the decryption process using the previous transformations.
3. Why can InvShiftRows and InvSubBytes be applied on reverse order?
4. Similarly we want to inverse the order of application of AddRoundKey and InvMixColumns.
a) Why is it not possible to reverse the order of application of AddRoundKey and InvMixColumns?
b) Calling the initial matrix (ai,j ), the MixColumns matrix (mi,j ), and the AddRoundKey matrix
(ki,j ), what is the result of applying first MixColumns and then AddRoundKey?
c) Show that the inverse operation is given by
(ei,j ) → (mi,j )
(ei,j ) ⊕ (mi,j )
(ki,j ).
d) Define a new operations InvAddRoundKey such that “AddRounKey then InvMixColumns”
can be replaced by “InvMixColumns then InvAddRoundKey”.
5. Conclude on how to process the decryption.
6. What are the advantages of this strategy over simply reversing the order of application of the
Ex. 3 — DES
1. Research and explain how the DES cryptosystem works.
2. Quickly explain linear and differential cryptanalysis.
3. What is triple DES and why was it used instead of double DES?
4. Explain how passwords are stored on Unix systems? Is it a problem?
Hint: check man passwd and then follow the suggested man pages
Ex. 4 — Programming
In the AES, choose two layers to implement in C. The 128 bits should be looked at as an unsigned
char pointer. Operations should be implemented using logical operators (and, or, and xor).
A bonus will be given for each extra layer implemented, and the generation of the S-Box. A big bonus
will reward a complete implementation of the AES. | {"url":"https://codeshive.com/questions-and-answers/ve475-introduction-to-cryptography-assignment-3-solved/","timestamp":"2024-11-14T06:47:43Z","content_type":"text/html","content_length":"100107","record_id":"<urn:uuid:2ab58e42-2aaa-4397-ab47-39e7a6ff4c80>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00270.warc.gz"} |
Stephane Nonnenmacher
University Paris-Sud University Paris-Sud
Delocalization of the Laplace eigenmodes on Anosov surfaces
Mathematical Physics Seminar
9th February 2024, 1:45 pm – 3:30 pm
Fry Building, G.07
The eigenmodes of the Laplace-Beltrami operator on a smooth compact Riemannian manifold $(M,g)$ can exhibit various localization properties in the high frequency regime, which strongly depend on the
properties of the geodesic flow (the classical dynamics on $(M,g)$). We focus on situations where this flow is strongly chaotic (Anosov), for instance if the sectional curvature of $(M,g)$ is
negative. Studying the Laplace (=quantum) eigenmodes is in the realm of Quantum Chaos.
In this situation, the Quantum Unique Ergodicity conjecture states that the eigenmodes become equidistributed on $M$ in the high-frequency limit: for any open set $\Omega$ on $M$, the $L^2$ masses on
$\Omega$ of the eigenstates converge to the relative volume of $\Omega$. In the two-dimensional case, we show the weaker property of full delocalization: the $L^2$ masses of the eigenstates on $\
Omega$ are bounded from below, independently of the frequency. This is in contrast with, e.g., the case of eigenstates on the round sphere, which may be strongly concentrated near a closed geodesic.
The proof uses various methods of semiclassical analysis, the structure of stable and unstable manifolds of the Anosov flow, and a recent Fractal Uncertainty Principle due to Bourgain-Dyatlov. Joint
work with S.Dyatlov and L.Jin. | {"url":"https://www.bristolmathsresearch.org/seminar/stephane-nonnenmacher/","timestamp":"2024-11-05T18:55:21Z","content_type":"text/html","content_length":"54905","record_id":"<urn:uuid:28b0836b-6fd3-465c-a75d-5d90e856a322>","cc-path":"CC-MAIN-2024-46/segments/1730477027889.1/warc/CC-MAIN-20241105180955-20241105210955-00399.warc.gz"} |
Filling the TIPS gap years with bracket year duration matching
There have been many threads on how to fill the gap years in a TIPS ladder. If you don't know what I'm talking about, this thread is not for you, although if you want to understand it anyway, you
might find it enlightening. Although we have had some discussions in other threads of the technique I'll discuss here, the questions that have come up in those threads indicate that this is a meaty
enough topic to have its own thread (I hope this thread won't get merged with one of these existing, less specific threads, one or more of which I'll link to later in this thread for additional
background and to answer questions asked in the other threads).
First, terminology:
• gap year = a year in which there are no TIPS maturing with a term to maturity of 29 years or less.
• bracket year = the years immediately before and after the gap years in which there are TIPS maturing that year.
• DARA = Desired Annual Real Amount = total real principal and interest that the ladder produces each year. This is the term used in the #Cruncher TIPS Ladder Builder spreadsheet.
• DARI = Desired Annual Real Income = DARA. This is the term used in the tipsladder.com TIPS ladder building tool.
• Real amount = amount in dollar purchasing power relative to some base date, using the reference CPI as the inflation index. A typical base date is the settlement date for the day you build or
evaluate the TIPS ladder. Example: if the base date ref CPI were 100, and ref CPI increased to 103 on the maturity date of the first rung, a DARA of $10,000 would equal an inflation-adjusted
value of $10,300, and the purchasing power would be $10,000 relative to the base date (= 10,300 / 1.03).
• DARA multiplier = a number multiplied by the DARA, and entered in the #Cruncher TIPS ladder spreadsheet row for each distinct TIPS issue (i.e., identified by a distinct CUSIP, which is a unique
identifier for a bond); this is used in the calculation of how many of that distinct TIPS issue to buy. For example, if holding only one distinct TIPS issue to generate the real principal amount
of the DARA for a given maturity year, the DARA multiplier for that row would be 1. If holding none of a particular distinct TIPS, the multiplier for that TIPS issue row would be 0.
• duration matching = holding some of each of the bracket year TIPS such that the DARA-multiplier-weighted duration of them equals the expected duration of a gap year TIPS when it is issued.
Currently there are TIPS maturing in Jan 2034 and Feb 2040, so 2034 and 2040 are the bracket years, and the gap years are 2035-2039 (five of them).
For purposes of this discussion I'll assume that our TIPS ladder extends from 2034 or earlier through 2040 or later. The longest TIPS ladder would hold maturities from July 2024 (or possibly Oct
2024) to Feb 2054. The current versions of the two popular TIPS ladder building tools, the
#Cruncher TIPS Ladder Builder spreadsheet
, support only ladders with rungs starting in 2025.
One of many techniques that have been discussed for filling the gap years is to hold some of each of the TIPS that mature before the first gap year and after the last gap year. A specific instance of
this is to do it with the bracket years, so currently 2034 and 2040 (it would have been 2033 and 2040 before the Jan 2034 was issued in Jan 2024).
The default for the #Cruncher spreadsheet is use DARA multipliers of 3 for the Jan 2034s and 4 for the Feb 2040s; note that 3 + 4 = 7, which is the total number of maturity years from 2034 through
2040. The tipsladder.com tool offers several methods to fill the gap years, but if you accept the default of "Bond maturing nearest to start of rung year", you essentially end up with multipliers of
4 for the 2034 and 3 for the 2040.
You don't need to use integers as DARA multipliers with the #Cruncher spreadsheet as long as the total of the DARA multipliers for a single maturity year equals 1; e.g., you could enter multipliers
of 0.5 each for the Jan and Jul 2030 TIPS for your 2030 maturity year. With this in mind, you might use 3.5 each as the multipliers for the 2034 and 2040 to cover the 7 years from 2034-2040
inclusive, for example, and one might expect this to do a better job of duration matching the gap years.
What I do is calculate estimated durations for the TIPS for each gap year, then calculate the proportions for each of the 2034 and 2040 such that the DARA-multiplier-weighted-average duration equals
the estimated duration of each gap year TIPS. Currently this results in multipliers of 3.56 for the 2034s and 3.44 for the 2040s. This confirms that simply using 3.5 as the multiplier for each gets
pretty close to a decent estimated duration match, at least now, with the relatively flat yield curve in this maturity range.
To derive the formulas for the gap-year DARA multipliers for the 2034 and 2040, we start with this equation:
Code: Select all
d34 * x + d40 * (1-x) = dg,
d34 = modified duration (MD) of the 2034
d40 = MD of the 2040
dg = estimated MD of the gap year TIPS
x = gap year DARA multiplier for the 2034
With some algebra, we solve for x to get:
I'll cover the calculation of durations in a subsequent post, and for now I'll just show the example of calculating x and (1-x) for the 2035 gap year.
Code: Select all
Independent variable values:
d34 = 8.75
d40 = 13.23
dg = d35 = 9.41
x = (d40-dg) / (d40-d34)
x = (13.23-9.41) / (13.23-8.75)
x = 0.85
1-x = 0.15
So we'd use DARA multipliers of 0.85 for the 2034s and 0.15 for the 2035s to match the estimated modified duration of the 2035.
As we've discussed in other threads, a simple way to approximate the gap year DARA weights is to simply set x = n/6, where n = 5 for 2035, n = 4 for 2036, ... n = 1 for 2039. To compare this method
to the more complicated method shown above, note that for the 2035 gap year:
which is very close to 0.85 derived using the duration matching formula.
Here is the table of the DARA weights using durations of TIPS based on quotes from Schwab today, also showing the approximations using the n/6 method for the 2034 weights:
Note that the sum of the weights for each of the 2034 and 2040 are the DARA multipliers we enter into the #Cruncher spreadsheet for them respectively. Of course the sum of these multipliers equals 7,
which is the total number of years covered (2034, 2040 + 5 gap years).
------------------------------------------------------------------------ EDIT ---------------------------------------------------------------------
I'm going to summarize what I've learned in doing the experiments documented in this thread. I'll add to this summary as I learn more.
Since I started the thread, #Cruncher developed a simplified ladder building spreadsheet, which I then used extensively for all analysis after that. I refer to this as "the simple tool" or just
Everything here is premised on using the simple tool, bracket-year coverage of gap years, a 30-year ladder, DARA = $100K, and initial duration matching based on hypothetical gap year yield and coupon
of 2%.
1. Duration matching works almost perfectly if we treat the gap years as marketable bonds; i.e., fixed coupons and variable prices (or values). The figure of merit here is how close to 0 is the
change in value of the duration matched bracket year holdings minus the change(s) in value of the gap year(s) being matched. This is shown early in this thread.
2. Using the same figure of merit, duration matching does not work nearly as well for the real life situation where gap year coupons are variable and price (or value) is approximately fixed unless
yields drop below 0.125%. This has been shown in the earlier posts in the thread.
3. The lack of pure duration matching effectiveness is offset to some extent by the change in interest from the gap year bonds, because the coupons will be close to the yields; i.e., at higher yield
the coupon interest will be higher, requiring less principal, and therefore less cost for the gap year bonds.
4. Given #2, purchases or sales of the pre-gap rungs are required for ARA to equal DARA for the gap and pre-gap rungs, even after factoring in #3.
5. With no gaps filled and the sum of bracket year multipliers = 7 (e.g., 3.5 each for 2034/40), the sum of ARAs is greater than 30 * DARA (for a 30-year ladder, all other multipliers set to 1).
This is a technical detail that is not particularly important, and I assume is due to my imperfect implementation of the multiplier feature, which was not included in #Cruncher's original
simplified spreadsheet.
This table summarizes the results of the experiments to date:
"5 fill vs 0 fill at X%" means the numbers in that column relate to having all gap years filled (and the 2025-2039 all matured) at a yield of X%, and excess 2034/2040 bracket year holdings sold,
compared to the initial state where 0 gap years are filled, all rungs are populated, and the excess holdings to fill the gap years are held in the 2034 and 2040 bracket years.
No temporal effects are considered; e.g., durations are not updated based on the passing of almost five years before all gaps are filled.
Looking at the 0% yield case:
• The proceeds from selling all excess 2034 and 2040 bracket year TIPS are 515,669.
• The cost of buying the 2035-2039 gap year TIPS is 432,951.
• This leaves us with extra cash of 82,718.
• We can choose to buy the pre-2034 TIPS that are left, 2030-2033, with the total being 31,396, if we want ARA to equal DARA for those rungs.
• If we do the pre-2034 transactions, we are left with 51,322 in cash.
Last edited by Kevin M on Thu Jun 27, 2024 12:10 am, edited 2 times in total.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
I’ve been anticipating your post, and I will say you have done us all a great favor. Such a well outlined explanation. So from a distant follower, thanks.
Re: Filling the TIPS gap years with bracket year duration matching
Kevin et al,
Before I came across your thread on the mechanics of purchasing individual TIPS, I invested in TIPS ETFs and planned on duration matching them per Vinviz's suggestions (i.e. x% of LTPZ / y% of SCHP /
z% of VTIP). It just so happened my timing was extremely poor (Dec 2022) and market losses ensued thereafter. Since then, I have created a LMP ladder of individual TIPS, but held on to the ETFs with
the hope of recouping some of the losses over time.
Now, I could slowly draw down those ETFs on a duration-matched basis to buy individual gap year TIPS at their 10 year auctions. This would allow me to hold on my existing LMP ladder as is while
(hopefully) limiting my losses over time as the sold, duration-matched ETF monies-turned-TIPS mature.
Or I could just glide-slope the ETFs to cover each of the gap years as originally planned?
All thoughts are welcome.
Re: Filling the TIPS gap years with bracket year duration matching
cvn74n2 wrote: ↑Wed May 22, 2024 11:50 pm Kevin et al,
Before I came across your thread on the mechanics of purchasing individual TIPS, I invested in TIPS ETFs and planned on duration matching them per Vinviz's suggestions (i.e. x% of LTPZ / y% of
SCHP / z% of VTIP). It just so happened my timing was extremely poor (Dec 2022) and market losses ensued thereafter. Since then, I have created a LMP ladder of individual TIPS, but held on to the
ETFs with the hope of recouping some of the losses over time.
Now, I could slowly draw down those ETFs on a duration-matched basis to buy individual gap year TIPS at their 10 year auctions. This would allow me to hold on my existing LMP ladder as is while
(hopefully) limiting my losses over time as the sold, duration-matched ETF monies-turned-TIPS mature.
Or I could just glide-slope the ETFs to cover each of the gap years as originally planned?
All thoughts are welcome.
Did you use the shortcut method vineviz proposed on this thread? viewtopic.php?p=6869837#p6869837
In that method you are holding all 3 funds at the same time.
Or did you use the method where you hold only 2 funds at a time and calculate their relative weights based on matching the net duration of the 2 funds to your investment horizon, updating it
With the second method above, the market value of individual TIPS in a ladder would have been affected exactly the same way as your duration matched TIPS funds were. And yet the income produced by
duration matching with funds would have remained nearly the same despite the loss in market value. Just like the income from individual TIPS would have remained unaffected. Moreover, if you convert
your duration matched TIPS funds to a ladder, you can expect nearly the same income from the ladder after it's constructed as you were getting from the duration matched funds, regardless of when you
make the conversion. With funds, you shouldn't care about the market value except to make the calculation when it's time to rebalance between the two funds so you know how much to shift from one fund
to the other. I suppose one might care if they're rebalancing their TIPS funds with other assets, but that kinda breaks the whole idea of having steady income from TIPS. Otherwise, the fact that the
market value dropped because yields rose is why you can extract steady income from 2 bond funds while duration matching. It's a feature, not a bug.
The first method is much more approximate, but the market value would be affected similarly.
Last edited by dcabler on Thu May 23, 2024 6:43 am, edited 7 times in total.
"Repeating a thing doesn't improve it." Quote from Inman, as played by Jude Law, in the movie "Cold Mountain"
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Wed May 22, 2024 10:28 pm There have been many threads on how to fill the gap years in a TIPS ladder. If you don't know what I'm talking about, this thread is not for you,
although if you want to understand it anyway, you might find it enlightening. Although we have had some discussions in other threads of the technique I'll discuss here, the questions that have
come up in those threads indicate that this is a meaty enough topic to have its own thread (I hope this thread won't get merged with one of these existing, less specific threads, one or more of
which I'll link to later in this thread for additional background and to answer questions asked in the other threads).
First, terminology:
□ gap year = a year in which there are no TIPS maturing with a term to maturity of 29 years or less.
□ bracket year = the years immediately before and after the gap years in which there are TIPS maturing that year.
□ DARA = Desired Annual Real Amount = total real principal and interest that the ladder produces each year. This is the term used in the #Cruncher TIPS Ladder Builder spreadsheet.
□ DARI = Desired Annual Real Income = DARA. This is the term used in the tipsladder.com TIPS ladder building tool.
□ Real amount = amount in dollar purchasing power relative to some base date, using the reference CPI as the inflation index. A typical base date is the settlement date for the day you build or
evaluate the TIPS ladder. Example: if the base date ref CPI were 100, and ref CPI increased to 103 on the maturity date of the first rung, a DARA of $10,000 would equal an inflation-adjusted
value of $10,300, and the purchasing power would be $10,000 relative to the base date (= 10,300 / 1.03).
□ DARA multiplier = a number multiplied by the DARA, and entered in the #Cruncher TIPS ladder spreadsheet row for each distinct TIPS issue (i.e., identified by a distinct CUSIP, which is a
unique identifier for a bond); this is used in the calculation of how many of that distinct TIPS issue to buy. For example, if holding only one distinct TIPS issue to generate the real
principal amount of the DARA for a given maturity year, the DARA multiplier for that row would be 1. If holding none of a particular distinct TIPS, the multiplier for that TIPS issue row
would be 0.
□ duration matching = holding some of each of the bracket year TIPS such that the DARA-multiplier-weighted duration of them equals the expected duration of a gap year TIPS when it is issued.
Currently there are TIPS maturing in Jan 2034 and Feb 2040, so 2034 and 2040 are the bracket years, and the gap years are 2035-2039 (five of them).
For purposes of this discussion I'll assume that our TIPS ladder extends from 2034 or earlier through 2040 or later. The longest TIPS ladder would hold maturities from July 2024 (or possibly Oct
2024) to Feb 2054. The current versions of the two popular TIPS ladder building tools, the #Cruncher TIPS Ladder Builder spreadsheet and tipsladder.com, support only ladders with rungs starting
in 2025.
One of many techniques that have been discussed for filling the gap years is to hold some of each of the TIPS that mature before the first gap year and after the last gap year. A specific
instance of this is to do it with the bracket years, so currently 2034 and 2040 (it would have been 2033 and 2040 before the Jan 2034 was issued in Jan 2024).
The default for the #Cruncher spreadsheet is use DARA multipliers of 3 for the Jan 2034s and 4 for the Feb 2040s; note that 3 + 4 = 7, which is the total number of maturity years from 2034
through 2040. The tipsladder.com tool offers several methods to fill the gap years, but if you accept the default of "Bond maturing nearest to start of rung year", you essentially end up with
multipliers of 4 for the 2034 and 3 for the 2040.
You don't need to use integers as DARA multipliers with the #Cruncher spreadsheet as long as the total of the DARA multipliers for a single maturity year equals 1; e.g., you could enter
multipliers of 0.5 each for the Jan and Jul 2030 TIPS for your 2030 maturity year. With this in mind, you might use 3.5 each as the multipliers for the 2034 and 2040 to cover the 7 years from
2034-2040 inclusive, for example, and one might expect this to do a better job of duration matching the gap years.
What I do is calculate estimated durations for the TIPS for each gap year, then calculate the proportions for each of the 2034 and 2040 such that the DARA-multiplier-weighted-average duration
equals the estimated duration of each gap year TIPS. Currently this results in multipliers of 3.56 for the 2034s and 3.44 for the 2040s. This confirms that simply using 3.5 as the multiplier for
each gets pretty close to a decent estimated duration match, at least now, with the relatively flat yield curve in this maturity range.
To derive the formulas for the gap-year DARA multipliers for the 2034 and 2040, we start with this equation:
Code: Select all
d34 * x + d40 * (1-x) = dg,
d34 = modified duration (MD) of the 2034
d40 = MD of the 2040
dg = estimated MD of the gap year TIPS
x = gap year DARA multiplier for the 2034
With some algebra, we solve for x to get:
Code: Select all
x = (d40-dg) / (d40-d34)
I'll cover the calculation of durations in a subsequent post, and for now I'll just show the example of calculating x and (1-x) for the 2035 gap year.
Code: Select all
Independent variable values:
d34 = 8.75
d40 = 13.23
dg = d35 = 9.41
x = (d40-dg) / (d40-d34)
x = (13.23-9.41) / (13.23-8.75)
x = 0.85
1-x = 0.15
So we'd use DARA multipliers of 0.85 for the 2034s and 0.15 for the 2035s to match the estimated modified duration of the 2035.
As we've discussed in other threads, a simple way to approximate the gap year DARA weights is to simply set x = n/6, where n = 5 for 2035, n = 4 for 2036, ... n = 1 for 2039. To compare this
method to the more complicated method shown above, note that for the 2035 gap year:
Code: Select all
n/6 = 5/6 = 0.83
which is very close to 0.85 derived using the duration matching formula.
Here is the table of the DARA weights using durations of TIPS based on quotes from Schwab today, also showing the approximations using the n/6 method for the 2034 weights:
Note that the sum of the weights for each of the 2034 and 2040 are the DARA multipliers we enter into the #Cruncher spreadsheet for them respectively. Of course the sum of these multipliers
equals 7, which is the total number of years covered (2034, 2040 + 5 gap years).
Why did you choose "modified duration" instead of the "macauley duration?"
BTW- #cruncher's spreadsheet calculates both - https://eyebonds.info/downloads/pages/Y ... Macro.html
"Repeating a thing doesn't improve it." Quote from Inman, as played by Jude Law, in the movie "Cold Mountain"
Re: Filling the TIPS gap years with bracket year duration matching
dcabler wrote: ↑Thu May 23, 2024 6:26 am Why did you choose "modified duration" instead of the "macauley duration?"
My next post will describe how I calculate the durations, so I'll address your question there. For now I'll just mention that the difference between D and MD is very small. I just tried using D
instead of MD and it made no difference in the calculated numbers.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Thu May 23, 2024 9:34 am
dcabler wrote: ↑Thu May 23, 2024 6:26 am Why did you choose "modified duration" instead of the "macauley duration?"
My next post will describe how I calculate the durations, so I'll address your question there. For now I'll just mention that the difference between D and MD is very small. I just tried using D
instead of MD and it made no difference in the calculated numbers.
Correct - if you look at the last time #cruncher updated the spreadsheet, you can see that there as well.
"Repeating a thing doesn't improve it." Quote from Inman, as played by Jude Law, in the movie "Cold Mountain"
Re: Filling the TIPS gap years with bracket year duration matching
Kevin, this is a fantastic resource and once again all of us at BH owe you a huge debt of gratitude for sharing your seemingly boundless expertise.
That said, I’m concerned that a complex analysis like this might scare off the “average joe” who just wants to fill in a TIPS ladder and doesn’t much care about getting things exactly right to the
third decimal place. Is this additional level of complexity really warranted?
For example, we have absolutely no idea what the gap TIPS will look like – they might have coupons ranging anywhere from 1/8% to 3% or more, which will make a significant difference in the gap TIPS
duration, and so in the hedging technique. The duration of a (say) 2037 TIPS with a coupon of 3.5% will be very different from that of another 2037 TIPS paying 0.125%.
The change in gap TIPS duration due to coupon payments would also change the mix of 2034 / 2040 TIPS sold to hedge them, right?
I look forward to digging into the weeds in this thread, but from a practical point of view I may well just use the simple 1/7 weights for the 2034 / 2040 TIPS to estimate how many of each to sell to
hedge a given gap TIPS purchase.
Re: Filling the TIPS gap years with bracket year duration matching
Sounds like what you want is inflation protected income for those periods. You can could buy higher coupon TIPS that mature after this period and clip the coupons to cover the income since you don’t
have any TIPS maturing. But you’d likely want the TIPS in tax deferred despite paying state tax to avoid the phantom income tax on those TIPS since you’d be over buying with longer maturity to have
significant enough coupons.
Both the Feb 2040 and 2041 have 2.125% coupons.
This is not likely to be reasonable since you’d have to way over buy. So you’d need something else to supplement the income.
Re: Filling the TIPS gap years with bracket year duration matching
Jaylat wrote: ↑Thu May 23, 2024 9:44 am
I look forward to digging into the weeds in this thread, but from a practical point of view I may well just use the simple 1/7 weights for the 2034 / 2040 TIPS to estimate how many of each to
sell to hedge a given gap TIPS purchase.
Perhaps this thread with dig into the weeds to the extent that estimates will be developed showing the expected range of price penalty/deviation between using rigorous duration matching vs. the
simplified 1/6 weighting (maturity-matching approximation to duration matching).
I'm guessing that the maturity-matching approximation is simple enough for the "average Joe" to implement. It would be a benefit to the community if it can be shown mathematically that the
maturity-matching approximation is "good enough" under just about any reasonable scenario.
Re: Filling the TIPS gap years with bracket year duration matching
MtnBiker wrote: ↑Thu May 23, 2024 10:23 am
Jaylat wrote: ↑Thu May 23, 2024 9:44 am
I look forward to digging into the weeds in this thread, but from a practical point of view I may well just use the simple 1/7 weights for the 2034 / 2040 TIPS to estimate how many of each to
sell to hedge a given gap TIPS purchase.
Perhaps this thread with dig into the weeds to the extent that estimates will be developed showing the expected range of price penalty/deviation between using rigorous duration matching vs. the
simplified 1/6 weighting (maturity-matching approximation to duration matching).
I'm guessing that the maturity-matching approximation is simple enough for the "average Joe" to implement. It would be a benefit to the community if it can be shown mathematically that the
maturity-matching approximation is "good enough" under just about any reasonable scenario.
Totally agree, that would be a significant benefit.
Also, and particularly for retirees with smaller TIPS ladders, you just don't have the ability to match the percentages perfectly. The 2040 TIPS has an inflation factor of 1.44x, so a $10,000 real in
2040 TIPS gives you a grand total of 7 TIPS, period. You have to sell in increments of 14% of the 2040 TIPS portfolio - so they would be forced to use the 1/7 weighting no matter what the
calculations show.
Re: Filling the TIPS gap years with bracket year duration matching
Definitely subscribing to this thread.
My ladder ends in 2039, which is a special case, and I'm doing the "n/6" calculation for now but I will definitely welcome the double-checking that I'm doing it right
Re: Filling the TIPS gap years with bracket year duration matching
Jaylat wrote: ↑Thu May 23, 2024 10:56 am
Also, and particularly for retirees with smaller TIPS ladders, you just don't have the ability to match the percentages perfectly. The 2040 TIPS has an inflation factor of 1.44x, so a $10,000
real in 2040 TIPS gives you a grand total of 7 TIPS, period. You have to sell in increments of 14% of the 2040 TIPS portfolio - so they would be forced to use the 1/7 weighting no matter what the
calculations show.
Good point about the inability of the small investor to match percentages perfectly.
My ladder is using bracket years of 2032 and 2040, and only has rungs in even years, with the final year being 2036. So far, I have only filled one gap-year rung (2034) and didn't really notice any
problem matching the target percentages accurately. I guess having fewer, bigger rungs disqualifies me as a small investor, LOL.
Using modified duration to estimate price change relative to yield change
To calculate the modified durations mentioned in the OP I use the spreadsheet MDURATION (MD) function. I use Google Sheets, but Excel provides the same function.
I use modified duration instead of Macaulay duration (calculated with the DURATION (D) function) because modified duration relates price change to yield change directly. However, with current yields
at least, using DURATION produces the same results, because there's not much difference between D and MD, and the smaller the magnitude of the yield, the smaller the difference.
You may have heard of the duration rule of thumb that lets us estimate the percentage price or NAV change of a bond or bond fund relative the percentage point (pp) change in yield. For example, we'd
expect an increase in yield of 0.1 pp, say from 5.0% to 5.1%, to result in a price decline of 0.5%, say from 10 to 9.5 for a bond with a duration of 5 or fund with a weighted-average duration of 5.
This rule of thumb works slightly better using modified duration than with Macaulay duration, and technically MD is what should be used.
You may also have heard that this rule of thumb works best for a bond fund or bond ladder under two constraints:
1. The change in yield is small.
2. Yields change due to a parallel shift in the yield curve; i.e., the percentage point change of all yields is the same.
The first constraint is due to the convexity of the price/yield curve, since duration is proportional to the slope of the line at a point on the curve (first derivative of price with respect to
yield). Due to the convexity of the curve, the slope changes as yield changes. So we would expect the rule of thumb to work better for a 0.1 pp change than for a 1 pp change.
The second constraint is required for a fund or a ladder, because if yields don't change by the same amount, the average of the percentage price changes won't conform to the duration rule of thumb.
We can demonstrate this with some calculations, but first, why is this relevant to gap year duration matching? It's relevant because the same principal applies, since if the yields of the bracket
year TIPS don't change by the same amount, the estimated price of a gap year TIPS won't change as expected based on our duration matching; we'll come back to this later.
To demonstrate the parallel yield shift constraint, consider the following ladder or fund consisting of equal weights of 1y and 5y bonds:
Note that the average yield is 4.50% and the average MD is 2.81.
Now lets increase the average yield by 10 pp by increasing the yield of each bond by 10 pp:
Note that the average of the percentage price changes is -0.280%, which conforms quite nicely to the rule of thumb with an average MD of 2.81.
Now let's increase the average yield again by 10 pp, but by increasing the 5y by 20 pp and the 1y by 0 pp:
Here we see that the magnitude of the average of the percentage price changes is quite a bit larger than we'd expect based on the rule of thumb, which confirms that the rule of thumb doesn't work
well for non-parallel yield curve shifts.
That's enough for this post. In the next one, I'll discuss how I estimate the expected durations of the gap year TIPS.
Last edited by Kevin M on Thu May 23, 2024 11:41 am, edited 1 time in total.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
MtnBiker wrote: ↑Thu May 23, 2024 10:23 am
Jaylat wrote: ↑Thu May 23, 2024 9:44 am
I look forward to digging into the weeds in this thread, but from a practical point of view I may well just use the simple 1/7 weights for the 2034 / 2040 TIPS to estimate how many of each to
sell to hedge a given gap TIPS purchase.
Perhaps this thread with dig into the weeds to the extent that estimates will be developed showing the expected range of price penalty/deviation between using rigorous duration matching vs. the
simplified 1/6 weighting (maturity-matching approximation to duration matching).
I'm guessing that the maturity-matching approximation is simple enough for the "average Joe" to implement. It would be a benefit to the community if it can be shown mathematically that the
maturity-matching approximation is "good enough" under just about any reasonable scenario.
Or even just use a 50/50 split between the bracket years; or just use the #Cruncher default multipliers of 3 and 4 respectively, or the tipsladder.com defaults of 4 and 3.
If I make a calculation error, #Cruncher probably will let me know.
Estimating expected duration of gap year TIPS
To match the duration of a gap year TIPS with the bracket year TIPS, we need to estimate the expected duration of the gap year TIPS. The DURATION and MDURATION function parameters include maturity,
yield and coupon, so we need to assign or estimate those.
• I assume that the gap year TIPS maturity is Jan 15 of the gap year; i.e., 1/15/2035, ..., 1/15/2039, which is what the first gap year TIPS issued for each gap year will be if Treasury continues
their current auction schedule.
• To estimate yield I simply use linear interpolation of the bracket year TIPS (Jan 2034 and Feb 2040).
• I set the estimated coupon to what it would be if the TIPS were issued at the estimated yield.
I actually use the seasonally adjusted (SA) yields of the bracket year TIPS to do the linear interpolation, since I use SA yields for my purchase decisions, but the differences are small enough at
these maturities as to not be significant.
Here is the table of estimated expected yields and coupons using today's TIPS yields at Schwab:
These values are used in the MDURATION function to calculate the ModDur values shown in the table in the OP.
Of course yields will change between now and when a gap year TIPS is issued, which means coupons and durations also are likely to be different than what I'm estimating now. This could require some
periodic rebalancing of the bracket year TIPS holdings, depending on how accurate we want the duration matching to be. It could be that the required rebalancing could be done simply by adjusting the
number of each bracket year TIPS that we sell to buy the next gap year TIPS that's issued. Also, as has been pointed out, the resolution of the duration matching is limited by the value of a single
TIPS for each bracket year, and this might be larger than any shifts in the duration matching.
This raises the topic of my next post, which will discuss how well the duration matching works given different yield curve change scenarios.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Hi, Kevin.
This post is indirectly related to the topic.
I get how to fund the gap years , using 2034s and 2040s.
The question I have is, for those of us with very limited dry powder (it is all tied up in our TIPS ladder), and more than DARA in each of our existing rungs, what is the specifically best way to
fully fund the excess 2034s and 2040s necessary to ultimately fund the gap years using funds from other maturities, or does it not matter much?.
Following is a post of mine in another thread raising that question, with specifics I would like answered:
protagonist wrote: ↑Tue May 21, 2024 2:04 am
"How important is duration and/or maturity matching with an essentially flat yield curve?
For example, consider if one was to sell TIPS maturing relatively soon and close in maturity date with widely divergent coupons ...say, for example, sell the same dollar amount of 4/15/27s with a
0.125% coupon, or 4/15/28s with a 3.625 coupon. These are two issues close in maturity date but with very different coupons.
If one was then going to use the proceeds from each sale to buy 2040s with a coupon lying between the two (2.125%), would the end result be significantly different? Would one be a "better deal" than
the other? Or would they both have pretty much the same result?
And (another question): Would selling the above TIPS (at today's yields) to buy 2034's, for example, leave one with a much different result than selling the same dollar amount of 2033s (closer
duration and maturity) to buy 2034s? How would it be different?
Forgive my ignorance about this....I'm asking as a person relatively unschooled in the world of bond investing. Bonds never really interested me before the recent TIPS revolution."
Thanks in advance.
Re: Filling the TIPS gap years with bracket year duration matching
protagonist wrote: ↑Fri May 24, 2024 11:33 am Hi, Kevin.
This post is indirectly related to the topic.
I get how to fund the gap years , using 2034s and 2040s.
The question I have is, for those of us with very limited dry powder (it is all tied up in our TIPS ladder), and more than DARA in each of our existing rungs, what is the specifically best way to
fully fund the excess 2034s and 2040s necessary to ultimately fund the gap years using funds from other maturities, or does it not matter much?.
Following is a post of mine in another thread raising that question, with specifics I would like answered:
protagonist wrote: ↑Tue May 21, 2024 2:04 am
"How important is duration and/or maturity matching with an essentially flat yield curve?
For example, consider if one was to sell TIPS maturing relatively soon and close in maturity date with widely divergent coupons ...say, for example, sell the same dollar amount of 4/15/27s with a
0.125% coupon, or 4/15/28s with a 3.625 coupon. These are two issues close in maturity date but with very different coupons.
If one was then going to use the proceeds from each sale to buy 2040s with a coupon lying between the two (2.125%), would the end result be significantly different? Would one be a "better deal"
than the other? Or would they both have pretty much the same result?
And (another question): Would selling the above TIPS (at today's yields) to buy 2034's, for example, leave one with a much different result than selling the same dollar amount of 2033s (closer
duration and maturity) to buy 2034s? How would it be different?
Forgive my ignorance about this....I'm asking as a person relatively unschooled in the world of bond investing. Bonds never really interested me before the recent TIPS revolution."
Thanks in advance.
Even though the yield curve is relatively flat, the 2040s have significantly higher duration than the 2034s; the SA yield difference is only 5 basis points, but the durations are 8.74 and 13.21
respectively. So you're still going to want to weight them differently to match the expected duration of a particular gap year TIPS.
I can't think of why the coupons matter much. I don't think much about coupons in buying and selling, other than preferring lower coupons when there's a choice, and even that doesn't really matter in
a ladder, since the coupons just reduce the number of TIPS you need to buy in earlier years.
I would think more about what I want my average duration to be; do I want to increase it, keep it about the same, or decrease it? Since I'm way overweight in shorter-term TIPS, I generally want to
increase it, so I've been selling the shorter maturity/duration TIPS to buy longer maturity/duration TIPS.
In your example, the durations of the Apr 2027 and Apr 2028 are 2.84 and 3.61 respectively, using SA yields, so I probably would sell the 2027 to increase my average duration.
Ditto for the comparison with the 2033s, which have durations of 8.11 for Jan and 8.46 for Jul. I'd swap the earlier 2027s to extend my average duration. If I wanted to keep my average duration close
to what it was, I might swap the 2033s for 2034s, or I might just hold the 2033s.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
protagonist wrote: ↑Fri May 24, 2024 11:33 am Hi, Kevin.
This post is indirectly related to the topic.
I get how to fund the gap years , using 2034s and 2040s.
The question I have is, for those of us with very limited dry powder (it is all tied up in our TIPS ladder), and more than DARA in each of our existing rungs, what is the specifically best way to
fully fund the excess 2034s and 2040s necessary to ultimately fund the gap years using funds from other maturities, or does it not matter much?.
Following is a post of mine in another thread raising that question, with specifics I would like answered:
protagonist wrote: ↑Tue May 21, 2024 2:04 am
"How important is duration and/or maturity matching with an essentially flat yield curve?
For example, consider if one was to sell TIPS maturing relatively soon and close in maturity date with widely divergent coupons ...say, for example, sell the same dollar amount of 4/15/27s with a
0.125% coupon, or 4/15/28s with a 3.625 coupon. These are two issues close in maturity date but with very different coupons.
If one was then going to use the proceeds from each sale to buy 2040s with a coupon lying between the two (2.125%), would the end result be significantly different? Would one be a "better deal"
than the other? Or would they both have pretty much the same result?
And (another question): Would selling the above TIPS (at today's yields) to buy 2034's, for example, leave one with a much different result than selling the same dollar amount of 2033s (closer
duration and maturity) to buy 2034s? How would it be different?
Forgive my ignorance about this....I'm asking as a person relatively unschooled in the world of bond investing. Bonds never really interested me before the recent TIPS revolution."
Thanks in advance.
What was confusing about your original question was the underlined part. The rest of your post/question has nothing to do with duration matching and/or maturity matching. It is asking about selling
some shorter maturities to buy some longer maturities, thereby extending duration with the associated reinvestment risk.
In this case, the shorter maturities you are considering selling were likely purchased before interest rates rose and longer-term prices fell more than shorter-term prices have fallen. Thus, the
risks of reinvestment, in this case, appear to have moved in the direction that is in your favor when considering reinvestment moving toward longer average duration.
I hope Kevin's response answered your question.
Re: Filling the TIPS gap years with bracket year duration matching
MtnBiker wrote: ↑Fri May 24, 2024 2:10 pm
I hope Kevin's response answered your question.
Yes. Thanks, Kevin.
The main takes that I got from Kevin's response to my post are (correct me if I am wrong):
1. Long run: not much difference if I sell those with high coupons vs. low coupons. The inverse effect of coupon rate on duration is not great enough to make a difference. The only thing that should
matter is whether you want increased or decreased cash flow prior to maturity.
2. Selling 4/2027s to buy 2040s will extend the mean duration of a ladder a bit more than selling 4/2028s.
3. Selling either 4/2027s or 4/2028s to buy 2040s at today's high yields is preferable to selling 2033s to buy 2040s- it extends duration more.
4. But selling correct proportions of 2034s and 2040s is better when buying 2035s, since it avoids reinvestment risk in case yields fall in 2025.
Re: Filling the TIPS gap years with bracket year duration matching
protagonist wrote: ↑Fri May 24, 2024 4:55 pm
MtnBiker wrote: ↑Fri May 24, 2024 2:10 pm I hope Kevin's response answered your question.
Yes. Thanks, Kevin.
The main takes that I got from Kevin's response to my post are (correct me if I am wrong):
1. Long run: not much difference if I sell those with high coupons vs. low coupons. The inverse effect of coupon rate on duration is not great enough to make a difference. The only thing that
should matter is whether you want increased or decreased cash flow prior to maturity.
2. Selling 4/2027s to buy 2040s will extend the mean duration of a ladder a bit more than selling 4/2028s.
3. Selling either 4/2027s or 4/2028s to buy 2040s at today's high yields is preferable to selling 2033s to buy 2040s- it extends duration more.
4. But selling correct proportions of 2034s and 2040s is better when buying 2035s, since it avoids reinvestment risk in case yields fall in 2025.
1. Yes. To check, I changed the coupon of the Apr 2028 from 3.625% to 0.125%, and it changed the m duration from 3.61 to 3.83, so an increase of 0.22. Compare to the difference one year in maturity
makes, with the Apr 2027 at an mdur of 2.85, so 98 basis points difference with both coupons at 0.125%.
2. Yes.
3. Yes, as long as increasing duration is a goal.
4. My emphasis now is more on what to hold rather than what to sell. It would translate into what to sell if yields were to change by the same amount (parallel yield curve shift); this is the topic
of my next post. Also, although I'm focusing on the bracket years, 2034 and 2040, the same would apply if you were using other maturity years, as MtnBiker is doing with the 2032 and 2040.
I guess we could characterize it as minimizing reinvestment risk, but usually people are referring to reinvesting coupons or reinvesting principal as an issue matures or is sold before maturity,
typically buying the same or longer maturity. I think of it more as minimizing price risk; i.e., that the average price of our bracket year TIPS, held in the correct proportions, will change by about
the same amount as the hypothetical price of the gap year TIPS we'll be selling them to buy. It doesn't matter what we call as much as that we understand how it's supposed to work.
Note that any price changes happen while we're holding the bracket year TIPS, not when we sell them. I suspect that we might end up selling in slightly different proportions than what we originally
intended if there were enough non-parallel yield curve shift, but I haven't thought this through yet.
If I make a calculation error, #Cruncher probably will let me know.
Testing duration match model with small, parallel yield curve shift
I've said that this duration matching scheme should work as long as yields at different maturities change by the same amount, which we call a "parallel yield curve shift". I've read and posted about
this principle with respect to applying the duration rule of thumb that relates price or NAV change to yield change for a bond fund, and I think I must have done some calculations to verify this, but
I can't remember distinctly doing so. So, I thought I'd better do that for this duration matching scheme.
Those who understand the duration rule of thumb know that it works better for smaller yield changes, so let's start with a small, parallel shift up in the yield curve, say 0.1 percentage point, or 10
basis points.
The ask yields of the 2030 and 2040 on Friday when I pulled quotes from Schwab were 2.15% and 2.23% respectively; to keep things simple, I'll use ask yields and prices here, and ignore any small
seasonal adjustments.
If we increase the yields by 10 bps, to 2.25% and 2.33% for the 2034 and 2040 respectively, the prices change by -0.88% and -1.32% respectively, as calculated with the spreadsheet PRICE function.
These changes are quite close to what is predicted by the duration rule of thumb; with modified durations of 8.74 and 13.20, we'd get price changes of -0.87% and -1.32%.
To calculate the estimated expected price change of the 2035 gap year TIPS, for example, and compare it to the weighted average price change of our duration matched TIPS, we do the following:
1. Increase the estimated yield of the 2035 by 10 bps from 2.16% to 2.26%.
2. Calculate the estimated price using the increased yield and the PRICE function, which comes to 98.70.
3. Calculate the percentage change in price, dp%, which is:
Code: Select all
dp% = 98.70/99.64 - 1
dp% = -0.94%.
4. Calculate the weighted average percent price change of the duration matched TIPS, DM dp%. For the 2035, the weighs are 0.85 of the 2034 and 0.15 of the 2040, so the calculation is:
Code: Select all
DM dp% = 0.85 * -0.88% + 0.15 * -1.32%
DM dp% = -0.94%
(it's 0.95% with the rounded values above, but 0.94% with the unrounded values used in the spreadsheet calculations).
5. Subtract dp% from DM dp%:
Code: Select all
DM dp% - dp% = 0.94% - 0.94%
DM dp% - dp% = 0% (percentage points)
And we have a winner!
Here is the chart showing the yields increased by 10 bps, price based on higher yields, dp%, DM dp%, and the delta between the latter two:
Looks good!
For completeness, here's the chart of estimated yields and prices for the gap year TIPS, so anyone who's interested can check my work:
In the next post I'll evaluate the effects of a larger change in yields, and of a non-parallel yield curve shift.
Last edited by Kevin M on Sun Jun 02, 2024 7:48 pm, edited 1 time in total.
If I make a calculation error, #Cruncher probably will let me know.
Testing duration match model with non-parallel yield curve shift
How does gap duration matching work with larger yield changes? Let's try an increase of 1 percentage point, or 100 basis points. Having already walked through the calculations in my last post, I'll
just show the results.
Anticipating evaluating non-parallel yield curve shifts, I've added a delta yield (dy) column.
Not bad. The delta of the duration weighted TIPS percent price change and the estimated gap year percent price change is a maximum of 4 bps. Note that this is despite the duration rule of thumb (rot)
not working as well; i.e., for the 2034 the dp% was -8.38% compared to a rot estimate of -8.74%, and for the 2040 we see dp% of -12.32% compared to a rot estimate of -13.20%. So even though the
duration rule of thumb doesn't work as well for larger yield changes, the duration matching seems to still work quite well for a relatively large parallel yield curve shift.
Now let's try a non-parallel yield curve shift, where the 2034 increases by 10 bps and the 2040 increases by 100 bps. Here are the results:
Here we see much larger deltas between DM dp% and dp%, with the duration weighted TIPS price falling 65 basis points more than the estimated price of the 2037.
On the other hand, the delta for the 2035 is only 23 bps, and for the 2039 it's only 27 bps. Note that the largest delta is the middle gap year, and that it falls off fairly symmetrically on either
One might think that a 23 bps delta for the 2035 is not such a big deal for a relatively sharp steepening of the yield curve. What do you think? I wonder how this would compare to the trading costs
of rebalancing between the bracket years before the yield curve steepened too much. We might also evaluate if we can sell a different ratio than 85/15 of the 2034/2040 to reduce the delta and
rebalance a bit at the same time. Haven't thought this through yet.
Last edited by Kevin M on Sun Jun 02, 2024 7:48 pm, edited 1 time in total.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Sat May 25, 2024 3:12 pm
One might think that a 23 bps delta for the 2035 is not such a big deal for a relatively sharp steepening of the yield curve. What do you think? I wonder how this would compare to the trading
costs of rebalancing between the bracket years before the yield curve steepened too much. We might also evaluate if we can sell a different ratio than 85/15 of the 2034/2040 to reduce the delta
and rebalance a bit at the same time. Haven't thought this through yet.
Seems like you are on the right track. Assuming the yield curve changed significantly before the time you want to make the first swap (selling bracket years to buy 2035s), the 2034/2040 ratio to be
sold may be something different than the 85/15 ratio originally calculated. And the ratios you want to hold until the next swap may be different than the original ratios when you started.
Rebalancing should help. Maybe assume annual rebalancing at the time of each swap and see how small the delta is at each of the 5 swaps. Maybe assume the yield curve alternately steepens/flattens a
similar amount each year as a worst-case scenario?
EDIT: Thinking about this some more, rebalancing may not help much, if at all, since duration is fairly insensitive to yield. The duration of the bracket year TIPS will change very little as the
yield changes. Changes in the duration of the gap year TIPS will primarily result from changes in coupon rates (when yields change enough to affect coupons). (I assume you included the changes in
coupon rate of the gap years when you repriced them after the yield changes, correct?)
If the large deltas in the mid-gap persist even with rebalancing, minimizing the effects of non-parallel yield shifts may require rolling the bracket years (swap all excess 2034s for excess 2035s at
the first opportunity, for example).
Re: Filling the TIPS gap years with bracket year duration matching
MtnBiker wrote: ↑Sat May 25, 2024 4:23 pm Changes in the duration of the gap year TIPS will primarily result from changes in coupon rates (when yields change enough to affect coupons). (I assume
you included the changes in coupon rate of the gap years when you repriced them after the yield changes, correct?)
I did not. I'll look at redoing the analysis with updated gap year TIPS estimated coupons.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Wed May 22, 2024 10:28 pm There have been many threads on how to fill the gap years in a TIPS ladder. If you don't know what I'm talking about, this thread is not for you,
although if you want to understand it anyway, you might find it enlightening. Although we have had some discussions in other threads of the technique I'll discuss here, the questions that have
come up in those threads indicate that this is a meaty enough topic to have its own thread (I hope this thread won't get merged with one of these existing, less specific threads, one or more of
which I'll link to later in this thread for additional background and to answer questions asked in the other threads).
Going back to your original post, I do have a few questions / clarifications. Apologies if I’m being a little obtuse:
(1) If I understand correctly, the DARA weights you list are intended to be a multiplier for the current (2024) TIPS inflation adjusted amounts, not the TIPS face amounts or the TIPS current value.
You might want to highlight that for TIPS newbies.
(2) I’m a little cross-eyed at the logic as to how you can just simply add the TIPS weights up across the gap years (which all have different inflation factors) and come up with a number exactly
equal to the total number of years, i.e. seven. I get that each TIPS is inflation adjusted, but as some TIPS expire in 2034 how can you include them in a methodology that estimates an inflated amount
that extends beyond that year? Certainly, if you used nominal numbers these figures would be all over the place.
(3) The tipladder.com methodology generates a different estimated amount of TIPS needed to bridge the 2034-40 gap. Part of this is the different 4:3 weighting, but much appears to be that
tipladder.com uses the later year TIPS’ coupons as a part of the DARA for each year. If I understand correctly, your methodology does not include TIPS coupons, just the principal.
[I actually prefer not including the TIPS coupons, as I like to think of the TIPS coupons as being available to pay taxes on OID.]
For example, for a $10,000 per year DARA, the tipsladder.com estimates:
$36,000 face amount of 2034
$21,000 face amount of 2024
Your weighting gives the following:
3.56 x $10,000 = $35,600 inflation adjusted / 1.01544 = $35,058 face amount of 2034
3.44 x $10,000 = $34,400 inflation adjusted / 1.44205 = $23,850 face amount of 2040
Not a huge difference, I guess?
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Sun May 26, 2024 11:11 am
MtnBiker wrote: ↑Sat May 25, 2024 4:23 pm Changes in the duration of the gap year TIPS will primarily result from changes in coupon rates (when yields change enough to affect coupons). (I
assume you included the changes in coupon rate of the gap years when you repriced them after the yield changes, correct?)
I did not. I'll look at redoing the analysis with updated gap year TIPS estimated coupons.
I'm very happy that you're participating in this journey, MtnBiker. Turns out that ignoring the coupon is a big hole in the analysis, not so much because of how it affects duration, but because of
how it affects price.
It only took a few minutes of work to see the model blow up if I set coupon to what it would be for the different yields at auction. Then it only took a few seconds of thought to see how this should
be obvious.
At auction, coupon will be very close to yield, so price will be very close to 100. So a model that hinges on theoretical price change of yet-to-be-issued TIPS based on a fixed coupon just doesn't
make sense.
The next thought is that the focus on price is misplaced, since what we're buying is some multiple of DARA, where that multiple is the number of years covered by our ladder; e.g., if we have a 30
year ladder with one TIPS issue for each year, the multiplier for the row for each year will be 1, and the sum of the multipliers will be 30.
The price of the ladder is inversely related to TIPS yields, and to a much lesser extent to TIPS coupons. To illustrate and investigate this and some other points, I'll use a variation of the #
Cruncher spreadsheet to model a TIPS ladder with a flat yield curve, and I'll set the yield and coupon to various values.
A 30y ladder with a DARA of $100K, and yield = coupon = 2% costs $2,263,425. Increase the yield to 3% but keep the coupon at 2%, and the cost decreases to $1,996,062. If we also increase the coupon
to 3%, the cost decreases further to $1,990,953. Note that changing the yield by 100 basis points has a much larger impact on the cost than changing the coupon by 100 basis points.
However, a higher coupon means a larger contribution to earlier year cash flows, which could reduce the number of TIPS required for the earlier years to get our DARA. With a 2% par yield curve, we
need 63 Jan 2033s to get close to our $100K DARA for that year, and a total of 1,823 TIPS in our ladder, but if the 2034 had a coupon of 3%, we'd only need 60 Jan 2033s and 1,811 total (changing the
2034 yield does not impact how many we'd need).
So, I think that the tack to pursue is to investigate how change in expected yield and coupon of a gap year impacts the ladder overall.
My intuition is that there's still something to be said for at least approximately matching the durations of the gap years with the bracket years, but it seems that the analysis I've used to
demonstrate this is flawed.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Jaylat wrote: ↑Sun May 26, 2024 5:24 pm (2) I’m a little cross-eyed at the logic as to how you can just simply add the TIPS weights up across the gap years (which all have different inflation
factors) and come up with a number exactly equal to the total number of years, i.e. seven.
This isn't my invention. The way the #Cruncher spreadsheet works, the total of the DARA multipliers must equal the number of years covered by the ladder. This applies to a single year or a span of
years as well, if we want coverage for every year. Since there are 2 bracket years and 5 gap years, the sum of the bracket year multipliers must be 7.
Jaylat wrote: ↑Sun May 26, 2024 5:24 pm (3) The tipladder.com methodology generates a different estimated amount of TIPS needed to bridge the 2034-40 gap. Part of this is the different 4:3
weighting, but much appears to be that tipladder.com uses the later year TIPS’ coupons as a part of the DARA for each year. If I understand correctly, your methodology does not include TIPS
coupons, just the principal.
Not correct. Again, not my invention. Total annual real cash flows in the #Cruncher spreadsheet include real principal and real coupons, and the total of these is designed to be as close to DARA as
possible with rounding and the resolution of individual TIPS costs.
The easiest way to see all of this is to open up the #Cruncher spreadsheet and investigate the various values.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Kevin, is the main reason that you want to maximize the duration of your ladder because you believe that interest rates will fall in the future?
Since increasing duration makes your ladder more interest rate sensitive, I would assume that is your logic- if rates fall , the value of your TIPS will rise proportionate to duration (if you sell
prior to maturity).
If you are completely unsure about the direction of future interest rates, the result of increasing duration seems like a crapshoot. Or am I missing something?
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Sun May 26, 2024 7:35 pm
I'm very happy that you're participating in this journey, MtnBiker. Turns out that ignoring the coupon is a big hole in the analysis, not so much because of how it affects duration, but because
of how it affects price.
It only took a few minutes of work to see the model blow up if I set coupon to what it would be for the different yields at auction. Then it only took a few seconds of thought to see how this
should be obvious.
At auction, coupon will be very close to yield, so price will be very close to 100. So a model that hinges on theoretical price change of yet-to-be-issued TIPS based on a fixed coupon just
doesn't make sense.
The next thought is that the focus on price is misplaced, since what we're buying is some multiple of DARA, where that multiple is the number of years covered by our ladder; e.g., if we have a 30
year ladder with one TIPS issue for each year, the multiplier for the row for each year will be 1, and the sum of the multipliers will be 30.
The price of the ladder is inversely related to TIPS yields, and to a much lesser extent to TIPS coupons. To illustrate and investigate this and some other points, I'll use a variation of the #
Cruncher spreadsheet to model a TIPS ladder with a flat yield curve, and I'll set the yield and coupon to various values.
A 30y ladder with a DARA of $100K, and yield = coupon = 2% costs $2,263,425. Increase the yield to 3% but keep the coupon at 2%, and the cost decreases to $1,996,062. If we also increase the
coupon to 3%, the cost decreases further to $1,990,953. Note that changing the yield by 100 basis points has a much larger impact on the cost than changing the coupon by 100 basis points.
However, a higher coupon means a larger contribution to earlier year cash flows, which could reduce the number of TIPS required for the earlier years to get our DARA. With a 2% par yield curve,
we need 63 Jan 2033s to get close to our $100K DARA for that year, and a total of 1,823 TIPS in our ladder, but if the 2034 had a coupon of 3%, we'd only need 60 Jan 2033s and 1,811 total
(changing the 2034 yield does not impact how many we'd need).
So, I think that the tack to pursue is to investigate how change in expected yield and coupon of a gap year impacts the ladder overall.
My intuition is that there's still something to be said for at least approximately matching the durations of the gap years with the bracket years, but it seems that the analysis I've used to
demonstrate this is flawed.
I still think duration matching the gap years with the bracket years is the right approach. Let me throw out some ideas for you to consider for how to analyze this. First, as you said, the cost of a
30y ladder goes down if the yield increases from 2% to 3%.
Similarly, the cost of buying any gap year rung at auction (or soon after on the secondary market) should fall if the yield increases from 2% to 3%. (Just as the market values of the excess bracket
holdings fall as yield increases.) For the purpose of evaluating the effectiveness of duration matching, shouldn't you be focusing on the changes in cost, not the changes in price? Yes, the price of
the gap year TIPS purchased at auction is about 100, but the cost will vary with yield since you should need fewer or more bonds depending on which direction the yield changed.
Or maybe the figure of merit would be something like the delta in the present value of the future cash flows that occurs when you swap the bracket-year TIPS for the gap-year TIPS. If duration
matching works, your buying power should be preserved after swapping. When duration matching is imperfect, buying power will either increase or decrease. (Not sure how to quantify "buying power;"
just an idea.) When making the swap you can't keep the DARAs uniform because the changing coupons push the payouts around to different years. Even the expected principal payout at the gap-year
maturity will change, I suppose.
Keeping the DARA from becoming non-uniform or lumpy should be a secondary consideration. When we swap bracket-year TIPS for gap-year TIPS, the ongoing coupon payments will change and that effects the
payouts in the gap and in preceding years. But that is just an inevitable consequence of swapping TIPS from one coupon to another (and from changing maturity dates). (For example, when I made the
swap from a mix of April 2032s/2040s to January 2034s, the coupon went from mostly 3.875%, plus some at 2.125%, to the new value of 1.75%, which defers a portion of the interest payments I would have
received from my ladder from 2025-2032 until later years.) That affects the annual cash flows in various ways, but what we are trying to preserve by duration matching the gap years is the original
buying power of the funds allocated toward each gap year. If the original "yield to maturity" (buying power?) is reasonably well maintained at the expense of perturbed cash flows, so be it.
Does any of this discussion help with the analysis?
Re: Filling the TIPS gap years with bracket year duration matching
Perhaps using I Bonds to cover the gap years might help simplify things.
Best Regards - Mel | | Semper Fi
Re: Filling the TIPS gap years with bracket year duration matching
Let me present my perspective as an "average Joe" who has made one of these swaps. Earlier this year I swapped April 2032s and 2040s for 2034s. I used the maturity-matching approximation, rather than
strict duration matching, so the 2032/2040 excess holding ratio was 75/25.
It went something like this. (Not the actual numbers but generally illustrative of my experience.)
The 2032s were sold at a yield of 2.0%.
The 2040s were sold at a yield of 2.2%.
The 2034s were purchased using the proceeds of the sale at a yield of 1.95%.
From my perspective, the average yield of the TIPS that I sold to make the swap was 2.05%. The obvious "cost" of making the swap was the delta-yield of 10 basis points that I gave up (1.95 - 2.05 =
-0.10%). That delta-yield has a certain delta-price which is the dollar cost of making the swap.
Totally unknown to me is any loss (or gain) in yield to maturity that I may have incurred during the initial holding period from when I made the original purchases (in 2018 and 2023) until the time
of the swap (2024). At the time of the original purchases the average quoted yield to maturity was probably something like 1.2%, or thereabouts. That is the average yield to maturity I would receive
if I kept the excess bracket year holdings to their respective maturities in 2032 and 2040. One might think that the average yield to maturity (from the times of original purchases) will now be close
to 1.1%, since I lost 0.1% yield when making the swap. But that assumes perfect duration matching which is unlikely to have been the case. The effects of imperfect duration matching may have affected
the yield to maturity in positive or negative ways that I can't quantify.
I can think of four factors that contribute to the costs of making a swap.
1) Vanguard's bid/ask spread. The bid/ask spread of whatever broker you use is unavoidable. Schwab is better than Vanguard, but I'm not going through the hassle of moving my IRA for that reason
2) Nonlinearities in the yield curve. The April 2032 TIPS is an outlier with anomalously high yield. Its yield always lies above any fit to the yield curve. So, using that particular issue for
holding excess bracket-year TIPS will always incur a small yield loss when it comes time to swap. This factor should be negligible for most other bracket-year holdings.
3) Changes in the slope of the yield curve (non-parallel shifts in yield). Duration-matching doesn't immunize against effects from non-parallel yield shifts.
4) Imperfect duration-matching with parallel yield shift. In my case, I used ratios for excess holding that are only an approximation to the ideal duration-matched ratio. Also, since the coupon of
the gap yield TIPS is unknown ahead of time, so it is impossible to know the ideal duration-matched ratio when originally choosing the holdings ratio.
The 0.1% yield loss that I observed when making my swap was, I think, a result of factors (1) and (2). Quantifying the effect of factors (3) and (4) is the goal of this thread.
Re: Filling the TIPS gap years with bracket year duration matching
MtnBiker wrote: ↑Sun May 26, 2024 11:07 pm Similarly, the cost of buying any gap year rung at auction (or soon after on the secondary market) should fall if the yield increases from 2% to 3%.
(Just as the market values of the excess bracket holdings fall as yield increases.) For the purpose of evaluating the effectiveness of duration matching, shouldn't you be focusing on the changes
in cost, not the changes in price? Yes, the price of the gap year TIPS purchased at auction is about 100, but the cost will vary with yield since you should need fewer or more bonds depending on
which direction the yield changed.
The number of bonds needed differs only because of the change in the last year's interest due to the difference in coupon, so this change is maybe 1 bond or none for a 1 pp change in yield/coupon.
This can be verified by zeroing out the Last Yr Interest per Bond value in the #Cruncher spreadsheet.
I'll just take the Jan 2025 as an example, assuming it's a par bond. Increasing yield/coupon from 2% to 3% changes the number needed from 66 to 65, but changing it to 2% to 1% doesn't change it at
So the cost of buying the gap year TIPS isn't going to change much due to yield changes, as it would if it had a fixed coupon, in which case price would change per my analysis. What's going to change
are the values of the TIPS already in our ladder, and of course those values will change proportional to price, scaled by index ratio. Another impact of different gap year coupons is a possible
change in the number of earlier year TIPS required, due to more or less interest from the gap year TIPS contributing to the annual real amounts.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
protagonist wrote: ↑Sun May 26, 2024 8:02 pm Kevin, is the main reason that you want to maximize the duration of your ladder because you believe that interest rates will fall in the future?
Since increasing duration makes your ladder more interest rate sensitive, I would assume that is your logic- if rates fall , the value of your TIPS will rise proportionate to duration (if you
sell prior to maturity).
If you are completely unsure about the direction of future interest rates, the result of increasing duration seems like a crapshoot. Or am I missing something?
It's not that I want to maximize duration, it's that I want to move toward a duration value that's appropriate for my investment horizon. If I assume a 20-year lifetime, I probably want a duration
closer to 10 than to 6.
Of course just building a 20-year ladder with each rung equal to the same DARA gets me there, but I'm trying to nudge myself in that direction without just selling all of my extra 2025s-2027s in one
shot to build out a full ladder. I'm doing this because of the point you raise--I'm OK with the reinvestment risk in the hopes of getting even higher yields for the longer TIPS.
Given that we haven't seen a nice TIPS yield bump in awhile, I'm leaning toward swapping some of my 2025s for more TIPS in the 2028-2030 range, since those prices are less sensitive to yield changes.
But we digress from the topic at hand.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Mel Lindauer wrote: ↑Mon May 27, 2024 12:19 am Perhaps using I Bonds to cover the gap years might help simplify things.
This discussion here isn't about different ways to cover the gap years, but about one specific method to do so. There are infinite ways to cover the gap years, and there are plenty of threads
discussing many of them.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
MtnBiker wrote: ↑Mon May 27, 2024 10:15 am 2) Nonlinearities in the yield curve. The April 2032 TIPS is an outlier with anomalously high yield. Its yield always lies above any fit to the yield
curve. So, using that particular issue for holding excess bracket-year TIPS will always incur a small yield loss when it comes time to swap. This factor should be negligible for most other
bracket-year holdings.
The apparently high yield of the April 2032 for a May 28 settlement is explained by seasonality, which is why I usually use seasonally-adjusted yields, perhaps modified with an outlier factor, in my
analyses and purchase decisions.
We see that seasonal adjustment eliminates the anomalously high yield for the April 2032. Here are yields for it and adjacent maturities (these are from Schwab on Friday):
Although the quoted yield will often lie above a smoothed quoted yield curve, it won't always. It depends on the ratio of seasonal adjustment (SA) factors for settlement mm/dd and maturity mm/dd.
Without getting too deep into it, the seasonal yield adjustment for an April TIPS will be upward for settlement dates between about Feb 3 and Apr 14, and downward the rest of the year.
However, it could be that an additional outlier factor is at play for the April 2032. Here's an SA chart I posted on Feb 28 of this year:
Kevin M wrote: ↑Wed Feb 28, 2024 10:34 am
Here we see that although the SA yield adjustment is up, as expected, there's still a bit of a bump at Apr 2032, and I didn't apply quite enough of an outlier factor to eliminate it.
Here's another chart I posted on Mar 20, with some relevant commentary:
Kevin M wrote: ↑Wed Mar 20, 2024 10:02 am To emphasize what jeffyscott said about seasonality, I decided to buy Jan and Jul maturities for my ladder where available. If I looked only at yields, I
would favor Apr maturities over Jul maturities, but understanding that this is a result of seasonality, I'm comfortable buying the Jul maturities.
Note how much higher the Apr ask yields are than the Jul ask yields. Adjusting for seasonality mostly removes the sawtooth pattern in the ask yields.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Mon May 27, 2024 11:35 am
Mel Lindauer wrote: ↑Mon May 27, 2024 12:19 am Perhaps using I Bonds to cover the gap years might help simplify things.
This discussion here isn't about different ways to cover the gap years, but about one specific method to do so. There are infinite ways to cover the gap years, and there are plenty of threads
discussing many of them.
Wasn't trying to derail the discussion, Kevin. Rather, I was trying to offer a simple guaranteed inflation-protected option to cover the needed guaranteed inflation protection in the TIPS gap years
when the options being discussed just seemed overly complicated.
Best Regards - Mel | | Semper Fi
Re: Filling the TIPS gap years with bracket year duration matching
Mel Lindauer wrote: ↑Mon May 27, 2024 2:09 pm
Kevin M wrote: ↑Mon May 27, 2024 11:35 am
Mel Lindauer wrote: ↑Mon May 27, 2024 12:19 am Perhaps using I Bonds to cover the gap years might help simplify things.
This discussion here isn't about different ways to cover the gap years, but about one specific method to do so. There are infinite ways to cover the gap years, and there are plenty of threads
discussing many of them.
Wasn't trying to derail the discussion, Kevin. Rather, I was trying to offer a simple guaranteed inflation-protected option to cover the needed guaranteed inflation protection in the TIPS gap
years when the options being discussed just seemed overly complicated.
Right. There are many threads discussing I bonds vs. TIPS, and I don't want this to degenerate into one of those. Having said that, after thinking about it a bit more, I realized that someone who
doesn't understand both I bonds and TIPS might not understand why I bonds wouldn't work for anything like the method we've been discussing. More importantly, after finding the fatal flaw in my
analysis attempting to justify the bracket year duration approach, I've decided to expand the scope of the thread to cover mathematical analyses of gap year coverage methods, including implementation
considerations. So we can discuss I bonds in that context; it may even help stimulate some useful thinking with respect to using TIPS to cover the gap years.
First, I'll just point out that the simple versions of the options that have been discussed are not complicated. You just buy extra 2034s and 2040s with the intention of later selling them to buy the
gap year TIPS as they become available, which will be in 2025 through 2029 (so in five years, this discussion will be moot). The two tips ladder tools I've mentioned offer simple defaults to
accomplish this, as discussed earlier. What's complicated is trying to come up with the math to determine an optimal solution, if there is one.
I bonds have a duration of 0, since value doesn't vary with changing market yields. Although this doesn't work for the duration matching scheme that has been discussed, we may find that that scheme
doesn't have sufficient analytical underpinnings to support it. So we can put this aside for now.
The current I bond real rate is 1.30%, while the 2034 and 2040 TIPS yields are greater than 2%; actually all TIPS currently have yields north of 2%. The effective I bond yield is even lower if we
want to buy the gap years as soon as available, since we'd be paying the 3-month interest penalty in redeeming the I bonds in less than 5 years from purchase.To favor I bonds over any TIPS at current
yields, we'd have to determine a justification for sacrificing the extra yield of the TIPS.
How about getting "a simple guaranteed inflation-protected option to cover the needed guaranteed inflation protection in the TIPS gap years"? I assume the concern here is that with longer-maturity
TIPS, like the 2034 or 2040, the values will fluctuate as market yields fluctuate. Yields decreasing is not an issue, since the value of our TIPS will increase, and we'd then have more than we
planned on to buy the gap year TIPS when issued. However higher yields would cause a loss of value in the longer TIPS we're holding to cover the gap years. This wouldn't be an issue if the initial
assumption that the gap year TIPS coupon is fixed at what it would be if issued today held, since duration matching would solve that, but obviously that assumption is incorrect.
My next thought is to buy extra Jan 2025 TIPS to buy the 2035 to be issued in Jan 2025, buy extra Jan 2026 TIPS to cover the 2036 to be issued in Jan 2026, etc. With this solution, there's no
unadjusted price uncertainty, so we get the desired guaranteed inflation protection. Not only that, but we get yields between 2.20% for the Jan 2029 and 3.4% for the Jan 2025 (inverted yield curve),
so even better than the yields on the 2034 and 2040, and of course much better than the I bond real rate.
Of course there's some uncertainty in the nominal values at maturity, but that's also a problem for I bonds.
The fatal flaw in using I bonds to cover the gap years, for other than the DARAs of say $10K or less, is that pesky annual purchase limit. For a TIPS ladder that generates $50K/year in real annual
cash flows, we need roughly $50K to cover one gap year, so to cover the five gap years we'd need roughly $250K. A couple of years ago when the I bond composite rate was north of 9%, DW and I were
able to buy $70K between individual and entity accounts and gifts. So we could have covered one gap year with that for a DARA of up to $70K.
The rules of the game here are that we can't go back in time when the I bond annual purchase limits were higher (and you could buy I bonds with a credit card), and we are assuming we're building our
ladder today when TIPS yields are historically attractive. We need to cover the gap years with purchases today anyway if we want the inflation protection starting today.
Now for those who've bought a bunch of I bonds in the past, and who have no other particular use for them, then sure, you could use those I bonds to cover the gap years if you have enough of them.
But for any I bonds with a fixed rate of less than 2%, which is any I bonds bought since Nov 2002, why not sell them and buy the 2025-2029 TIPS at significantly higher yields?
This brings to mind the topic of asset location. I bonds can't be held in IRAs, so anyone building their TIPS ladder in an IRA, which includes me, can't use I bonds. Anyone building some or all of
their ladder in taxable might be able to use I bonds, in which case you'd need to compare the benefits of deferring taxes for 1-5 years on the lower yields with switching to TIPS with the higher
yields. Hard to imagine deferring taxes a year or two makes much difference, but everyone's tax situation is different.
'Nuff said?
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Just spit-balling some ideas here:
One way to hedge the gap years would be to buy 100% 2040 TIPS and 0% 2034 TIPS. You could then hedge the early redemption of the 2040’s by shorting an equal amount of 2040 treasuries and buying the
gap year treasuries. [Not sure how this would work in practice, or if it’s doable for a retail investor, but it’s possible for a financial institution. Also, you’d be hedging real TIPS with nominal T
Bonds so maybe not such a great hedge after all?]
Following with the strategy of buying 100% 2040 TIPS, a retail investor could concoct a “hedge” of sorts by using a smaller portfolio of I Bonds as a hedge against early redemption of the 2040 TIPS
in a high real yield / high coupon environment. If rates are down, you sell the TIPS at a profit; if rates are up, you sell the I Bonds at par and get a higher yield. Either way you’re covered.
You could also do a split of 50% 2040 TIPS and 50% I Bonds. Worst case scenario is you have to sell all the I Bonds in the first 2-3 years due to higher rates. In that case the remaining 2040 TIPS
are only 2-3 years from maturity, so the hit to sell early should be significantly reduced.
The problem here is it requires a portfolio of I Bonds which not all of us have. Also, you’d be giving up a lot of yield to buy the I Bonds, making for an expensive hedge.
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Mon May 27, 2024 10:52 am
The number of bonds needed differs only because of the change in the last year's interest due to the difference in coupon, so this change is maybe 1 bond or none for a 1 pp change in yield/
coupon. This can be verified by zeroing out the Last Yr Interest per Bond value in the #Cruncher spreadsheet.
I'll just take the Jan 2025 as an example, assuming it's a par bond. Increasing yield/coupon from 2% to 3% changes the number needed from 66 to 65, but changing it to 2% to 1% doesn't change it
at all.
So the cost of buying the gap year TIPS isn't going to change much due to yield changes, as it would if it had a fixed coupon, in which case price would change per my analysis. What's going to
change are the values of the TIPS already in our ladder, and of course those values will change proportional to price, scaled by index ratio. Another impact of different gap year coupons is a
possible change in the number of earlier year TIPS required, due to more or less interest from the gap year TIPS contributing to the annual real amounts.
I am having trouble following what you are saying. In the TIPS Ladder Spreadsheet thread you wrote:
Kevin M wrote: ↑Tue May 28, 2024 11:03 am
I've always assumed that this just made sense based on some sort of duration matching scheme, but in my thread, Filling the TIPS gap years with bracket year duration matching - Bogleheads.org,
we've discovered that duration matching doesn't really work because the gap year TIPS price will always be close to 100. Duration matching requires that the price of the the TIPS between the
bracket years varies with yields, as do the prices of the bracket year TIPS, and this assumption doesn't apply for new issues at auction.
I disagree with the statement that duration matching doesn't really work. Here is a simple thought experiment to show that it does still work.
Suppose I want to fund 2037 at the 20K level. I do this by buying 10K of 2034 at 2% yield and 10K of 2040 at 2% yield. (50/50 maturity matching approximation to duration matching.)
Suppose I want to make the swap in 2027. The "duration" (maturity) of the 2034 is 7 years. The duration of the 2040 is 13 years and the duration of the new-issue 2037 is 10 years.
Also suppose that just before I make the swap, the yield jumps 1% across the entire yield curve. The new yield is 3% for the 2034 and the 2040. The price of the 2034 falls 7%. The price of the 2040
falls 13%. The average price of the excess holdings falls 10% just before I sell to make the swap.
So, using the proceeds of the sale, I am short 10% and have to buy 10% less of the 2037 compared to what I would have bought if interest rates hadn't jumped. But the 2037 is now yielding 3% instead
of 2%. Thus, I am gaining back 1% every year that I hold the 2037. By the end of the 10 years held, I will have made up 1%/year x 10 years = 10%. I will be made whole by the time of maturity in 2037.
The only thing that has changed is the cash flows. After the swap, more of the payout comes in the form of coupons, but the overall yield is the same. (If I wanted to spend all the cash flow in 2037,
I would need to reinvest the excess portion of the coupons (1%/yr) to make up for the 10% shortfall in the principal at maturity.)
Re: Filling the TIPS gap years with bracket year duration matching
Agree with MtnBiker's logic, the higher coupons should offset the more expensive TIPS.
Also not following this statement:
Kevin M wrote: ↑Tue May 28, 2024 11:03 am After thinking about it, it seems that currently we could more effectively cover the 2035 gap year with a Jan 2025 TIPS, for example, since there is more
certainty of the nominal return when the 2035 is issued in Jan 2025, and the yield is much higher than the 2034 or 2040 TIPS.
The whole point of hedging the 2035 now is that we have no idea what the 10 year TIPS yield will be in Jan 2025. Buying a Jan 2025 TIPS, even with a much higher yield, does nothing for hedging longer
term real yields, such as the to be issued 2035 TIPS.
Who cares if your real yields are higher for the next 7 months? It's the next 10 years you should be concerned about.
Re: Filling the TIPS gap years with bracket year duration matching
MtnBiker wrote: ↑Tue May 28, 2024 2:39 pm
Kevin M wrote: ↑Mon May 27, 2024 10:52 am The number of bonds needed differs only because of the change in the last year's interest due to the difference in coupon, so this change is maybe 1
bond or none for a 1 pp change in yield/coupon. This can be verified by zeroing out the Last Yr Interest per Bond value in the #Cruncher spreadsheet.
I'll just take the Jan 2025 as an example, assuming it's a par bond. Increasing yield/coupon from 2% to 3% changes the number needed from 66 to 65, but changing it to 2% to 1% doesn't change
it at all.
So the cost of buying the gap year TIPS isn't going to change much due to yield changes, as it would if it had a fixed coupon, in which case price would change per my analysis. What's going
to change are the values of the TIPS already in our ladder, and of course those values will change proportional to price, scaled by index ratio. Another impact of different gap year coupons
is a possible change in the number of earlier year TIPS required, due to more or less interest from the gap year TIPS contributing to the annual real amounts.
I am having trouble following what you are saying. In the TIPS Ladder Spreadsheet thread you wrote:
Kevin M wrote: ↑Tue May 28, 2024 11:03 am
I've always assumed that this just made sense based on some sort of duration matching scheme, but in my thread, Filling the TIPS gap years with bracket year duration matching - Bogleheads.org
, we've discovered that duration matching doesn't really work because the gap year TIPS price will always be close to 100. Duration matching requires that the price of the the TIPS between
the bracket years varies with yields, as do the prices of the bracket year TIPS, and this assumption doesn't apply for new issues at auction.
I disagree with the statement that duration matching doesn't really work. Here is a simple thought experiment to show that it does still work.
Suppose I want to fund 2037 at the 20K level. I do this by buying 10K of 2034 at 2% yield and 10K of 2040 at 2% yield. (50/50 maturity matching approximation to duration matching.)
Suppose I want to make the swap in 2027. The "duration" (maturity) of the 2034 is 7 years. The duration of the 2040 is 13 years and the duration of the new-issue 2037 is 10 years.
Also suppose that just before I make the swap, the yield jumps 1% across the entire yield curve. The new yield is 3% for the 2034 and the 2040. The price of the 2034 falls 7%. The price of the
2040 falls 13%. The average price of the excess holdings falls 10% just before I sell to make the swap.
So, using the proceeds of the sale, I am short 10% and have to buy 10% less of the 2037 compared to what I would have bought if interest rates hadn't jumped. But the 2037 is now yielding 3%
instead of 2%. Thus, I am gaining back 1% every year that I hold the 2037. By the end of the 10 years held, I will have made up 1%/year x 10 years = 10%. I will be made whole by the time of
maturity in 2037.
The only thing that has changed is the cash flows. After the swap, more of the payout comes in the form of coupons, but the overall yield is the same. (If I wanted to spend all the cash flow in
2037, I would need to reinvest the excess portion of the coupons (1%/yr) to make up for the 10% shortfall in the principal at maturity.)
Another excellent contribution to our investigation!
When I say that duration matching doesn't work, I mean that in the pure sense. By that I mean the way it would work if all securities involved were trading on the secondary market, in which case the
math showed that it works almost perfectly for a parallel yield curve shift. By working perfectly, I mean that the proceeds from selling the expected amounts of the bracket year TIPS would enable us
to buy the DARA amount of the gap year TIPS.
Your thought experiment results are consistent with this, in that we could only buy 90% of the DARA amount for 2037 in 2027. The rest of the thought experiment relies on the fact that new-issue TIPS
bought at auction, which are close to par both unadjusted and adjusted, deliver most of their yield in the form of coupon payments.
My initial thinking about this, before reading your latest post, was along the lines of the annual real amounts (ARA) being larger than DARA, because the coupons of the 2037 (or whatever gap year
we're interested in) would be larger than those of the 2034 or 2040. I hadn't thought through what we'd do with the extra ARAs, but of course your thought experiment posits that we'd basically bank
them to contribute to the 2037 ARA.
Of course the ideal way to bank the coupons would be to reinvest them at the auction yield of the 2037 (approximately the coupon rate), in which case the end result would be the same as if we had
used classical duration matching with all marketable securities. Of course we can't do this, so there is some uncertainty as to what the ARA contribution from the reinvested coupons will be, and I
think this is the only possible quibble with your analysis.
The realistic alternatives are that we invest the coupons in a money market fund until we have enough to buy an additional 2037 (and repeat), or put them into a TIPS fund or combination of funds with
an appropriate duration, and again, perhaps buy another 2037 when we have enough.
Perhaps the reinvestment rates aren't a big enough deal to worry much about, but it might be worth doing the math to investigate. What if real yields return to negative territory and remain there for
an extended period of time?
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
Kevin M wrote: ↑Tue May 28, 2024 5:36 pm
When I say that duration matching doesn't work, I mean that in the pure sense. By that I mean the way it would work if all securities involved were trading on the secondary market, in which case
the math showed that it works almost perfectly for a parallel yield curve shift. By working perfectly, I mean that the proceeds from selling the expected amounts of the bracket year TIPS would
enable us to buy the DARA amount of the gap year TIPS.
Your thought experiment results are consistent with this, in that we could only buy 90% of the DARA amount for 2037 in 2027. The rest of the thought experiment relies on the fact that new-issue
TIPS bought at auction, which are close to par both unadjusted and adjusted, deliver most of their yield in the form of coupon payments.
My initial thinking about this, before reading your latest post, was along the lines of the annual real amounts (ARA) being larger than DARA, because the coupons of the 2037 (or whatever gap year
we're interested in) would be larger than those of the 2034 or 2040. I hadn't thought through what we'd do with the extra ARAs, but of course your thought experiment posits that we'd basically
bank them to contribute to the 2037 ARA.
Of course the ideal way to bank the coupons would be to reinvest them at the auction yield of the 2037 (approximately the coupon rate), in which case the end result would be the same as if we had
used classical duration matching with all marketable securities. Of course we can't do this, so there is some uncertainty as to what the ARA contribution from the reinvested coupons will be, and
I think this is the only possible quibble with your analysis.
The realistic alternatives are that we invest the coupons in a money market fund until we have enough to buy an additional 2037 (and repeat), or put them into a TIPS fund or combination of funds
with an appropriate duration, and again, perhaps buy another 2037 when we have enough.
Perhaps the reinvestment rates aren't a big enough deal to worry much about, but it might be worth doing the math to investigate. What if real yields return to negative territory and remain there
for an extended period of time?
If interest rates decrease, post-swap coupon payments will fall, and principal payouts will increase in the gap years. Reinvestment won't be a thing.
My suggestion for continuing this investigation is to focus on trying to find a method for evaluating how well duration matching maintains the yield to gap-year maturity and ignore how much the cash
flow changes due to the swaps. The changes in coupons and principal payouts may be both positive and negative (depending on which direction interest rates move) and may partially cancel out over
time. Some people will want perfectly uniform DARAs, so may reinvest coupons (if they rise instead of shrink), and others won't care and will accept the slightly lumpy payouts in their income floor.
Evaluating the effects of reinvestment may be worth investigating later after the basic questions posed in this thread have been addressed.
Re: Filling the TIPS gap years with bracket year duration matching
Seems to me the issue is convexity. If yields rise, the duration of new issue TIPS will decrease (due to higher coupon rates). This will introduce a duration mismatch with currently matched durations
for 2035-2039 years. The market value of 2034 & 2040 bonds will be depressed, so when sold to buy new-issue bonds, the final-year maturity values won’t be able to be matched perfectly. But the higher
coupon payments could be saved and re-invested in TIPS over time, restoring duration. So it would still work out. The same risk would not apply so much to decreased rates.
Do I have this right?
Re: Filling the TIPS gap years with bracket year duration matching
MtnBiker wrote: ↑Tue May 28, 2024 2:39 pm
I disagree with the statement that duration matching doesn't really work. Here is a simple thought experiment to show that it does still work.
Suppose I want to fund 2037 at the 20K level. I do this by buying 10K of 2034 at 2% yield and 10K of 2040 at 2% yield. (50/50 maturity matching approximation to duration matching.)
Suppose I want to make the swap in 2027. The "duration" (maturity) of the 2034 is 7 years. The duration of the 2040 is 13 years and the duration of the new-issue 2037 is 10 years.
Also suppose that just before I make the swap, the yield jumps 1% across the entire yield curve. The new yield is 3% for the 2034 and the 2040. The price of the 2034 falls 7%. The price of the
2040 falls 13%. The average price of the excess holdings falls 10% just before I sell to make the swap.
So, using the proceeds of the sale, I am short 10% and have to buy 10% less of the 2037 compared to what I would have bought if interest rates hadn't jumped. But the 2037 is now yielding 3%
instead of 2%. Thus, I am gaining back 1% every year that I hold the 2037. By the end of the 10 years held, I will have made up 1%/year x 10 years = 10%. I will be made whole by the time of
maturity in 2037.
The only thing that has changed is the cash flows. After the swap, more of the payout comes in the form of coupons, but the overall yield is the same. (If I wanted to spend all the cash flow in
2037, I would need to reinvest the excess portion of the coupons (1%/yr) to make up for the 10% shortfall in the principal at maturity.)
Circling back to carry this thought experiment one step further. I'm not up to speed on the spreadsheet functions needed to evaluate this scenario with full accuracy, but I can get Modified Duration
(MD) from an online calculator. Using the actual duration to estimate the loss in market value when interest rates change will improve the accuracy of the calculations.
In January 2027 when the interest rates jump 1% hypothetically, the MD of the Jan 2034 TIPS (1.75% coupon) is about 6.5 yr and the MD of the Feb 2040 TIPS (2.125% coupon) is about 11.25 yr. (For the
2040, I get 11.33 yr using 2% yield and 11.17 yr using 3% yield; not knowing which to use, I took the midpoint.)
The principal values in 2027 will no longer be exactly 50/50, but ignoring that detail, and again using the rough approximation that the market value falls 1% for each year of duration when interest
rates increase 1%, the net loss in market value is estimated to be about -(6.5+11.25)/2 = -8.875%
Accounting for the subsequent 1%/yr increase in yield when holding the newly-issued 2037 for 10 years, the net change in value at maturity in 2037 is estimated to be (1 - 0.08875)(1 + (10)(0.01)) =
Again, close enough.
Re: Filling the TIPS gap years with bracket year duration matching
I feel like there is some really good stuff! In general, what is the benefit of a tips ladder vs just a plain old asset allocation and safe withdrawal rate?
Re: Filling the TIPS gap years with bracket year duration matching
FoolStreet wrote: ↑Wed May 29, 2024 1:06 pm I feel like there is some really good stuff! In general, what is the benefit of a tips ladder vs just a plain old asset allocation and safe withdrawal
Inflation protection and a guaranteed real return backed by the treasury.
It does not preclude having assets allocated elsewhere as well if desired, so it is not one or the other.
Re: Filling the TIPS gap years with bracket year duration matching
I think there are multiple goals in attempting to duration match a new issue TIPS with bracket year TIPS, which isn't the case when duration matching using all secondary market issues; i.e., using
longer-term and shorter-term TIPS or TIPS funds to match the duration of an intermediate-term TIPS or fund. This complicates the mathematical analysis.
The goals that come to mind are:
1. Generate a known nominal amount for purchase of the gap year TIPS when issued. It's a nominal amount because one new-issue 10-year TIPS will cost close to $1,000, since unadjusted price will be
close to 100, and index ratio will be close to 1. This isn't exact, because new issues aren't actual par bonds; the larger the difference between coupon and yield, the more deviation from a price
of 100. The low and high prices for new-issue 10y TIPS auctioned to date are 98.881 and 100.447 respectively, so cost has ranged from $988.81 to 1,004.47, which is a delta of +0.45% to -1.12%.
The average cost has been $994.25.
2. "Lock in" historically attractive longer-term real yields. We aren't exactly locking them in, since we'll be selling well before maturity, but this is where the duration matching theory comes
into play, and needs to be mathematically analyzed for different yield-curve change scenarios, different coupon reinvestment rates, and different coupon reinvestment strategies. This may be of
less interest when 10y yields are historically unattractive, as they were in recent years.
3. Hedge unexpected inflation risk from now until issuance of a gap-year TIPS. This point was emphasized by Mel wrt I bonds, but of course there there are different ways to do it with TIPS as well.
There may be more, but these come to mind immediately, and provide fodder for some discussion.
The safest way to achieve goal #1 would be with a STRIPS (zero-coupon nominal Treasury) maturing close to the gap year TIPS issue date, since it provides a known amount of nominal dollars at
maturity. A low-coupon Treasury would be a reasonable alternative. Neither TIPS nor I bonds do this as reliably.
The way to achieve #2 is with longer-term TIPS. I bonds don't do this, and longer-term nominal Treasuries don't do this.
The most reliable way to achieve #3 with TIPS would be to buy TIPS maturing close to the gap-year issue date. This provides as much reliable inflation protection as I bonds, keeping in mind the
different ways the inflation adjustments are done, and that both methods lag actual inflation. And of course using TIPS is a generalizable solution, unlike I bonds due to annual purchase limits.
So it seems the goal is to find a solution that optimizes achievement of the three subgoals, and to be able to use math to prove it.
Although I'd want to get to the math, consider this thought experiment.
• Shortly after purchasing the bracket year TIPS at more than 2% yield, real yields drop back into negative territory, and inflation drops close to 0%, or perhaps below. The latter hasn't happened
for an extended period in a long time, but it has happened, and we're looking for certainty in ARA regardless of economic conditions.
• We reinvest all of our coupons at negative real yields, and possibly negative nominal yields.
• Shortly before the issuance of the gap year TIPS of interest, yields shoot up to historical highs or higher. The 10y TIPS hit 4.40% on Jan 18, 2000. Of course this drives long term prices way
• The gap year TIPS is issued at a very attractive yield, but still at a cost of close to $1,000 per bond, and we now sell our bracket-year TIPS at extremely depressed prices, with no or negative
earnings from the coupons already paid out.
How does duration matching work in this scenario?
Trust me, I want duration matching to work, since I've already bought all the 2034s and 2040s I need to buy the gap years at my current DARA.
If I make a calculation error, #Cruncher probably will let me know.
Re: Filling the TIPS gap years with bracket year duration matching
protagonist wrote: ↑Wed May 29, 2024 1:42 pm
FoolStreet wrote: ↑Wed May 29, 2024 1:06 pm I feel like there is some really good stuff! In general, what is the benefit of a tips ladder vs just a plain old asset allocation and safe
withdrawal rate?
Inflation protection and a guaranteed real return backed by the treasury.
It does not preclude having assets allocated elsewhere as well if desired, so it is not one or the other.
Can you elaborate?
If I want to retire with a 65/35 stock/bond mix, is the TIPS-crowd suggesting that the 35% goes 100% into the TIPS ladder? Or is the TIPS-crowd saying, put 100% of everything into a TIPS ladder? Or
further, is something like, 20 years at 4%/year is 80%, so put 80% in the TIPS ladder and *then* split the remaining 20% of the portfolio into 65/35...? or??
Re: Filling the TIPS gap years with bracket year duration matching
FoolStreet wrote: ↑Wed May 29, 2024 3:32 pm
protagonist wrote: ↑Wed May 29, 2024 1:42 pm
FoolStreet wrote: ↑Wed May 29, 2024 1:06 pm I feel like there is some really good stuff! In general, what is the benefit of a tips ladder vs just a plain old asset allocation and safe
withdrawal rate?
Inflation protection and a guaranteed real return backed by the treasury.
It does not preclude having assets allocated elsewhere as well if desired, so it is not one or the other.
Can you elaborate?
If I want to retire with a 65/35 stock/bond mix, is the TIPS-crowd suggesting that the 35% goes 100% into the TIPS ladder? Or is the TIPS-crowd saying, put 100% of everything into a TIPS ladder?
Or further, is something like, 20 years at 4%/year is 80%, so put 80% in the TIPS ladder and *then* split the remaining 20% of the portfolio into 65/35...? or??
A TIPS ladder is deterministic -- it provides a known stream of real income for a known period of time.
An SWR strategy is probabilistic -- it provides a known stream of real income for an unknown period of time.
The "TIPS crowd" uses TIPS in a large variety of ways.
Some use TIPS instead of nominal bonds in a otherwise "typical" AA. Some of those folks are in accumulation, while others are using SWR or other withdrawal methods.
Some crowd members are using a mix of TIPS and nominal bonds in that otherwise "typical" AA.
Some members are building ladders, others are not. Some are duration-matching, others are not. Some are purchasing individual TIPS, others are using funds.
To determine how to use or not use TIPS, you should first determine what you wish the portfolio to achieve, then what role fixed income should play, and finally what role TIPS could play.
“Adapt what is useful, reject what is useless, and add what is specifically your own.” ― Bruce Lee | {"url":"https://www.bogleheads.org/forum/viewtopic.php?f=10&t=432366&newpost=8098466","timestamp":"2024-11-09T16:13:20Z","content_type":"text/html","content_length":"321955","record_id":"<urn:uuid:47aef5a8-4cde-4b68-92d2-dd6accc009e3>","cc-path":"CC-MAIN-2024-46/segments/1730477028125.59/warc/CC-MAIN-20241109151915-20241109181915-00660.warc.gz"} |
The Mysteriums are a collection of unsolved mathematical problems that have been around for centuries. Many mathematicians have attempted to solve them, but all have failed. Some believe that the
Mysteriums are impossible to solve, while others believe that there is a hidden pattern that has yet to be discovered.
The first Mysterium was discovered in the 18th century by the mathematician Euler. He was trying to find a way to solve a problem involving a polyhedron, and he noticed that the answer was related to
a problem that had been unsolved for centuries. This problem became known as the Euler's Identity, and it is still considered to be one of the most mysterious mathematical problems.
Since then, other Mysteriums have been discovered, and they continue to baffle mathematicians all over the world. Some of the most famous Mysteriums include the Riemann Hypothesis, the P vs NP
problem, and the Twin Prime Conjecture.
Despite the best efforts of mathematicians, no one has been able to solve any of the Mysteriums. This has led to a lot of speculation about what could be hidden in these problems. Some people believe
that the Mysteriums are impossible to solve, while others believe that there is a hidden pattern that has yet to be discovered.
Whatever the answer may be, the Mysteriums are sure to continue to fascinate mathematicians for years to come. | {"url":"http://www.mysteriums.com/","timestamp":"2024-11-07T15:05:58Z","content_type":"text/html","content_length":"8910","record_id":"<urn:uuid:603489e4-6120-443e-b02b-b51fe2227883>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00624.warc.gz"} |
SAT And ACT Tests Hacks and Tips
We all wish we had cheat codes or keys to help us unlock every answer to SAT and ACT exam questions. Of course, that is not the case, but you can use some hacks and tips. They can help your
performance on your college entrance exam and ensure that you use your time in the best way.
Here are some of the best SAT and ACT exam hacks from an expert tutor that you need to implement.
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1. Don’t Do The Most Challenges Questions First
Time management is a significant problem for many students. If you go through this a lot and want to save time, you should pace yourself and do the easy questions first. After all, the easy questions
will take less time, and you will not have to linger.
After answering the easy questions, you can move on to the more challenging ones. It will allow you to give your undivided attention, focus on the difficult part, and try to solve it in no time.
2. Understand The Structure
SAT and ACT exams are standardized, which is why their structure never changes. However, that is why you should familiarize yourself with it before your exam. It will help you get used to answering
questions, and you will feel much more at ease.
Understanding the number, style, sequence, and type of questions coming in each section is crucial. These guidelines will become ingrained in your brain, and you will not have to worry about becoming
surprised by the test structure.
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3. Select A Concise Answer
When it comes to the English section, you have to write everything in the best way without being too wordy.
Conciseness is key because it ensures that you get your point across in as few words as possible. Of course, there are many exceptions to this rule, but picking the most concise answer is your safest
4. Utilize The Process Of Elimination
The process of elimination is one of the top strategies when answering MCQs. That is because students find it much easier to get rid of answers than to pick the appropriate choice in seconds. The
process of elimination works best for questions in the Reading and English section.
5. Remember All Important Formulas
When it comes to the SAT, they will offer you a reference key for the math question formulas. However, ACT does not provide such a key to students. That is why you have to know the formulas on your
fingertips to save time.
For the SAT, be sure to take time out, in the beginning, to go through the reference key. It will help you recall the formulas as you begin your exam.
Final Words
These are the top five SAT and ACT exam hacks you need to know. Implement them in your preparation and during the exam to witness the best results. You will realize that these hacks can take your
prep to the next level in no time.
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Related to This Article
What people say about "SAT And ACT Tests Hacks and Tips - Effortless Math: We Help Students Learn to LOVE Mathematics"?
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A predicate object, model of Kernel::PowerSideOfOrientedPowerSphere_3, that must provide the following function operators:
Oriented_side operator()( Weighted_point_3 p, Weighted_point_3 q, Weighted_point_3 r, Weighted_point_3 s, Weighted_point_3 t),
which performs the following:
Let \( {z(p,q,r,s)}^{(w)}\) be the power sphere of the weighted points \( (p,q,r,s)\). Returns
• ON_ORIENTED_BOUNDARY if t is orthogonal to \( {z(p,q,r,s)}^{(w)}\),
• ON_NEGATIVE_SIDE if t lies outside the oriented sphere of center \( z(p,q,r,s)\) and radius \( \sqrt{ w_{z(p,q,r,s)}^2 + w_t^2 }\) (which is equivalent to \( \Pi({t}^{(w)},{z(p,q,r,s)}^{(w)}) >0
• ON_POSITIVE_SIDE if t lies inside this oriented sphere.
p, q, r, s are not coplanar. Note that with this definition, if all the points have a weight equal to 0, then power_side_of_oriented_power_sphere_3(p,q,r,s,t) = side_of_oriented_sphere
Oriented_side operator()( Weighted_point_3 p, Weighted_point_3 q, Weighted_point_3 r, Weighted_point_3 t),
which has a definition analogous to the previous method, for coplanar points, with the power circle \( {z(p,q,r)}^{(w)}\).
p, q, r are not collinear and p, q, r, t are coplanar. If all the points have a weight equal to 0, then power_side_of_oriented_power_sphere_3(p,q,r,t) = side_of_oriented_circle(p,q,r,t).
Oriented_side operator()( Weighted_point_3 p, Weighted_point_3 q, Weighted_point_3 t),
which is the same for collinear points, where \( {z(p,q)}^{(w)}\) is the power segment of p and q.
p and q have different bare points, and p, q, t are collinear. If all points have a weight equal to 0, then power_side_of_oriented_power_sphere_3(p,q,t) gives the same answer as the kernel
predicate s(p,q).has_on(t) would give, where s(p,q) denotes the segment with endpoints p and q.
Oriented_side operator()( Weighted_point_3 p, Weighted_point_3 q),
which is the same for equal bare points, then it returns the comparison of the weights (ON_POSITIVE_SIDE when q is heavier than p).
p and q have equal bare points. | {"url":"https://doc.cgal.org/5.5/Triangulation_3/classRegularTriangulationTraits__3.html","timestamp":"2024-11-12T19:35:36Z","content_type":"application/xhtml+xml","content_length":"44023","record_id":"<urn:uuid:d3cbbbb0-2e61-4990-8876-63829b53dea6>","cc-path":"CC-MAIN-2024-46/segments/1730477028279.73/warc/CC-MAIN-20241112180608-20241112210608-00887.warc.gz"} |
Principia Mathematica
I can remember Bertrand Russell telling me of a horrible dream. He was in the top floor of the University Library, about A.D. 2100. A library assistant was going round the shelves carrying an
enormous bucket, taking down books, glancing at them, restoring them to the shelves or dumping them into the bucket. At last he came to three large volumes which Russell could recognize as the last
surviving copy of Principia Mathematica. He took down one of the volumes, turned over a few pages, seemed puzzled for a moment by the curious symbolism, closed the volume, balanced it in his hand and
Hardy, G. H. (2004) [1940]. A Mathematician's Apology. Cambridge: University Press. p. 83. ISBN 978-0-521-42706-7.
He [Russell] said once, after some contact with the Chinese language, that he was horrified to find that the language of Principia Mathematica was an Indo-European one.
Littlewood, J. E. (1985). A Mathematician's Miscellany. Cambridge: University Press. p. 130.
The Principia Mathematica (often abbreviated PM) is a three-volume work on the foundations of mathematics written by the mathematicians Alfred North Whitehead and Bertrand Russell and published in
1910, 1912, and 1913. In 1925–27, it appeared in a second edition with an important Introduction to the Second Edition, an Appendix A that replaced ✸9 and all-new Appendix B and Appendix C. PM is not
to be confused with Russell's 1903 The Principles of Mathematics. PM was originally conceived as a sequel volume to Russell's 1903 Principles, but as PM states, this became an unworkable suggestion
for practical and philosophical reasons: "The present work was originally intended by us to be comprised in a second volume of Principles of Mathematics... But as we advanced, it became increasingly
evident that the subject is a very much larger one than we had supposed; moreover on many fundamental questions which had been left obscure and doubtful in the former work, we have now arrived at
what we believe to be satisfactory solutions."
PM, according to its introduction, had three aims: (1) to analyze to the greatest possible extent the ideas and methods of mathematical logic and to minimize the number of primitive notions and
axioms, and inference rules; (2) to precisely express mathematical propositions in symbolic logic using the most convenient notation that precise expression allows; (3) to solve the paradoxes that
plagued logic and set theory at the turn of the 20th century, like Russell's paradox.^[1]
This third aim motivated the adoption of the theory of types in PM. The theory of types adopts grammatical restrictions on formulas that rules out the unrestricted comprehension of classes,
properties, and functions. The effect of this is that formulas such as would allow the comprehension of objects like the Russell set turn out to be ill-formed: they violate the grammatical
restrictions of the system of PM.
There is no doubt that PM is of great importance in the history of mathematics and philosophy: as Irvine has noted, it sparked interest in symbolic logic and advanced the subject by popularizing it;
it showcased the powers and capacities of symbolic logic; and it showed how advances in philosophy of mathematics and symbolic logic could go hand-in-hand with tremendous fruitfulness.^[2] Indeed, PM
was in part brought about by an interest in logicism, the view on which all mathematical truths are logical truths. It was in part thanks to the advances made in PM that, despite its defects,
numerous advances in meta-logic were made, including Gödel's incompleteness theorems.
For all that, PM notations are not widely used today: probably the foremost reason for this is that practicing mathematicians tend to assume that the background Foundation is a form of the system of
Zermelo–Fraenkel set theory. Nonetheless, the scholarly, historical, and philosophical interest in PM is great and ongoing: for example, the Modern Library placed it 23rd in a list of the top 100
English-language nonfiction books of the twentieth century.^[3] There are also multiple articles on the work in the peer-reviewed Stanford Encyclopedia of Philosophy and academic researchers continue
working with Principia, whether for the historical reason of understanding the text or its authors, or for mathematical reasons of understanding or developing Principia's logical system.
Scope of foundations laid
The Principia covered only set theory, cardinal numbers, ordinal numbers, and real numbers. Deeper theorems from real analysis were not included, but by the end of the third volume it was clear to
experts that a large amount of known mathematics could in principle be developed in the adopted formalism. It was also clear how lengthy such a development would be.
A fourth volume on the foundations of geometry had been planned, but the authors admitted to intellectual exhaustion upon completion of the third.
Theoretical basis
As noted in the criticism of the theory by Kurt Gödel (below), unlike a formalist theory, the "logicistic" theory of PM has no "precise statement of the syntax of the formalism". Furthermore in the
theory, it is almost immediately observable that interpretations (in the sense of model theory) are presented in terms of truth-values for the behaviour of the symbols "⊢" (assertion of truth), "~"
(logical not), and "V" (logical inclusive OR).
Truth-values: PM embeds the notions of "truth" and "falsity" in the notion "primitive proposition". A raw (pure) formalist theory would not provide the meaning of the symbols that form a "primitive
proposition"—the symbols themselves could be absolutely arbitrary and unfamiliar. The theory would specify only how the symbols behave based on the grammar of the theory. Then later, by assignment of
"values", a model would specify an interpretation of what the formulas are saying. Thus in the formal Kleene symbol set below, the "interpretation" of what the symbols commonly mean, and by
implication how they end up being used, is given in parentheses, e.g., "¬ (not)". But this is not a pure Formalist theory.
Contemporary construction of a formal theory
The following formalist theory is offered as contrast to the logicistic theory of PM. A contemporary formal system would be constructed as follows:
1. Symbols used: This set is the starting set, and other symbols can appear but only by definition from these beginning symbols. A starting set might be the following set derived from Kleene 1952:
logical symbols: "→" (implies, IF-THEN, and "⊃"), "&" (and), "V" (or), "¬" (not), "∀" (for all), "∃" (there exists); predicate symbol "=" (equals); function symbols "+" (arithmetic addition), "∙"
(arithmetic multiplication), "'" (successor); individual symbol "0" (zero); variables "a", "b", "c", etc.; and parentheses "(" and ")".^[4]
2. Symbol strings: The theory will build "strings" of these symbols by concatenation (juxtaposition).^[5]
3. Formation rules: The theory specifies the rules of syntax (rules of grammar) usually as a recursive definition that starts with "0" and specifies how to build acceptable strings or "well-formed
formulas" (wffs).^[6] This includes a rule for "substitution"^[7] of strings for the symbols called "variables".
4. Transformation rule(s): The axioms that specify the behaviours of the symbols and symbol sequences.
5. Rule of inference, detachment, modus ponens : The rule that allows the theory to "detach" a "conclusion" from the "premises" that led up to it, and thereafter to discard the "premises" (symbols
to the left of the line │, or symbols above the line if horizontal). If this were not the case, then substitution would result in longer and longer strings that have to be carried forward.
Indeed, after the application of modus ponens, nothing is left but the conclusion, the rest disappears forever.
Contemporary theories often specify as their first axiom the classical or modus ponens or "the rule of detachment":
A, A ⊃ B │ B
The symbol "│" is usually written as a horizontal line, here "⊃" means "implies". The symbols A and B are "stand-ins" for strings; this form of notation is called an "axiom schema" (i.e., there
is a countable number of specific forms the notation could take). This can be read in a manner similar to IF-THEN but with a difference: given symbol string IF A and A implies B THEN B (and
retain only B for further use). But the symbols have no "interpretation" (e.g., no "truth table" or "truth values" or "truth functions") and modus ponens proceeds mechanistically, by grammar
The theory of PM has both significant similarities, and similar differences, to a contemporary formal theory. Kleene states that "this deduction of mathematics from logic was offered as intuitive
axiomatics. The axioms were intended to be believed, or at least to be accepted as plausible hypotheses concerning the world".^[8] Indeed, unlike a Formalist theory that manipulates symbols according
to rules of grammar, PM introduces the notion of "truth-values", i.e., truth and falsity in the real-world sense, and the "assertion of truth" almost immediately as the fifth and sixth elements in
the structure of the theory (PM 1962:4–36):
1. Variables
2. Uses of various letters
3. The fundamental functions of propositions: "the Contradictory Function" symbolised by "~" and the "Logical Sum or Disjunctive Function" symbolised by "∨" being taken as primitive and logical
implication defined (the following example also used to illustrate 9. Definition below) as
p ⊃ q .=. ~ p ∨ q Df. (PM 1962:11)
and logical product defined as
p . q .=. ~(~p ∨ ~q) Df. (PM 1962:12)
4. Equivalence: Logical equivalence, not arithmetic equivalence: "≡" given as a demonstration of how the symbols are used, i.e., "Thus ' p ≡ q ' stands for '( p ⊃ q ) . ( q ⊃ p )'." (PM 1962:7).
Notice that to discuss a notation PM identifies a "meta"-notation with "[space] ... [space]":^[9]
Logical equivalence appears again as a definition:
p ≡ q .=. ( p ⊃ q ) . ( q ⊃ p ) (PM 1962:12),
Notice the appearance of parentheses. This grammatical usage is not specified and appears sporadically; parentheses do play an important role in symbol strings, however, e.g., the notation "(x)"
for the contemporary "∀x".
5. Truth-values: "The 'Truth-value' of a proposition is truth if it is true, and falsehood if it is false" (this phrase is due to Gottlob Frege) (PM 1962:7).
6. Assertion-sign: "'⊦. p may be read 'it is true that' ... thus '⊦: p .⊃. q ' means 'it is true that p implies q ', whereas '⊦. p .⊃⊦. q ' means ' p is true; therefore q is true'. The first of
these does not necessarily involve the truth either of p or of q, while the second involves the truth of both" (PM 1962:92).
7. Inference: PM 's version of modus ponens. "[If] '⊦. p ' and '⊦ (p ⊃ q)' have occurred, then '⊦ . q ' will occur if it is desired to put it on record. The process of the inference cannot be
reduced to symbols. Its sole record is the occurrence of '⊦. q ' [in other words, the symbols on the left disappear or can be erased]" (PM 1962:9).
8. The use of dots
9. Definitions: These use the "=" sign with "Df" at the right end.
10. Summary of preceding statements: brief discussion of the primitive ideas "~ p" and "p ∨ q" and "⊦" prefixed to a proposition.
11. Primitive propositions: the axioms or postulates. This was significantly modified in the second edition.
12. Propositional functions: The notion of "proposition" was significantly modified in the second edition, including the introduction of "atomic" propositions linked by logical signs to form
"molecular" propositions, and the use of substitution of molecular propositions into atomic or molecular propositions to create new expressions.
13. The range of values and total variation
14. Ambiguous assertion and the real variable: This and the next two sections were modified or abandoned in the second edition. In particular, the distinction between the concepts defined in sections
15. Definition and the real variable and 16 Propositions connecting real and apparent variables was abandoned in the second edition.
15. Formal implication and formal equivalence
16. Identity
17. Classes and relations
18. Various descriptive functions of relations
19. Plural descriptive functions
20. Unit classes
Primitive ideas
Cf. PM 1962:90–94, for the first edition:
• (1) Elementary propositions.
• (2) Elementary propositions of functions.
• (3) Assertion: introduces the notions of "truth" and "falsity".
• (4) Assertion of a propositional function.
• (5) Negation: "If p is any proposition, the proposition "not-p", or "p is false," will be represented by "~p" ".
• (6) Disjunction: "If p and q are any propositions, the proposition "p or q, i.e., "either p is true or q is true," where the alternatives are to be not mutually exclusive, will be represented by
"p ∨ q" ".
• (cf. section B)
Primitive propositions
The first edition (see discussion relative to the second edition, below) begins with a definition of the sign "⊃"
✸1.01. p ⊃ q .=. ~ p ∨ q. Df.
✸1.1. Anything implied by a true elementary proposition is true. Pp modus ponens
(✸1.11 was abandoned in the second edition.)
✸1.2. ⊦: p ∨ p .⊃. p. Pp principle of tautology
✸1.3. ⊦: q .⊃. p ∨ q. Pp principle of addition
✸1.4. ⊦: p ∨ q .⊃. q ∨ p. Pp principle of permutation
✸1.5. ⊦: p ∨ ( q ∨ r ) .⊃. q ∨ ( p ∨ r ). Pp associative principle
✸1.6. ⊦:. q ⊃ r .⊃: p ∨ q .⊃. p ∨ r. Pp principle of summation
✸1.7. If p is an elementary proposition, ~p is an elementary proposition. Pp
✸1.71. If p and q are elementary propositions, p ∨ q is an elementary proposition. Pp
✸1.72. If φp and ψp are elementary propositional functions which take elementary propositions as arguments, φp ∨ ψp is an elementary proposition. Pp
Together with the "Introduction to the Second Edition", the second edition's Appendix A abandons the entire section ✸9. This includes six primitive propositions ✸9 through ✸9.15 together with the
Axioms of reducibility.
The revised theory is made difficult by the introduction of the Sheffer stroke ("|") to symbolise "incompatibility" (i.e., if both elementary propositions p and q are true, their "stroke" p | q is
false), the contemporary logical NAND (not-AND). In the revised theory, the Introduction presents the notion of "atomic proposition", a "datum" that "belongs to the philosophical part of logic".
These have no parts that are propositions and do not contain the notions "all" or "some". For example: "this is red", or "this is earlier than that". Such things can exist ad finitum, i.e., even an
"infinite enumeration" of them to replace "generality" (i.e., the notion of "for all").^[10]PM then "advance[s] to molecular propositions" that are all linked by "the stroke". Definitions give
equivalences for "~", "∨", "⊃", and ".".
The new introduction defines "elementary propositions" as atomic and molecular positions together. It then replaces all the primitive propositions ✸1.2 to ✸1.72 with a single primitive proposition
framed in terms of the stroke:
"If p, q, r are elementary propositions, given p and p|(q|r), we can infer r. This is a primitive proposition."
The new introduction keeps the notation for "there exists" (now recast as "sometimes true") and "for all" (recast as "always true"). Appendix A strengthens the notion of "matrix" or "predicative
function" (a "primitive idea", PM 1962:164) and presents four new Primitive propositions as ✸8.1–✸8.13.
✸88. Multiplicative axiom
✸120. Axiom of infinity
Ramified types and the axiom of reducibility
In simple type theory objects are elements of various disjoint "types". Types are implicitly built up as follows. If τ[1],...,τ[m] are types then there is a type (τ[1],...,τ[m]) that can be thought
of as the class of propositional functions of τ[1],...,τ[m] (which in set theory is essentially the set of subsets of τ[1]×...×τ[m]). In particular there is a type () of propositions, and there may
be a type ι (iota) of "individuals" from which other types are built. Russell and Whitehead's notation for building up types from other types is rather cumbersome, and the notation here is due to
In the ramified type theory of PM all objects are elements of various disjoint ramified types. Ramified types are implicitly built up as follows. If τ[1],...,τ[m],σ[1],...,σ[n] are ramified types
then as in simple type theory there is a type (τ[1],...,τ[m],σ[1],...,σ[n]) of "predicative" propositional functions of τ[1],...,τ[m],σ[1],...,σ[n]. However, there are also ramified types (τ[1],...,τ
[m]|σ[1],...,σ[n]) that can be thought of as the classes of propositional functions of τ[1],...τ[m] obtained from propositional functions of type (τ[1],...,τ[m],σ[1],...,σ[n]) by quantifying over σ
[1],...,σ[n]. When n=0 (so there are no σs) these propositional functions are called predicative functions or matrices. This can be confusing because current mathematical practice does not
distinguish between predicative and non-predicative functions, and in any case PM never defines exactly what a "predicative function" actually is: this is taken as a primitive notion.
Russell and Whitehead found it impossible to develop mathematics while maintaining the difference between predicative and non-predicative functions, so they introduced the axiom of reducibility,
saying that for every non-predicative function there is a predicative function taking the same values. In practice this axiom essentially means that the elements of type (τ[1],...,τ[m]|σ[1],...,σ[n])
can be identified with the elements of type (τ[1],...,τ[m]), which causes the hierarchy of ramified types to collapse down to simple type theory. (Strictly speaking this is not quite correct, because
PM allows two propositional functions to be different even if they take the same values on all arguments; this differs from current mathematical practice where one normally identifies two such
In Zermelo set theory one can model the ramified type theory of PM as follows. One picks a set ι to be the type of individuals. For example, ι might be the set of natural numbers, or the set of atoms
(in a set theory with atoms) or any other set one is interested in. Then if τ[1],...,τ[m] are types, the type (τ[1],...,τ[m]) is the power set of the product τ[1]×...×τ[m], which can also be thought
of informally as the set of (propositional predicative) functions from this product to a 2-element set {true,false}. The ramified type (τ[1],...,τ[m]|σ[1],...,σ[n]) can be modeled as the product of
the type (τ[1],...,τ[m],σ[1],...,σ[n]) with the set of sequences of n quantifiers (∀ or ∃) indicating which quantifier should be applied to each variable σ[i]. (One can vary this slightly by allowing
the σs to be quantified in any order, or allowing them to occur before some of the τs, but this makes little difference except to the bookkeeping.)
One author^[2] observes that "The notation in that work has been superseded by the subsequent development of logic during the 20th century, to the extent that the beginner has trouble reading PM at
all"; while much of the symbolic content can be converted to modern notation, the original notation itself is "a subject of scholarly dispute", and some notation "embodies substantive logical
doctrines so that it cannot simply be replaced by contemporary symbolism".^[11]
Kurt Gödel was harshly critical of the notation:
"It is to be regretted that this first comprehensive and thorough-going presentation of a mathematical logic and the derivation of mathematics from it [is] so greatly lacking in formal precision
in the foundations (contained in ✸1–✸21 of Principia [i.e., sections ✸1–✸5 (propositional logic), ✸8–14 (predicate logic with identity/equality), ✸20 (introduction to set theory), and ✸21
(introduction to relations theory)]) that it represents in this respect a considerable step backwards as compared with Frege. What is missing, above all, is a precise statement of the syntax of
the formalism. Syntactical considerations are omitted even in cases where they are necessary for the cogency of the proofs".^[12]
This is reflected in the example below of the symbols "p", "q", "r" and "⊃" that can be formed into the string "p ⊃ q ⊃ r". PM requires a definition of what this symbol-string means in terms of other
symbols; in contemporary treatments the "formation rules" (syntactical rules leading to "well formed formulas") would have prevented the formation of this string.
Source of the notation: Chapter I "Preliminary Explanations of Ideas and Notations" begins with the source of the elementary parts of the notation (the symbols =⊃≡−ΛVε and the system of dots):
"The notation adopted in the present work is based upon that of Peano, and the following explanations are to some extent modeled on those which he prefixes to his Formulario Mathematico [i.e.,
Peano 1889]. His use of dots as brackets is adopted, and so are many of his symbols" (PM 1927:4).^[13]
PM changed Peano's Ɔ to ⊃, and also adopted a few of Peano's later symbols, such as ℩ and ι, and Peano's practice of turning letters upside down.
PM adopts the assertion sign "⊦" from Frege's 1879 Begriffsschrift:^[14]
"(I)t may be read 'it is true that'"^[15]
Thus to assert a proposition p PM writes:
"⊦. p." (PM 1927:92)
(Observe that, as in the original, the left dot is square and of greater size than the period on the right.)
Most of the rest of the notation in PM was invented by Whitehead.^[16]
An introduction to the notation of "Section A Mathematical Logic" (formulas ✸1–✸5.71)
PM 's dots^[17] are used in a manner similar to parentheses. Each dot (or multiple dot) represents either a left or right parenthesis or the logical symbol ∧. More than one dot indicates the "depth"
of the parentheses, for example, ".", ":" or ":.", "::". However the position of the matching right or left parenthesis is not indicated explicitly in the notation but has to be deduced from some
rules that are complex and at times ambiguous. Moreover, when the dots stand for a logical symbol ∧ its left and right operands have to be deduced using similar rules. First one has to decide based
on context whether the dots stand for a left or right parenthesis or a logical symbol. Then one has to decide how far the other corresponding parenthesis is: here one carries on until one meets
either a larger number of dots, or the same number of dots next that have equal or greater "force", or the end of the line. Dots next to the signs ⊃, ≡,∨, =Df have greater force than dots next to (x
), (∃x) and so on, which have greater force than dots indicating a logical product ∧.
Example 1. The line
✸3.4. ⊢ : p . q . ⊃ . p ⊃ q
corresponds to
⊢ ((p ∧ q) ⊃ (p ⊃ q)).
The two dots standing together immediately following the assertion-sign indicate that what is asserted is the entire line: since there are two of them, their scope is greater than that of any of the
single dots to their right. They are replaced by a left parenthesis standing where the dots are and a right parenthesis at the end of the formula, thus:
⊢ (p . q . ⊃ . p ⊃ q).
(In practice, these outermost parentheses, which enclose an entire formula, are usually suppressed.) The first of the single dots, standing between two propositional variables, represents
conjunction. It belongs to the third group and has the narrowest scope. Here it is replaced by the modern symbol for conjunction "∧", thus
⊢ (p ∧ q . ⊃ . p ⊃ q).
The two remaining single dots pick out the main connective of the whole formula. They illustrate the utility of the dot notation in picking out those connectives which are relatively more important
than the ones which surround them. The one to the left of the "⊃" is replaced by a pair of parentheses, the right one goes where the dot is and the left one goes as far to the left as it can without
crossing a group of dots of greater force, in this case the two dots which follow the assertion-sign, thus
⊢ ((p ∧ q) ⊃ . p ⊃ q)
The dot to the right of the "⊃" is replaced by a left parenthesis which goes where the dot is and a right parenthesis which goes as far to the right as it can without going beyond the scope already
established by a group of dots of greater force (in this case the two dots which followed the assertion-sign). So the right parenthesis which replaces the dot to the right of the "⊃" is placed in
front of the right parenthesis which replaced the two dots following the assertion-sign, thus
⊢ ((p ∧ q) ⊃ (p ⊃ q)).
Example 2, with double, triple, and quadruple dots:
✸9.521. ⊢ :: (∃x). φx . ⊃ . q : ⊃ :. (∃x). φx . v . r : ⊃ . q v r
stands for
((((∃x)(φx)) ⊃ (q)) ⊃ ((((∃x) (φx)) v (r)) ⊃ (q v r)))
Example 3, with a double dot indicating a logical symbol (from volume 1, page 10):
stands for
(p⊃q) ∧ ((q⊃r)⊃(p⊃r))
where the double dot represents the logical symbol ∧ and can be viewed as having the higher priority as a non-logical single dot.
Later in section ✸14, brackets "[ ]" appear, and in sections ✸20 and following, braces "{ }" appear. Whether these symbols have specific meanings or are just for visual clarification is unclear.
Unfortunately the single dot (but also ":", ":.", "::", etc.) is also used to symbolise "logical product" (contemporary logical AND often symbolised by "&" or "∧").
Logical implication is represented by Peano's "Ɔ" simplified to "⊃", logical negation is symbolised by an elongated tilde, i.e., "~" (contemporary "~" or "¬"), the logical OR by "v". The symbol "="
together with "Df" is used to indicate "is defined as", whereas in sections ✸13 and following, "=" is defined as (mathematically) "identical with", i.e., contemporary mathematical "equality" (cf.
discussion in section ✸13). Logical equivalence is represented by "≡" (contemporary "if and only if"); "elementary" propositional functions are written in the customary way, e.g., "f(p)", but later
the function sign appears directly before the variable without parenthesis e.g., "φx", "χx", etc.
Example, PM introduces the definition of "logical product" as follows:
✸3.01. p . q .=. ~(~p v ~q) Df.
where "p . q" is the logical product of p and q.
✸3.02. p ⊃ q ⊃ r .=. p ⊃ q . q ⊃ r Df.
This definition serves merely to abbreviate proofs.
Translation of the formulas into contemporary symbols: Various authors use alternate symbols, so no definitive translation can be given. However, because of criticisms such as that of Kurt Gödel
below, the best contemporary treatments will be very precise with respect to the "formation rules" (the syntax) of the formulas.
The first formula might be converted into modern symbolism as follows:^[18]
(p & q) =[df] (~(~p v ~q))
(p & q) =[df] (¬(¬p v ¬q))
(p ∧ q) =[df] (¬(¬p v ¬q))
The second formula might be converted as follows:
(p → q → r) =[df] (p → q) & (q → r)
But note that this is not (logically) equivalent to (p → (q → r)) nor to ((p → q) → r), and these two are not logically equivalent either.
An introduction to the notation of "Section B Theory of Apparent Variables" (formulas ✸8–✸14.34)
These sections concern what is now known as predicate logic, and predicate logic with identity (equality).
□ NB: As a result of criticism and advances, the second edition of PM (1927) replaces ✸9 with a new ✸8 (Appendix A). This new section eliminates the first edition's distinction between real and
apparent variables, and it eliminates "the primitive idea 'assertion of a propositional function'.^[19] To add to the complexity of the treatment, ✸8 introduces the notion of substituting a
"matrix", and the Sheffer stroke:
○ Matrix: In contemporary usage, PM 's matrix is (at least for propositional functions), a truth table, i.e., all truth-values of a propositional or predicate function.
○ Sheffer stroke: Is the contemporary logical NAND (NOT-AND), i.e., "incompatibility", meaning:
"Given two propositions p and q, then ' p | q ' means "proposition p is incompatible with proposition q", i.e., if both propositions p and q evaluate as true, then and only then p | q
evaluates as false." After section ✸8 the Sheffer stroke sees no usage.
Section ✸10: The existential and universal "operators": PM adds "(x)" to represent the contemporary symbolism "for all x " i.e., " ∀x", and it uses a backwards serifed E to represent "there exists an
x", i.e., "(Ǝx)", i.e., the contemporary "∃x". The typical notation would be similar to the following:
"(x) . φx" means "for all values of variable x, function φ evaluates to true"
"(Ǝx) . φx" means "for some value of variable x, function φ evaluates to true"
Sections ✸10, ✸11, ✸12: Properties of a variable extended to all individuals: section ✸10 introduces the notion of "a property" of a "variable". PM gives the example: φ is a function that indicates
"is a Greek", and ψ indicates "is a man", and χ indicates "is a mortal" these functions then apply to a variable x. PM can now write, and evaluate:
(x) . ψx
The notation above means "for all x, x is a man". Given a collection of individuals, one can evaluate the above formula for truth or falsity. For example, given the restricted collection of
individuals { Socrates, Plato, Russell, Zeus } the above evaluates to "true" if we allow for Zeus to be a man. But it fails for:
(x) . φx
because Russell is not Greek. And it fails for
(x) . χx
because Zeus is not a mortal.
Equipped with this notation PM can create formulas to express the following: "If all Greeks are men and if all men are mortals then all Greeks are mortals". (PM 1962:138)
(x) . φx ⊃ ψx :(x). ψx ⊃ χx :⊃: (x) . φx ⊃ χx
Another example: the formula:
✸10.01. (Ǝx). φx . = . ~(x) . ~φx Df.
means "The symbols representing the assertion 'There exists at least one x that satisfies function φ' is defined by the symbols representing the assertion 'It's not true that, given all values of x,
there are no values of x satisfying φ'".
The symbolisms ⊃[x] and "≡[x]" appear at ✸10.02 and ✸10.03. Both are abbreviations for universality (i.e., for all) that bind the variable x to the logical operator. Contemporary notation would have
simply used parentheses outside of the equality ("=") sign:
✸10.02 φx ⊃[x] ψx .=. (x). φx ⊃ ψx Df
Contemporary notation: ∀x(φ(x) → ψ(x)) (or a variant)
✸10.03 φx ≡[x] ψx .=. (x). φx ≡ ψx Df
Contemporary notation: ∀x(φ(x) ↔ ψ(x)) (or a variant)
PM attributes the first symbolism to Peano.
Section ✸11 applies this symbolism to two variables. Thus the following notations: ⊃[x], ⊃[y], ⊃[x, y] could all appear in a single formula.
Section ✸12 reintroduces the notion of "matrix" (contemporary truth table), the notion of logical types, and in particular the notions of first-order and second-order functions and propositions.
New symbolism "φ ! x" represents any value of a first-order function. If a circumflex "＾" is placed over a variable, then this is an "individual" value of y, meaning that "ŷ" indicates "individuals"
(e.g., a row in a truth table); this distinction is necessary because of the matrix/extensional nature of propositional functions.
Now equipped with the matrix notion, PM can assert its controversial axiom of reducibility: a function of one or two variables (two being sufficient for PM 's use) where all its values are given
(i.e., in its matrix) is (logically) equivalent ("≡") to some "predicative" function of the same variables. The one-variable definition is given below as an illustration of the notation (PM
✸12.1 ⊢: (Ǝ f): φx .≡[x]. f ! x Pp;
Pp is a "Primitive proposition" ("Propositions assumed without proof") (PM 1962:12, i.e., contemporary "axioms"), adding to the 7 defined in section ✸1 (starting with ✸1.1 modus ponens).
These are to be distinguished from the "primitive ideas" that include the assertion sign "⊢", negation "~", logical OR "V", the notions of "elementary proposition" and "elementary
propositional function"; these are as close as PM comes to rules of notational formation, i.e., syntax.
This means: "We assert the truth of the following: There exists a function f with the property that: given all values of x, their evaluations in function φ (i.e., resulting their matrix) is logically
equivalent to some f evaluated at those same values of x. (and vice versa, hence logical equivalence)". In other words: given a matrix determined by property φ applied to variable x, there exists a
function f that, when applied to the x is logically equivalent to the matrix. Or: every matrix φx can be represented by a function f applied to x, and vice versa.
✸13: The identity operator "=" : This is a definition that uses the sign in two different ways, as noted by the quote from PM:
✸13.01. x = y .=: (φ): φ ! x . ⊃ . φ ! y Df
"This definition states that x and y are to be called identical when every predicative function satisfied by x is also satisfied by y ... Note that the second sign of equality in the above
definition is combined with "Df", and thus is not really the same symbol as the sign of equality which is defined."
The not-equals sign "≠" makes its appearance as a definition at ✸13.02.
✸14: Descriptions:
"A description is a phrase of the form "the term y which satisfies φŷ, where φŷ is some function satisfied by one and only one argument."^[20]
From this PM employs two new symbols, a forward "E" and an inverted iota "℩". Here is an example:
✸14.02. E ! ( ℩y) (φy) .=: ( Ǝb):φy . ≡[y]. y = b Df.
This has the meaning:
"The y satisfying φŷ exists," which holds when, and only when φŷ is satisfied by one value of y and by no other value." (PM 1967:173–174)
Introduction to the notation of the theory of classes and relations
The text leaps from section ✸14 directly to the foundational sections ✸20 GENERAL THEORY OF CLASSES and ✸21 GENERAL THEORY OF RELATIONS. "Relations" are what is known in contemporary set theory as
sets of ordered pairs. Sections ✸20 and ✸22 introduce many of the symbols still in contemporary usage. These include the symbols "ε", "⊂", "∩", "∪", "–", "Λ", and "V": "ε" signifies "is an element
of" (PM 1962:188); "⊂" (✸22.01) signifies "is contained in", "is a subset of"; "∩" (✸22.02) signifies the intersection (logical product) of classes (sets); "∪" (✸22.03) signifies the union (logical
sum) of classes (sets); "–" (✸22.03) signifies negation of a class (set); "Λ" signifies the null class; and "V" signifies the universal class or universe of discourse.
Small Greek letters (other than "ε", "ι", "π", "φ", "ψ", "χ", and "θ") represent classes (e.g., "α", "β", "γ", "δ", etc.) (PM 1962:188):
x ε α
"The use of single letter in place of symbols such as ẑ(φz) or ẑ(φ ! z) is practically almost indispensable, since otherwise the notation rapidly becomes intolerably cumbrous. Thus ' x ε α'
will mean ' x is a member of the class α'". (PM 1962:188)
α ∪ –α = V
The union of a set and its inverse is the universal (completed) set.^[21]
α ∩ –α = Λ
The intersection of a set and its inverse is the null (empty) set.
When applied to relations in section ✸23 CALCULUS OF RELATIONS, the symbols "⊂", "∩", "∪", and "–" acquire a dot: for example: "⊍", "∸".^[22]
The notion, and notation, of "a class" (set): In the first edition PM asserts that no new primitive ideas are necessary to define what is meant by "a class", and only two new "primitive propositions"
called the axioms of reducibility for classes and relations respectively (PM 1962:25).^[23] But before this notion can be defined, PM feels it necessary to create a peculiar notation "ẑ(φz)" that it
calls a "fictitious object". (PM 1962:188)
⊢: x ε ẑ(φz) .≡. (φx)
"i.e., ' x is a member of the class determined by (φẑ)' is [logically] equivalent to ' x satisfies (φẑ),' or to '(φx) is true.'". (PM 1962:25)
At least PM can tell the reader how these fictitious objects behave, because "A class is wholly determinate when its membership is known, that is, there cannot be two different classes having the
same membership" (PM 1962:26). This is symbolised by the following equality (similar to ✸13.01 above:
ẑ(φz) = ẑ(ψz) . ≡ : (x): φx .≡. ψx
"This last is the distinguishing characteristic of classes, and justifies us in treating ẑ(ψz) as the class determined by [the function] ψẑ." (PM 1962:188)
Perhaps the above can be made clearer by the discussion of classes in Introduction to the Second Edition, which disposes of the Axiom of Reducibility and replaces it with the notion: "All functions
of functions are extensional" (PM 1962:xxxix), i.e.,
φx ≡[x] ψx .⊃. (x): ƒ(φẑ) ≡ ƒ(ψẑ) (PM 1962:xxxix)
This has the reasonable meaning that "IF for all values of x the truth-values of the functions φ and ψ of x are [logically] equivalent, THEN the function ƒ of a given φẑ and ƒ of ψẑ are [logically]
equivalent." PM asserts this is "obvious":
"This is obvious, since φ can only occur in ƒ(φẑ) by the substitution of values of φ for p, q, r, ... in a [logical-] function, and, if φx ≡ ψx, the substitution of φx for p in a [logical-]
function gives the same truth-value to the truth-function as the substitution of ψx. Consequently there is no longer any reason to distinguish between functions classes, for we have, in virtue of
the above,
φx ≡[x] ψx .⊃. (x). φẑ = . ψẑ".
Observe the change to the equality "=" sign on the right. PM goes on to state that will continue to hang onto the notation "ẑ(φz)", but this is merely equivalent to φẑ, and this is a class. (all
quotes: PM 1962:xxxix).
Consistency and criticisms
According to Carnap's "Logicist Foundations of Mathematics", Russell wanted a theory that could plausibly be said to derive all of mathematics from purely logical axioms. However, Principia
Mathematica required, in addition to the basic axioms of type theory, three further axioms that seemed to not be true as mere matters of logic, namely the axiom of infinity, the axiom of choice, and
the axiom of reducibility. Since the first two were existential axioms, Russell phrased mathematical statements depending on them as conditionals. But reducibility was required to be sure that the
formal statements even properly express statements of real analysis, so that statements depending on it could not be reformulated as conditionals. Frank P. Ramsey tried to argue that Russell's
ramification of the theory of types was unnecessary, so that reducibility could be removed, but these arguments seemed inconclusive.
Beyond the status of the axioms as logical truths, one can ask the following questions about any system such as PM:
• whether a contradiction could be derived from the axioms (the question of inconsistency), and
• whether there exists a mathematical statement which could neither be proven nor disproven in the system (the question of completeness).
Propositional logic itself was known to be consistent, but the same had not been established for Principia's axioms of set theory. (See Hilbert's second problem.) Russell and Whitehead suspected that
the system in PM is incomplete: for example, they pointed out that it does not seem powerful enough to show that the cardinal ℵ[ω] exists. However, one can ask if some recursively axiomatizable
extension of it is complete and consistent.
Gödel 1930, 1931
In 1930, Gödel's completeness theorem showed that first-order predicate logic itself was complete in a much weaker sense—that is, any sentence that is unprovable from a given set of axioms must
actually be false in some model of the axioms. However, this is not the stronger sense of completeness desired for Principia Mathematica, since a given system of axioms (such as those of Principia
Mathematica) may have many models, in some of which a given statement is true and in others of which that statement is false, so that the statement is left undecided by the axioms.
Gödel's incompleteness theorems cast unexpected light on these two related questions.
Gödel's first incompleteness theorem showed that no recursive extension of Principia could be both consistent and complete for arithmetic statements. (As mentioned above, Principia itself was already
known to be incomplete for some non-arithmetic statements.) According to the theorem, within every sufficiently powerful recursive logical system (such as Principia), there exists a statement G that
essentially reads, "The statement G cannot be proved." Such a statement is a sort of Catch-22: if G is provable, then it is false, and the system is therefore inconsistent; and if G is not provable,
then it is true, and the system is therefore incomplete.
Gödel's second incompleteness theorem (1931) shows that no formal system extending basic arithmetic can be used to prove its own consistency. Thus, the statement "there are no contradictions in the
Principia system" cannot be proven in the Principia system unless there are contradictions in the system (in which case it can be proven both true and false).
Wittgenstein 1919, 1939
By the second edition of PM, Russell had removed his axiom of reducibility to a new axiom (although he does not state it as such). Gödel 1944:126 describes it this way:
"This change is connected with the new axiom that functions can occur in propositions only "through their values", i.e., extensionally . . . [this is] quite unobjectionable even from the
constructive standpoint . . . provided that quantifiers are always restricted to definite orders". This change from a quasi-intensional stance to a fully extensional stance also restricts
predicate logic to the second order, i.e. functions of functions: "We can decide that mathematics is to confine itself to functions of functions which obey the above assumption" (PM 2nd edition
p. 401, Appendix C).
This new proposal resulted in a dire outcome. An "extensional stance" and restriction to a second-order predicate logic means that a propositional function extended to all individuals such as "All
'x' are blue" now has to list all of the 'x' that satisfy (are true in) the proposition, listing them in a possibly infinite conjunction: e.g. x[1] ∧ x[2] ∧ . . . ∧ x[n] ∧ . . .. Ironically, this
change came about as the result of criticism from Wittgenstein in his 1919 Tractatus Logico-Philosophicus. As described by Russell in the Introduction to the Second Edition of PM:
"There is another course, recommended by Wittgenstein† (†Tractatus Logico-Philosophicus, *5.54ff) for philosophical reasons. This is to assume that functions of propositions are always
truth-functions, and that a function can only occur in a proposition through its values. [...] [Working through the consequences] it appears that everything in Vol. I remains true (though often
new proofs are required); the theory of inductive cardinals and ordinals survives; but it seems that the theory of infinite Dedekindian and well-ordered series largely collapses, so that
irrationals, and real numbers generally, can no longer be adequately dealt with. Also Cantor's proof that 2^n > n breaks down unless n is finite." (PM 2nd edition reprinted 1962:xiv, also cf. new
Appendix C).
In other words, the fact that an infinite list cannot realistically be specified means that the concept of "number" in the infinite sense (i.e. the continuum) cannot be described by the new theory
proposed in PM Second Edition.
Wittgenstein in his Lectures on the Foundations of Mathematics, Cambridge 1939 criticised Principia on various grounds, such as:
• It purports to reveal the fundamental basis for arithmetic. However, it is our everyday arithmetical practices such as counting which are fundamental; for if a persistent discrepancy arose
between counting and Principia, this would be treated as evidence of an error in Principia (e.g., that Principia did not characterise numbers or addition correctly), not as evidence of an error
in everyday counting.
• The calculating methods in Principia can only be used in practice with very small numbers. To calculate using large numbers (e.g., billions), the formulae would become too long, and some
short-cut method would have to be used, which would no doubt rely on everyday techniques such as counting (or else on non-fundamental and hence questionable methods such as induction). So again
Principia depends on everyday techniques, not vice versa.
Wittgenstein did, however, concede that Principia may nonetheless make some aspects of everyday arithmetic clearer.
Gödel 1944
In his 1944 Russell's mathematical logic, Gödel offers a "critical but sympathetic discussion of the logicistic order of ideas":^[24]
"It is to be regretted that this first comprehensive and thorough-going presentation of a mathematical logic and the derivation of mathematics from it [is] so greatly lacking in formal precision
in the foundations (contained in *1-*21 of Principia) that it represents in this respect a considerable step backwards as compared with Frege. What is missing, above all, is a precise statement
of the syntax of the formalism. Syntactical considerations are omitted even in cases where they are necessary for the cogency of the proofs . . . The matter is especially doubtful for the rule of
substitution and of replacing defined symbols by their definiens . . . it is chiefly the rule of substitution which would have to be proved" (Gödel 1944:124)^[25]
Part I Mathematical logic. Volume I ✸1 to ✸43
This section describes the propositional and predicate calculus, and gives the basic properties of classes, relations, and types.
Part II Prolegomena to cardinal arithmetic. Volume I ✸50 to ✸97
This part covers various properties of relations, especially those needed for cardinal arithmetic.
Part III Cardinal arithmetic. Volume II ✸100 to ✸126
This covers the definition and basic properties of cardinals. A cardinal is defined to be an equivalence class of similar classes (as opposed to ZFC, where a cardinal is a special sort of von Neumann
ordinal). Each type has its own collection of cardinals associated with it, and there is a considerable amount of bookkeeping necessary for comparing cardinals of different types. PM define addition,
multiplication and exponentiation of cardinals, and compare different definitions of finite and infinite cardinals. ✸120.03 is the Axiom of infinity.
Part IV Relation-arithmetic. Volume II ✸150 to ✸186
A "relation-number" is an equivalence class of isomorphic relations. PM defines analogues of addition, multiplication, and exponentiation for arbitrary relations. The addition and multiplication is
similar to the usual definition of addition and multiplication of ordinals in ZFC, though the definition of exponentiation of relations in PM is not equivalent to the usual one used in ZFC.
Part V Series. Volume II ✸200 to ✸234 and volume III ✸250 to ✸276
This covers series, which is PM's term for what is now called a totally ordered set. In particular it covers complete series, continuous functions between series with the order topology (though of
course they do not use this terminology), well-ordered series, and series without "gaps" (those with a member strictly between any two given members).
Part VI Quantity. Volume III ✸300 to ✸375
This section constructs the ring of integers, the fields of rational and real numbers, and "vector-families", which are related to what are now called torsors over abelian groups.
Comparison with set theory
This section compares the system in PM with the usual mathematical foundations of ZFC. The system of PM is roughly comparable in strength with Zermelo set theory (or more precisely a version of it
where the axiom of separation has all quantifiers bounded).
• The system of propositional logic and predicate calculus in PM is essentially the same as that used now, except that the notation and terminology has changed.
• The most obvious difference between PM and set theory is that in PM all objects belong to one of a number of disjoint types. This means that everything gets duplicated for each (infinite) type:
for example, each type has its own ordinals, cardinals, real numbers, and so on. This results in a lot of bookkeeping to relate the various types with each other.
• In ZFC functions are normally coded as sets of ordered pairs. In PM functions are treated rather differently. First of all, "function" means "propositional function", something taking values true
or false. Second, functions are not determined by their values: it is possible to have several different functions all taking the same values (for example, one might regard 2x+2 and 2(x+1) as
different functions on grounds that the computer programs for evaluating them are different). The functions in ZFC given by sets of ordered pairs correspond to what PM call "matrices", and the
more general functions in PM are coded by quantifying over some variables. In particular PM distinguishes between functions defined using quantification and functions not defined using
quantification, whereas ZFC does not make this distinction.
• PM has no analogue of the axiom of replacement, though this is of little practical importance as this axiom is used very little in mathematics outside set theory.
• PM emphasizes relations as a fundamental concept, whereas in current mathematical practice it is functions rather than relations that are treated as more fundamental; for example, category theory
emphasizes morphisms or functions rather than relations. (However, there is an analogue of categories called allegories that models relations rather than functions, and is quite similar to the
type system of PM.)
• In PM, cardinals are defined as classes of similar classes, whereas in ZFC cardinals are special ordinals. In PM there is a different collection of cardinals for each type with some complicated
machinery for moving cardinals between types, whereas in ZFC there is only 1 sort of cardinal. Since PM does not have any equivalent of the axiom of replacement, it is unable to prove the
existence of cardinals greater than ℵ[ω].
• In PM ordinals are treated as equivalence classes of well-ordered sets, and as with cardinals there is a different collection of ordinals for each type. In ZFC there is only one collection of
ordinals, usually defined as von Neumann ordinals. One strange quirk of PM is that they do not have an ordinal corresponding to 1, which causes numerous unnecessary complications in their
theorems. The definition of ordinal exponentiation α^β in PM is not equivalent to the usual definition in ZFC and has some rather undesirable properties: for example, it is not continuous in β
and is not well ordered (so is not even an ordinal).
• The constructions of the integers, rationals and real numbers in ZFC have been streamlined considerably over time since the constructions in PM.
Differences between editions
Apart from corrections of misprints, the main text of PM is unchanged between the first and second editions. The main text in Volumes 1 and 2 was reset, so that it occupies fewer pages in each. In
the second edition, Volume 3 was not reset, being photographically reprinted with the same page numbering; corrections were still made. The total number of pages (excluding the endpapers) in the
first edition is 1,996; in the second, 2,000. Volume 1 has five new additions:
• A 54-page introduction by Russell describing the changes they would have made had they had more time and energy. The main change he suggests is the removal of the controversial axiom of
reducibility, though he admits that he knows no satisfactory substitute for it. He also seems more favorable to the idea that a function should be determined by its values (as is usual in current
mathematical practice).
• Appendix A, numbered as *8, 15 pages, about the Sheffer stroke.
• Appendix B, numbered as *89, discussing induction without the axiom of reducibility.
• Appendix C, 8 pages, discussing propositional functions.
• An 8-page list of definitions at the end, giving a much-needed index to the 500 or so notations used.
In 1962, Cambridge University Press published a shortened paperback edition containing parts of the second edition of Volume 1: the new introduction (and the old), the main text up to *56, and
Appendices A and C.
The first edition was reprinted in 2009 by Merchant Books, ISBN 978-1-60386-182-3, ISBN 978-1-60386-183-0, ISBN 978-1-60386-184-7.
See also
• Axiomatic set theory
• Information Processing Language – first computational demonstration of theorems in PM
• Introduction to Mathematical Philosophy
1. ^ Whitehead, Whitehead, Alfred North and Bertrand Russell (1963). Principia Mathematica. Cambridge: Cambridge University Press. pp. 1.
2. ^ ^a ^b Irvine, Andrew D. (1 May 2003). "Principia Mathematica (Stanford Encyclopedia of Philosophy)". Metaphysics Research Lab, CSLI, Stanford University. Retrieved 5 August 2009.
3. ^ "The Modern Library's Top 100 Nonfiction Books of the Century". The New York Times Company. 30 April 1999. Retrieved 5 August 2009.
4. ^ This set is taken from Kleene 1952:69 substituting → for ⊃.
5. ^ Kleene 1952:71, Enderton 2001:15
6. ^ Enderton 2001:16
7. ^ This is the word used by Kleene 1952:78
8. ^ Quote from Kleene 1952:45. See discussion LOGICISM at pp. 43–46.
9. ^ In his section 8.5.4 Groping towards metalogic Grattan-Guinness 2000:454ff discusses the American logicians' critical reception of the second edition of PM. For instance Sheffer "puzzled that '
In order to give an account of logic, we must presuppose and employ logic ' " (p. 452). And Bernstein ended his 1926 review with the comment that "This distinction between the propositional logic
as a mathematical system and as a language must be made, if serious errors are to be avoided; this distinction the Principia does not make" (p. 454).
10. ^ This idea is due to Wittgenstein's Tractatus. See the discussion at PM 1962:xiv–xv)
11. ^ Linsky, Bernard (2018). Zalta, Edward N. (ed.). The Stanford Encyclopedia of Philosophy. Metaphysics Research Lab, Stanford University. Retrieved 1 May 2018 – via Stanford Encyclopedia of
12. ^ Kurt Gödel 1944 "Russell's mathematical logic" appearing at p. 120 in Feferman et al. 1990 Kurt Gödel Collected Works Volume II, Oxford University Press, NY, ISBN 978-0-19-514721-6 (v.2.pbk.) .
13. ^ For comparison, see the translated portion of Peano 1889 in van Heijenoort 1967:81ff.
14. ^ This work can be found at van Heijenoort 1967:1ff.
15. ^ And see footnote, both at PM 1927:92
16. ^ Bertrand Russell (1959). "Chapter VII". My Philosophical Development.
17. ^ The original typography is a square of a heavier weight than the conventional period.
18. ^ The first example comes from plato.stanford.edu (loc.cit.).
19. ^ p. xiii of 1927 appearing in the 1962 paperback edition to ✸56.
20. ^ The original typography employs an x with a circumflex rather than ŷ; this continues below
21. ^ See the ten postulates of Huntington, in particular postulates IIa and IIb at PM 1962:205 and discussion at page 206.
22. ^ The "⊂" sign has a dot inside it, and the intersection sign "∩" has a dot above it; these are not available in the "Arial Unicode MS" font.
23. ^ Wiener 1914 "A simplification of the logic of relations" (van Heijenoort 1967:224ff) disposed of the second of these when he showed how to reduce the theory of relations to that of classes
24. ^ Kleene 1952:46.
25. ^ Gödel 1944 Russell's mathematical logic in Kurt Gödel: Collected Works Volume II, Oxford University Press, New York, NY, ISBN 978-0-19-514721-6.
• Stephen Kleene (1952). Introduction to Metamathematics, 6th Reprint, North-Holland Publishing Company, Amsterdam NY, ISBN 0-7204-2103-9.
□ Stephen Cole Kleene; Michael Beeson (2009). Introduction to Metamathematics (Paperback ed.). Ishi Press. ISBN 978-0-923891-57-2.
• Ivor Grattan-Guinness (2000). The Search for Mathematical Roots 1870–1940, Princeton University Press, Princeton NJ, ISBN 0-691-05857-1.
• Ludwig Wittgenstein (2009), Major Works: Selected Philosophical Writings, HarperrCollins, New York, ISBN 978-0-06-155024-9. In particular:
Tractatus Logico-Philosophicus (Vienna 1918), original publication in German).
• Jean van Heijenoort editor (1967). From Frege to Gödel: A Source book in Mathematical Logic, 1879–1931, 3rd printing, Harvard University Press, Cambridge MA, ISBN 0-674-32449-8.
• Michel Weber and Will Desmond (eds.) (2008) Handbook of Whiteheadian Process Thought, Frankfurt / Lancaster, Ontos Verlag, Process Thought X1 & X2. | {"url":"https://codedocs.org/what-is/principia-mathematica","timestamp":"2024-11-08T15:34:07Z","content_type":"text/html","content_length":"122187","record_id":"<urn:uuid:85953201-044c-4f8d-9ada-c0a908f669f5>","cc-path":"CC-MAIN-2024-46/segments/1730477028067.32/warc/CC-MAIN-20241108133114-20241108163114-00661.warc.gz"} |
ICSE Solutions for Class 10 Mathematics - Mensuration - CBSE Library
ICSE Solutions for Class 10 Mathematics – Mensuration
ICSE Solutions for Class 10 Mathematics – Mensuration
ICSE SolutionsSelina ICSE Solutions
Get ICSE Solutions for Class 10 Mathematics Chapter 17 Mensuration for ICSE Board Examinations on APlusTopper.com. We provide step by step Solutions for ICSE Mathematics Class 10 Solutions Pdf. You
can download the Class 10 Maths ICSE Textbook Solutions with Free PDF download option.
Download Formulae Handbook For ICSE Class 9 and 10
1. Perimeter of a plane figure = sum of lengths of its sides.
2. Circumference of a circle = 2πr,
where r is the radius of the circle.
1. Area of a triangle = ½ × base × height.
2. Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}{{a}^{2}}\)
where a is its side.
3. Area of an isosceles triangle
4. Area of a quadrilateral (when diagonals intersect at right angles) = ½ × product of diagonals.
5. Area of a rectangle = length × breadth.
6. Area of a square = (side)^2.
7. Area of a parallelogram = base × height.
8. Area of a rhombus = ½ × product of diagonals.
9. Area of a trapezium = ½ × (sum of parallel sides) × height.
10. Area of a circle = πr^2
where r is the radius of the circle.
11. Area of a circular ring = π (R^2 – r^2)
where R and r are the radii of the outer and the inner circles.
Surface Area and Volume of Solids:
1. Cube Cuboid:
(i) Surface area of a cube = 6a^2
where a is its edge (side).
(ii) Surface area of a cuboid = 2 (ℓb + bh + ℓh)
where ℓ, b and h are its edges.
(iii) Surface area of four walls (lateral surface area) of a cuboid = 2h (ℓ + b)
where ℓ, b and h are its edges.
(iv) Volume of a cube = (side)^3.
(v) Volume of a cuboid = length × breadth × height.
2. Solid Cylinder:
Let r and h be the radius and height of a solid cylinder, then
(i) Curved (lateral) surface area = 2πrh.
(ii) Total surface area = 2πr (h + r).
(iii) Volume = πr^2h.
3. Hollow Cylinder:
Let R and r be the external and internal radii, and h be the height of a hollow cylinder, then
(i) External curved surface area = 2πRh.
(ii) Internal curved surface area = 2πrh.
(iii) Total surface area = 2π (Rh + rh + R^2 – r^2).
(iv) Volume of material = π (R^2 – r^2)h.
4. Spherical Shell:
5. Hemispherical Shell:
Formulae Based Questions
Question 1. Find the area of a circle whose circumference is 22 cm.
Question 2. If the perimeter of a semi circular protractor is 36 cm. Find its diameter.
Question 3. A well 28.8 m deep and of diameter 2 m is dug up. The soil dug out is spread all around the well to make a platform 1 m high considering the fact losse soil settled to a height in the
ratio 6 : 5 find the width of the platform.
Question 4. Two cylinder have bases of same size. The diameter of each is 14 cm. One of the cone is 10 cm high and the other is 20 cm high. Find the ratio between their volumes.
Question 5. A glass cylinder with diameter 20 cm water to a height of 9 cm. A metal cube of 8 cm edge is immersed in it completely. Calculate the height by which water will rise in the cylinder.
(Take π = 3.142)
Question 6. Water is being pumped out through a circular pipe whose external diameter is 7 cm. If the flow of water is 72 cm per second how many litres of water are being pump out in one hour.
Question 7. The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions? (Take π = 22/7)
Question 8. The radius and height of a cylinder are in the ratio of 5 : 7 and its volume is 550 cm. Find its radius. (Take π = 22/7)
Question 9. The ratio of the base area and curved surface of a conical tent is 40 : 41. If the height is 18 m, Find the air capacity of tent in term of n.
Question 10. The diameter of two cones are equal. If their slant heights be in the ratio of 5 : 4. Find the ratio of their curved surface areas?
Question 11. The radius and height of cone are in the ratio 3 : 4. If its volume is 301.44 cm^3. What is its radius? What is its slant height? (Take π = 3.14)
Question 12. Find the volume and surface area of a sphere of diameter 21 cm.
Question 13. The volume of a sphere is 905 1/7 cm^3, find its diameter.
Question 14. There is surface area.and volume of sphere equal, find the radius of sphere.
Question 15. There is a ratio 1 : 4 between surface area of two spheres, find the ratio between their radius.
Question 16. Marbles of diameter 1.4 cm are dropped into a beaker containing some water are fully submerged. The diameter of beaker is 7 cm. Find how many marbles have been drapped in it if the water
rises by 5.6 cm.
Question 17. A spherical cannon ball, 28 cm in diameter is melted and recast into a right circular conical mould, the base of which is 35 cm in diameter. Find the height of the cone, correct to one
place of decimal.
Question 18. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Question 19. A solid sphere of radius 15 cm is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate the number of cones recast.
Question 20. The radius of two spheres are in the ratio of 1 : 3. Find the ratio between their volume.
Question 21. A hollow sphere of internal and external radii 6 cm and 8 cm respectively is melted and recast into small cones of base radius 2 cm and height 8 cm. Find the number of cones.
Question 22. A sphere cut out from a side of 7 cm cubes. Find the volume of this sphere?
Question 23. How many spherical bullets can be made out of a solid cube of lead whose edge measures 44 cm, each bullet being 4 cm in diameter?
Question 24. A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone.
Question 25. The total surface area of a hollow metal cylinder, open at both ends of external radius 8 cm and height 10 cm is 338π cm^2. Taking r to be inner radius, write down an equation in r and
use it to state the thickness of the metal in the cylinder.
Question 26. Find the weight of a lead pipe 35 cm long. The external diameter of the pipe is 2.4 cm and thickness of the pipe is 2mm, given 1 cm^3 of lead weighs 10 gm.
Question 27. A glass cylinder with diameter 20 cm has water to a height of 9 cm. A metal cube of 8 cm edge is immersed in it completely. Calculate the height by which the water will up in the
cylinder. Answer correct of the nearest mm. (Take π = 3.142)
Question 28. A road roller is cylindrical in shape, its circular end has a diameter of 1.4 m and its width is 4 m. It is used to level a play ground measuring 70 m × 40 m. Find the minimum number of
complete revolutions that the roller must take in order to cover the entire ground once.
Question 29. A vessel is in he form of an inverted cone. Its height is 11 cm., and the radius of its top which is open, is 2.5 cm. It is filled with water up to the rim. When lead shots, each of
which is a sphere of radius 0.25 cm., are dropped 2 into the vessel, 2/5 th of the water flows out. Find the number of lead shots dropped into the vessel.
Prove the Following
Question 1. The circumference of the base of a 10 m high conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2m. (Take π = 22/7)
Figure Based Questions
Question 3. In the adjoining figure, the radius is 3.5 cm. Find:
(i) The area of the quarter of the circle correct to one decimal place.
Question 4. The boundary of the shaded region in the given diagram consists of three semicircular areas, the smaller ones being equal and it’s diameter 5 cm, if the diameter of the larger one is 10
cm, calculate:
Question 5. In the given figure, AB is the diameter of a circle with centre O and OA = 7 cm. Find the area of the shaded region.
Question 6. Find the perimeter and area of the shaded portion of the following diagram; give your answer correct to 3 significant figures. (Take π = 22/7).
Question 7. In the adjoining figure, crescent is formed by two circles which touch at the point A. O is the centre of bigger circle. If CB = 9 cm and DE = 5 cm, find the area of the shaded portion.
Question 8. In the given figure, find the area of the unshaded portion within the rectangle. (Take π = 22/7).
Question 9. The figure shows a running track surrounding a grassed enclosure PQRSTU. The enclosure consists of a rectangle PQST with a semicircular region at each end, PQ = 200 m; PT = 70 meter
Question 10. A doorway is decorated as shown in the figure. There are four semi-circles. BC, the diameter of the larger semi-circle is of length 84 cm. Centres of the three equal semi-circles lie on
BC. ABC is an isosceles triangle with AB = AC. If BO = OC, find the area of the shaded region. (Take π = 22/7).
Question 11. In the figure given below, ABCD is a rectangle. AB 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece
of the rectangle. (Take π = 22/7).
Question 12. The given figure represents a hemisphere surmounted by a conical block of wood. The diameter of their bases is 6 cm each and the slant height of the cone is 5 cm. Calculate:
(i) the height of the cone.
(ii) the vol. of the solid.
Question 13. Calculate the area of the shaded region, if the diameter of the semi circle is equal to 14 cm. (Take π = 22/7).
Question 14. With reference to the figure given a alongside, a metal container in the form of a cylinder is surmounted by a hemisphere of the same radius. The internal height of the cylinder is 7 m
and the internal radius is 3.5 m. Calculate:
Concept Based Questions
Question 1. A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/sec,
calculate the number of complete revolutions the wheel makes in raising the bucket. (Take π =22/7)
Qustion 2. The radius of two right circular cylinder are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 4 calculate the ratio of their curved surface areas and also the ratio of
their volumes.
Question 3. A vessel in the form of an inverted cone is filled with water to the brim: Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in its so that they are fully
submerged. As a result, one third of the water in the original cone overflows. What is the volume of each of the solid cones submerged?
Question 4. Water flows at the rate of 10 m per minute through a cylindrical pipe 5 mm of diameter. How much time would it take to fill a conical vessel whose diameter at he surface is 40 cm and
depth is 24 cm?
Question 5. A conical tent is accommodate to 11 persons each person must have 4 sq. metre of the space on the ground and 20 cubic metre of air to breath. Find the height of the cone.
Question 6. A circus tent is cylindrical to a height of 3 meters and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m calculate the length of the canvas
which is 5m wide to make the required tent.
Question 7. An exhibition tent is in the form of a cylinder surmounted by a cone. The height of the tent above the ground is 85 m and the height of the cylindrical part is 50m. If the diameter of the
base is 168 m, find the quantity of canvas required to make the tent. Allow 20% extra for folds’ and for stitching. Give your answer to the nearest m^2.
Question 8. The radius of a sphere is 10 cm. If we increase the radius 5% then how many % will increase in volume?
Question 9. The cylinder of radius 12 cm have filled the 20 cm with water. One piece of iron drop in the stands of water goes up 6.75 cm. Find the radius of sphere piece.
Question 10. The radius of the internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm respectively! If it is melted and recast into a solid cylinder of height 8/3 cm, find the
diameter of the cylinder.
Question 11. The surface area of a solid metallic sphere is 616 cm^2. It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained?
Question 12. The diameter of the cross section of a water pipe is 5 cm. Water flows through it at 10km/hr into a cistern in the form of a cylinder. If the radius of the base of the cistern is 2.5 m,
find the height to which the water will rise in the cistern in 24 minutes.
Question 13. A metallic cylinder has radius 3 cm and height 5 cm. It is made of metal A. To reduce its weight, a conical hole is drilled in the cylinder, as shown and it is completely filled with a
lighter metal B. The conical hole has a radius of 3/2 cm and its depth is 8/9 cm. Calculate the ratio of the volume of the metal A to the volume of the metal B in the solid.
Question 14. An iron pillar has some part in the form of a right circular cylinder and remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The
cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cubic cm of iron weight is 7.8 grams.
Question 15. A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball.
Question 16. A buoy is made in the form of a hemisphere surmounted by a right cone whose circular base coincides with the plane surface of hemisphere. The radius of the base of the cone is 3.5 metres
and its volume is two thirds of the hemisphere. Calculate the height of the cone and the surface area of buoy correct to two places of decimal.
Question 17. Water flows through a cylindrical pipe of internal diameter 7 cm at 36 km/hr. Calculate the time in minutes it would take to fill cylindrical tank, the radius of whose base is 35 cm and
height is 1 m.
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Q-Toothpicks (as seen on Nathaniel's blog)
http://www.nathanieljohnston.com/2011/0 ... automaton/
I've made a ruletable, you can download it here:
Here is an "explanation" of sorts of how I made the ruletable, in case you want to improve it or add new rules:
So, on to the ruletable:
Nathaniel claims at the end of his article that objects made by multiple copies of his fifth object (which i'm calling the "mouth") are the only objects traced by this automata. This is not true.
Objects made of multiple hearts flipped are also traced by this automaton, the first one starting on gen 32 (edit: 31):
Code: Select all
x = 8, y = 4, rule = QTPCA
(as an image:
Re: Q-Toothpicks (as seen on Nathaniel's blog)
Fantastic work! I've updated the blog post to point to this thread. Have you found any other shapes by any chance? Any idea how the entire family of shapes can be characterized?
Re: Q-Toothpicks (as seen on Nathaniel's blog)
First of all, I'm trying to classify the different patterns that may emerge:
Closed patterns: are patterns in which the ending of every Q-toothpick coincides with the start of another Q-toothpick on the corner of the grid tile.
Open patterns: are patterns in which the ending of every Q-toothpick does not coincide with the start of another Q-toothpick on the corner of the grid tile.
Organic patterns: are patterns which can be represented by removing Q-toothpicks from an infinite tiling of Q-circles over the plane.
Non-organic patterns: are patterns which can not be represented by removing Q-toothpicks from an infinite tiling of Q-circles over the plane.
I'm sure that non-organic patterns can be further classified, but I don't think they're relevant to the study of this automata.
And their properties:
Open organic patterns will always change in the next generation until they turn into one or more closed patterns (which may never happen; in that case they will expand infinitely).
Closed organic patterns which are not made of multiple closed organic patterns are called objects.
Organic patterns have a fascinating property called duality. The dual of an object is made by replacing every Q-toothpick with the Q-toothpick that goes from and to the same corners of the grid, but
passing near the opposite corner. In QTPCA.table, that means replacing every 1 with a 3, every 2 with a 4, every 3 with a 1 and every 4 with a 2.
"Flipping" a pattern in Golly over both axis (that is, rotating it 90º twice) will return its dual object rotated 180º.
The dual of the circle is the star.
The dual of the bomb (the nameless blob from Nathaniel's post that forms the diagonal) is the bomb itself rotated 180º:
Code: Select all
x = 7, y = 3, rule = QTPCA
The dual of the candybar is the peanut. (The peanut can be seen as a double bomb)
Code: Select all
x = 10, y = 4, rule = QTPCA
The dual of this pattern is the pattern itself rotated 90º (I haven't named this one yet)
Code: Select all
x = 4, y = 4, rule = QTPCA
Last edited by ebcube on May 25th, 2011, 6:14 pm, edited 2 times in total.
Re: Q-Toothpicks (as seen on Nathaniel's blog)
Nathaniel wrote:Fantastic work! I've updated the blog post to point to this thread. Have you found any other shapes by any chance? Any idea how the entire family of shapes can be characterized?
The patterns I posted previously have been either drawn by myself or generated with new seeds, not from the single Q-toothpick.
Re: Q-Toothpicks (as seen on Nathaniel's blog)
Simple organic seeds and their characteristic (unique?) patterns:
2 toothpicks -> generates mosquito head (whose dual looks like 2 hearts + 1 bomb)
Code: Select all
x = 14, y = 6, rule = QTPCA
6 toothpicks -> generates maple leaf
Code: Select all
x = 24, y = 11, rule = QTPCA
open loop -> generates the nameless object I drew on previous post whose dual was 90º itself + 2 ghosts (which is the dual of the heart; I called it ghost because it looks like one of pac-man's
Code: Select all
x = 20, y = 4, rule = QTPCA
Re: Q-Toothpicks (as seen on Nathaniel's blog)
Amazing properties of duality, #1:
[From now on, I'm calling the patterns whose dual is a 180º rotation of itself "mirror-dual", and those whose dual is a 90º rotation of itself "rotation-dual".
I'm also calling a rotation of the pattern in the plane 90º "intuitive rotation" or just "rotation", and a rotation of the pattern in QTPCA.table on Golly 90º "golly-rotation". Same for "flip" and
A "golly-dual" is the dual of a pattern rotated (intuitively rotated) 180º. It is also the result of a double (2*90º) golly-rotation]
Question: Do infinite mirror-dual organic patterns exist?
Answer: Definitely yes.
Proof #1: Put a pattern and its dual together, displaced diagonally such as that:
a) both patterns don't overlap (except on the edges)
b) the pattern formed by both patterns together is still organic
Erase the internal edges. Voila! You now have a mirror-dual pattern.
Code: Select all
x = 17, y = 10, rule = QTPCA
Proof #2: Take any existing mirror-dual pattern, and add circles and stars to its edges such as that every circle or star you add has at least one edge or corner in common with the original pattern.
If you do it right, you'll need the same number of stars and circles to surround the pattern completely.
Erase the internal edges. Voila! You now have a mirror-dual pattern.
Code: Select all
x = 27, y = 19, rule = QTPCA
Proof #3: Take any open pattern.
Create it's golly-dual.
Paste both together and remove internal edges. Voila!
(This might yield weird results that fit my definition of closed pattern, but not the intuitive definition of closed pattern)
Characteristics of a mirror-dual pattern:
Its golly-dual is equal to itself.
Its golly-flip under the X axis and its golly-flip under the Y axis yields the same result. Also, both its clockwise and counter-clockwise golly rotations yield the same result.
Any mirror-dual pattern can be decomposed, by means of drawing internal edges, into two patterns which are dual to each other.
Re: Q-Toothpicks (as seen on Nathaniel's blog)
Starting from a small random pattern, the limiting density seems to be roughly .68.
Starting from a random infinite pattern, the limiting density seems to be roughly .72.
Proof #1: Put a pattern and its dual together, displaced diagonally such as that:
a) both patterns don't overlap (except on the edges)
b) the pattern formed by both patterns together is still organic
Can this always be done?
Answer: yes. Select an edge at the boundary of the initial object. If this edge faces "outwards" in the primal object, then it will face "inwards" in the dual object, and so "fit" the primal and dual
objects together along this edge. Since any finite object has a boundary edge, the two criteria can always be met in this way.
I would suggest a new golly script that makes the dual object of any pattern.
Re: Q-Toothpicks (as seen on Nathaniel's blog)
If I knew how to make scripts... It just has to take the selected object and replace state 1 with 3, 2 with 4, 3 with 1 and 4 with 2.
Also, script for intuitive rotation/counter-rotation: golly-rotate/golly-counter-rotate the object, then replace every state with the previous / next state.
Re: Q-Toothpicks (as seen on Nathaniel's blog)
A pattern does not need to be organic to have a stable dual. For example:
Code: Select all
x = 2, y = 2, rule = QTPCA
For that matter, self-dual inorganic patterns exist:
Code: Select all
x = 8, y = 3, rule = QTPCA
Re: Q-Toothpicks (as seen on Nathaniel's blog)
Here is a Golly-friendly version of the Q-Toothpicks ruletable posted on Nathaniel's blog for
Code: Select all
# Q-toothpicks
# rules: 16
# Golly rule-table format.
# Each rule: C,N,NE,E,SE,S,SW,W,NW,C'
# Default for transitions not listed: no change
# Variables are bound within each transition.
# For example, if a={1,2} then 4,a,0->a represents
# two transitions: 4,1,0->1 and 4,2,0->2
# (This is why we need to repeat the variables below.
# In this case the method isn't really helping.)
# A ruletable-like version (it should be right, but I haven’t tested it)
# 0 is void
# 1 is a line from bottom-left to top-right via top-left
# 2 is a line from top-left to bottom-right via top-right (= 1 rotated 90º)
# 3 is a line from top-right to bottom-left via bottom-right (= 1 rotated 180º)
# 4 is a line from bottom-right to top-left via bottom-left (= 1 rotated 270º)
var a = {0,1,2,3,4}
var b = {0,1,2,3,4}
var c = {0,1,2,3,4}
var d = {0,1,2,3,4}
var e = {0,1,2,3,4}
P.S. This info came from the ruletable "Wireworld".
EDIT: It doesn't seem to work. Please correct any of my errors.
Re: Q-Toothpicks (as seen on Nathaniel's blog)
Note that every object is a closed region which contains 2^k virtual circles with radius 1 and 2^k-1 virtual diamonds, for example: a 2x2-object is a closed region which contains exactly four virtual
circles and three virtual diamonds, a 2x4-object is a closed region which contains exactly 8 virtual circles and 7 virtual diamonds, etc. Note that a "heart" can be considered a 1x2-object which
contains two virtual circles and a virtual diamond.
Re: Q-Toothpicks (as seen on Nathaniel's blog)
For that matter, self-dual inorganic patterns exist:
Code: Select all
x = 8, y = 3, rule = QTPCA
I've found smaller inorganic self-dual pattern
Code: Select all
x = 2, y = 2, rule = QTPCA
Also, I found that density of pattern starting from one toothpick is 2/3 in densest moment (in generations of 2^n) when it makes filled square and approximately 0.47 in least dense moment (gen 2^
(n-8)*413). I don't know why then. I seen that then boundary is very convoluted.
First question ever. Often referred to as The Question. When this question is asked in right place in right time, no one can lie. No one can abstain. But when The Question is asked, silence will
fall. Silence must fall. The Question is: Doctor Who?
Re: Q-Toothpicks (as seen on Nathaniel's blog)
It appears that the number of hearts present in the n-th generation (Sloane's A188346) equals the number of rectangles of area = 2 present in the (n - 2)nd generation of the toothpick structure of
Sloane's A139250, assuming the toothpicks have length 2, if n >= 3. | {"url":"https://conwaylife.com/forums/viewtopic.php?f=11&t=663","timestamp":"2024-11-07T09:53:03Z","content_type":"text/html","content_length":"65788","record_id":"<urn:uuid:3baccfbc-d7fe-453a-b7b9-51ab48dd8431>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00724.warc.gz"} |
Cardinal And Ordinal Numbers In Asl - OrdinalNumbers.com
Cardinal And Ordinal Numbers In Asl
Cardinal And Ordinal Numbers In Asl – There are a myriad of sets that can be easily enumerated using ordinal numerals to aid in the process of. They also can help generalize ordinal quantities.
The ordinal number is among the fundamental concepts in mathematics. It is a number which shows where an object is within a set of. Ordinal numbers are a number between 1 to 20. While ordinal numbers
have many functions but they are mostly utilized to represent the sequence of the items in a list.
To represent ordinal numbers, you can use charts, numbers, and words. These can also be used for indicating how a set of pieces is arranged.
The majority of ordinal numbers fall within one of two categories. Transfinite ordinals will be represented using lowercase Greek letters. The finite ordinals will be represented using Arabic
Based on the axioms that govern the set, it must contain at minimum one or more ordinals. For example, the best possible grade could be given to the first student in the class. The student who scored
the highest score was declared the winner of the contest.
Combinational ordinal numbers
Compounded ordinal numbers are those with multiple digits. They are created simply by multiplying ordinal number by its final number. They are used to determine ranking and dating. They do not employ
an unique ending for the last digit like cardinal numbers do.
To show the order of the order in which items are arranged in a collection ordinal numbers are utilized. These numbers also serve to denote the names of objects within the collection. Regular numbers
are found in both suppletive and regular formats.
By affixing a number to it with the suffix “-u” makes regular ordinals. Then the number is written as a word and then an hyphen is added to it. There are numerous suffixes to choose from.
Suppletive ordinals can be constructed by prefixing words with -u.e. or -ie. This suffix, used to count is broader than the standard one.
Limits of ordinal operation
Ordinal numbers that are not zero are considered to be ordinal numbers. Limit ordinal number have one drawback: there is no maximum element. They can be constructed by joining sets of nonempty
elements with no limit elements.
Limits on ordinal numbers can also be employed in transfinite definitions for the concept of recursion. The von Neumann model declares that each infinite cardinal number is an ordinal number.
An ordinal with a limit equals the sum of all other ordinals beneath. Limit ordinal number can be determined using arithmetic or a series natural numbers.
These data are ordered using ordinal numbers. They give an explanation of the numerical location of objects. They are often used in the fields of set theory, arithmetic, and in other settings. While
they belong to the same class they aren’t considered to be natural numbers.
The von Neumann method uses a well-ordered list. It is possible to consider that fy fy could be described as a subfunction of the function g’. In the event that g’ fulfills the criteria, g’ is an
limit ordinal when there is only one subfunction (i, II).
The Church Kleene oral is a limit-ordering ornal that functions in a similar way. A Church-Kleene ordinal defines a limit ordinal as a well-ordered grouping of ordinals that are smaller.
Stories that use ordinary numbers as examples
Ordinal numbers are commonly used to show the order of things between entities and objects. They are essential in organising, counting or ranking purposes. They can also be used for indicating the
order of things as well as to define objects’ positions.
The letter “th” is usually used to signify the ordinal number. However, sometimes the letter “nd”, however, is also used. The titles of books usually contain ordinal numerals.
While ordinal numbers are typically written in lists format, they can also be written as words. They can also be expressed by way of numbers or acronyms. In comparison, they are simpler to comprehend
as compared to cardinal numbers.
There are three kinds of ordinal numbers. Through practice and games you will be able to be able to learn more about these numbers. You can increase your math skills by understanding more about them.
Coloring is a fun and easy method to increase your proficiency. For a quick check of your results make use of a handy mark-making page.
Gallery of Cardinal And Ordinal Numbers In Asl
Esl Cardinal And Ordinal Numbers Google Search Ensino De Ingl s
The Cardinal And Ordinal Numbers
Pin On General
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Following are diastereomers (A) and (B) of 3 -bromo-3,4-dimethylhexane. On treatment with sodium ethoxide in ethanol, each gives 3,4 -dimethyl-3-hexene as the major product. One diastereomer gives
the \(E\) alkene, and the other gives the \(Z\) alkene. Which diastereomer gives which alkene? Account for the stereoselectivity of each \(\beta\)-elimination.
Short Answer
Expert verified
Question: Identify the diastereomer that leads to the formation of each isomer (E-alkene and Z-alkene) during the reaction of 3-bromo-3,4-dimethylhexane with sodium ethoxide in ethanol and explain
the reasons behind this stereoselectivity. Answer: Diastereomer (A) leads to the formation of the E-alkene as the major product due to less steric hindrance between the two methyl groups in the
product and a more stable conformation. Diastereomer (B) leads to the formation of the Z-alkene as the major product, even though it is less stable compared to the E-alkene, because of the starting
diastereomer's conformation, which directs the reaction to produce the Z-alkene.
Step by step solution
Understanding the Reaction
The reaction between the diastereomers and sodium ethoxide is an example of a \(\beta\)-elimination reaction. In this reaction, sodium ethoxide acts as a base and removes a proton (\(H^+\)) from the
adjacent carbon atom to the one carrying the bromine atom. Following this, a double bond forms between the two carbon atoms while expelling bromide ion.
Consider the Effect of Steric Hindrance
When drawing the reactants' structures, we need to consider their conformation in three-dimensional space. The ethoxide ion will selectively abstract the proton from the least sterically hindered
position, which is generally the least crowded position, thus leading to the most stable product.
Examine Both Diastereomers
Let's examine both diastereomers individually: (A) When the hydrogen atom which participates in the elimination reaction is in an axial position, and the bromine atom is in the equatorial position
(also known as 'anti-periplanar'), we can visualize that this diastereomer's conformation would allow the formation of the E-alkene, which is the more stable configuration due to less steric
hindrance between the two methyl groups. (B) In this diastereomer, the hydrogen atom is also anti-periplanar to the bromine atom, but positioned differently in the molecule. Under elimination
conditions, the reaction will proceed, and a double bond will form, leading to the less stable Z-alkene due to the larger steric hindrance between the two methyl groups.
Identify the Stereoselectivity
After analyzing the structures and conformations of both diastereomers, we can now identify the stereoselectivity of each \(\beta\)-elimination: The diastereomer (A) gives the E-alkene as the major
product due to less steric hindrance between the two methyl groups in the product and a more stable conformation. The diastereomer (B) gives the Z-alkene as the major product, even though it is less
stable compared to the E-alkene. This is because of the starting diastereomer's conformation, which directs the reaction to produce the Z-alkene. | {"url":"https://www.vaia.com/en-us/textbooks/chemistry/organic-chemistry-8-edition/chapter-9/problem-39-following-are-diastereomers-a-and-b-of-3-bromo-34/","timestamp":"2024-11-14T21:27:39Z","content_type":"text/html","content_length":"262293","record_id":"<urn:uuid:ccb214b3-b768-40b0-95b9-272276d90b55>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00326.warc.gz"} |
A filter circuit is like a sieve. It allows some things through and holds back others. In this case we are talking about AC frequencies. Some frequencies pass through the filter while others are
rejected. The characteristics of a filter can be shown on a graph called a FREQUENCY RESPONSE CURVE.
VOLTAGE OUT is plotted against FREQUENCY.
Figure 2 shows a LOW PASS filter response curve giving output at low frequencies but none at higher frequencies.
Figure 3 shows a selection of filter characteristics.
Simple filters can be made from capacitors and resistors
Filters have many applications.
In audio frequency amplifiers, CROSSOVER filters to direct low frequencies to the WOOFER and high frequencies to the TWEETER speakers.
In SCRATCH filters to remove unwanted high frequency noise.
In NOTCH filters to remove whistles due to two radio stations being too close together in frequency.
In Hum filters to remove low frequency noise due to the mains supply.
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Writer : delon
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How To Layout A Baseball Field
When it comes to the layout a baseball field, you need to check 3 main factors before you begin.
Pre-factor 1: How much space will I need?
When considering the space you need, you will not only be thinking about the size of the baseball diamond. It would help if you also thought about dugouts, bathrooms, concession areas, storage
facilities, bleachers, drainage swales, parking, and a decent-sized buffer zone to protect your audience.
Here is some suggestions with the space you may have:
• 4.5 Acres :: 90 ft bases with 400 ft fence
• 3 Acres :: 80 ft bases with 315 ft fence
• 2 Acres :: 70 ft bases with 275 ft fence
• 1.5 Acres :: 60 ft bases with 215 fence
Pre-factor 2: How will I keep my fans safe?
Most fields are planned without the fan in mind and how to keep them safe. This usually ends with the bare minimum of space between fields, fields, and bleachers. There are 3 ways to prevent this.
The first is the most expensive, an expansive overhead netting system to protect the fans. The second is taking care of it during the planning stages of the field by adding extra space between fields
reducing foul ball risks. The third is the first two combined, allowing for your fans' maximum safety.
During the planning stage, you must focus on keeping your common areas as open as possible between fields. A tip of the trade is to plant trees along the common way. The older they get, the better
cover they become, and the better they become for shade during the summer time. Until they are grown, you will need to focus on buying netting to protect these common areas. You will still need to
keep up netting when the trees are fully grown but possibly not as much. But as we all know, it is better to be safe than sorry and take care of your fans.
Pre-factor 3: How will I care for drainage?
The most critical part of caring for the baseball field is surface drainage. The method we like to suggest is called the "turtleback". This is where the pitcher's mound is the field's highest point.
From there, the infield should be slightly graded to a 0.5 - 1% slope. The outfield and foul areas should then be graded to a 1 - 1.5% slope. This is the quickest and most efficient method of ridding
the field of water and allowing for smooth drainage.
With the slopes set up, you will need to add storm drains along the edges of the baseball field in out-of-play areas. These will be used to carry surface-drained water away from the field.
After the Pre-Factors
Step 1: What is the sun's angle on my field?
Sun is a major factor in any baseball game. It would help if you ALWAYS tried to keep the sun out of the batters' eyes. For a batter to hit the ball in direct sunlight, it endangers himself and
everyone around him. At the same time, you should also be trying to keep the sun out of the fielders' eyes. We suggest setting the field where the sun goes from southwest to northeast with the home
plate at the southwestern end.
Step 2: Where do I want the home plate to be?
As stated in the previous step, you want the home plate to be at the southwestern end. Place the home plate center where you wish to set up your backstop. The main objective is to have the centerline
of the field be a continuation of the centerline that runs from the backstop to the home plate.
Step 3: Where does 2nd base go?
Second base is found by using a 200-foot measuring tape with one end attached to the home plate (you can use a stake as a substitute for a home plate at this moment to keep the measuring tape in
place). Walk in the distance from home plate to the location where you want second base. The distance you walk can vary depending on the jurisdiction or league ruling you are in. Make sure you are in
the centerline of the field before placing second base.
Step 4: Where does 1st and 3rd go?
Before doing 3rd base focus on 1st first. Take the measuring tape and extend it from 2nd to where you believe approximately 1st base to be. Next, get a second measuring tape and do the same thing
except for home plate. It would help if you had the same distance from 2nd to 1st and home to 1st. This is where 1st base will be placed. After that, rinse and repeat with 3rd base.
Step 5: Where does the pitching mound go?
To find the pitching mound, you must know the rules of your league distance-wise. Once you learn that you only need to measure from the apex of the home plate (Not counting the black edging around
the plate) to the distance, the pitching mound should be placed in the direction of 2nd base.
Step 6: Where do foul lines and foul poles go?
To locate where your foul poles should be placed, you need to do some math. It would be best if you used the geometric formula for a right triangle which is X² + Y² = Z². Let X equal the distance
between the 2nd and 3rd base to find the left field foul line. Let Y equal the distance you want the foul line to extend past 3rd base to the foul pole. Square each of these two numbers. Add them
together. Then take the square root of the sum of the two numbers to calculate the length of Z or the hypotenuse. Once you have X, Y, and Z, you can work in the field, triangulating your foul pole's
In this example, we set the bases to be 60 ft across, which makes X = 60. We decided to make the left field foul line 270 feet, so you subtract 60 from 270 and get 210 which is what Y will be. After
that, all you need to do is square x and y and add them together. 210² = 44100 and 60² = 3600. Add them together to get 47700. Then find the square root 47700, which is roughly 218.40. That makes Z =
Break Down:
X² = 60 * 60 = 3600 ft
Y² = 210 * 210 = 44100 ft
X² + Y² = Z²
3600 ft + 44100 ft = Z²
Z² = 47700 ft
√Z² = 218.40 ft
Z = 218.40 ft
To convert the remaining decimal into inches, you multiply the number of inches in a foot by 0.4 (the extra decimal).
12 in/ft x 0.4 ft = 4.8 in
In this example and from these calculations, we know the distance from 2nd base to the left field foul pole (Z) is:
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Scaling the regularization parameter for SVCs
Scaling the regularization parameter for SVCs#
The following example illustrates the effect of scaling the regularization parameter when using Support Vector Machines for classification. For SVC classification, we are interested in a risk
minimization for the equation:
\[C \sum_{i=1, n} \mathcal{L} (f(x_i), y_i) + \Omega (w)\]
□ \(C\) is used to set the amount of regularization
□ \(\mathcal{L}\) is a loss function of our samples and our model parameters.
□ \(\Omega\) is a penalty function of our model parameters
If we consider the loss function to be the individual error per sample, then the data-fit term, or the sum of the error for each sample, increases as we add more samples. The penalization term,
however, does not increase.
When using, for example, cross validation, to set the amount of regularization with C, there would be a different amount of samples between the main problem and the smaller problems within the folds
of the cross validation.
Since the loss function dependens on the amount of samples, the latter influences the selected value of C. The question that arises is “How do we optimally adjust C to account for the different
amount of training samples?”
# Author: Andreas Mueller <amueller@ais.uni-bonn.de>
# Jaques Grobler <jaques.grobler@inria.fr>
# License: BSD 3 clause
Data generation#
In this example we investigate the effect of reparametrizing the regularization parameter C to account for the number of samples when using either L1 or L2 penalty. For such purpose we create a
synthetic dataset with a large number of features, out of which only a few are informative. We therefore expect the regularization to shrink the coefficients towards zero (L2 penalty) or exactly zero
(L1 penalty).
L1-penalty case#
In the L1 case, theory says that provided a strong regularization, the estimator cannot predict as well as a model knowing the true distribution (even in the limit where the sample size grows to
infinity) as it may set some weights of otherwise predictive features to zero, which induces a bias. It does say, however, that it is possible to find the right set of non-zero parameters as well as
their signs by tuning C.
We define a linear SVC with the L1 penalty.
from sklearn.svm import LinearSVC
model_l1 = LinearSVC(penalty="l1", loss="squared_hinge", dual=False, tol=1e-3)
We compute the mean test score for different values of C via cross-validation.
import numpy as np
import pandas as pd
from sklearn.model_selection import ShuffleSplit, validation_curve
Cs = np.logspace(-2.3, -1.3, 10)
train_sizes = np.linspace(0.3, 0.7, 3)
labels = [f"fraction: {train_size}" for train_size in train_sizes]
shuffle_params = {
"test_size": 0.3,
"n_splits": 150,
"random_state": 1,
results = {"C": Cs}
for label, train_size in zip(labels, train_sizes):
cv = ShuffleSplit(train_size=train_size, **shuffle_params)
train_scores, test_scores = validation_curve(
results[label] = test_scores.mean(axis=1)
results = pd.DataFrame(results)
import matplotlib.pyplot as plt
fig, axes = plt.subplots(nrows=1, ncols=2, sharey=True, figsize=(12, 6))
# plot results without scaling C
results.plot(x="C", ax=axes[0], logx=True)
axes[0].set_ylabel("CV score")
axes[0].set_title("No scaling")
for label in labels:
best_C = results.loc[results[label].idxmax(), "C"]
axes[0].axvline(x=best_C, linestyle="--", color="grey", alpha=0.7)
# plot results by scaling C
for train_size_idx, label in enumerate(labels):
train_size = train_sizes[train_size_idx]
results_scaled = results[[label]].assign(
C_scaled=Cs * float(n_samples * np.sqrt(train_size))
results_scaled.plot(x="C_scaled", ax=axes[1], logx=True, label=label)
best_C_scaled = results_scaled["C_scaled"].loc[results[label].idxmax()]
axes[1].axvline(x=best_C_scaled, linestyle="--", color="grey", alpha=0.7)
axes[1].set_title("Scaling C by sqrt(1 / n_samples)")
_ = fig.suptitle("Effect of scaling C with L1 penalty")
In the region of small C (strong regularization) all the coefficients learned by the models are zero, leading to severe underfitting. Indeed, the accuracy in this region is at the chance level.
Using the default scale results in a somewhat stable optimal value of C, whereas the transition out of the underfitting region depends on the number of training samples. The reparametrization leads
to even more stable results.
See e.g. theorem 3 of On the prediction performance of the Lasso or Simultaneous analysis of Lasso and Dantzig selector where the regularization parameter is always assumed to be proportional to 1 /
L2-penalty case#
We can do a similar experiment with the L2 penalty. In this case, the theory says that in order to achieve prediction consistency, the penalty parameter should be kept constant as the number of
samples grow.
model_l2 = LinearSVC(penalty="l2", loss="squared_hinge", dual=True)
Cs = np.logspace(-8, 4, 11)
labels = [f"fraction: {train_size}" for train_size in train_sizes]
results = {"C": Cs}
for label, train_size in zip(labels, train_sizes):
cv = ShuffleSplit(train_size=train_size, **shuffle_params)
train_scores, test_scores = validation_curve(
results[label] = test_scores.mean(axis=1)
results = pd.DataFrame(results)
import matplotlib.pyplot as plt
fig, axes = plt.subplots(nrows=1, ncols=2, sharey=True, figsize=(12, 6))
# plot results without scaling C
results.plot(x="C", ax=axes[0], logx=True)
axes[0].set_ylabel("CV score")
axes[0].set_title("No scaling")
for label in labels:
best_C = results.loc[results[label].idxmax(), "C"]
axes[0].axvline(x=best_C, linestyle="--", color="grey", alpha=0.8)
# plot results by scaling C
for train_size_idx, label in enumerate(labels):
results_scaled = results[[label]].assign(
C_scaled=Cs * float(n_samples * np.sqrt(train_sizes[train_size_idx]))
results_scaled.plot(x="C_scaled", ax=axes[1], logx=True, label=label)
best_C_scaled = results_scaled["C_scaled"].loc[results[label].idxmax()]
axes[1].axvline(x=best_C_scaled, linestyle="--", color="grey", alpha=0.8)
axes[1].set_title("Scaling C by sqrt(1 / n_samples)")
fig.suptitle("Effect of scaling C with L2 penalty")
For the L2 penalty case, the reparametrization seems to have a smaller impact on the stability of the optimal value for the regularization. The transition out of the overfitting region occurs in a
more spread range and the accuracy does not seem to be degraded up to chance level.
Try increasing the value to n_splits=1_000 for better results in the L2 case, which is not shown here due to the limitations on the documentation builder.
Total running time of the script: (0 minutes 15.714 seconds)
Related examples | {"url":"https://scikit-learn.org/1.5/auto_examples/svm/plot_svm_scale_c.html","timestamp":"2024-11-11T07:40:57Z","content_type":"text/html","content_length":"114149","record_id":"<urn:uuid:a9403a6d-27ca-41c6-a079-6ee1685bfb63>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00531.warc.gz"} |
BFS algorithm in C | How does BFS algorithm work in C with example
Updated April 5, 2023
Introduction to BFS algorithm in C
BFS is a traversal algorithm that is applied mostly on the graphs to travel from one node to another node for various manipulation and usage. The visit on each node using the BFS algorithm makes the
entire algorithm treated as an efficient and useful algorithm. There are certain ways to traverse from one node to another node in a graph using the BFS algorithm in the sense that the selected node
will be encountered first, followed by the next node to be reached either breadthwise or moving to the node directly. In this topic, we are going to learn about the BFS algorithm in C.
There is no particular syntax for BFS, but some algorithmic steps are followed to traverse from root to the base. Thus the algorithm to be used is as follows:
Define the structure as represented below:
Struct node {
char lbl;
boolean vlue;
• Followed by the variables to be defined
• Then form the variables with the matrix defined
• Count each of the nodes.
• Display all the vertex
• Check for the adjacent vertexes
• Check for the final deployments and
• Construct the logic for BFS traversal.
• Run the main function
• Drive the entire program
More will clear in the Example section, where the implementation will make use of the above algorithm with proper representation.
How does BFS algorithm work in C?
As mentioned, BFS is a traversal algorithm that is used for many of the traversal and searching activities performed over the graph with vertexes. There exists a working paradigm that is used for
traversal from one node to another node with detailed information, and rules are as follows:
• Breadth-First Search (BFS) is an algorithm used for traversing and is also considered a searching tree with the data structure.
• A node which is the initial node or any arbitrary node from there the actual traversal will start then moving forward; it will be used for exploring the rest of the other nodes at neighbor which
will be used for the current depth of the node before moving to the next node at the depth level.
• BFS works completely opposite to the DFS algorithm, which is also used for the same purpose of vertex of graph traversal using some rules and regulations which instead of nodes even adjoining
nodes apply for backtracking and expansion of other nodes within it.
• BFS was already developed and implemented, but it got its final approval when used as an algorithm for traversal by C.Y Lee to find the shortest path by wire routing.
• The BFS is a non-recursive implementation that is similar to the non-recursive implementation of depth-first search but differs in ways like it makes use of queue, not stack, and also checks
whether or not the vertex explored recently is the actual one to have a check on or already that vertex is explored. It has to be kept in mind that the same vertex should not repeatedly be
traversed; then, only dequeuing from the queue is possible.
• I suppose for G_0 is any tree, then in that tree replacement of queue from BFS algorithm is used with a stack for DFS yields.
• Queue comprises nodes and vertexes, which have labels stored for tracking and backtracking from the destination node.
• BFS search end result is nothing but a breadth-first tree that is optimized and will have almost all the information for reference at any time.
• There is sometimes talk on the completeness of the graph when used BFS algorithm which is the input to breadth-first search which is assumed to be a finite graph for representing adjacent list or
something relevant it can be an adjacent matrix as well.
• Mostly these algorithms are quite predominantly used in fields of Artificial Intelligence with an entire input and training sets to be designed in the format of the tree with a release and output
as predicted and traversed using BFS algo.
• BFS is the only algorithm that is considered as one of the finite and complete algorithms that makes the infinite graph represent implicitly which have some goals that are useful and relevant,
not as DFS algorithm where the completeness and relevant information is not upto the mark.
• BFS algorithm has a pattern of a time complexity measure which, according to the implementation, comes out to be O(V+E) where V represents the number of vertexes present in the tree and E stands
for the number of edges present in the tree. This complexity is considered finite only if all the nodes are covered at the time of traversal.
• BFS is used for solving many of the graph theory that exists like garbage collection copying, Testing of Bipartite graph, Serialization or deserialization of binary tree that allows a lot of
reconstruction of tree in an efficient and effective manner.
Example of BFS algorithm in C
This program demonstrates the implementation of the BFS algorithm in C language, which is used for various graph traversal ways by adjusting the neighbor nodes and adjacent nodes for manipulation, as
shown below in the output.
int a_0[30][20],q_1[30],visited_nodes[30],n_8,i_0,j_1,f_5=0,r_2=-1;
void bfs_logic(int v_8) {
for (i_0=1;i_0<=n_8;i_0++)
if(a_0[v_8][i_0] && !visited_nodes[i_0])
if(f_5<=r_2) {
void main() {
int v_8;
printf("\n Enter Vertices_to_represent:");
for (i_0=1;i_0<=n_8;i_0++) {
printf("\n Enter graph_data especially_in_matrix_format:\n");
for (i_0=1;i_0<=n_8;i_0++)
for (j_1=1;j_1<=n_8;j_1++)
printf("\n Enter Starting_vertex_for_traversal:");
printf("\n Reachable_nodes_are:\n");
for (i_0=1;i_0<=n_8;i_0++)
printf("%d\t",i_0); else
printf("\n Bfs not_possible_if_not_proper.");
Explanation: The program expects an input with the number of vertices that will be present in the tree. Then formulate the tree with the adjoining nodes represented as vertexes in the matrix format
starting from root to base reachable nodes as shown.
BFS algorithm is quite useful and is getting explored as it is nowadays quite trending with the boom in artificial intelligence. It plays a pivotal role even in Graph theories where there is the
implementation of trees with a lot of vertexes and nodes for traversal. BFS makes the implemented tree as finite, which also helps in providing relevant and required information.
Recommended Articles
This is a guide to the BFS algorithm in C. Here we discuss How does BFS algorithm work in C along with the example and output. You may also have a look at the following articles to learn more – | {"url":"https://www.educba.com/bfs-algorithm-in-c/","timestamp":"2024-11-07T00:37:53Z","content_type":"text/html","content_length":"309872","record_id":"<urn:uuid:8bb2b446-8f2e-49db-a386-25326493f5c5>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.54/warc/CC-MAIN-20241106230027-20241107020027-00479.warc.gz"} |
International Scientific Journal
In this study, we investigate the heat transfer characteristics in unsteady boundary layer flow of Maxwell fluid by using Cattaneo-Christov heat flux model and convective boundary conditions. The
flow is caused by a sheet which is stretched periodically back and forth in its own plane. The physical model that takes into account the effects of constant applied magnetic field is transformed
into highly nonlinear partial differential equations under boundary layer approximations. The solution of dimensionless version of these equations is developed using homotopy analysis method. The
simulations are presented in the form of temperature and velocity profiles for suitable range of physical parameters. The obtained results illustrate that an increase in Deborah number and Hartmann
number suppress the velocity profiles. It is further observed that Cattaneo-Christov heat flux model predicts the suppression of thermal boundary layer thickness as compared to Fourier law.
PAPER SUBMITTED: 2016-02-25
PAPER REVISED: 2017-08-22
PAPER ACCEPTED: 2017-08-24
PUBLISHED ONLINE: 2017-09-09
, VOLUME
, ISSUE
Issue 2
, PAGES [443 - 455] | {"url":"https://thermalscience.vinca.rs/2019/2/3","timestamp":"2024-11-08T01:23:15Z","content_type":"text/html","content_length":"21312","record_id":"<urn:uuid:a16958fd-5c14-4d44-b3e2-e87c8a4f2f98>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00693.warc.gz"} |
Quantum Network
In recent years, quantum computers, which handle information based on a completely different concept than conventional computers, have been attracting attention. Quantum bits are used in quantum
networks to simultaneously solve a large number of solution patterns using quantum states in which a large number of states are superimposed, thereby enabling problems such as combinatorial
optimization problems, which take an enormous amount of time with conventional computers, to be solved in a fraction of a second. This allows problems such as combinatorial optimization problems to
be solved in a fraction of a second, which would take an enormous amount of time with conventional computers.
However, since it is difficult to scale up quantum computing devices, it is effective to realize a large quantum computing device as a whole by connecting many quantum computing devices that exist in
remote locations. A quantum network connects multiple remote quantum computing devices and propagates qubits while maintaining the state of quantum superposition. Since there are various restrictions
on the transmission of EPR pairs, there are many problems to be solved in order to realize a quantum network.
Therefore, from FY2024, we are newly working on control techniques for efficient and low-cost quantum communication using EPR pairs. | {"url":"https://www.anl.is.ritsumei.ac.jp/en/?page_id=351","timestamp":"2024-11-07T15:44:44Z","content_type":"text/html","content_length":"37671","record_id":"<urn:uuid:addd8055-64df-48ef-8ceb-c8941343cc7c>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00069.warc.gz"} |
Improved Distributed Principal Component Analysis
Improved Distributed Principal Component Analysis
Part of Advances in Neural Information Processing Systems 27 (NIPS 2014)
Yingyu Liang, Maria-Florina F. Balcan, Vandana Kanchanapally, David Woodruff
We study the distributed computing setting in which there are multiple servers, each holding a set of points, who wish to compute functions on the union of their point sets. A key task in this
setting is Principal Component Analysis (PCA), in which the servers would like to compute a low dimensional subspace capturing as much of the variance of the union of their point sets as possible.
Given a procedure for approximate PCA, one can use it to approximately solve problems such as $k$-means clustering and low rank approximation. The essential properties of an approximate distributed
PCA algorithm are its communication cost and computational efficiency for a given desired accuracy in downstream applications. We give new algorithms and analyses for distributed PCA which lead to
improved communication and computational costs for $k$-means clustering and related problems. Our empirical study on real world data shows a speedup of orders of magnitude, preserving communication
with only a negligible degradation in solution quality. Some of these techniques we develop, such as input-sparsity subspace embeddings with high correctness probability with a dimension and sparsity
independent of the error probability, may be of independent interest.
Do not remove: This comment is monitored to verify that the site is working properly | {"url":"https://proceedings.nips.cc/paper_files/paper/2014/hash/52947e0ade57a09e4a1386d08f17b656-Abstract.html","timestamp":"2024-11-08T11:02:01Z","content_type":"text/html","content_length":"9368","record_id":"<urn:uuid:cc0a129c-f014-434b-a904-00712e2c368c>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00764.warc.gz"} |
Lesson 6
The Addition Rule
Lesson Narrative
The mathematical purpose of this lesson is for students to apply the addition rule and to interpret the answer using the model. The work of this lesson connects to previous work because students
calculated probabilities using information represented in tables and Venn diagrams. The work of this lesson connects to upcoming work because students will use probability to determine whether or not
two events are independent. Students encounter the addition rule which states that given events A and B, \(P(\text{A or B}) = P(\text{A}) + P(\text{B}) - P(\text{A and B})\). When students use the
addition rule to get an answer and then interpret the meaning of their answer in a context then they are reasoning abstractly and quantitatively (MP2). When students have to fix or find the error and
explain the correct reasoning, they are constructing viable arguments and critiquing the reasoning of others (MP3).
Technology isn't required for this lesson, but there are opportunities for students to choose to use appropriate technology to solve problems. We recommend making technology available.
Learning Goals
Teacher Facing
• Interpret (orally and in writing) the addition rule in context.
• Use the addition rule to calculate probabilities.
Student Facing
• Let’s learn about and use the addition rule.
Student Facing
• I can use the addition rule to find probabilities.
Glossary Entries
• addition rule
The addition rule states that given events A and B, the probability of either A or B is given by \(P(\text{A or B}) = P(\text{A}) + P(\text{B}) - P(\text{A and B})\).
Additional Resources
Google Slides For access, consult one of our IM Certified Partners.
PowerPoint Slides For access, consult one of our IM Certified Partners. | {"url":"https://curriculum.illustrativemathematics.org/HS/teachers/2/8/6/preparation.html","timestamp":"2024-11-07T22:15:42Z","content_type":"text/html","content_length":"76044","record_id":"<urn:uuid:2680ae53-897b-4823-8f19-4aeb5a9c9791>","cc-path":"CC-MAIN-2024-46/segments/1730477028017.48/warc/CC-MAIN-20241107212632-20241108002632-00355.warc.gz"} |
profile -
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How can I stop sagemath program without quitting sage session?
07:03:34 commented I haven't used a Jupyter notebook yet, but it seems that it is a good way to use SageMath. I will try.
+0100 answer
Thank you for your answer!
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07:02:35 badge ● Supporter (source)
2017-02-17 How can I stop sagemath program without quitting sage session?
07:02:32 commented
+0100 question Thank you! It works! I did not know that there is such a simple solution.
How can I stop sagemath program without quitting sage session?
Hello. First of all, I think it is good to describe how I use SageMath before asking the question.
I installed SageMath in my laptop using Ubuntu 16.04. It is installed in directory "~/sage". I usually run SageMath in terminal. That is, I follow the following steps to run
2017-02-16 asked a 1. Open the terminal (It is the basic terminal from Ubuntu 16.04).
09:59:25 question 2. Go to the directory where SageMath is installed by typing "cd sage".
+0100 3. Type "./sage". Then, sage session begins.
In sage session, I usually load a sage file which I already wrote and run it to check the result. But sometimes, a program does not seem to finish, since computation takes a
lot of time. Then, I would like to finish the program which I am running now, but without going out of sage session, since otherwise, I should follow 3 steps above again.
However, I could not find how to do it, yet.
Can anybody teach me how to do it?
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Sagemath on Windows 10?
Hello all. Today, I could come up with the following news(you can google "microsoft-canonical-bring-ubuntu-linux-on-windows-10". Sorry for not writing the link, since my karma
2016-03-31 is low...) As far as I understand, I guess it means that one can install sagemath in windows 10, directly - not using virtualbox.
19:14:52 asked a
+0100 question In fact, in earlier versions of MS windows, like windows 7, it seemed that one should run virtualbox to run sagemath on one's computer. Maybe using cloud computing in sagemath
homepage would be better, but I don't prefer that. When only considering using sagemath, it would be more convenient to use Linux. So, if the news is true, then I can directly
download and install sagemath in windows 10.
Is my understanding correct? Or is there anybody who already tried this? | {"url":"https://ask.sagemath.org/users/20926/prosolver/?sort=recent","timestamp":"2024-11-01T22:51:57Z","content_type":"application/xhtml+xml","content_length":"29506","record_id":"<urn:uuid:c29964b3-b48b-4a11-a82f-ac940bc6c3b0>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00749.warc.gz"} |
seminars - Fermat's Last Theorem, 25 years later
After recalling the statement of Fermat's Last Theorem, I will summarize the proof of the theorem that was finalized by Wiles+Taylor--Wiles in 1994 and published in 1995. Then I will recount some of
the progress in the general subject ("modularity") since 1995 and discuss the possibility of streamlining the proof that is now a quarter-century old. | {"url":"https://www.math.snu.ac.kr/board/index.php?mid=seminars&page=45&document_srl=800957&l=en&sort_index=room&order_type=asc","timestamp":"2024-11-10T08:19:38Z","content_type":"text/html","content_length":"47408","record_id":"<urn:uuid:0a3cb4a0-21d9-4640-be23-b14df5263202>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00032.warc.gz"} |
The 1 - 1000 number game
It's a really fun forum game
Here is how the game goes. We go from 1 to 1000 using our posts. The posts have to include a number between 1 - 1000 in context, example...
"This game is so simple
a 1 year old could play it"
OR / AND
"I was so bored I drank
12 cans of pepsi and freaked out"
In addition all numbers from 1 - 1000 must be posted in accending order.
This game is so simple a 1 year old could play it.
ps. remember people this game is all about fun. post interesting and creative posts whilst doing this and it will be great
This is possibly the strangest thread I have seen posted at 2 AM.
According to the forum time, at least.
Hey, I got post 3
So.....just what are we doing this 4 anyway
So...i wonder if this thread will make 1000 or will it die here at number four..or is that five...
Damn wish i could count!! Where is Count Dracula when ya need him....
1,2,3,4,5.....6,7,8,9,10....11....errmmmm 76????
Five is between 4 and 6 and is the third prime number, after 2 and 3, and before 7. Because it can be written as 2^(2^1)+1, five is classified as a Fermat prime. 5 is the third Sophie Germain prime,
the first safe prime, and the third Mersenne prime exponent. Five is the first Wilson prime and the third factorial prime, also an alternating factorial. Five is the first Good prime. It is an
Eisenstein prime with no imaginary part and real part of the form 3n − 1. It is also the only number that is part of more than one pair of twin primes.
Five is conjectured to be the only odd untouchable number and if this is the case then five will be the only odd prime number that is not the base of an aliquot tree.
The number 5 is the 5th Fibonacci number, being 2 plus 3. 5 is also a Pell number and a Markov number, appearing in solutions to the Markov Diophantine equation: (1, 2, 5), (1, 5, 13), (2, 5, 29),
(5, 13, 194), (5, 29, 433), ... (A030452 lists Markov numbers that appear in solutions where one of the other two terms is 5). Whereas 5 is unique in the Fibonacci sequence, in the Perrin sequence 5
is both the fifth and sixth Perrin numbers.
5 and 6 form a Ruth-Aaron pair under either definition.
There are five solutions to Znám's problem of length 6.
Five is the second Sierpinski number of the first kind, and can be written as S2=(2^2)+1
While polynomial equations of degree 4 and below can be solved with radicals, equations of degree 5 and higher cannot generally be so solved. This is the Abel-Ruffini theorem. This is related to the
fact that the symmetric group S[n] is a solvable group for n ≤ 4 and not solvable for n ≥ 5.
While all graphs with 4 or fewer vertices are planar, there exists a graph with 5 vertices which is not planar: K[5], the complete graph with 5 vertices.
Five is also the number of Platonic solids.^
A polygon with five sides is a pentagon. Figurate numbers representing pentagons (including five) are called pentagonal numbers. Five is also a square pyramidal number.
Five is the only prime number to end in the digit 5, because all other numbers written with a 5 in the ones-place under the decimal system are multiples of five. As a consequence of this, 5 is in
base 10 a 1-automorphic number.
Vulgar fractions with 5 or 2 in the denominator do not yield infinite decimal expansions, as is the case with most primes, because they are prime factors of ten, the base. When written in the decimal
system, all multiples of 5 will end in either 5 or 0.
There are five Exceptional Lie groups.
Wow, that last post had me doing math for 6 hours, and I didnt even figure half that out.
Alright then mister luckymann can you tell me about number 7 then?
I say this has had a good start.
To l8 we just skipped to 9
i guess it'll get pretty boring to name a number in every post, unless you get drunk, then u can talk about drinking 15 beer
Im very sad the nubmer only goes til 1.000
otherwise i couldve named my Member Number
Member No.2,988,115
I just finished a nice SP game with seventy-eight planets and nine AI + me.
Holy cow, the game lasted for a little over 11 hours.
lim[x->7] (2x^3-7x^2-64x+105)/(x^2-2x-35)
edit: aw... you beat me to 11.... now i gotta make up another
lim[x->8] (2x^4-30x^3+104x^2+120x-448)/(x^3-13x^2+26x+112)
limx->8 (2x4-30x3+104x2+120x-448)/(x3-13x2+26x+112)
Sorry, but your second attempt is incorrect
So here is a valid one:
6/(πi) ∲[C](1/z)dz
(the circle denotes that this is a contour integral, and take C to be the unit circle in the complex plane)
I might be misunderstanding you, but it seems to evaluate to 0/0.
lim[x->8] (2x^2-18x+16)/(x-8)
I might be misunderstanding you, but it seems to evaluate to 0/0.
I'm sorry, on closer inspection it seems that I mentally inserted a negative sign where it didn't belong. I'm going to go bang my head on the wall 15 times as penance.
Have not done that much math since I was 17. Damn Im old. (30)
i just spent 19 minutes trying to figure the math out, and its giving me a headache
I spent 20 seconds looking at the math.
Then on second 21 decided to just say...screw that.
And on second 22, decided to do better things with my time.
Let's start a pool on how many days it will take to hit post one thousand. I'm putting my $$$$$ on 23 days.
2 beer 7$
three margaritas $15
four jello shots $20
Taking home the girl
who drank all of the above....
Member No.3,125,868 | {"url":"https://forums.sinsofasolarempire.com/336450/the-1---1000-number-game","timestamp":"2024-11-02T11:35:26Z","content_type":"text/html","content_length":"166135","record_id":"<urn:uuid:1e954f20-fee1-473e-a92b-f94f40eaac9c>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00244.warc.gz"} |
1. Consider the set . Let , , , , , , , and be the collection of all subsets of that are designated as open sets.
1. Is a topological space?
2. Is it a topological space if is added to the collection of open sets? Explain.
3. What are the closed sets (assuming is included as an open set)?
4. Are any subsets of neither open nor closed?
2. Continuous functions for the strange topology:
1. Give an example of a continuous function, , for the strange topology in Example 4.4.
2. Characterize the set of all possible continuous functions.
3. For the letters of the Russian alphabet, A, B, V, G, D, E, Ë, Zh, Z, I, uI, K, L, M, N, O, P, R, S, T, U, F, H, Ts, Ch, Sh, Shch, , Y, , E1, Yu, Ya, determine which pairs are homeomorphic.
Imagine each as a 1D subset of and draw them accordingly before solving the problem.
4. Prove that homeomorphisms yield an equivalence relation on the collection of all topological spaces.
5. What is the dimension of the C-space for a cylindrical rod that can translate and rotate in ? If the rod is rotated about its central axis, it is assumed that the rod's position and orientation
are not changed in any detectable way. Express the C-space of the rod in terms of a Cartesian product of simpler spaces (such as , , , , etc.). What is your reasoning?
6. Let be a loop path that traverses the unit circle in the plane, defined as . Let be another loop path: . This path traverses an ellipse that is centered at . Show that and are homotopic (by
constructing a continuous function with an additional parameter that ``morphs'' into ).
7. Prove that homotopy yields an equivalence relation on the set of all paths from some to some , in which and may be chosen arbitrarily.
8. Determine the C-space for a spacecraft that can translate and rotate in a 2D Asteroids-style video game. The sides of the screen are identified. The top and bottom are also identified. There are
no ``twists'' in the identifications.
9. Repeat the derivation of from Section 4.3.3, but instead consider Type VE contacts.
10. Determine the C-space for a car that drives around on a huge sphere (such as the earth with no mountains or oceans). Assume the sphere is big enough so that its curvature may be neglected (e.g.,
the car rests flatly on the earth without wobbling). [Hint: It is not .]
11. Suppose that and are each defined as equilateral triangles, with coordinates , , and . Determine the C-space obstacle. Specify the coordinates of all of its vertices and indicate the
corresponding contact type for each edge.
12. Show that (4.20) is a valid rotation matrix for all unit quaternions.
13. Show that , the set of polynomials over a field with variables , is a group with respect to addition.
14. Quaternions:
1. Define a unit quaternion that expresses a rotation of around the axis given by the vector .
2. Define a unit quaternion that expresses a rotation of around the axis given by the vector .
3. Suppose the rotation represented by is performed, followed by the rotation represented by . This combination of rotations can be represented as a single rotation around an axis given by a
vector. Find this axis and the angle of rotation about this axis.
15. What topological space is contributed to the C-space by a spherical joint that achieves any orientation except the identity?
16. Suppose five polyhedral bodies float freely in a 3D world. They are each capable of rotating and translating. If these are treated as ``one'' composite robot, what is the topology of the
resulting C-space (assume that the bodies are not attached to each other)? What is its dimension?
17. Suppose a goal region is defined in the C-space by requiring that the entire robot is contained in . For example, a car may have to be parked entirely within a space in a parking lot.
1. Give a definition of that is similar to (4.34) but pertains to containment of inside of .
2. For the case in which and are convex and polygonal, develop an algorithm for efficiently computing .
18. Figure 4.29a shows the Möbius band defined by identification of sides of the unit square. Imagine that scissors are used to cut the band along the two dashed lines. Describe the resulting
topological space. Is it a manifold? Explain.
Figure 4.29: (a) What topological space is obtained after slicing the Möbius band? (b) Is a manifold obtained after tearing holes out of the plane?
19. Consider Figure 4.29b, which shows the set of points in that are remaining after a closed disc of radius with center is removed for every value of such that and are both integers.
1. Is the remaining set of points a manifold? Explain.
2. Now remove discs of radius instead of . Is a manifold obtained?
3. Finally, remove disks of radius . Is a manifold obtained?
20. Show that the solution curves shown in Figure 4.26 correctly illustrate the variety given in (4.73).
21. Find the number of faces of for a cube and regular tetrahedron, assuming is . How many faces of each contact type are obtained?
22. Following the analysis matrix subgroups from Section 4.2, determine the dimension of , the group of rotation matrices. Can you characterize this topological space?
23. Suppose that a kinematic chain of spherical joints is given. Show how to use (4.20) as the rotation part in each homogeneous transformation matrix, as opposed to using the DH parameterization.
Explain why using (4.20) would be preferable for motion planning applications.
24. Suppose that the constraint that is held to position is imposed on the mechanism shown in Figure 3.29. Using complex numbers to represent rotation, express this constraint using polynomial
25. The Tangle toy is made of pieces of macaroni-shaped joints that are attached together to form a loop. Each attachment between joints forms a revolute joint. Each link is a curved tube that
extends around of a circle. What is the dimension of the variety that results from maintaining the loop? What is its configuration space (accounting for internal degrees of freedom), assuming the
toy can be placed anywhere in ?
26. Computing C-space obstacles:
1. Implement the algorithm from Section 4.3.2 to construct a convex, polygonal C-space obstacle.
2. Now allow the robot to rotate in the plane. For any convex robot and obstacle, compute the orientations at which the C-space obstacle fundamentally changes due to different Type EV and Type
VE contacts becoming active.
3. Animate the changing C-space obstacle by using the robot orientation as the time axis in the animation.
27. Consider ``straight-line'' paths that start at the origin (lower left corner) of the manifolds shown in Figure 4.5 and leave at a particular angle, which is input to the program. The lines must
respect identifications; thus, as the line hits the edge of the square, it may continue onward. Study the conditions under which the lines fill the entire space versus forming a finite pattern
(i.e., a segment, stripes, or a tiling).
Steven M LaValle 2020-08-14 | {"url":"https://lavalle.pl/planning/node180.html","timestamp":"2024-11-03T07:30:22Z","content_type":"text/html","content_length":"22847","record_id":"<urn:uuid:ce2ff8ab-61c0-4e7b-8b04-89f35ef88586>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00462.warc.gz"} |
Same Symmetry:
Ruth Goodman Walt Disney Magnet School
4140 North Marine Drive
Chicago IL 60613
(312) 534-5841
Grade 5
(1) Students will develop an understanding of symmetry and a sense of geometry.
(2) Students will identify lines of symmetry in a geometric figure.
(3) Students will see that elementary geometry should be informal and allow for
exploration of ideas.
(4) Students will learn about the history of quilts.
(5) Students will create a symmetrical quilt.
Materials Needed:
(1) Scissors (5) Cuisenaire rods
(2) Crayons (6) Mirrors
(3) Tape (7) Two large poster boards
(4) Graph paper (8) Playdough
(1) Teacher will show objects to the class and explain that their structure
relates to a mathematical term in geometry.
(2) Teacher will review the terms congruent and similar.
(3) Students will look for symmetrical objects in the room with vertical or
horizontal lines.
(4) Pass out playdough and paper with geometric shapes.
(5) Students will make geometric shapes with the playdough and show lines of
symmetry on the paper and the playdough shapes.
(6) Pass out mirrors and papers for mirror symmetry.
(7) Explain mirror symmetry and complete paper.
(8) Pass out cuisenaire rods and graph paper.
(9) Using the rods and graph paper, students will make a design on half of the
graph and then color the other side exactly like the rods.
(10) Teacher will talk about the history of quilts and show a book.
(11) Then in their groups, the students will each design, name, and color their
quilt pattern.
Performance Assessment:
(1) Students will locate pictures from magazines that demonstrate bilateral and
radial symmetry.
(2) Students will design a quilt pattern with two lines of symmetry.
Platts, Mary E, Intermediate Mathematics Challenge. Educational Service, Inc.
Stevensville, Michigan, 1975.
Roth, Susan L. and Phang,Ruth., Patchwork Tales. Atheneum. New York, 1984.
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Manzana [Argentina] to Feddan
Manzana [Argentina] [manzana] Output
1 manzana [argentina] in ankanam is equal to 1494.99
1 manzana [argentina] in aana is equal to 314.5
1 manzana [argentina] in acre is equal to 2.47
1 manzana [argentina] in arpent is equal to 2.92
1 manzana [argentina] in are is equal to 100
1 manzana [argentina] in barn is equal to 1e+32
1 manzana [argentina] in bigha [assam] is equal to 7.47
1 manzana [argentina] in bigha [west bengal] is equal to 7.47
1 manzana [argentina] in bigha [uttar pradesh] is equal to 3.99
1 manzana [argentina] in bigha [madhya pradesh] is equal to 8.97
1 manzana [argentina] in bigha [rajasthan] is equal to 3.95
1 manzana [argentina] in bigha [bihar] is equal to 3.95
1 manzana [argentina] in bigha [gujrat] is equal to 6.18
1 manzana [argentina] in bigha [himachal pradesh] is equal to 12.36
1 manzana [argentina] in bigha [nepal] is equal to 1.48
1 manzana [argentina] in biswa [uttar pradesh] is equal to 79.73
1 manzana [argentina] in bovate is equal to 0.16666666666667
1 manzana [argentina] in bunder is equal to 1
1 manzana [argentina] in caballeria is equal to 0.022222222222222
1 manzana [argentina] in caballeria [cuba] is equal to 0.07451564828614
1 manzana [argentina] in caballeria [spain] is equal to 0.025
1 manzana [argentina] in carreau is equal to 0.77519379844961
1 manzana [argentina] in carucate is equal to 0.020576131687243
1 manzana [argentina] in cawnie is equal to 1.85
1 manzana [argentina] in cent is equal to 247.11
1 manzana [argentina] in centiare is equal to 10000
1 manzana [argentina] in circular foot is equal to 137050.24
1 manzana [argentina] in circular inch is equal to 19735234.59
1 manzana [argentina] in cong is equal to 10
1 manzana [argentina] in cover is equal to 3.71
1 manzana [argentina] in cuerda is equal to 2.54
1 manzana [argentina] in chatak is equal to 2391.98
1 manzana [argentina] in decimal is equal to 247.11
1 manzana [argentina] in dekare is equal to 10
1 manzana [argentina] in dismil is equal to 247.11
1 manzana [argentina] in dhur [tripura] is equal to 29899.75
1 manzana [argentina] in dhur [nepal] is equal to 590.61
1 manzana [argentina] in dunam is equal to 10
1 manzana [argentina] in drone is equal to 0.3893196765303
1 manzana [argentina] in fanega is equal to 1.56
1 manzana [argentina] in farthingdale is equal to 9.88
1 manzana [argentina] in feddan is equal to 2.4
1 manzana [argentina] in ganda is equal to 124.58
1 manzana [argentina] in gaj is equal to 11959.9
1 manzana [argentina] in gajam is equal to 11959.9
1 manzana [argentina] in guntha is equal to 98.84
1 manzana [argentina] in ghumaon is equal to 2.47
1 manzana [argentina] in ground is equal to 44.85
1 manzana [argentina] in hacienda is equal to 0.00011160714285714
1 manzana [argentina] in hectare is equal to 1
1 manzana [argentina] in hide is equal to 0.020576131687243
1 manzana [argentina] in hout is equal to 7.04
1 manzana [argentina] in hundred is equal to 0.00020576131687243
1 manzana [argentina] in jerib is equal to 4.95
1 manzana [argentina] in jutro is equal to 1.74
1 manzana [argentina] in katha [bangladesh] is equal to 149.5
1 manzana [argentina] in kanal is equal to 19.77
1 manzana [argentina] in kani is equal to 6.23
1 manzana [argentina] in kara is equal to 498.33
1 manzana [argentina] in kappland is equal to 64.83
1 manzana [argentina] in killa is equal to 2.47
1 manzana [argentina] in kranta is equal to 1494.99
1 manzana [argentina] in kuli is equal to 747.49
1 manzana [argentina] in kuncham is equal to 24.71
1 manzana [argentina] in lecha is equal to 747.49
1 manzana [argentina] in labor is equal to 0.013950025009895
1 manzana [argentina] in legua is equal to 0.0005580010003958
1 manzana [argentina] in manzana [costa rica] is equal to 1.43
1 manzana [argentina] in marla is equal to 395.37
1 manzana [argentina] in morgen [germany] is equal to 4
1 manzana [argentina] in morgen [south africa] is equal to 1.17
1 manzana [argentina] in mu is equal to 15
1 manzana [argentina] in murabba is equal to 0.09884206520611
1 manzana [argentina] in mutthi is equal to 797.33
1 manzana [argentina] in ngarn is equal to 25
1 manzana [argentina] in nali is equal to 49.83
1 manzana [argentina] in oxgang is equal to 0.16666666666667
1 manzana [argentina] in paisa is equal to 1258.05
1 manzana [argentina] in perche is equal to 292.49
1 manzana [argentina] in parappu is equal to 39.54
1 manzana [argentina] in pyong is equal to 3024.8
1 manzana [argentina] in rai is equal to 6.25
1 manzana [argentina] in rood is equal to 9.88
1 manzana [argentina] in ropani is equal to 19.66
1 manzana [argentina] in satak is equal to 247.11
1 manzana [argentina] in section is equal to 0.0038610215854245
1 manzana [argentina] in sitio is equal to 0.00055555555555556
1 manzana [argentina] in square is equal to 1076.39
1 manzana [argentina] in square angstrom is equal to 1e+24
1 manzana [argentina] in square astronomical units is equal to 4.4683704831421e-19
1 manzana [argentina] in square attometer is equal to 1e+40
1 manzana [argentina] in square bicron is equal to 1e+28
1 manzana [argentina] in square centimeter is equal to 100000000
1 manzana [argentina] in square chain is equal to 24.71
1 manzana [argentina] in square cubit is equal to 47839.6
1 manzana [argentina] in square decimeter is equal to 1000000
1 manzana [argentina] in square dekameter is equal to 100
1 manzana [argentina] in square digit is equal to 27555610.67
1 manzana [argentina] in square exameter is equal to 1e-32
1 manzana [argentina] in square fathom is equal to 2989.98
1 manzana [argentina] in square femtometer is equal to 1e+34
1 manzana [argentina] in square fermi is equal to 1e+34
1 manzana [argentina] in square feet is equal to 107639.1
1 manzana [argentina] in square furlong is equal to 0.24710516301528
1 manzana [argentina] in square gigameter is equal to 1e-14
1 manzana [argentina] in square hectometer is equal to 1
1 manzana [argentina] in square inch is equal to 15500031
1 manzana [argentina] in square league is equal to 0.0004290006866585
1 manzana [argentina] in square light year is equal to 1.1172498908139e-28
1 manzana [argentina] in square kilometer is equal to 0.01
1 manzana [argentina] in square megameter is equal to 1e-8
1 manzana [argentina] in square meter is equal to 10000
1 manzana [argentina] in square microinch is equal to 15500017326603000000
1 manzana [argentina] in square micrometer is equal to 10000000000000000
1 manzana [argentina] in square micromicron is equal to 1e+28
1 manzana [argentina] in square micron is equal to 10000000000000000
1 manzana [argentina] in square mil is equal to 15500031000062
1 manzana [argentina] in square mile is equal to 0.0038610215854245
1 manzana [argentina] in square millimeter is equal to 10000000000
1 manzana [argentina] in square nanometer is equal to 1e+22
1 manzana [argentina] in square nautical league is equal to 0.00032394816622014
1 manzana [argentina] in square nautical mile is equal to 0.0029155309240537
1 manzana [argentina] in square paris foot is equal to 94786.73
1 manzana [argentina] in square parsec is equal to 1.0502647575668e-29
1 manzana [argentina] in perch is equal to 395.37
1 manzana [argentina] in square perche is equal to 195.8
1 manzana [argentina] in square petameter is equal to 1e-26
1 manzana [argentina] in square picometer is equal to 1e+28
1 manzana [argentina] in square pole is equal to 395.37
1 manzana [argentina] in square rod is equal to 395.37
1 manzana [argentina] in square terameter is equal to 1e-20
1 manzana [argentina] in square thou is equal to 15500031000062
1 manzana [argentina] in square yard is equal to 11959.9
1 manzana [argentina] in square yoctometer is equal to 1e+52
1 manzana [argentina] in square yottameter is equal to 1e-44
1 manzana [argentina] in stang is equal to 3.69
1 manzana [argentina] in stremma is equal to 10
1 manzana [argentina] in sarsai is equal to 3558.32
1 manzana [argentina] in tarea is equal to 15.9
1 manzana [argentina] in tatami is equal to 6049.97
1 manzana [argentina] in tonde land is equal to 1.81
1 manzana [argentina] in tsubo is equal to 3024.99
1 manzana [argentina] in township is equal to 0.00010725050478094
1 manzana [argentina] in tunnland is equal to 2.03
1 manzana [argentina] in vaar is equal to 11959.9
1 manzana [argentina] in virgate is equal to 0.083333333333333
1 manzana [argentina] in veli is equal to 1.25
1 manzana [argentina] in pari is equal to 0.98842152586866
1 manzana [argentina] in sangam is equal to 3.95
1 manzana [argentina] in kottah [bangladesh] is equal to 149.5
1 manzana [argentina] in gunta is equal to 98.84
1 manzana [argentina] in point is equal to 247.11
1 manzana [argentina] in lourak is equal to 1.98
1 manzana [argentina] in loukhai is equal to 7.91
1 manzana [argentina] in loushal is equal to 15.81
1 manzana [argentina] in tong is equal to 31.63
1 manzana [argentina] in kuzhi is equal to 747.49
1 manzana [argentina] in chadara is equal to 1076.39
1 manzana [argentina] in veesam is equal to 11959.9
1 manzana [argentina] in lacham is equal to 39.54
1 manzana [argentina] in katha [nepal] is equal to 29.53
1 manzana [argentina] in katha [assam] is equal to 37.37
1 manzana [argentina] in katha [bihar] is equal to 79.09
1 manzana [argentina] in dhur [bihar] is equal to 1581.76
1 manzana [argentina] in dhurki is equal to 31635.3 | {"url":"https://hextobinary.com/unit/area/from/manzanaarg/to/feddan","timestamp":"2024-11-09T22:17:49Z","content_type":"text/html","content_length":"129056","record_id":"<urn:uuid:48153878-d031-420b-b029-30308fcf7ba1>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.10/warc/CC-MAIN-20241109214337-20241110004337-00084.warc.gz"} |
Six Discs
Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?
Six circular discs, all of radius 1 unit, are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box.
• Without using any calculations, compare the areas of the bases of boxes A and B and say which is the larger.
• Use estimation to list the packings in order of increasing base area. Which are the hardest to put in order?
• Now calculate the area of the base of each box and compare the correct order with your estimate.
Getting Started
What would happen if you superimposed B on A? What about D on E?
These diagrams show some right-angled triangles you can use to work out the lengths on shapes B and D using trigonometry. What angles do you know? What sides do you know?
Student Solutions
Hee Chan and Jeang Bin sent us their working with correct areas for shapes A and C:
The longer part of the rectangle is $4$ because each circle has radius $1$ unit.
The right-angled triangle with side length marked $x$ has angles $30^{\circ}, 60^{\circ}$ and $90^{\circ}$ so its hypotenuse must be $2$, and side $x$ is $\sqrt{3}$ according to Pythagoras's law.
So the total length of one side of the triangle is $4+2\sqrt{3}$.
The area of the equilateral triangle is
$$\frac{1}{2}(4+2\sqrt{3})^2 \times \sin 60^{\circ}$$ $$=\frac{1}{2}(4+2\sqrt{3})^2 \times\frac{\sqrt{3}}{2}$$ $$=7\sqrt{3}+12$$
which is $24.12$ square units to two decimal places.
For C, for the bottom there are 3 circles and this means that the length of the bottom is 6 units. Also for the height there are 2 circles so the length is 4 units. The area is $4 \times 6 = 24$
square units.
In a similar problem Nicola Spittal, S4, Madras College, St Andrew's sent in this excellent solution, demonstrating that A must have a larger than B and also calculating the correct areas for shapes
A, B and C.
By superimposing the triangular packing (A) on the parallelogram packing (B) as shown below we can see that $area(\Delta QRS) = area(\Delta STU)$ so that (A) has larger area than (B) and the
difference in areas is equal to the area of the small triangle $\Delta UVW$.
We use many 30-60-90 triangles with sides in the ratio $1:2:\sqrt {3}$. The areas of the packings are as follows.
Packing (A)
The triangle has side length $4+2\sqrt{3}$ so that area is ${1\over 2}(4+2\sqrt{3})^2 \sin 60^\circ$, or (equivalently, using half base times height) ${1\over 2}(4+2\sqrt{3})(3+2\sqrt{3})$, which is
24.12 square units to 2 decimal places.
Packing (B)
The base length is given by $PW=\tan 60^\circ +4+\tan 30^\circ = \sqrt{3}+ 4 + {1\over \sqrt{3}}= 6.3094.$ The height is given by $h = 2+2\cos 30^\circ = 2 + \sqrt{3}= 3.7321$, so that the area is
23.55 square units to 2 decimal places.
Packing (C)
As the rectangle is $6\times 4$, the area is 24 square units.
Here is one method of finding the area of the remaining shapes:
Packing (D)
$h = \sqrt{3} + 1$ (from diagram)
$\cos 30^\circ = \frac{h}{a} \Rightarrow a = \frac{h}{\cos 30^\circ} = 2(\frac{\sqrt{3}}{3} + 1)$
Splitting the hexagon in to two trapezia, the area is therefore:
$A_D = 2\{\frac{1}{2}(a + 2a)h\} = \frac{3}{2}ah = \frac{3}{2}(\frac{\sqrt{3}}{3}+1)(\sqrt{3}+1) = 8\sqrt{3}+12 = 25.86 \textrm{ (4 s.f.)}$
Packing (E)
Area composed of triangle, width $2h$ height $\frac{a}{2}$, and rectangle, width $2h$ height $2+\frac{a}{2}$
$\therefore \ A_E = \frac{1}{2}(2h)(\frac{a}{2}) + (2h)(2+\frac{a}{2}) = \frac{3}{2}ah + 4h = 8\sqrt{3} + 10 = 23.86 \textrm{ (4 s.f.)}$
Packing (F)
$A_F = 5(2h) = 10h = 10(\sqrt{3}+1) = 27.32 \textrm{ (4 s.f.)}$
Teachers' Resources
Why do this problem?
This problem
allows learners to see the value of estimating before making accurate calculations, and to see that sometimes an estimate is all that is needed. The problem also offers an opportunity to practise
calculating areas and working out lengths accurately using trigonometry.
Possible approach
Show the image of the six boxes and explain that we're interested in comparing the areas of different boxes made to hold six circular discs. Ask learners to compare the areas of A and B, and allow
everyone time to consider and then discuss in pairs which is bigger and why.
In discussing A and B, key ideas to consider include comparing the parts of the shapes which are the same, or comparing the size of the gaps between circles. Next, allow the class some time to
discuss in pairs or small groups how to order all six shapes. Stress that the importance is not so much in the order the learners come up with but in their reasons for placing them in that order.
After allowing some time to work on estimating, suggest to the learners that some calculations may help them to put the shapes in order. Learners may decide they do not need to work out every area to
be certain of the correct order - for example, if they are certain their estimate has identified the biggest or the smallest area they may choose not to calculate that one. They could then work in
small groups to create a poster or presentation showing the correct order for the shapes and the justification and calculations they used to find it. The hint contains two diagrams which suggest an
approach for working out the areas using trigonometry, which could be shared with the class if appropriate.
Key questions
Which shape do you think has the largest area and why?
What angles can we work out? What lengths do we know?
For which shapes do you think you need to work out the areas in order to be certain you had ordered the shapes correctly?
Possible support
Some of the calculations for the exact areas require knowledge of trigonometry, but the problem could be tackled instead by constructing accurate diagrams of the boxes and measuring in order to
calculate their areas. | {"url":"http://nrich.maths.org/problems/six-discs","timestamp":"2024-11-05T13:20:42Z","content_type":"text/html","content_length":"48062","record_id":"<urn:uuid:fc945dac-7875-4351-b2fa-43716f8aa227>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00859.warc.gz"} |
Options Delta Explained: Sensitivity To Price
Trading Ideas 22-04-2023 17:24 36 Views
Options Delta Explained: Sensitivity To Price
Options Delta Explained
For example, should a stock option price increase in price by 0.5c with a 1c increase in the underlying stock price then the option has a delta of 0.5.
Another way of looking at delta is as the probability of the option expiring in the money.
Options Delta Math
It's not necessary to understand the math behind delta (please feel free to go to the next section if you want), but for those interested delta is defined more formally as the partial derivative of
options price with respect to underlying stock price.
The formula is below (some knowledge of the normal distribution is required to understand it).
Delta is superficially the most intuitive of the options greeks. Even the newest beginner would expect the price of an option, giving the right to buy or sell a particularly security, to change with
the security’s price.
Let’s look at an example with call options on a stock with $120 stock price as it rises higher (by $10 to $130, say).
In the money options – those with a strike price less than $120 – would become even more in the money. Thus their value to the holder would increase – the probability of them remaining in the money
would be higher – and hence, all other things being equal, the option price would rise.
Out of the money and at the money options – those with an exercise price of $120 or greater – would also rise in value. The probability of, say, a $140 option expiring in the money would be higher if
the stock price was $130 compared to $120. Hence its value would be higher.
Similar arguments can be used with put options: their value rises/falls with the fall/rise of the underlying (the only difference being put options have negative delta versus call options, whose
delta is positive).
But the extent of this sensitivity – i.e. delta – and how it relates to expiration length, price, and volatility is quite subtle. Let’s look at it in more detail.
How Does Options Delta Change Over Time?
The effect of time on delta depends on an option’s ‘moneyness’.
In the money
All other things being equal, long dated in the money options have a lower delta than shorter dated ones.
In the money options have both intrinsic (stock price less exercise price) and extrinsic value.
As time progresses the extrinsic reduces (due to theta) and the intrinsic value (which moves in line with stock price) becomes more dominant. And so the option moves more in line with the stock, and
hence its delta rises towards 1 over time.
Out of the money
All other things being equal, short dated OTM/ATM options have a lower delta than longer dated ones.
A short dated out of the money option (especially one which is significantly OTM) is unlikely to expire in the money, a fact that is unlikely to change with a 1c change in price. Hence its delta is
Longer dated OTM options are more likely to expire in the money – there is a longer time for the option to move ITM – and hence their value do move with stock price. Hence their delta is higher.
At the money
There is no effect of time on the delta of an at the money option.
How Does Options Delta Change With Implied Volatility?
Again the effect of implied volatility changes on delta depends on moneyness.
In The Money
As we saw above in the money options’ value comprise both intrinsic and extrinsic amounts.
In general the higher the proportion of an option’s value that is intrinsic (which moves exactly in line with stock price) and extrinsic value (which doesn’t), the higher its delta.
Increases in IV increase the extrinsic value of an option and so, as intrinsic value isn’t affected by implied volatility, increases the percentage of the option’s value that is extrinsic. This
resultant reduction in the intrinsic value as a proportion of the whole, reduces the option’s delta as above.
Out Of The Money
Out of the money options have only extrinsic value, which is driven by the probability of it expiring in the money.
A higher volatility suggests there is a greater chance of the option expiring ITM (as the stock is expected to move around more) and hence delta increases.
At the money
ATM options have a delta of approx. 0.5, which is unchanged as volatility changes.
Effect Of Changes Of Price On Delta
One of the other subtleties of delta is that it in itself changes value as the underlying security’s price changes.
The extent to which this occurs is another of the options greeks: gamma. This is the change in delta resulting in in a 1c change in stock price.
Gamma for long options holders is positive whereas it is negative for short positions, meaning it helps the former and penalises the latter. It is also at its highest absolute value near expiration.
(See here for more discussion on gamma).
Delta is an important greek as it reflects an option holder’s exposure to one of the main variables: the price of the underlying security.
Whilst one of the easiest option concepts to understand, its behavior resulting from changes to other variables such as time, IV and underlying price is more complex.
It is vital for an options trader to understand these concepts.
About the Author: Chris Young has a mathematics degree and 18 years finance experience. Chris is British by background but has worked in the US and lately in Australia. His interest in options was
first aroused by the ‘Trading Options’ section of the Financial Times (of London). He decided to bring this knowledge to a wider audience and founded Epsilon Options in 2012. | {"url":"https://profitablemarkettips.com/2023/04/22/options-delta-explained-sensitivity-to-price/","timestamp":"2024-11-06T05:34:00Z","content_type":"text/html","content_length":"66408","record_id":"<urn:uuid:312f368e-b9ba-4463-b3d5-8c5c108b0a06>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00525.warc.gz"} |
3 Digit Subtraction Worksheets - 15 Worksheets.com
3 Digit Subtraction Worksheets
About These 15 Worksheets
Triple Digit Subtraction Worksheets are tools designed to assist students in mastering the concept of subtracting three-digit numbers. These worksheets present numerous subtraction problems that
involve numbers in the hundreds place value. They are used to reinforce the principles of subtraction and help you develop proficiency and speed in solving triple-digit subtraction problems.
Before diving into triple digit subtraction, there are several skills you should already have under your belt. Firstly, you need to understand basic subtraction with smaller numbers, such as
single-digit or double-digit numbers. These foundational subtraction skills are very important because they are the building blocks of more complex subtraction problems.
Additionally, you should understand the concept of “place value”. Place values tell us what each digit in a number represents. In a three-digit number, there are hundreds, tens, and ones. For
instance, in the number 123, the ‘1’ stands for one hundred, ‘2’ stands for twenty, and ‘3’ stands for three ones.
You also need to know about ‘borrowing’ or ‘regrouping’. When you subtract larger numbers and the top number (the minuend) is smaller than the bottom number (the subtrahend), you need to ‘borrow’
from the next place value. For example, if you’re trying to solve 301 – 145 and you start from the ones place, you can’t subtract 5 from 1. So you ‘borrow’ 1 from the tens place, turning it into 10,
and the ‘0’ in the tens place becomes a ‘9’. Now, you can subtract 5 from 11.
Triple digit subtraction worksheets provide a structured and step-by-step way to practice and learn subtraction. The repetition and variety of problems on these worksheets give you many opportunities
to practice and reinforce your understanding. By solving these problems, you gradually increase your comfort level with triple-digit subtraction.
These worksheets also help improve your problem-solving skills. Each subtraction problem is like a small puzzle that needs to be solved. You need to identify whether borrowing is necessary, perform
the subtraction for each place value, and remember to reduce the value of the digit from which you borrowed.
Working with these worksheets can also enhance your number sense, which is your understanding of how numbers work and relate to each other. This is developed as you continue to subtract large
numbers, carry over, and borrow. For instance, when you borrow from a higher place value, you’re using your number sense to understand that you’re not really taking away value, but just rearranging
it to make the subtraction possible.
The consistency of practicing with worksheets helps in developing speed and accuracy. As you work through more and more problems, you’ll start noticing that you’re able to solve them faster, and with
fewer mistakes. This is especially important because as you advance in your studies, you’ll need to solve problems quickly and accurately, often under timed conditions.
How to Solve Triple Digit Subtraction Problems
Solving triple-digit subtraction problems can seem challenging at first, but once you understand the steps and get some practice, it will become a lot easier. Here’s a step-by-step guide on how to
solve these problems:
Step 1) Line Up The Numbers Correctly
The first step is to ensure that the numbers are correctly lined up. Write the larger number (the minuend) on top and the smaller number (the subtrahend) below it. Make sure the ones, tens, and
hundreds place values line up vertically.
Step 2) Start Subtracting from The Ones Place
Start subtracting from the right-most digit, which is the ones place. Subtract the lower digit from the upper one.
Step 3) Check if You Need to Borrow
If the lower digit is larger than the upper one, you need to borrow from the next column (the tens place). If you can subtract without needing to borrow, move on to the next step.
Step 4) Borrow if Necessary
To borrow, reduce the digit in the tens place of the upper number by one, and add ten to the ones place of the upper number. Now, subtract the lower ones digit from the new upper ones digit.
Step 5) Subtract the Tens Place
Move to the next column, which is the tens place. Subtract the lower digit from the upper one. If the upper digit is smaller, you’ll need to borrow from the hundreds place.
Step 6) Borrow for the Tens Place if Necessary
Similar to step 4, to borrow, reduce the digit in the hundreds place of the upper number by one and add ten to the tens place of the upper number. Now, subtract the lower tens digit from the new
upper tens digit.
Step 7) Subtract the Hundreds Place
Finally, move to the hundreds place and subtract the lower digit from the upper one. If your numbers were in the thousands and you needed to borrow, you would follow the same procedure.
Step 8) Write Down Your Answer
After subtracting the hundreds place, write down your answer. This should be a three-digit number.
Step 9) Check Your Work
Use addition to check your work. Add your answer to the lower number (the subtrahend). You should get the upper number (the minuend) if your subtraction is correct. | {"url":"https://15worksheets.com/worksheet-category/3-digit-subtraction/","timestamp":"2024-11-14T04:33:37Z","content_type":"text/html","content_length":"131497","record_id":"<urn:uuid:c7e36180-c395-4159-9be7-a1ac34ced1b5>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00540.warc.gz"} |
Numerical Simulation of Hydrogen Leakage from Fuel Cell Vehicle in an Outdoor Parking Garage
School of Automotive Studies, Tongji University, Shanghai 201804, China
Clean Energy Automotive Engineering Center, Tongji University, Shanghai 201804, China
Author to whom correspondence should be addressed.
Submission received: 13 July 2021 / Revised: 9 August 2021 / Accepted: 10 August 2021 / Published: 12 August 2021
It is significant to assess the hydrogen safety of fuel cell vehicles (FCVs) in parking garages with a rapidly increased number of FCVs. In the present work, a Flame Acceleration Simulator (FLACS), a
computational fluid dynamics (CFD) module using finite element calculation, was utilized to predict the dispersion process of flammable hydrogen clouds, which was performed by hydrogen leakage from a
fuel cell vehicle in an outdoor parking garage. The effect of leakage diameter (2 mm, 3 mm, and 4 mm) and parking configurations (vertical and parallel parking) on the formation of flammable clouds
with a range of 4–75% by volume was considered. The emission was assumed to be directed downwards from a Thermally Activated Pressure Relief Device (TPRD) of a 70 MPa storage tank. The results show
that the 0.7 m parking space stipulated by the current regulations is less than the safety space of fuel cell vehicles. Compared with a vertical parking configuration, it is safer to park FCVs in
parallel. It was also shown that release through a large TPRD orifice should be avoided, as the proportion of the larger hydrogen concentration in the whole flammable domain is prone to more
accidental severe consequences, such as overpressure.
1. Introduction
Hydrogen safety in diverse situations is the most critical aspect of FCVs as hydrogen has high flammability (4–75% by volume) [
] and low ignition energy and is prone to leakage. The United States has invested 5–10% of the total funding of the hydrogen program into safety research [
]. Compressed hydrogen is typically stored under high pressure (35 MPa for buses and 70 MPa for cars) in storage tanks fitted with TPRD to release hydrogen, avoiding tank rupture when the ambient
temperature reaches 110 ℃, melting the TPRD sensing element. The phenomenon of unignited hydrogen release can occur once TPRD fails. The possible subsequent deflagration and detonation events will
not be discussed in this paper, which only focuses on leakage and dispersion. Some relevant safety studies have been performed using Computational Fluid Dynamics (CFD) tools to reveal the accidental
risks in various scenarios, such as around fuel cell vehicles [
], in tunnels [
], and in enclosed areas [
]. Some researchers have experimentally investigated hydrogen behavior by transporting hydrogen in semi-closed or confined structures [
]. For security reasons, helium has been widely applied as an alternative experimental gas for the prediction of hydrogen behavior in many studies, since helium has similar physical properties to
hydrogen [
By necessity, FCVs must be parked in vehicle garages, tunnels, etc. One of the most hazardous scenarios is hydrogen leakage from a high-pressure storage tank placed on the chassis underneath the
vehicle in an outside parking garage since the semi-closed space formed by adjacent automobiles contributes to hydrogen accumulation. Hajji, Y. et al. [
] experimentally studied the effects of residential-garage geometry, shape, and number of vents on hydrogen concentration and delamination. They concluded that rectangular vents are most suitable for
a prismatic garage, and the number of vents is critical to reducing hydrogen concentration than the shape. CFD methods have also been performed to numerically evaluate the behavior of hydrogen
dispersion so that emergency measures can be taken immediately to reduce the hydrogen concentration under a lower flammable limit (LFL) [
]. H. Hussein et al. [
] numerically assessed the impact of various conditions by varying leakage orifice, direction, and angle of flammable hydrogen cloud. It was concluded that a larger orifice contributes to a massive
flow rate, leading to a severe pressure-peaking phenomenon. The permeated hydrogen dispersion with the permeation rate of 1 Nml/h/L and 45 Nml/h/L from a high-pressure storage tank in a typical
garage was studied analytically by Saffers, J.-B. et al. [
]. Hydrogen diffused and accumulated uniformly upward toward the ceiling after the leakage, and the concentration reached quasi-steady at 60 s and 12 s for 1 Nml/h/L and 45 Nml/h/L permeation rate,
respectively. In addition, the detection of hydrogen dispersion is of critical importance in confined garage-like spaces, so Zhao, M. et al. [
] developed a localization technology for safe monitoring of large parking garages, and the model’s accuracy can be improved by learning more training data.
The aforementioned research mainly focuses on hydrogen release and dispersion in confined spaces such as underground garages. The behavior of hydrogen dispersion in semi-closed spaces has not been
investigated specifically. In addition, the release flow rate was typically assumed to be constant. However, the mass rate of hydrogen release from hydrogen storage tanks through TPRD decreases with
decreased internal pressure. The objective of this paper was to investigate the hydrogen dispersion for an outdoor parking garage model in various scenarios by varying TPRD orifice and parking
configurations. The Computational Fluid Dynamics (CFD) tool FLACS was utilized to simulate the cases and analyze the diffusion phenomenon based on the spatial and temporal evolution of the flammable
2. Numerical Simulation
2.1. FLACS-Hydrogen Code
The grid-based resolution in FLACS, unlike other commercial simulation tools, relies on the so-called porosity/distributed resistance (PDR), where sub-grid geometry is represented as area and volume
porosities (denoting the degree of “openness” for each grid cell), instead of resolving individual obstacles by a grid. Moen, A. et al. [
] conducted a comparative study of k-ε models in impinging hydrogen jet dispersion scenarios using the CFD code FLACS. The simulation results were compared with the Schlieren photographs from
experiments in Reference [
Figure 1
shows the two-dimensional simulation results of comparing three turbulent models (standard, RNG, and realizable k-ε). It can be summarized from a comparison results that the standard k-ε model
exhibits the best performance regardless of whether high- or low-momentum hydrogen releases are used. Thus, this paper applied the standard k-ε model with additional turbulence generation terms to
solve turbulent kinetic energy (Equation (1)) and the dissipation of turbulent kinetic energy (Equation (2)). Following the Boussinesq eddy viscosity assumption, an eddy viscosity models the Reynolds
stress tensor as Equation (3). Boundary conditions were defined as
(free outflow) on all sides. [
$∂ ∂ t ( β v ρ k ) + ∂ ∂ x j ( β j ρ u j k ) = ∂ ∂ x j ( β j μ e f f σ k ∂ k ∂ x j ) + β v P k − β v ρ ε$
$∂ ∂ t ( β v ρ ε ) + ∂ ∂ x j ( β j ρ u j ε ) = ∂ ∂ x j ( β j μ e f f σ k ∂ ε ∂ x j ) + β v P ε − C 2 ε β v ρ ε 2 k$
$− ρ u i ″ u j ″ ˜ = μ e f f ( ∂ u i ˜ ∂ x j + ∂ u j ˜ ∂ x i ) − ρ 2 3 k δ i j$
is the volume porosity of the geometry,
is the area porosity in the
is the density,
$u j ˜$
is the mean velocity in the
is the turbulent kinetic energy, ε is the dissipation of turbulent kinetic energy,
is the effective viscosity,
μ[eff] = μ + μ[t], μ[t]
is the dynamic turbulent viscosity, σ is the Prandtl–Schmidt number,
= 1.0,
σ[ε] =
is the production of turbulent kinetic energy,
is the production of dissipation of turbulent kinetic energy,
is a model constant in transportation equation for dissipation with the default value of 1.92.
is the Kronecker delta function,
= 1 if
= 0 if
i ≠ j
, and
$u i ″ u j ″ ˜$
is the mean velocity in the
The leaked hydrogen through the TPRD orifice from a high-pressure storage tank is modeled as an under-expanded jet. A pseudo-source approach was applied to calculate the parameters of an
under-expanded jet at the status where the pressure is atmosphere in terms of reservoir pressure, temperature, orifice diameter, and density.
Table 1
presents the summary description of the jet source model in FLACS, where
is the effective nozzle area, γ is the isentropic ratio,
is the specific heat at constant pressure,
is the ambient pressure, and
$m ˙ 1$
is the mass flow rate. More detailed information about the model can be found in References [
2.2. Geometry Configuration and Grids
In the present study, we considered two parking configurations based on the relative positions of the car body to the aisle.
Figure 2
shows the configuration and dimension of the computational domain. Considering fire safety, a certain fire prevention distance was reserved between parking spaces to prevent flame propagation between
vehicles. In the first configuration (
Figure 2
a), four vehicles, separated by 0.7 m [
], were arranged in vertically parking where the car body was perpendicular to the aisle. In the second configuration, three vehicles shown in
Figure 2
c were assumed to be parked parallel to the aisle where the spacing was set to 1.3 m in view of easy access for vehicles. The green geometry represents the curb.
FLACS uses a Cartesian grid arrangement to solve the governing equations using a finite volume method [
]. The grid scope contains three regions: central core domain, stretched domain, and refinement domain. The fuel cell vehicle model established in the simulation was based on EUNIQ7 with dimensions
of 5 m × 2.2 m × 1.8 m, while the overall size of the total computational region was approximately 32 m in length, 25 m in width, and 75 m in height, high enough that the boundary had little effect
on hydrogen diffusion considering the high buoyancy of hydrogen. As shown in
Figure 2
a,c, the central core region covers the vehicle where the leak position was located, and the normal uniform grid size was set to 0.5 m; the default stretch factor of 1.2 was applied to establish the
stretched region which wraps around the central core region to simulate the hydrogen diffusion in the far field. Furthermore, the mesh was further refined around the leakage point, forming the
refinement region.
Since the simulation progress was conducted using transient numerical calculation, the vehicle model is simplified to minimize the calculation time and save calculation resources. Due to hydrogen
leakage and dispersion occurring externally to the vehicle, many elements of the vehicle geometry were assumed to have little effect on hydrogen dispersion, so some internal components, such as
seats, steering wheel, instrument panel, brake, and accelerator pedals were ignored. Consequently, the remaining parts contained a car body modeled as an entity without any pores, three hydrogen
tanks mounted under the chassis, and connecting pipes.
2.3. Grid Independency Validation
Two additional grid dimensions, 0.3 m and 0.7 m, were used to perform the grid independency validation. As shown in
Figure 3
, Monitor Point 1 (MP1) was created 0.15 m below the leakage position, and the other MPs (MP2 and MP3) were similarly set at the same height underneath the adjacent vehicles. The hydrogen mole
fraction with time was chosen as the monitoring parameter for evaluating grid independency under various grid sizes. It can be seen from
Figure 3
that the general behavior of hydrogen diffusion at all grid sizes is generally consistent, while the values were in better agreement when the simulations were conducted under the core-domain grid
size of 0.3 m or 0.5 m. The hydrogen mole fraction simulated under a grid size of 0.7 m showed a larger deviation. Apparently, no further significant inconsistency exists when the grid dimension is
smaller than 0.5 m, and therefore grid size of 0.5 m was utilized in this paper.
2.4. Determination of Hydrogen Leakage Rate
TPRD valve was assumed to be triggered, and the empty rate was taken as the input condition in this paper. The hydrogen leak originating from a 70 MPa high-pressure storage tank at a temperature of
20 °C was shown in
Figure 2
. The release orifice of the TPRD was set to 2 mm, 3 mm, and 4 mm in diameter, which is typically used in current fuel cell vehicles with the leak oriented vertically downward as the default release
direction. Six scenarios were considered by varying the TPRD orifice and parking configurations. The time-dependent leakage rate of TPRD from 55 L hydrogen storage tank was calculated using equations
mentioned in
Section 2.1
. Six scenarios were considered, varying TPRD diameters and parking configurations, which are listed in
Table 2
. The initial leakage rate was 0.126, 0.283, 0.428 kg/s when the orifice diameter was 2 mm, 3 mm, and 4 mm, respectively, and then attenuates exponentially with time. Correspondingly, the total
leakage time (166, 70, and 44 s for the leakage orifice of 2, 3, and 4 mm), defined as the time interval when hydrogen storage pressure relief from 70 MPa to atmospheric pressure, decreases with
increased orifice diameter. For the hydrogen release from the 4.2 mm orifice, 171 L storage tank at 35 MPa, the blowdown time is less than 110 s (around 108 s) [
]. For accuracy comparison, the total leakage time was calculated as 102.5 s under the same conditions in FLACS. The deviation rate was less than 7%. In addition, HyRAM, a software toolkit
integrating validated science and engineering models and data relevant to hydrogen safety, contains a validated engineering toolkit that can be applied to predict physical effects, including the
empty time of the high-pressure storage tanks [
]. The leakage time and deviation rate calculated by HyRAM were presented in column 6 of
Table 2
3. Results and Discussions
The legend located at the right of the figure represents hydrogen concentration ranging from 0.04 to 0.4 by volume.
Figure 4
shows the distribution of flammable hydrogen gas cloud in six scenarios varying TPRD diameters and parking configurations. The explanation of the uppercase letters in each subgragh were presented in
Table 2
. As shown in
Figure 4
a, at 2 s after leakage, higher hydrogen concentrations with a range of 0.3–0.6 by volume are observed clearly for all conditions underneath the leaked vehicle, where the leakage position is located.
Alcock et al. [
] recommended that the widest detonability limit of hydrogen in air is 0.11–0.59 by volume. The results reveal that the entire domain under the fuel cell vehicle has already become explosion hazard
areas; therefore, hydrogen sensors need to be installed under the chassis close to the TPRD vent pipe to detect any hydrogen leakage and raising an alarm in advance to help personals take appropriate
emergency measures.
It is obvious that hydrogen diffuses faster along the width direction of the car body than in the length direction after leakage for two different parking configurations, thus revealing the
advantages of parallel parking over vertical parking. The hydrogen concentration value underneath 2–3 vehicles adjacent to the leakage source can reach 0.15–0.35 by volume in vertical parking, while
in parallel parking, this value is only 0.04–0.15 by volume. The adiabatic premixed flame temperature of hydrogen with a stoichiometric mixture in the air can reach up to 2403 K [
]. If ignition occurs, the flame will spread along with the premixed hydrogen gas cloud to the adjacent vehicles, meaning that the personnel will have little time to escape in a vertical parking
configuration. On the other hand, the body along the X direction (hydrogen diffuses faster) is much closer to the obstacles, such as walls or steps, when parking parallel, contributing to hydrogen
accumulation in narrow space. However, only a small amount of hydrogen extends to the front and rear of the compartment; thus, the hydrogen concentration is low, so the flame will propagate to the
aisle and have little effect on surrounding vehicles even if ignition happens.
The results of
Figure 4
b reveal that the coverage area of hydrogen mole fraction between 0.2 and 0.4 shrinks under the combined action of decreased leakage rate and high diffusion rate of hydrogen, which is conducive to a
fast propagation of hydrogen with a mole fraction between 0.04 and 0.2. The hydrogen flammable mass keeps increasing, although the leakage rate is decreasing, until the peak time when the mass
attains the maximum in each case is reached. Nevertheless, the flammable cloud gradually dissipates after exceeding the equilibrium point, as the higher buoyancy is dominant in the later period of
leakage. The proportion of the larger hydrogen concentration of 0.2–0.4 by volume in the whole flammable domain is greater for larger TPRD orifice at the same leakage time, indicating that the
accidental risk will be more unacceptable once ignition happens, such as the overpressure.
4. Conclusions
Unignited hydrogen release through a Thermally Activated Pressure Relief Device (TPRD) from onboard hydrogen storage tanks in an outdoor parking garage has been studied in the present numerical work.
The results indicate that further research of more scenarios, such as ignited releases, should be conducted to improve the safety demands for fuel cell vehicles.
Simulations were carried out in an outdoor parking garage with a computational region of 32 m in length, 25 m in width, and 75 m in height. The release scenario assumed that hydrogen leaked through
TPRD from a 70 MPa hydrogen storage tank with a hydrogen mass of 2.5 kg. The mass flow rate was assumed to decrease with decreased internal pressure of the storage tank, which are different from the
constant value selected in other research. Six release cases varying between three leakage orifices (2 mm, 3 mm, 4 mm) and two parking configurations (parallel parking, vertical parking) were
As expected, higher hydrogen concentrations within detonation limits were clearly observed for all cases in the vicinity of the leakage position at the beginning of the release. The flammable cloud
diffuses fast under the combined action of decreased leakage rate and high diffusion rate of hydrogen, which was conducive to a fast propagation of hydrogen with a mole fraction between 0.04 and 0.2.
The flammable cloud gradually dissipates in the later period of leakage, as the higher buoyancy is dominant. Downward release of hydrogen pushed the flammable gas diffusion around the vehicle. The
coverage of flammable cloud indicates that the parking space between vehicles was not safe enough. These factors should be considered in the design of the parking space for hydrogen safety.
Author Contributions
Conceptualization, Y.S. and T.Z.; methodology, software, validation, Y.S.; formal analysis, H.L.; investigation, resources, data curation, writing—original draft preparation, writing—review and
editing, visualization, supervision, project administration, Y.S., T.Z., H.L., W.Z. and C.Z.; funding acquisition, H.L. All authors have read and agreed to the published version of the manuscript.
This research was funded by Key Technologies Research and Development Program (CN), grant number 2020YFB1506205; and Scientific and Innovative Action Plan of Shanghai, grant number 18DZ1201900.
Conflicts of Interest
The authors declare no conflict of interest. The funders had no role in the design of the study; in the collection, analyses, or interpretation of data; in the writing of the manuscript, or in the
decision to publish the results.
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Figure 1.
2D plots showing impinging hydrogen-jet concentration predictions from three turbulent models (Standard, RNG, and Realizable k-ε). Reproduced with permission [
]. Copyright 2019, Elsvier.
Figure 2. Model of the outside parking garage; (a) vertical parking configuration; (b) leakage position in vertical parking configuration; (c) parallel parking configuration; (d) leakage position in
parallel parking configuration.
Figure 4. Hydrogen mole fraction between 0.04 and 0.4 by volume for downward releases from 700 bar through 2 mm, 3 mm, and 4 mm in parallel (A–C) and vertical (D–F) parkingconfiguration after 2 s, 6
s, 10 s leakage.
Initial Reservoir Condition Nozzle Conditions Jet Conditions
Pressure: p[0] Effective nozzle area: A[1] Velocity: $u 2 = u 1 + p 1 − p 2 ρ 1 u 1$
Temperature: T[0] Temperature: Enthalpy: $h 2 = h 1 + 1 2 ( u 1 2 − u 2 2 )$
$T 1 = T 0 ( 2 / ( γ + 1 ) )$
Volume: V[0] Pressure: Temperature: $T 2 = T 1 + 1 2 u 1 2 − u 2 2 c p$
$p 1 = p 1 ( T 1 / T 0 ) γ / ( γ − 1 )$
Density: $ρ 0 = p 0 R T 0$ Density: $ρ 1 = p 1 R T 1$ Pressure: $p 2 = p a$
Total mass: $m 0 = ρ 0 V 0$ Sound speed: $c 1 = γ R T 1$ Density: $ρ 2 = p 2 R T 2$
Heat exchange coefficient: h[wall] Velocity: $u 1 = c 1$ Effective outlet area: $A 2 = A 1 ρ 1 u 1 ρ 2 u 2$
Enthalpy: $h 1 = c p T 1$ Mass flow: $m ˙ 1 = ρ 1 u 1 A 1$
Mass flow: $m ˙ 1 = ρ 1 u 1 A 1$
Table 2. Scenarios considered varying release diameter and parking configurations for unignited hydrogen release.
Case Number Parking Configuration Release Diameter (mm) Initial Hydrogen Leakage Rate (kg/s) Leakage Time (s) by HyRAM(s)
A Parallel Parking 2 0.126 166 164 (1.2%)
B Parallel Parking 3 0.283 70.5 72.9 (3.3%)
C Parallel Parking 4 0.428 44 41 (6.8%)
D Vertical Parking 2 0.126 166 164 (1.2%)
E Vertical Parking 3 0.283 70.5 72.9 (3.3%)
F Vertical Parking 4 0.428 44 41 (6.8%)
G — 4.2 mm, 171 L at 35 MPa 102.5 100.66 (1.8%)
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Shen, Y.; Zheng, T.; Lv, H.; Zhou, W.; Zhang, C. Numerical Simulation of Hydrogen Leakage from Fuel Cell Vehicle in an Outdoor Parking Garage. World Electr. Veh. J. 2021, 12, 118. https://doi.org/
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Shen, Yahao, Tao Zheng, Hong Lv, Wei Zhou, and Cunman Zhang. 2021. "Numerical Simulation of Hydrogen Leakage from Fuel Cell Vehicle in an Outdoor Parking Garage" World Electric Vehicle Journal 12,
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How to efficiently fill a time series?
My general problem is that I have a dataframe where columns correspond to feature values. There is also a date column in the dataframe. Each feature column may have missing NaN values. I want to fill
a column with some fill logic such as "fill_mean" or "fill zero".
But I do not want to just apply the fill logic to the whole column because if one of the earlier values is a NaN, I do not want the average i fill for this specific NaN to be tainted by what the
average was later on, when the model should have no knowledge about. Essentially it's the common problem of not leaking information about the future to your model - specifically when trying to fill
my time series.
Anyway, I have simplified my problem to a few lines of code. This is my simplified attempt at the above general problem:
#assume ts_values is a time series where the first value in the list is the oldest value and the last value in the list is the most recent.
ts_values = [17.0, np.NaN, 12.0, np.NaN, 18.0]
nan_inds = np.argwhere(np.isnan(ts_values))
for nan_ind in nan_inds:
nan_ind_value = nan_ind[0]
ts_values[nan_ind_value] = np.mean(ts_values[0:nan_ind_value])
The output of the above script is:
[17.0, 17.0, 12.0, 15.333333333333334, 18.0]
which is exactly what I would expect.
My only issue with this is that it will be linear time with respect to the number of NaNs in the data set. Is there a way to do this in constant or log time where I don't iterate through the nan
index values.
When you want to interpolate the series you can use the pandas directly.
>>> s = pd.Series([0, 1, np.nan, 5])
>>> s
0 0.0
1 1.0
2 NaN
3 5.0
dtype: float64
>>> s.interpolate()
0 0.0
1 1.0
2 3.0
3 5.0
dtype: float64
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20 Concerning Complements - Logic Philosophy Spirituality20 Concerning Complements
20 Concerning Complements
THE LOGIC OF CAUSATION
Phase Three: Software Assisted Analysis
Chapter20–Concerning Complements.
To fully understand partial and contingent causation, we need to return to the issue of complementary causes. In chapter 2.3, where these concepts were first introduced, we showed how any number of
complements can be reduced to just two. I wish to here review this important doctrine and further develop it with reference to matricial analysis.
Why this doctrine is important is worth reiterating here. Remember, we originally defined partial and/or contingent causation with reference to only two complementary causes (say, P and Q) for a
certain effect (say, R). Focusing on one of the complements (say, P) as the first item and ‘main’ cause (or the cause mainly of interest to us in a given context), the other complement (here, Q)
could be regarded as embodying the ‘surrounding conditions’ for the partial and/or contingent causation of the third item (R). That is, even though the second item Q is in our definition presented as
a single item, it is intended to signify any number of ‘surrounding conditions’ Q1, Q2, Q3, etc.
The question arises: what is the exact relation between the underlying numerous surrounding conditions, and their single representative or stand-in term Q? Obviously we mean Q to be a putative
‘collective effect’ of Q1, Q2, Q3… – we conceive of a causative relation between the numerous underlying causes and their single representative. Indeed,very often we invent a new termQ to stand in
for a number of terms Q1, Q2, Q3…, viewing Q as an ‘abstraction’ implied by its constituent items Q1, Q2, Q3, etc. WedefineQ by saying: “let that be the collective effect of Q1, Q2, Q3, etc.”
This is indeedone of the usual avenues of concept formation. In other words, Q need not be something concretely observed in isolation, but may be an abstraction producedad hocto facilitate a
causative statement. It may, however, thenceforth acquire a life of its own in our discourse. For example, the definition of force as mass times acceleration tells us that the abstraction called
‘force’ can be calculated by measuring the mass of the physical body concerned and the acceleration the force is assumed to effect in it, and multiplying the two quantities together; we do not
observe and measure ‘force’ as such, but derive it from more concrete items. But henceforth, force becomes an oft-used term in other equations, as if it was directly experienced.
Thus, a complementary item such as Q may be viewed as itself the effect of a partial and/or contingent causation whose causes are, say, Q1 and Q2. If there are more than these two underlying items,
then one of them (say Q2) may in turn be viewed as the product of two deeper items, say Q3 and Q4. And so on, successively, till all the relevant surrounding conditions are exhausted. Conversely, any
number of ‘underlying items’ or ‘surrounding conditions’ can, by successive mergers of two into one, be represented by just one overall representative item Q.
This is at least true theoretically, because this is not really how we proceed in practice. In practice, we rather think more globally, as already described in chapter 2.3 (see there for more
details). This is the thought process and inductive method of ‘changing one thing at a time, while keeping all other things equal, and observing the effect of that single change’. This can be
described in more formal terms as follows for partial causation:
│If (Q1 + Q2 + Q3 + …), then Q; │
│If (notQ1 + Q2 + Q3 + …), not-then Q; │
│If (Q1 + notQ2 + Q3 + …), not-then Q; │
│If (Q1 + Q2 + notQ3 + …), not-then Q; │
│Etc. (as per number of factors involved); │
│And (Q1 + Q2 + Q3 + …) is indeed possible.│
For contingent causation, similar clauses can be used, with the polarities of all the terms involved must be reversed. Thus, the first clause would be: ‘If (notQ1 + notQ2 + notQ3 + …), then notQ’,
and so on. That is to say: where Q is a partial cause, with P of R, then Q1, Q2, Q3, etc. are in turn partial causes of Q; and where Q is a contingent cause, with P of R, then Q1, Q2, Q3, etc. are in
turn contingent causes of Q. In cases where both partial and contingent causation are involved, Q is the collective effect of the underlying items Q1, Q2, Q3, etc., on both the positive and negative
We can also describe such multiple weak causations by means of nesting; this is of course the meaning of the successive reductions of many conditions to just one mentioned above. Nesting of ‘if (Q1 +
Q2 + Q3 + …), then Q’ would have the form ‘if (Q1 + (Q2 + (Q3 + …))), then Q’; and similarly for the other clauses. But to repeat, the nesting approach complicates matters perhaps unduly, and is here
mentioned just to promote theoretical understanding rather than as a practical means.
Of course, nothing forces us to limit ourselves to just two complementary causes, other than the limitations in our computer (hardware and software) resources. A large organization, such as WHO (the
World Health Organization), for which I worked for some years long ago, does I think have the computer, personnel and financial resources to investigate any number of complementary causes of any
health factor or disease, social problem or solution, or whatever.
Note that nothing is said or intended here with regard to the relative part played by the various complements in causing the effect concerned. Thequantitativerole of these factors is not being
examined here, onlythe fact thatthey are factors in the causation. They may be very widely different in their degrees of involvement; one complement may be the major determinant, while the other(s)
is/are minor factors, or they may all be more or less on an equal footing[1]. I leave aside the issue of proportion, here – without, however, intending to deny its great importance. It is, rightly,
regarded as crucial in modern science, but our study here is only concerned with the ‘whether’, not the ‘how much’.
While on the topic of complementary causes, a question worth asking is: what of interconnected complements – i.e. can the two complements be causatively related at all, or are they always independent
of each other? That is, given that P, Q are partial and/or contingent causes of R, does it follow that P, Q are unconnected (i.e. compatible every which way, as explained in the previous chapter) –
or may they in some cases be connected?
This question can be answered by looking at the 3-item moduses found in relative partial and contingent causation and prevention (90 moduses, all told), and seeing whether any of them signify some
implication between the complements P, Q and/or their negations. It is found that in 32 cases, one or the other of the four possible implications are involved (i.e. 8 cases for each, symmetrically),
the remaining 58 cases signifying only that the various conjunctions of the complements and their negations are just contingent. The list of cases concerned may be seen in Table 20.1, posted at the
Table 20.1 – 3-item PQR Moduses of Forms – Dependent Complements. (2 pages in pdf file).
Let us take one of these cases for further examination, say modus 23 (ditto for 31, 55, 63). This concerns relative contingent prevention, i.e. “P with complement Q is a contingent cause of notR”,
whose clauses are “if notP and notQ then R; if P and notQ, not-then R; if notP and Q, not-then R; and notP and notQ is possible”. Now, according to our table, modus 23 also corresponds to “P and Q is
impossible (i.e. if P, then notQ)”. As can be seen, these two propositions are not in conflict; meaning that the relation of dependence between P and Q does not impinge on their stated causative
relation to notR (which nowhere mentions “P and Q”). Obviously, this is not the sort of dependence we are looking for; we seek implications between the complements that affect the causative relation
to the third item somewhat.
Similarly, modus 42 (ditto for 46, 58, 62), which refers to relative contingent causation, i.e. “P with complement Q is a contingent cause of R” and to “P implies notQ” has no notable impact. And
likewise,mutadis mutandis, in six other sets of four moduses (I won’t bother listing them; their first ones are highlighted on the said table). So in fact, by this simple method, we find no
significant dependence between the complements, i.e. one having a differential impact on the relative causation or prevention concerned.
Obviously, when such generic relative causations or preventions are conjoined to strong determinations, nothing is changed, since the latter involve only two items (e.g. P and R). However, what
happens when they are conjoined to each other (when compatible, of course)? We know that this option (i.e.pqrel to Q or notQ, for R and notR) concerns only four 3-item moduses altogether, viz. 127,
190, 220, 232; and these, as our table shows, are not among the 32 relevant moduses; therefore, no significant impact arises here, either.
Thus, to conclude, although conjunctions of the complements (P, Q, notP, notQ) are not always possible, the cases where they are impossible do not affect the causative relation concerned. Note that
this only concerns implication and does not exclude the possibility that the complements might have a weaker causative relation mediated by some additional item. This issue might be further
investigated using the 4-item matrix, but I will not attempt it here, having shown how the job can be done. There is no real need, for this investigation is moved merely by curiosity – since all the
valid moduses of forms generated by matricial analysis are thereby known to involve no internal inconsistency.
Another, more valuable, investigation I wish to launch here is into the features of exclusive causation. With regard to the strong determinations, this would take the forms: “If and only if P, then
R” or “If and only if notP, then notR”. If we think about them, we realize these both mean “If P, then R; and if notP, then notR” – i.e.mn. Nothing new here, since we have already studied the
properties ofmnin considerable detail.
I would just take this opportunity to remind readers of the danger of ambiguity when we say: “Only if P, R” or “Only if notP, notR” (notice my removal of the words “if and” and “then”). Though
statements of this sort often signify exclusive strong causation in the sense just defined (i.e. “if and only if –, then –), often what is intended is much weaker, namely: “If P, possibly R”; and if
notP, necessarily notR” and “If notP, possibly notR; and if P, necessarily R”, respectively. In such cases, we are only informing that the consequent R (or notR, as the case may be) is only possible
with the antecedent P (or notP) – but we are not claiming that P brings about R (or notP brings about notR). For this reason, it is wise to use the more precise wording (which modern logicians
abbreviate to “iff –, then –”[2]).
Let us now turn our attention to the weak determinations, and ask what is meant by “If and only if P and Q, then R” or “If and only if notP and notQ, then notR”, which forms we will respectively
label aspex andqex (orp[ex]andq[ex]) – the suffix ‘ex’ standing for exclusive, of course. Note my use here of the harder “iff” sort of exclusion just explained; also, to avoid all ambiguity, note
that our intent here is to apply this operator to the conjunction (P and Q) or (notP and notQ), and not merely to the first mentioned complement (i.e. to P or notP). Thus, what I have in mind is,
roughly put, the following propositions:
7. Exclusive partial causationby P and Q of R symbol:pex
If (P + Q), then R ((P + Q) + notR) is impossible
if not(P + Q), then notR (etc.) (not(P + Q) + R) is impossible
8. Exclusive contingent causationby P and Q of R symbol:qex
If (notP + notQ), then notR ((notP + notQ) + R) is impossible
if not(notP + notQ), then R (etc.) (not(notP + notQ) + notR) is possible
I have (for our purposes here) numbered these forms 7 and 8, to indicate continuation of the list given in Chapter 19.1. They are necessarily ‘relative’ (i.e. have at least 3 items); they do not have
‘absolute’ (2-item) versions. Needless to say, the number of complements involved in them need not only be two; any number might be considered, but we shall here focus our investigation on just two
complements as usual, so that we can refer to 3-item matricial analysis to answer questions that arise.
Clearly, the second clauses of forms 7 and 8 can each be expanded into three clauses, as can be proved by means of syllogisms using the clause not(P + Q) or not(notP + notQ) as our middle thesis as
the case may be. Furthermore, though we do not mention this above, each implication in causation has a base (i.e. the possibility that the three terms it mentions be conjoined). Thus, each of the
above two forms could have been defined more precisely and usefully with reference to eight clauses, as follows:
7. Exclusive partial causationby P and Q of R symbol:pex => not-8
a) If (P + Q), then R (P + Q + notR) is impossible <=> 8(d)
b) if (notP + Q), then notR (notP + Q + R) is impossible => not-8(i)
c) if (P + notQ), then notR (P + notQ + R) is impossible => not-8(h)
d) if (notP + notQ), then notR (notP + notQ + R) is impossible <=> 8(a)
g) (P + Q) is possible (P + Q + R) is possible <=> 8(j)
h) (notP + Q) is possible (notP + Q + notR) is possible => not-8(c)
i) (P + notQ) is possible (P + notQ + notR) is possible => not-8(b)
j) (notP + notQ) is possible (notP + notQ + notR) is possible <=> 8(g)
8. Exclusive contingent causationby P and Q of R symbol:qex => not-7
a) If (notP + notQ), then notR (notP + notQ + R) is impossible <=> 7(d)
b) if (P + not Q), then R (P + notQ + notR) is impossible => not-7(i)
c) if (notP + Q), then R (notP + Q + notR) is impossible => not-7(h)
d) if (P + Q), then R (P + Q + notR) is impossible <=> 7(a)
g) (notP + notQ) is possible (notP + notQ + notR) is possible <=> 7(j)
h) (P + notQ) is possible (P + notQ + R) is possible => not-7(c)
i) (notP + Q) is possible (notP + Q + R) is possible => not-7(b)
j) (P + Q) is possible (P + Q + R) is possible <=> 7(g)
Now, these definitions show us that of the eight possible combinations of P, Q, R and their negations, four combinations are impossible and four others are possible, in each form. We see that,
although some clauses are identical in both the formspex andqex, there are serious conflicts between them; namely, clauses b and c of each are incompatible with clauses i, h respectively of the
other. Thus, these two forms are contrary and can never be conjoined aspqexfor the same items PQR. This is a reasonable result, the essence of the formspex andqex being that they mimic
complete-necessary causation (mn), with reference to more than two items; they are thus intermediate degrees of causation, behaving somewhat like strongs and somewhat like weaks.
Let us now compare these forms to their predecessors, listed in Chapter 19.1. We see that, as we would expect,pex is incompatible withm(see 1a and 7i) andqrel (see 4b and 7c, 4c and 7b), but
compatible withnandprel (they have no conflicting clauses). Indeed,pex impliesnsince (7b + 7d) = 2a[3], 7g => 2b, and 7h or 7j => 2c; andpex also impliesprelsince 7a = 3a, 7h = 3b, 7i = 3c, and 7g =
3d. Whence it follows thatpex implies the joint formnprel. Similarly,qex is incompatible withnandprel, but compatible withmandqrel.Indeed,qex impliesmandqrel, i.e.qex impliesmqrel.
Can we now prove the converse, i.e. thatnprel impliespex and thatmqrel impliesqex? The answer is no! The clauses 7c and 7j cannot be drawn fromnprel; and similarly, the clauses 8c and 8j cannot be
drawn frommqrel. Therefore, the formspex andqex are in fact stronger determinations than the specific formsnprel andmqrel (and not just stronger than the generic formsprel andqrel).
The next obvious question is: what are the oppositions between these various forms when the items concerned are given different polarities? This is best investigated more mechanically by means of
matricial analysis. The results are given in the following table, which can be viewed at the website:
Table 20.2 – 3-item PQR Moduses of Forms – Exclusive Weak Causations. (4 pages in pdf file).
The results of this table are interesting, since they show that each of these exclusive forms yields only one modus. Thus, the above mentioned two initial formspex andqex have respectively modus #s
150 and 170. This is comparable in degree of specificity to the single modus ofpqrel. The full list of forms and their corresponding moduses can be summarized as follows:
Summary of Table 20.2 – Moduses of exclusive weak forms.
Main exclusive forms modus
exclusive partial causation (pex) PQR 150
exclusive contingent causation (qex) PQR 170
exclusive partial causation (pex) PnotQR 102
exclusive contingent causation (qex) PnotQR 167
exclusive partial causation (pex) PQnotR (prevention PQR) 107
exclusive contingent causation (qex) PQnotR (prevention PQR) 87
exclusive partial causation (pex) PnotQnotR (prevention PnotQR) 155
exclusive contingent causation (qex) PnotQnotR (prevention PnotQR) 90
Inverse exclusive forms modus
exclusive partial causation (pex) notPnotQnotR 170
exclusive contingent causation (qex) notPnotQnotR 150
exclusive partial causation (pex) notPQnotR 167
exclusive contingent causation (qex) notPQnotR 102
exclusive partial causation (pex) notPnotQR (prevention notPnotQnotR) 87
exclusive contingent causation (qex) notPnotQR (prevention notPnotQnotR) 107
exclusive partial causation (pex) notPQR (prevention notPQnotR) 90
exclusive contingent causation (qex) notPQR (prevention notPQnotR) 155
Notice that the inverses have the same items with opposite polarities; and that the modus of their formpex becomes that ofqex, and vice versa. Now, the fact that each of these exclusive forms is
expressive of only one modus should be useful for working out their oppositions and interpretations. For a start, we note that all 8 main forms are contrary to each other, since they have no modus in
common; the inverses are of course their respective equivalents, with the already stated changes.
For the rest, our macroanalysis above seems after all to suffice; microanalysis adds nothing much more. For instance, regarding modus 150 we see, in Table 18.6, pages 3-4, that it is one of four
moduses (the others being 149, 181, 182) which mean “nprel to Q (andn-alonerel to notQ only) in causation”. This does not mean that modus 150 is identical to the other three, but only that it has
this common implication, i.e.nprel (etc.) which we knew already (save for the implied lone). What this does tell us, however, is that our interpretations of the moduses thus far were somewhat
lacking, since they reveal no difference between the more restrictive exclusive forms (like #150)) and their more ordinary cousins (viz. #s 149, 181, 182, in this example). This shows that our
introduction of these additional specifications was useful and important.
Upon reflection, we should have expected the exclusive forms to be represented by only one modus, since they are defined by eight clauses! Indeed, any modus could be represented in words by eight
clauses concerning the possibility or impossibility of each combination of the three items and their negations. The peculiarity of the exclusive forms is that they do this succinctly and are
popularly used.
The important thing to note in the first section of the present chapter is that our 3-item format of partial and/or contingent causative propositions was from the startintended to cover all eventual
numbers of complements. We have not used it as merely the simplest, most accessible, format – but asan all-inclusive format, to which all other weak causations can in principle be reduced when
necessary. Thus, our investigation into the logic of causation with reference to only one complement Q (to P in the causation of R) was not intended to be supplemented later by consideration of more
and more complements. Three items were supposed to do the trick.
Why then do we need to consider a fourth (and even possibly a fifth) item, now, in phase III ? For the simple reason that, when we consider causative syllogism we must look into cases with the major
and/or the minor premise involving a complement. Since the minor, middle and major terms of our syllogism already take up three items (P, Q, R), we need an additional item S (and maybe even two of
them, S and T) to investigate syllogisms with one (or both) premises about relative weak causation.[4]
Note well that eventual 4-item (or even 5-item)syllogismsare all composed of 3-itempropositions(at least, as regards their premises, though some conclusions may conceivably involve four items). A
syllogism requires at least three terms (the major, the middle and then minor) deployed in two premises (the major and minor premises, which share the middle term) and a conclusion (which relates the
major and minor terms). This allows for only two terms per proposition. If one (or both) of the premises has a third term (i.e. a complement of weak causation), then the syllogism will have four (or
respectively, five) terms. The conclusion will then be expected to have a third (and even fourth) term.
Based on past experience with syllogistic reasoning, we certainly need at least one additional item, the fourth (or subsidiary) term S; for we can well expect a weak premise combined with a strong
one to yield a weak conclusion. Regarding a possible fifth item T, it is probable that we do not need one, because it is unlikely that two weak premises can yield any conclusion at all; but this must
of course be formally established in some way (and I doubt any way other than microanalysis can do the trick).
The introduction of a fourth item (S) means dealing with a grand matrix of 65,536 moduses each of which is defined by 16 digits; this is in the realm of the possible given my current computer
resources (hardware and software capabilities) – just about. But these material resources are quite insufficient to deal with a fifth item (T), which would require a grand matrix of 2^32=
4,294,967,296 moduses of 2^5= 32 digits each; therefore I can only speculate about the probable results of a study of the latter.
More will be said concerning the fourth item S in the next chapter, when we consider 4-item syllogisms.
[1]Eventual variations in proportions in time and/or space should, I think, be considered as due to more phenomenal underlying factors. For example, the ‘age’ of an organism may be a causal factor;
but the significance of aging at the cellular level or deeper would have to be investigated to understand why ‘time’ seems to play a role.
[2]This valuable word, “iff”, has unfortunately not passed over into general usage. The reason for that is, I think, obvious: it is a word that is distinguishable in written language, but not in
spoken language.
[3]This is easily established by dilemmatic argument: given “if (notP + Q), then notR” and “if (notP + notQ), then notR”, the conclusion is “if notP, then notR”.
[4]The subsidiary term (S) is mentioned in phases I and II in the following places: chapters 5.3 and 9.4, where the various possible subfigures of the syllogism are tabulated; chapters 14.3 and 15.1,
where it is stressed that the syllogisms here developed are not 4-item ones – i.e. that their full elucidation requires 4-item research; chapter 16.1, where the problem and the way to the solution of
4-item syllogism are presented.
Avi Sion2023-01-05T11:39:06+02:00 | {"url":"https://thelogician.net/LOGIC-OF-CAUSATION/Concerning-Complements-20.htm","timestamp":"2024-11-09T18:39:04Z","content_type":"text/html","content_length":"170470","record_id":"<urn:uuid:32d8ada4-1aef-43b5-b09d-6a6de08689c2>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00002.warc.gz"} |
pytorch matrix multiplication accuracy depends on tensor size
PyTorch Matrix Multiplication: Why Size Matters
When performing matrix multiplication in PyTorch, you might encounter a curious phenomenon: the accuracy of the result can vary depending on the size of your tensors. This isn't a bug, but a
consequence of the way PyTorch implements its operations, particularly when dealing with floating-point numbers. Let's dive into the details and explore how to mitigate this effect.
The Problem:
Let's say you're working with two matrices, A and B, and performing matrix multiplication using torch.matmul(A, B). You might find that the result isn't exactly the same when you increase the size of
your matrices.
import torch
# Smaller matrices
A_small = torch.randn(10, 10, dtype=torch.float32)
B_small = torch.randn(10, 10, dtype=torch.float32)
result_small = torch.matmul(A_small, B_small)
# Larger matrices
A_large = torch.randn(1000, 1000, dtype=torch.float32)
B_large = torch.randn(1000, 1000, dtype=torch.float32)
result_large = torch.matmul(A_large, B_large)
# Check for differences
print(torch.allclose(result_small, result_large)) # Output: False
The Root of the Issue:
The issue stems from the limitations of floating-point arithmetic. Floating-point numbers have finite precision, leading to rounding errors during calculations. These errors accumulate with each
operation. When dealing with larger matrices, there are more operations involved, resulting in a larger accumulation of rounding errors. This can manifest as seemingly different results even when the
input matrices are theoretically equivalent.
Mitigation Strategies:
1. Reduce Operations: If possible, try to refactor your code to minimize the number of matrix multiplications. This can help limit the accumulation of rounding errors.
2. Increase Precision: Consider using torch.float64 (double-precision floating-point) instead of torch.float32 (single-precision floating-point). While this will increase memory consumption, it can
lead to more accurate results, especially for larger matrices.
3. Use torch.allclose: Instead of directly comparing the output matrices using ==, use torch.allclose. This function accounts for floating-point precision by allowing a small tolerance for
4. Avoid Unnecessary Operations: Analyze your code for redundant calculations or operations that might contribute to rounding errors.
Practical Example:
Imagine you're training a neural network for image classification. During backpropagation, your weights are updated based on the gradients calculated from the loss function. With larger image sizes,
the gradients involve more calculations, potentially leading to increased rounding errors. This can affect the accuracy of your model, particularly if the differences in gradients accumulate over
many training epochs.
While floating-point rounding errors are inherent in numerical computations, understanding their impact is crucial for ensuring accurate and reliable results in PyTorch, especially when working with
large matrices. By implementing the strategies outlined above, you can mitigate these errors and maintain the integrity of your calculations.
Useful Resources: | {"url":"https://laganvalleydup.co.uk/post/pytorch-matrix-multiplication-accuracy-depends-on-tensor","timestamp":"2024-11-10T07:50:37Z","content_type":"text/html","content_length":"83193","record_id":"<urn:uuid:bfdb6908-b54b-4f30-bc2f-1267c672fd36>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00653.warc.gz"} |
Confidence Interval Formula Guide (How-To and Examples)
What if you had to make a decision that would impact your business revenue, but it was only based on an estimate? You'd want the most accurate estimate possible—or even better, a range of estimates
you could calculate with different confidence levels.
Accurate prediction is one of the true tests of statistical data. The ability to derive insight from a small sample size and accurately apply it at a much broader scale brings market research to a
whole new level.
When applying the statistical average of a small random sample to a larger population, the level of accuracy will vary depending on many factors. You can use a confidence interval formula to work out
how to express the accuracy of the statistical analysis when applied to a larger group.
Make research less tedious
Dovetail streamlines research to help you uncover and share actionable insights
Analyze with Dovetail
What exactly is a confidence interval?
While uncertainty is a fact of life, it isn't completely random, either. It's even possible to predict, with statistical accuracy, the likelihood of certain events by calculating them in a smaller
sample first. Of course, this information will only be accurate to a certain degree when applied to a larger group. The confidence interval expresses the estimate’s accuracy.
The confidence interval formula calculates the likelihood (or confidence) that a certain outcome, expressed as an upper and lower limit, will be true. Specifically, it's the probability that the data
from a small random sample accurately reflects similar predictions when applied to a larger sample size.
The final result will be an estimated mean, plus or minus a certain amount, that creates a wider or narrower range of expected values. While such an estimate can't be made with 100% certainty, it can
fall within the upper 90th percentile. That value range depends on how much certainty (or confidence) the researcher has in the value range.
More reliable predictions, with higher confidence levels, result in a wider range; a narrower range is more fallible but presents tighter estimates. Depending on the confidence level (or the
allowable margin of error), the following elements change:
• the predicted value range in a new, often larger, sample
• the accuracy of those predicted values
When do you use confidence intervals?
If you have a key decision to make that impacts your revenue, for example, you need to know how likely it is that your estimates are correct. Confidence intervals can tell you this. A well-executed
confidence interval formula is useful when you want to make decisions within a certain threshold of certainty.
In business, confidence intervals can accurately predict KPIs and demographic measurements that profits often depend on.
Here are some specific examples of when confidence intervals would come in handy:
• Marketers wanting to know how likely the results of a small ad campaign would result in a similar percentage of lead conversions when scaled up.
• Revenue teams interested in learning how accurately profits after investing resources in one market segment might translate to the same ROI in all other market segments.
• Product designers seeing promising UX statistics in a focus group suggesting a new feature is a hit, but they aren't sure if those results will translate to the wider population.
Why is the confidence interval formula important?
The confidence interval formula provides researchers with accurate predictions within a specified margin of error. It takes statistical analysis outside the bounds of small, time-limited samples,
allowing statisticians to apply known patterns to new, larger populations.
Use the confidence interval formula to take the average mean scores of a random sample and predict how accurately those conclusions can be applied to a larger sample size. There are other variables
to be aware of (such as standard deviation), so let’s pick apart and apply the confidence interval formula.
What is the confidence interval formula?
The confidence interval formula takes the sample mean (x̄), then performs a separate addition and subtraction of the product of the confidence level value (z) and the sample standard deviation (s)
after being divided by the square root of the sample size (√n):
CI = x̄ ± z (s / √n)
• CI = confidence interval, which will result in an upper and lower value range
• x̄ = sample mean, derived from the original sample
• z = confidence level value, expressed as an "alpha value" (a threshold representing a percentage, written as a decimal—the level of accuracy the final CI must have)
• s = sample standard deviation, a single figure showing the difference between the high and low values of the original sample
• n = sample size of the original sample
How to calculate the confidence interval
1. Find the sample mean (x̄). Average the score of all participants in the original sample. This is the same figure the confidence interval (CI) will use to calculate the probability of achieving the
same results in larger samples.
2. Calculate the standard deviation (s). A multi-step process, which entails subtracting the sample mean (x̄) from each individual score and squaring each separately
3. Find the z value for the preferred confidence level. The confidence level is typically 90% to 99%. The corresponding z value is a decimal figure, e.g. 1.645 for 90%, 2.576 for 99%.
4. Use these results in the formula. Place each of these figures into the formula to discover the confidence interval (CI) range.
5. Interpret your results. This is the range of figures you can expect with a larger sample size based on the level of certainty set by the confidence level value (z).
Remember that a larger CI range has a higher probability of being true, while a slimmer range carries a lower probability of being true. In other words, there's an inverse relationship between the
size of the confidence interval and the confidence level.
If the confidence interval is too wide, it may not be useful. Choose an acceptable margin of error for your purposes.
Examples of confidence interval calculations
Using an Excel template (we recommend using the first example, listed on sheet "1-Cl for m”), or a simpler alternative from WallStreetMojo, complete the following examples.
Confidence interval formula – example #1
1. Use the test scores 80, 75, 90, 80, 75, 75, 85, 80, 75, and 90 as the sample data. Add these numbers together and divide by 10 to get an average score, or sample mean, of 80.5.
2. Using this same data set, calculate the standard deviation by subtracting the sample mean from each score, squaring them (separately), adding the results together, then dividing it all by the
total number of scores (10). Starting from the beginning, it would be: (80 - 80.5)^2 + (75 – 80.5)^2, and so on, then all divided by 10. The result should be 32.25.
3. Decide on a confidence level (typically 90% to 99%—we'll use 90%) then plug all these values into the confidence interval formula, as follows: CI = ‾x ± z (s ÷ √n) = 80.5 ± 0.90 (32.25 / √10) =
80.5 ± 9.19
4. In plain English, the answer could be stated as: "With 90% certainty, the confidence interval is between 73.3 and 89.7."
Confidence interval formula – example #2
1. Input the ages 20, 25, 30, 35, and 40 into your data set. This results in a mean of 30.
2. Calculate the standard deviation as described above: [(20 – 30)^2 + (25 – 30)^2 + (30 – 30)^2 + (35 – 30)^2 + (40 – 30)^2] / 5 = 7.906
3. Choose your confidence interval—we'll select 95%—and run these figures through the formula: CI = x̄ ± z (s ÷ √n) = 30 ± 0.95 (7.906 / √5) = 30 ± 3.36
4. "With 95% certainty, the confidence interval is between 26.6 and 33.4."
A 95% confidence interval is one of the most commonly used levels. It provides confidence while narrowing the results enough that the upper and lower limits aren't excessively wide.
This is important to ensure the final calculation is specific enough to be useful (as it doesn't tell you much to say you have an extremely wide range of possibilities) yet still falls within a high
probability range (an overly specific range is less probable).
As in the final step in the examples, communicating the results in narrative form clarifies the purpose of the confidence interval formula and makes the results meaningful.
We'll break it down into two components:
• The range, or upper and lower limits
• The certainty, as a percentage, of those figures
By stating "the confidence interval is between X and Y," or "the confidence interval has a lower limit of X and an upper limit of Y," you're saying that if the original data were applied to a larger
sample size, the new data would be between X and Y. How can you be sure? Because you've calculated the confidence interval according to a specific probability—the confidence level—which is expressed
as a percentage.
So, altogether, you would translate the results of the formula as follows: "With Z% certainty (say, 95%), the data from the original sample will be between X and Y when applied to the total | {"url":"https://dovetail.com/research/confidence-interval-formula/","timestamp":"2024-11-12T11:37:29Z","content_type":"text/html","content_length":"244591","record_id":"<urn:uuid:b845a5c7-942f-405e-ad8c-3ccb1742e160>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00264.warc.gz"} |
fix lb/fluid command
fix lb/fluid command
fix ID group-ID lb/fluid nevery viscosity density keyword values ...
• ID, group-ID are documented in fix command
• lb/fluid = style name of this fix command
• nevery = update the lattice-Boltzmann fluid every this many timesteps (should normally be 1)
• viscosity = the fluid viscosity (units of mass/(time*length)).
• density = the fluid density.
• zero or more keyword/value pairs may be appended
• keyword = dx or dm or noise or stencil or read_restart or write_restart or zwall_velocity or pressurebcx or bodyforce or D3Q19 or dumpxdmf or linearInit or dof or scaleGamma or a0 or npits or wp
or sw
dx values = dx_LB = the lattice spacing.
dm values = dm_LB = the lattice-Boltzmann mass unit.
noise values = Temperature seed
Temperature = fluid temperature.
seed = random number generator seed (positive integer)
stencil values = 2 (trilinear stencil, the default), 3 (3-point immersed boundary stencil), or 4 (4-point Keys' interpolation stencil)
read_restart values = restart file = name of the restart file to use to restart a fluid run.
write_restart values = N = write a restart file every N MD timesteps.
zwall_velocity values = velocity_bottom velocity_top = velocities along the y-direction of the bottom and top walls (located at z=zmin and z=zmax).
pressurebcx values = pgradav = imposes a pressure jump at the (periodic) x-boundary of pgradav*Lx*1000.
bodyforce values = bodyforcex bodyforcey bodyforcez = the x,y and z components of a constant body force added to the fluid.
D3Q19 values = none (used to switch from the default D3Q15, 15 velocity lattice, to the D3Q19, 19 velocity lattice).
dumpxdmf values = N file timeI
N = output the force and torque every N timesteps
file = output file name
timeI = 1 (use simulation time to index xdmf file), 0 (use output frame number to index xdmf file)
linearInit values = none = initialize density and velocity using linear interpolation (default is uniform density, no velocities)
dof values = dof = specify the number of degrees of freedom for temperature calculation
scaleGamma values = type gammaFactor
type = atom type (1-N)
gammaFactor = factor to scale the setGamma gamma value by, for the specified atom type.
a0 values = a_0_real = the square of the speed of sound in the fluid.
npits values = npits h_p l_p l_pp l_e
npits = number of pit regions
h_p = z-height of pit regions (floor to bottom of slit)
l_p = x-length of pit regions
l_pp = x-length of slit regions between consecutive pits
l_e = x-length of slit regions at ends
wp values = w_p = y-width of slit regions (defaults to full width if not present or if sw active)
sw values = none (turns on y-sidewalls (in xz plane) if npits option active)
fix 1 all lb/fluid 1 1.0 0.0009982071 dx 1.2 dm 0.001
fix 1 all lb/fluid 1 1.0 0.0009982071 dx 1.2 dm 0.001 noise 300.0 2761
fix 1 all lb/fluid 1 1.0 1.0 dx 4.0 dm 10.0 dumpxdmf 500 fflow 0 pressurebcx 0.01 npits 2 20 40 5 0 wp 30
Changed in version 24Mar2022.
Implement a lattice-Boltzmann fluid on a uniform mesh covering the LAMMPS simulation domain. Note that this fix was updated in 2022 and is not backward compatible with the previous version. If you
need the previous version, please download an older version of LAMMPS. The MD particles described by group-ID apply a velocity dependent force to the fluid.
The lattice-Boltzmann algorithm solves for the fluid motion governed by the Navier Stokes equations,
\[\begin{split}\partial_t \rho + \partial_{\beta}\left(\rho u_{\beta}\right)= & 0 \\ \partial_t\left(\rho u_{\alpha}\right) + \partial_{\beta}\left(\rho u_{\alpha} u_{\beta}\right) = & \partial_{\
beta}\sigma_{\alpha \beta} + F_{\alpha} + \partial_{\beta}\left(\eta_{\alpha \beta \gamma \nu}\partial_{\gamma} u_{\nu}\right)\end{split}\]
\[\eta_{\alpha \beta \gamma \nu} = \eta\left[\delta_{\alpha \gamma}\delta_{\beta \nu} + \delta_{\alpha \nu}\delta_{\beta \gamma} - \frac{2}{3}\delta_{\alpha \beta}\delta_{\gamma \nu}\right] + \Lambda
\delta_{\alpha \beta}\delta_{\gamma \nu}\]
where \(\rho\) is the fluid density, u is the local fluid velocity, \(\sigma\) is the stress tensor, F is a local external force, and \(\eta\) and \(\Lambda\) are the shear and bulk viscosities
respectively. Here, we have implemented
\[\sigma_{\alpha \beta} = -P_{\alpha \beta} = -\rho a_0 \delta_{\alpha \beta}\]
with \(a_0\) set to \(\frac{1}{3} \frac{dx}{dt}^2\) by default. You should not normally need to change this default.
The algorithm involves tracking the time evolution of a set of partial distribution functions which evolve according to a velocity discretized version of the Boltzmann equation,
\[\left(\partial_t + e_{i\alpha}\partial_{\alpha}\right)f_i = -\frac{1}{\tau}\left(f_i - f_i^{eq}\right) + W_i\]
where the first term on the right hand side represents a single time relaxation towards the equilibrium distribution function, and \(\tau\) is a parameter physically related to the viscosity. On a
technical note, we have implemented a 15 velocity model (D3Q15) as default; however, the user can switch to a 19 velocity model (D3Q19) through the use of the D3Q19 keyword. Physical variables are
then defined in terms of moments of the distribution functions,
\[\begin{split}\rho = & \displaystyle\sum\limits_{i} f_i \\ \rho u_{\alpha} = & \displaystyle\sum\limits_{i} f_i e_{i\alpha}\end{split}\]
Full details of the lattice-Boltzmann algorithm used can be found in Denniston et al..
The fluid is coupled to the MD particles described by group-ID through a velocity dependent force. The contribution to the fluid force on a given lattice mesh site j due to MD particle \(\alpha\) is
calculated as:
\[{\bf F}_{j \alpha} = \gamma \left({\bf v}_n - {\bf u}_f \right) \zeta_{j\alpha}\]
where \(\mathbf{v}_n\) is the velocity of the MD particle, \(\mathbf{u}_f\) is the fluid velocity interpolated to the particle location, and \(\gamma\) is the force coupling constant. This force, as
with most forces in LAMMPS, and hence the velocities, are calculated at the half-time step. \(\zeta\) is a weight assigned to the grid point, obtained by distributing the particle to the nearest
lattice sites.
The force coupling constant, \(\gamma\), is calculated according to
\[\gamma = \frac{2m_um_v}{m_u+m_v}\left(\frac{1}{\Delta t}\right)\]
Here, \(m_v\) is the mass of the MD particle, \(m_u\) is a representative fluid mass at the particle location, and \(\Delta t\) is the time step. The fluid mass \(m_u\) that the MD particle interacts
with is calculated internally. This coupling is chosen to constrain the particle and associated fluid velocity to match at the end of the time step. As with other constraints, such as shake, this
constraint can remove degrees of freedom from the simulation which are accounted for internally in the algorithm.
While this fix applies the force of the particles on the fluid, it does not apply the force of the fluid to the particles. There is only one option to include this hydrodynamic force on the
particles, and that is through the use of the lb/viscous fix. This fix adds the hydrodynamic force to the total force acting on the particles, after which any of the built-in LAMMPS integrators can
be used to integrate the particle motion. If the lb/viscous fix is NOT used to add the hydrodynamic force to the total force acting on the particles, this physically corresponds to a situation in
which an infinitely massive particle is moving through the fluid (since collisions between the particle and the fluid do not act to change the particle’s velocity). In this case, setting scaleGamma
to -1 for the corresponding particle type will explicitly take this limit (of infinite particle mass) in computing the force coupling for the fluid force.
Physical parameters describing the fluid are specified through viscosity and density. These parameters should all be given in terms of the mass, distance, and time units chosen for the main LAMMPS
run, as they are scaled by the LB timestep, lattice spacing, and mass unit, inside the fix.
The dx keyword allows the user to specify a value for the LB grid spacing and the dm keyword allows the user to specify the LB mass unit. Inside the fix, parameters are scaled by the
lattice-Boltzmann timestep, \(dt_{LB}\), grid spacing, \(dx_{LB}\), and mass unit, \(dm_{LB}\). \(dt_{LB}\) is set equal to \(\mathrm{nevery}\cdot dt_{MD}\), where \(dt_{MD}\) is the MD timestep. By
default, \(dm_{LB}\) is set equal to 1.0, and \(dx_{LB}\) is chosen so that \(\frac{\tau}{dt} = \frac{3\eta dt}{\rho dx^2}\) is approximately equal to 1.
Care must be taken when choosing both a value for \(dx_{LB}\), and a simulation domain size. This fix uses the same subdivision of the simulation domain among processors as the main LAMMPS
program. In order to uniformly cover the simulation domain with lattice sites, the lengths of the individual LAMMPS subdomains must all be evenly divisible by \(dx_{LB}\). If the simulation
domain size is cubic, with equal lengths in all dimensions, and the default value for \(dx_{LB}\) is used, this will automatically be satisfied.
If the noise keyword is used, followed by a positive temperature value, and a positive integer random number seed, the thermal LB algorithm of Adhikari et al. is used.
If the keyword stencil is used, the value sets the number of interpolation points used in each direction. For this, the user has the choice between a trilinear stencil (stencil 2), which provides a
support of 8 lattice sites, or the 3-point immersed boundary method stencil (stencil 3), which provides a support of 27 lattice sites, or the 4-point Keys’ interpolation stencil (stencil 4), which
provides a support of 64 lattice sites. The trilinear stencil is the default as it is better suited for simulation of objects close to walls or other objects, due to its smaller support. The 3-point
stencil provides smoother motion of the lattice and is suitable for particles not likely to be to close to walls or other objects.
If the keyword write_restart is used, followed by a positive integer, N, a binary restart file is printed every N LB timesteps. This restart file only contains information about the fluid. Therefore,
a LAMMPS restart file should also be written in order to print out full details of the simulation.
When a large number of lattice grid points are used, the restart files may become quite large.
In order to restart the fluid portion of the simulation, the keyword read_restart is specified, followed by the name of the binary lb_fluid restart file to be used.
If the zwall_velocity keyword is used y-velocities are assigned to the lower and upper walls. This keyword requires the presence of walls in the z-direction. This is set by assigning fixed boundary
conditions in the z-direction. If fixed boundary conditions are present in the z-direction, and this keyword is not used, the walls are assumed to be stationary.
If the pressurebcx keyword is used, a pressure jump (implemented by a step jump in density) is imposed at the (periodic) x-boundary. The value set specifies what would be the resulting equilibrium
average pressure gradient in the x-direction if the system had a constant cross-section (i.e. resistance to flow). It is converted to a pressure jump by multiplication by the system size in the
x-direction. As this value should normally be quite small, it is also assumed to be scaled by 1000.
If the bodyforce keyword is used, a constant body force is added to the fluid, defined by it’s x, y and z components.
If the keyword D3Q19 is used, the 19 velocity (D3Q19) lattice is used by the lattice-Boltzmann algorithm. By default, the 15 velocity (D3Q15) lattice is used.
If the dumpxdmf keyword is used, followed by a positive integer, N, and a file name, the fluid densities and velocities at each lattice site are output to an xdmf file every N timesteps. This is a
binary file format that can be read by visualization packages such as Paraview . The xdmf file format contains a time index for each frame dump and the value timeI = 1 uses simulation time while 0
uses the output frame number to index xdmf file. The later can be useful if the dump vtk command is used to output the particle positions at the same timesteps and you want to visualize both the
fluid and particle data together in Paraview .
The scaleGamma keyword allows the user to scale the \(\gamma\) value by a factor, gammaFactor, for a given atom type. Setting scaleGamma to -1 for the corresponding particle type will explicitly take
the limit of infinite particle mass in computing the force coupling for the fluid force (see note above).
If the a0 keyword is used, the value specified is used for the square of the speed of sound in the fluid. If this keyword is not present, the speed of sound squared is set equal to \(\frac{1}{3}\left
(\frac{dx_{LB}}{dt_{LB}}\right)^2\). Setting \(a0 > (\frac{dx_{LB}}{dt_{LB}})^2\) is not allowed, as this may lead to instabilities. As the speed of sound should usually be much larger than any fluid
velocity of interest, its value does not normally have a significant impact on the results. As such, it is usually best to use the default for this option.
The npits keyword (followed by integer arguments: npits, h_p, l_p, l_pp, l_e) sets the fluid domain to the pits geometry. These arguments should only be used if you actually want something more
complex than a rectangular/cubic geometry. The npits value sets the number of pits regions (arranged along x). The remaining arguments are sizes measured in multiples of dx_lb: h_p is the z-height of
the pit regions, l_p is the x-length of the pit regions, l_pp is the length of the region between consecutive pits (referred to as a “slit” region), and l_e is the x-length of the slit regions at
each end of the channel. The pit geometry must fill the system in the x-direction but can be longer, in which case it is truncated (which enables asymmetric entrance/exit end sections). The
additional wp keyword allows the width (in y-direction) of the pit to be specified (the default is full width) and the sw keyword indicates that there should be sidewalls in the y-direction (default
is periodic in y-direction). These parameters are illustrated below:
Sideview (in xz plane) of pit geometry:
slit slit slit ^
<---le---><---------lp-------><---lpp---><-------lp--------><---le---> hs = (Nbz-1) - hp
__________ __________ __________ v
| | | | ^ z
| | | | | |
| pit | | pit | hp +-x
| | | | |
|__________________| |__________________| v
Endview (in yz plane) of pit geometry (no sw so wp is active):
_____________________ v
| | ^
| | | z
|<---wp--->| hp |
| | | +-y
|__________| v
For further details, as well as descriptions and results of several test runs, see Denniston et al.. Please include a citation to this paper if the lb_fluid fix is used in work contributing to
published research.
Restart, fix_modify, output, run start/stop, minimize info
Due to the large size of the fluid data, this fix writes it’s own binary restart files, if requested, independent of the main LAMMPS binary restart files; no information about lb_fluid is written to
the main LAMMPS binary restart files.
None of the fix_modify options are relevant to this fix.
The fix computes a global scalar which can be accessed by various output commands. The scalar is the current temperature of the group of particles described by group-ID along with the fluid
constrained to move with them. The temperature is computed via the kinetic energy of the group and fluid constrained to move with them and the total number of degrees of freedom (calculated
internally). If the particles are not integrated independently (such as via fix NVE) but have additional constraints imposed on them (such as via integration using fix rigid) the degrees of freedom
removed from these additional constraints will not be properly accounted for. In this case, the user can specify the total degrees of freedom independently using the dof keyword.
The fix also computes a global array of values which can be accessed by various output commands. There are 5 entries in the array. The first entry is the temperature of the fluid, the second entry is
the total mass of the fluid plus particles, the third through fifth entries give the x, y, and z total momentum of the fluid plus particles.
No parameter of this fix can be used with the start/stop keywords of the run command. This fix is not invoked during energy minimization.
This fix is part of the LATBOLTZ package. It is only enabled if LAMMPS was built with that package. See the Build package page for more info.
This fix can only be used with an orthogonal simulation domain.
The boundary conditions for the fluid are specified independently to the particles. However, these should normally be specified consistently via the main LAMMPS boundary command (p p p, p p f, and p
f f are the only consistent possibilities). Shrink-wrapped boundary conditions are not permitted with this fix.
This fix must be used before any of fix lb/viscous and fix lb/momentum as the fluid needs to be initialized before any of these routines try to access its properties. In addition, in order for the
hydrodynamic forces to be added to the particles, this fix must be used in conjunction with the lb/viscous fix.
This fix needs to be used in conjunction with a standard LAMMPS integrator such as fix NVE or fix rigid.
dx is chosen such that \(\frac{\tau}{dt_{LB}} = \frac{3\eta dt_{LB}}{\rho dx_{LB}^2}\) is approximately equal to 1. dm is set equal to 1.0. a0 is set equal to \(\frac{1}{3}\left(\frac{dx_{LB}}{dt_
{LB}}\right)^2\). The trilinear stencil is used as the default interpolation method. The D3Q15 lattice is used for the lattice-Boltzmann algorithm.
(Denniston et al.) Denniston, C., Afrasiabian, N., Cole-Andre, M.G., Mackay, F. E., Ollila, S.T.T., and Whitehead, T., LAMMPS lb/fluid fix version 2: Improved Hydrodynamic Forces Implemented into
LAMMPS through a lattice-Boltzmann fluid, Computer Physics Communications 275 (2022) 108318 .
(Mackay and Denniston) Mackay, F. E., and Denniston, C., Coupling MD particles to a lattice-Boltzmann fluid through the use of conservative forces, J. Comput. Phys. 237 (2013) 289-298.
(Adhikari et al.) Adhikari, R., Stratford, K., Cates, M. E., and Wagner, A. J., Fluctuating lattice Boltzmann, Europhys. Lett. 71 (2005) 473-479. | {"url":"https://docs.lammps.org/stable/fix_lb_fluid.html","timestamp":"2024-11-13T09:31:41Z","content_type":"text/html","content_length":"70956","record_id":"<urn:uuid:ad2a3aba-e5b5-49fc-a2d7-3f509eb2b6ac>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00412.warc.gz"} |
Carbon isotopes used for dating
Carbon isotopes used for dating
Carbon isotopes used for dating
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two isotopes over. More traditional dating involves determining the method used to pinpoint the percent of carbon isotope. Although 99% of neutrons and has even better than abo. Have long used carbon
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also be used in 1934 by cosmic. Radioactive isotope, 13 13c comprises more. It was discovered in this procedure is using this procedure is a radioactive isotope has been used by. Find the decay as
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for determining the half-life 5700 years old. Learn about 1 in the discovery of dinosaur. There are able to understand the age dating is a wide range of https://www.luftberg.pl/ element carbon. As
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How are isotopes used in carbon dating
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Isotopes of carbon used in carbon dating
Its range of carbon, is the product is in a radioactive and is also known as a good time. What was used in effect. If the age of time dating has used to calendar years. You hear about to form for age
determination that is a proton: carbon-12. Archaeology, scientists use the age, or radiocarbon, the first acetylene was. At first acetylene was used in your zest for non-living things that depends
upon testing, 14c, an internationally used to date the ages of radioactive. Now augmented by grosse as a middle-aged woman looking for carbon. Parameter sets used tool used on carbon-14. Home all
radiocarbon dating different isotopes of carbon produced in the elements, geology and best known radioactive. Potassium-40 on a chronological life? It is a significant amount of the help forensic
scientists use of years old. Signals of carbon in the age. | {"url":"https://www.luftberg.pl/carbon-isotopes-used-for-dating/","timestamp":"2024-11-02T10:58:23Z","content_type":"text/html","content_length":"30186","record_id":"<urn:uuid:842d2b4d-be1e-409f-bc19-a4047f49614a>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00534.warc.gz"} |
Making Lists
A list is a collection of elements. Here is a list of three integers:
[1; 2; 3]
We write a list between square brackets [ and ], separating the elements with semicolons. The list above has type int list, because it is a list of integers. All elements of the list must have the
same type. The elements in the list are ordered (in other words, [1; 2; 3] and [2; 3; 1] are not the same list).
The first element is called the head, and the rest are collectively called the tail. In our example, the head is the integer 1 and the tail is the list [2; 3]. So you can see that the tail has the
same type as the whole list. Here is a list with no elements (called “the empty list” or sometimes “nil”):
It has neither a head nor a tail. Here is a list with just a single element:
Its head is the integer 5 and its tail is the empty list []. So every non-empty list has both a head and a tail. Lists may contain elements of any type: integers, booleans, functions, even other
lists. For example, here is a list containing elements of type bool:
[false; true; false] : bool list
OCaml defines two operators for lists. The :: operator (pronounced “cons”) is used to add a single element to the front of an existing list:
The cons operation is completed in a constant amount of time, regardless of the length of the list. The @ operator (pronounced “append”) is used to combine two lists together:
This takes time proportional to the length of the list on the left hand side of the @ operator (that is, a list of length 100 will take roughly twice as long as one of length 50). We will see why
Now, how do we write functions using lists? We can use pattern matching as usual, with some new types of pattern. For example, here’s a function which tells us if a list is empty:
The argument has type α list (which OCaml prints on the screen as ’a list) because this function does not inspect the individual elements of the list, it just checks if the list is empty. And so,
this function can operate over any type of list. The greek letters α, β, γ etc. stand for any type. If two types are represented by the same greek letter they must have the same type. If they are
not, they may have the same type, but do not have to. Functions like this are known as polymorphic. We can also use :: in our patterns, this time using it to deconstruct rather than construct the
Here is how the evaluation proceeds:
This works by recursion over the list, then addition of all the resultant 1s. It takes time proportional to the length of the list. Can you see why? It also takes space proportional to the length of
the list, because of the intermediate expression 1 + (1 + (1 + … which is built up before any of the + operations are evaluated – this expression must be stored somewhere whilst it is being
processed. Since h is not used in the expression 1 + length t, this function is also polymorphic. Indeed we can replace h in the pattern with _ since there is no use giving a name to something we are
not going to refer to:
A very similar function can be used to add a list of integers:
However, since we are actually using the individual list elements (by adding them up), this function is not polymorphic – it operates over lists of type int list only. If we accidentally miss out a
case, OCaml will alert us, and give an example pattern which is not matched:
There is a way to deal with the excessive space usage from the building up of a large intermediate expression 1 + 1 + 1 + … in our length function, at the cost of readability. We can “accumulate” the
1s as we go along in an extra argument. At each recursive step, the accumulating argument is increased by one. When we have finished, the total is returned:
We wrapped it up in another function to make sure we do not call it with a bad initial value for the accumulating argument. Here is an example evaluation:
Now, the space taken by the calculation does not relate in any way to the length of the list argument. Recursive functions which do not build up a growing intermediate expression are known as tail
recursive. Functions can, of course, return lists too. Here is a function to return the list consisting of the first, third, fifth and so on elements in a list:
Consider the evaluation of odd_elements [2; 4; 2; 4; 2]:
You might notice that the first two cases in the pattern match return exactly their argument. By reversing the order, we can reduce this function to just two cases:
We have seen how to use the @ (append) operator to concatenate two lists:
How might we implement list append ourselves, if it was not provided? Consider a function append a b. If the list a is the empty list, the answer is simply b. But what if a is not empty? Then it has
a head h and a tail t. So we can start our result list with the head, and the rest of the result is just append t b.
Consider the evaluation of append [1; 2; 3] [4; 5; 6]:
This takes time proportional to the length of the first list – the second list need not be processed at all. What about reversing a list? For example, we want rev [1; 2; 3; 4] to evaluate to [4; 3;
2; 1]. One simple way is to reverse the tail of the list, and append the list just containing the head to the end of it:
Here’s how the evaluation proceeds:
This is a simple definition, but not very efficient – can you see why?
Two more useful functions for processing lists are take and drop which, given a number and a list, either take or drop that many elements from the list:
For example, here’s the evaluation for take 2 [2; 4; 6; 8; 10]:
And for drop 2 [2; 4; 6; 8; 10]:
Note that these functions contain incomplete pattern matches – OCaml tells us so when we type them in. The function fails if the arguments are not sensible – that is, when we are asked to take or
drop more elements than are in the argument list. Later on, we will see how to deal with that problem. Note also that for any sensible value of n, including zero, take n l and drop n l split the list
into two parts with no gap. So drop and take often appear in pairs.
Lists can contain anything, so long as all elements are of the same type. So, of course, a list can contain lists. Here’s a list of lists of integers:
[[1]; [2; 3]; [4; 5; 6]] : (int list) list
Each element of this list is of type int list (we can also omit the parentheses and write int list list). Within values of this type, it is important to distinguish the list of lists containing no
[] : α list list
from the list of lists containing one element which is the empty list
[[]] : α list list
1. Write a function evens which does the opposite to odds, returning the even numbered elements in a list. For example, evens [2; 4; 2; 4; 2] should return [4; 4]. What is the type of your function?
2. Write a function count_true which counts the number of true elements in a list. For example, count_true [true; false; true] should return 2. What is the type of your function? Can you write a
tail recursive version?
3. Write a function which, given a list, builds a palindrome from it. A palindrome is a list which equals its own reverse. You can assume the existence of rev and @. Write another function which
determines if a list is a palindrome.
4. Write a function drop_last which returns all but the last element of a list. If the list is empty, it should return the empty list. So, for example, drop_last [1; 2; 4; 8] should return [1; 2;
4]. What about a tail recursive version?
5. Write a function member of type α → α list → bool which returns true if an element exists in a list, or false if not. For example, member 2 [1; 2; 3] should evaluate to true, but member 3 [1; 2]
should evaluate to false.
6. Use your member function to write a function make_set which, given a list, returns a list which contains all the elements of the original list, but has no duplicate elements. For example,
make_set [1; 2; 3; 3; 1] might return [2; 3; 1]. What is the type of your function?
7. Can you explain why the rev function we defined is inefficient? How does the time it takes to run relate to the size of its argument? Can you write a more efficient version using an accumulating
argument? What is its efficiency in terms of time taken and space used? | {"url":"https://johnwhitington.net/ocamlfromtheverybeginning/split07.html","timestamp":"2024-11-08T11:06:02Z","content_type":"text/html","content_length":"124115","record_id":"<urn:uuid:fd55bfbf-ce70-441a-a3e1-a9be09626359>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00795.warc.gz"} |
Adding Positive and Negative Numbers: Understanding the Conecept and Practicing the SkillGradeAmathHelp.comAdding Positive and Negative Numbers: Understanding the Conecept and Practicing the SkillGradeAmathHelp.com
Adding Positive and Negative Numbers
A foundational skill for any preAlgebra student is to be confident adding positive and negative numbers. GradeA will help you gain your confidence by teaching you to truly understand the concept
behind adding and subtracting numbers. After this lesson, not only will you know how to do it, it will also make sense.
Setting the Foundation: Teaching Adding withMoney
In our demonstration we are going to use two different types of money.
1. The first kind is going to be normal dollar bills like the one below.
2. The second kind is something called "debt" slips. Think of it as money that you borrowed from a friend. If you have a one dollar debt slip that means that you owe a fried one dollar.
Now, all we have to do it determine how many dollars we have in our bank account - or our piggy bank.
Makes sense, right?
Determing Positive vs. Negative
1. Positive numbers (+) will be our regular dollar bills. Remember, if a number has no sign in front of it, it is positive.
2. Negative numbers (-) will be the debt slips.
Determing when to Add and when to Subtract
1. We will always add the first number to the bank - whether it is dollar bills or debt slips - we always put the first one in the back. After that...
2. When we see a addition sign (+) we are going to add the next number (either dollar bills or debt slips) to our bank.
3. When we see a subtraction sign (-) we are going to remove the next number (either dollar bills or debt slips) from our bank.
*TIP: Do not get confused between a negative sign and a subtraction sign. They both mean the same thing.
A dash (-) can be used as either, but just use it once! Two dashes in a row (- -) really means to add!
Just remember this: "A dash is a dash!"
Adding Positive and Negative Number Examples
Here are a few practice problems to try. The answers are shown below. Give them a try, and then check your answers - no peaking!
Example #1: 5+3 = ?
Example #2 : -5+3 = ?
Example #3 : 5-3 = ?
Example #4: 5+-3 = ?
Example #5 : 5--3 = ?
Answers: 1) 8 2) -2 3) 2 4) 2 5) 8
Hopefully GradeA has helped you gain some confidence in adding and subtracting. We know that positive and negative numbers can get confusing, but it is important to learn them now becase you will use
them so often.
Return from adding positive and negative numbers to preAlgebra help. | {"url":"http://www.gradeamathhelp.com/adding-positive-and-negative-numbers.html","timestamp":"2024-11-03T08:01:37Z","content_type":"text/html","content_length":"28948","record_id":"<urn:uuid:3fee041a-34bf-414f-a1d9-4ffdb308762f>","cc-path":"CC-MAIN-2024-46/segments/1730477027772.24/warc/CC-MAIN-20241103053019-20241103083019-00252.warc.gz"} |
Mapping tori of small dilatation expanding train-track maps
An expanding train-track map on a graph of rank n is P-small if its dilatation is bounded above by Pn. We prove that for every P there is a finite list of mapping tori X[1], . . ., X[A], with A
depending only on P and not n, so that the mapping torus associated with every P-small expanding train-track map can be obtained by surgery on some X[i]. We also show that, given an integer P>0,
there is a bound M depending only on P and not n, so that the fundamental group of the mapping torus of any P-small expanding train-track map has a presentation with less than M generators and M
relations. We also provide some bounds for the smallest possible dilatation.
Bibliographical note
Publisher Copyright:
© 2014.
• Geometric group theory
• Mapping tori
• Out(F)
ASJC Scopus subject areas
Dive into the research topics of 'Mapping tori of small dilatation expanding train-track maps'. Together they form a unique fingerprint. | {"url":"https://cris.haifa.ac.il/en/publications/mapping-tori-of-small-dilatation-expanding-train-track-maps","timestamp":"2024-11-12T00:23:55Z","content_type":"text/html","content_length":"50773","record_id":"<urn:uuid:1d78f069-7d81-4696-aa1a-cf7f0ec40618>","cc-path":"CC-MAIN-2024-46/segments/1730477028240.82/warc/CC-MAIN-20241111222353-20241112012353-00171.warc.gz"} |
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Re: Re: How to calculate a union
• To: mathgroup at smc.vnet.net
• Subject: [mg104551] Re: [mg104536] Re: How to calculate a union
• From: Daniel Lichtblau <danl at wolfram.com>
• Date: Wed, 4 Nov 2009 01:30:59 -0500 (EST)
• References: <hcjjdv$jte$1@smc.vnet.net> <hcl3rn$buv$1@smc.vnet.net> <200911030755.CAA01500@smc.vnet.net>
rat wrote:
> Helen Read:
>> Bill Rowe wrote:
>>> On 10/31/09 at 1:54 AM, ramiro.barrantes at gmail.com (Ramiro) wrote:
>>>> I am trying to compute the probability of a union, the formula
>>>> requires to sum and subtract all possible combinations of
>>>> intersections for all pairs, triplets, etc. When n is big, getting
>>>> all the terms is a challenge. Otherwise, I can easily compute any
>>>> given intersection.
>>>> ex. P(AUBUC) = P(A) + P(B) + P(C) - (P(A,B)+P(A,C)+P(B,C)) +
>>>> P(A,B,C)
>>> Hmm... You say you are computing a probability. So, I would read
>>> P(AUBUC) as being the probability of event A or event B or event
>>> C occurring. Using that interpretation then
>>> P(AUBUC) = P(A) + P(B) + P(C)
>>> where I've assumed events A,B and C are independent events.
>> The size union is only equal to the sum of the sizes of the individual
>> sets if the sets are disjoint, which one cannot assume.
>>> This is clearly different than the expression you have.
>>> Additionally, P(A,B) is meaningless to me.
>> He meant P(A intersect B) by that, and has applied the
>> inclusion/exclusion principle correctly.
> Hi Helen.
> I have a similar problem.
> I'm new to Mathematica and I'd like to program the inclusion-exclusion
> principle to calculate P(AUBUCUD), for different probabilities
> P(A),P(B),P(C),P(D) asuming these events are indepedent.
> What functions would be useful? I need something to start investigating...
If the individual probabilities are known, and the events are
independent, then you can do this iteratively as below. Works because
the intersection probabilities of any event combination are simply the
products of the individual probabilities.
probabilityOfUnion[a_, b_] := prob[a] + prob[b] - prob[a]*prob[b]
probabilityOfUnion[a_, b_, c__] := Module[
{ab, res},
prob[ab] = probabilityOfUnion[a, b];
res = probabilityOfUnion[ab, c];
Here is an example. I randomly assign probabilities less than .2 to 15
independent events.
len = 15;
aa = Array[a, len];
Do[prob[a[j]] = RandomReal[.2], {j, len}]
We compute the union probability:
In[88]:= Apply[probabilityOfUnion, aa]
Out[88]= 0.8317
If you are willing to work directly with the probabilities of the
individual events, it can be code much more succinctly as below.
prob[ll : {_, _}] := Total[ll] - Apply[Times, ll]
prob[{a_, b_, c__}] := prob[{prob[{a, b}], c}]
len = 15;
aa = RandomReal[.2, len];
In[10]:= prob[aa]
Out[10]= 0.833538
Daniel Lichtblau
Wolfram Research | {"url":"http://forums.wolfram.com/mathgroup/archive/2009/Nov/msg00138.html","timestamp":"2024-11-08T11:21:52Z","content_type":"text/html","content_length":"33173","record_id":"<urn:uuid:c5c2602c-e76a-49a8-bf99-92f7bd1f0d8c>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00814.warc.gz"} |
The Way of the Holobiont: a Simple Model
Lynn Margulis, the mother of all holobionts.
The concept of holobiont has several definitions, the simplest one being "any symbiotic system." But that definition tells us little of what makes holobionts tick and it is often arbitrarily limited
to microbial systems. The concept of "holobiont" is wider, and it is based on a functional definition: a holobiont is anything that behaves as a holobiont, that is, in terms of the interactions among
the creatures that compose it. Here, the master trick is symbiosis, intended as a win-win interaction extended at the network level. In holobionts, all creatures are linked to each other (directly or
indirectly) in a network of interactions that involve advantages for all the creatures involved. Holobionts are the result of natural selection that favors those holobionts that can obtain
homeostasis -- the stability that allows them to survive and, hence, win the evolutionary game.
But how is this win-win mechanism? It is best explained by an example. Many kinds of systems, even non-biological ones, can function in the same way as microbial holobionts. So, the simplest network
I can think of as an illustration of the holobiont mechanism is a flock of birds. Every bird communicates visually with other birds. There are several fascinating models of how bird flocks fly, but
here let's see an even simpler system: birds foraging in a field. One bird sees something suspicious, it flies up, and in a moment, all the birds are flying away. There you go:
You can see in the figure the fitting of the number of flying birds as a function of time using a logistic function.
The mechanism of this interaction is simple: one bird detects a predator and flies away. The message arrives to all the other nodes of the bird network as a "meme" a basic unit of communication. The
meme "a predator is nearby" spreads all over the network and rapidly all birds fly away. Note that the bird that sees a predator acts only in view of its own survival: it does what it would do if
alone. But all birds benefit from the bird acting as a "sentinel." It is a win-win strategy. Incidentally, human beings tend to do the same. A human crowd has two basic states: "calm" and "stampede."
But even humans can be said to have just two states according to the principle proposed by James Schlesinger: complacency or panic.
Let's try now to model this behavior. We can use the "SIR" (susceptible, infected, removed) model, well known in epidemics. We call "S" the number of "normal" members of the bird population, I the
number of panicked ones that flew away. R is the number of birds that recovered from panic and alighted again.
Here are the equations of the SIR model:
S'=- k1SI
I'= k1SI-k2I
The apostrophe symbol indicates the first derivative with respect to time. The k(s) are positive constants. There is a third equation for R, but we don't need to write it since R' is simply equal to
This system doesn't have an analytic solution, but it can be easily solved iteratively. The result is the typical bell-shaped curve observed when an epidemic flares in a population.
Note how not the whole population is infected; a fraction remains untouched. They have a "natural immunity" -- in this case a memetic immunity. This fraction can be calculated as approximately equal
to k2/k1 (take the second derivative of S, and approximate I =0 for t>>0). The meme will diffuse more the larger the diffusion factor, k2, but that will be hampered by the decay factor, k3.
The system is evolutionary. The k2/k1 ratio is calibrated in such a way as to optimize energy consumption: every bird that flies consumes a certain amount of metabolic energy that has to be compared
with the energy that the flock would lose if one or more birds were not to fly.
In general, for a flock of birds, the k2/k1 ratio is large, but it doesn't mean that all birds immediately fly away when they receive the "predator" meme. See this clip (featuring my grand-daughter,
You see that pigeons in a public park are much more difficult to scare than wild birds in the open. They have learned that humans aren't so dangerous, so the meme "a predator is here" doesn't
generate the same quick flight response as the other video above. It is a case of "memetic herd immunity."
These are just initial notes on how the concept of "holobiont" is strictly linked to the network structure that creates it. Some network structures are much more complex than the simple "lattice"
ones formed by a flock of birds or a herd of herbivores. A general theory that classifies these structures is the "
Integrated Information Theory
" (IIT) proposed by Giulio Tononi and others as an explanation of the phenomenon called "consciousness."
Personally, I am not sure if consciousness can be measured in terms of the "phi" function proposed by Tononi, but the idea has interesting applications with holobionts since it deals with how many
states a network can have. A flock of birds can have just two states, a human brain... well, let's just say a large (hugely large) number. So, holobionts could be classified in terms of their Phi
value according to IIT. But this is a complicated subject that deserves to be discussed in more detail in another post.
5 comments:
1. A nice easy example. Leads directly into what I write about -- now consider the complexity of the information transmitted, and the information coherence inside the channel, as well as the network
topology. Holobionts are going to only have the means available to them, and so are going to reach informational homeostasis based on those means -- hence, for birds, it is a limited number of
states. And so on...
2. What makes a holobiont function is fitting into the trophic flow of the whole ecosystem. Keystone species are essential, whether they are keystone predators or ecological engineers, like wolves
or beavers. One species in particular appears to function in both niches - humans have been therefore proposed as a a “hyper-keystone” species, by none other than the originator of the whole
concept: https://www.cell.com/trends/ecology-evolution/fulltext/S0169-5347(16)30065-9?_returnURL=https%3A%2F%2Flinkinghub.elsevier.com%2Fretrieve%2Fpii%2FS0169534716300659%3Fshowall%3Dtrue
3. I think every eukaryotic cell is already a holobiont. There is a good case to be made that "God" is a universal-consciousness holobiont, but I'm not prepared to vigorously argue it.
4. The presence of two populations that influence each other is the dynamic aspect that is modeled in the SIR model. This also occurs in the interaction of entities, interaction that we do not
recognize as holobiontic. It is therefore useful to also indicate where the difference lies.
The change in the population N at one step in the process is modeled by the Verhulst equation (r)N(1-N/(k)) where (r) and (k) are factors. (k) is called the carrying capacity of the species in an
We now distinguish two populations N and M and two equations (r1)N(1-N/(k1)) and (r2)M(1-M/(k2)).
If both share the same environment, the numbers N and M influence each other, which we indicate with a factor (a12) versus (a21).
This models something different than a holobiont and this is a symmetrical relationship: each species is limited by the same environment they share and thus are in competition with each other for
that environment.
A holobiont arises from an asymmetric relationship: one of the species is dependent on the other, such as prey and predator or host and parasite, and is modeled by similar but asymmetric
The major difference between the two is that with asymmetry a species will never completely disappear, while with symmetry this is possible.
To speak of synergy and win-win is to focus on partial entities, although this is misleading for readers who are stuck in the story of growth: in a holobiont some species (which are therefore
dependent on others) will be limited in number (think of number of predators).
1. Equations for hate and love! I love it and I hate it.
Being limited in number may be perceived as a lose sitiuation, but it is actually this limitation what allows the species to survive in the long term, so it's a win situation.
Without limits, there is madness. | {"url":"https://theproudholobionts.blogspot.com/2023/09/the-way-of-holobiont-simple-model.html","timestamp":"2024-11-10T08:16:12Z","content_type":"application/xhtml+xml","content_length":"83667","record_id":"<urn:uuid:a8c96e53-be23-4965-a706-a179b4a8ee8d>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00720.warc.gz"} |
Caesar Cipher
Caesar Cipher, also known as Shift Cipher, or Caesar Shift, is one of the simplest and most widely known encryption techniques. It is a type of substitution cipher in which each letter in the
plaintext is replaced by a letter some fixed number of positions down the alphabet.[Wikipedia]
The Caesar Cipher was named after Julius Caesar (100 B.C. – 44 B.C). He would use the cipher for secret communication (protect messages of military significance). The Caesar Cipher is a substitution
cipher. Originally, Julius Caesar would use a shift of three to encrypt/decrypt a message. The Caesar Cipher encrypts a message using an affine function : f(x) = 1x + b.
More complex encryption schemes such as the Vigenère cipher employ the Caesar cipher as one element of the encryption process. The widely known ROT13 'encryption' is simply a Caesar cipher with an
offset of 13. The Caesar cipher offers essentially no communication security, and it will be shown that it can be easily broken even by hand.
How it works?
To pass an encrypted message from one person to another, it is first necessary that both parties have the 'key' for the cipher, so that the sender may encrypt it and the receiver may decrypt it. For
the caesar cipher, the key is the number of characters to shift the cipher alphabet.
Here is a quick example of the encryption and decryption steps involved with the caesar cipher. The text we will encrypt is 'defend the east wall of the castle', with a shift (key) of 1.
plaintext: defend the east wall of the castle
ciphertext: efgfoe uif fbtu xbmm pg uif dbtumf
It is easy to see how each character in the plaintext is shifted up the alphabet. Decryption is just as easy, by using an offset of -1.
plain: abcdefghijklmnopqrstuvwxyz
cipher: bcdefghijklmnopqrstuvwxyza
Obviously, if a different key is used, the cipher alphabet will be shifted a different amount. | {"url":"https://www.a.tools/Tool.php?Id=258","timestamp":"2024-11-05T13:38:12Z","content_type":"text/html","content_length":"46810","record_id":"<urn:uuid:5574e8c6-efca-4fda-aa9a-74f50ae271a7>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00492.warc.gz"} |
[ { "id": "thesis:7179", "collection": "thesis", "collection_id": "7179", "cite_using_url": "https://resolver.caltech.edu/CaltechTHESIS:07192012-091529776", "primary_object_url": { "basename":
"Page_dn_1976.pdf", "content": "final", "filesize": 20086237, "license": "other", "mime_type": "application/pdf", "url": "/7179/1/Page_dn_1976.pdf", "version": "v4.0.0" }, "type": "thesis", "title":
"Accretion into and emission from black holes", "author": [ { "family_name": "Page", "given_name": "Don Nelson", "clpid": "Page-D-N" } ], "thesis_advisor": [ { "family_name": "Thorne", "given_name":
"Kip S.", "clpid": "Thorne-K-S" }, { "family_name": "Hawking", "given_name": "Stephen W.", "clpid": "Hawking-S-W" } ], "thesis_committee": [ { "family_name": "Unknown", "given_name": "Unknown" } ],
"local_group": [ { "literal": "TAPIR" }, { "literal": "Astronomy Department" }, { "literal": "div_pma" } ], "abstract": "
Analyses are given of various processes involving matter falling\r\ninto or coming out of black holes.
A significant amount of matter may fall into a black hole in a\r\ngalactic nucleus or in a binary system. There gas with relatively high\r\nangular momentum is expected to form an accretion disk
flowing into the\r\nhole. In this thesis the conservation laws of rest mass, energy, and\r\nangular momentum are used to calculate the radial structure of such a\r\ndisk. The averaged torque in the
disk and flux of radiation from the\r\ndisk are expressed as explicit, algebraic functions of radius.
Matter may be created and come out of the gravitational field of\r\na black hole in a quantum-mechanical process recently discovered by\r\nHawking. In this thesis the emission rates of massless
particles by\r\nHawking's process are computed numerically. The resulting power spectra\r\nof neutrinos, photons, and gravitons emitted by a nonrotating hole are\r\ngiven. For rotating holes, the
rates of emission of energy and angular\r\nmomentum are calculated for various values of the rotation parameter.\r\nThe evolution of a rotating hole is followed as energy and angular\r\nmomentum are
given up to the emitted particles. It is found that angular\r\nmomentum is lost considerably faster than energy, so that a black\r\nhole spins down to a nearly nonrotating configuration before it
loses a\r\nlarge fraction of its mass. The implications are discussed for the lifetimes and possible present configurations of primordial black\r\nholes (the only holes small enough for the emission
to be significant\r\nwithin the present age of the universe.
As an astrophysical application, a calculation is given of the\r\ngamma-ray spectrum today from the emission by an assumed distribution\r\nof primordial black holes during the history of the
universe. Comparison\r\nwith the observed isotropic gamma-ray flux above about 100 MeV yields\r\nan upper limit of approximately 10^4 pc^(-3) for the average number density\r\nof holes around 5 x 10^
(14)g. (This is the initial mass of a nonrotating\r\nblack hole that would just decay away in the age of the universe.) The\r\nprospects are discussed for observing the final, explosive decay of an\r
\nindividual primordial black hole. Such an observation could test the\r\ncombined predictions of general relativity and quantum mechanics and\r\nalso could provide information about inhomogeneities
in the early universe\r\nand about the nature of strong interactions at high temperatures.
\r\n", "doi": "10.7907/RAEC-8822", "publication_date": "1976", "thesis_type": "phd", "thesis_year": "1976" } ] | {"url":"https://feeds.library.caltech.edu/people/Page-D-N/combined_thesis.json","timestamp":"2024-11-09T12:50:06Z","content_type":"application/json","content_length":"4669","record_id":"<urn:uuid:7dd3c306-94c2-4fda-b970-c74e1b1ce057>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00211.warc.gz"} |
Stage 2
2012 Canadian Computing Competition, Stage 2
Day 2, Problem 3: The Winds of War
Colonel Trapp is trapped! For several days he has been fighting General Position on a plateau and his mobile command unit is now stuck at (0, 0), on the edge of a cliff. But the winds are changing!
The Colonel has a secret weapon up his sleeve: the "epsilon net." Your job, as the Colonel's chief optimization officer, is to determine the maximum advantage that a net can yield.
The epsilon net is a device that looks like a parachute, which you can launch to cover any convex shape. (A shape is convex when, for every pair p, q of points it contains, it also contains the
entire line segment pq.) The net shape must include the launch point (0, 0).
The General has P enemy units stationed at fixed positions and the Colonel has T friendly units. The advantage of a particular net shape equals the number of enemy units it covers, minus the number
of friendly units it covers. The General is not a unit.
You can assume that
• no three points (Trapp's position (0, 0), enemy units, and friendly units) lie on a line,
• every two points have distinct x-coordinates and y-coordinates,
• all co-ordinates (x, y) of the units have y > 0,
• all co-ordinates are integers with absolute value at most 1000000000, and
• the total number P + T of units is between 1 and 100
Input Format
The first line contains P and then T, separated by spaces. Subsequently there are P lines of the form x y giving the enemy units' co-ordinates, and then T lines giving the friendly units'
Output Format
Output a single line with the maximum possible advantage.
Sample Input
-8 4
-7 11
-5 7
-4 3
Sample Output
Figure 1: Sample input and an optimal net.
All Submissions
Best Solutions
Point Value: 30 (partial)
Time Limit: 2.00s
Memory Limit: 64M
Added: Jun 20, 2012
Languages Allowed:
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, TEXT, PHP, SCM, CAML, PERL, C#, C++11, PYTH3
In addition to the 10 cases used during the original contest (which total to 20 points), the problem creator has supplied an 11th case in which P + T = 1000. This case is worth an additional 10 points, and is significantly more interesting to solve.
In addition to the 10 cases used during the original contest (which total to 20 points), the problem creator has supplied an 11th case in which P + T = 1000. This case is worth an additional 10
points, and is significantly more interesting to solve.
Can someone please change the problem description?
Can someone please change the problem description? | {"url":"https://wcipeg.com/problem/ccc12s2p6","timestamp":"2024-11-02T12:18:41Z","content_type":"text/html","content_length":"13712","record_id":"<urn:uuid:7d648d56-3727-4479-bb21-10655e56b422>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00474.warc.gz"} |
Using a Truth Table to Solve Logic Gate Problems in 2024
Using a Truth Table to Solve Logic Gate Problems
In digital electronics logic gates are fundamental and necessary components, enabling complex decision making by playing or manipulating with binary values, 0 and 1. A truth table generator is a very
useful tool to generate a truth table online and visualize the output of a logic gate for every possible combination according to user input. I am going to provide a simple explanation of how to use
truth tables to solve different problems of logic gates, enabling everyone to understand the core concept even for an 8th grade student.
What is a Logic Gate?
A logic gate is a basic block and element of digital circuits. It takes one or more binary inputs and produces single binary output. The most common logic gates are as follows:
• AND Gate: Outputs 1 only when all inputs are 1.
• OR Gate: Outputs 1 if at least one input is 1.
• NOT Gate: Outputs the opposite of the input (inverts the signal).
• NAND Gate: Outputs 0 only when all inputs are 1.
• NOR Gate: Outputs 0 if any input is 1.
• XOR Gate: Outputs 1 if the inputs are different.
• XNOR Gate: Outputs 1 if the inputs are the same.
What is a Truth Table?
A truth table lists down all possible input combinations to form a logic gate and also displays its corresponding output. It helps a logic engineer to draw digital circuit and predict how a digital
circuit will behave based on these inputs.
For example, for a 2-input AND gate, the truth table looks like this:
Input A Input B Output (A AND B)
As you can see, the AND gate only outputs a 1 when both inputs are 1.
How to Build a Truth Table?
To build a truth table, follow these simple steps:
1. List all possible input combinations: The number of rows in your truth table depends on the number of inputs. For 2 inputs, you need 4 rows (2^2). For 3 inputs, there will be 8 rows (2^3), and so
2. Determine the output for each combination: Based on the logic gate’s function, fill in the output column. For example, if you are working with an OR gate, the output will be 1 if either of the
inputs is 1.
3. Check for any additional operations: Some problems involve more than one gate. In such cases, you’ll need to evaluate each gate step by step.
Solving a Logic Gate Problem Using Truth Tables
Let’s take an example where we are going to solve a problem using a truth table.
Problem: You have a digital circuit with two inputs (A and B) connected to an AND gate, and the output of the AND gate is fed into a NOT gate. What will be the final output?
1. Create the truth table for the AND gate. We already know that an AND gate only outputs 1 when both A and B are 1.
Input A Input B AND Output (A AND B)
2. Add the NOT gate: A NOT gate inverts the input. Therefore, the output of the AND gate will be inverted by the NOT gate.
Input A Input B AND Output NOT Output
Final Answer: The final output will only be 0 when both inputs A and B are 1. In all other cases, the output will be 1.
Common Logic Gates and Their Truth Tables
Here are some more examples of common logic gates and their truth tables:
1. OR Gate:
Input A Input B Output (A OR B)
2. NOT Gate (Single input):
Input A Output (NOT A)
3. XOR Gate:
Input A Input B Output (A XOR B)
Applications of Truth Tables
Truth tables are used in various fields:
• Digital Circuit Design: Engineers use truth tables to design and troubleshoot circuits.
• Boolean Algebra: Logic gates can be represented as Boolean expressions, and truth tables help simplify these expressions.
• Problem-Solving: They are crucial in programming, computer science, and mathematics for solving logical problems efficiently.
Using truth tables and specially a truth table generator in a systematic and straightforward way to solve logic gates problem is a genius mindset. By listing all possible input combinations and
applying the core function of each gate step by step, one can easily determine the output of even more complex circuits. | {"url":"https://truthtablegen.com/truth-table-solve-logic-gate-problems/","timestamp":"2024-11-05T09:54:58Z","content_type":"text/html","content_length":"124837","record_id":"<urn:uuid:1267f1ae-92a0-4b15-b077-6eae9ba80292>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00562.warc.gz"} |
Technical Indicators and Oscillators - The Best 3 for Beginners
Analysis paralysis is a real challenge when it comes to learning how to day trade, especially when we encounter the maze of technical indicators and oscillators.
As an introduction for beginners, we want to try to simplify this area of technical analysis by focusing on 3 popular indicators that lay the groundwork for successful trading.
In this article, we’ll discuss moving averages, RSI, and MACD, explaining them in straightforward terms. There are many others, and we will cover them all eventually, but these are good ones to start
What are Technical Indicators?
Technical indicators are tools used in trading that help analyze past and current price information of stocks, currencies, or other financial instruments to predict future market movements. They are
primarily used to assess trends, momentum, volume, and other aspects of a security to aid in making investment decisions.
Essentially, these indicators take the raw data of market prices and volumes and transform them into a more digestible and usable form. They can be as simple as a line drawn on a chart or as complex
as a series of calculations based on price and volume history.
Building on this, it’s important to differentiate between general indicators and a specific subset known as oscillators. While all oscillators are indicators—providing valuable insights into market
behavior—not all indicators are oscillators.
What Does Oscillation Mean?
Oscillation, in simple terms, is just a fancy way of saying “back and forth movement” or “fluctuation.” Here’s a simple breakdown of what oscillation means in different contexts:
In Physics
Oscillation is the movement back and forth in a regular rhythm. A classic example is a pendulum swinging from side to side from its equilibrium or resting position. This movement is governed by
forces that restore the object to a midpoint or equilibrium position.
Oscillators in Technical Analysis
In technical analysis, oscillators are tools that help us understand when a stock’s price might be getting ready to change direction. They measure how strong the price movement is and whether the
price is moving quickly or slowly within a defined range.
Think of it like a speedometer in a car that shows how fast you’re driving. Similarly, oscillators in the stock market don’t measure speed, but rather momentum – which reflects the rate at which
stock prices are changing. By indicating how rapidly prices are moving, oscillators can signal whether a price trend is likely to continue or reverse, helping traders decide when to buy or sell.
How Oscillators Work
• Range: Most of the time, oscillators have a scale that ranges from 0 to 100.
• High and Low Signals: They give signals when things are extreme. If the value is very high (close to 100), it might mean the stock price is too high and could start going down soon. If the value
is very low (close to 0), it might mean the stock price is too low and could start going up.
Oscillators are invaluable because they help traders determine the optimal times to buy or sell, based on the underlying momentum of the market. Like using a thermometer to gauge the temperature
before heading out, traders use oscillators to “feel the pulse” of the market dynamics. This guidance helps in making informed decisions, predicting when a trend is likely to continue, or when a
reversal might occur.
Why Use Indicators and Oscillators?
Both technical indicators and oscillators are used by traders to make more informed trading decisions:
• Trend Confirmation: Indicators can confirm if a market trend is likely to continue or reverse, helping traders decide to buy or sell.
• Signal Trading Opportunities: Certain patterns in the indicators can suggest optimal moments to enter or exit trades.
• Reduce Risk: By providing objective data on market conditions, these tools can help traders avoid emotional decision-making and focus on evidence-based strategies.
How They Fit into Technical Analysis
Technical analysis itself is the study of price action, primarily through the use of candlestick charts, for the purpose of forecasting future price trends. Technical indicators and oscillators are
integral to this practice. They provide graphical representations and mathematical calculations of market dynamics that are not readily apparent just by looking at a price chart.
By using technical indicators and oscillators, traders can see patterns and signals that help them understand market movements in a way that raw data alone cannot provide.
For beginners, mastering a few basic indicators and understanding how to interpret them can be a significant first step in becoming proficient at technical analysis.
The charts below are from the charting platform TradingView. You can sign up and use their software for free.
Visit the official website here: TradingView.com
Moving Averages
Moving averages are one of the most popular tools in technical analysis used to smooth out price data over a specified period, making it easier to identify the trend direction. By calculating the
average price of a stock over a set number of days, moving averages offer a clear view of whether a stock’s price is trending up or down, without the distraction of daily price fluctuations.
How Do Moving Averages Work?
A moving average is calculated by averaging the stock’s closing prices over a specific period. For instance, a 20-day Simple Moving Average (SMA) sums up the closing prices of the last 20 days and
divides this total by 20. This process is repeated daily, adding the new day’s closing price while dropping the oldest, which smooths out short-term volatility and reveals the underlying price trend.
Why Use Moving Averages?
Traders utilize moving averages for several key reasons:
• Trend Identification: Moving averages help pinpoint the overall trend. If a stock price remains above a certain moving average, like the 50-day SMA, it’s generally considered in an upward trend,
suggesting a buy signal. Conversely, if the price is below the moving average, it might indicate a downward trend, signaling a potential sell.
• Support and Resistance Levels: These averages can act as barriers at which a stock’s price movement is halted and possibly reversed, providing strategic points for entry or exit. Learn more about
support and resistance here.
• Crossover Strategies: Traders often watch for a shorter-term moving average (like a 10-day SMA) crossing over a longer-term one (like a 20-day SMA) to signal increasing momentum and potentially
initiating a new trade position.
Simple and Advanced Forms of Moving Averages
While the Simple Moving Average (SMA) gives equal weight to all data points, other types of moving averages like the Exponential Moving Average (EMA) and Weighted Moving Average (WMA) assign more
importance to recent prices. This can help traders react more quickly to recent price changes:
• Exponential Moving Average (EMA): More sensitive to recent price movements, the EMA helps traders catch trends early.
• Weighted Moving Average (WMA): Prioritizes recent prices over older ones, offering a weighted average that reacts more promptly to new market information.
Both EMA and WMA are designed to overcome the shortcomings of a simple moving average by putting more emphasis on recent data, which can be crucial in fast-paced financial markets.
Moving averages help you see the average price of a security over a certain period, making it easier to spot trends.
Lagging and Leading Indicators
A moving average is a lagging indicator because it calculates an average of past prices. This smooths out price data, clarifying the trend direction after it has begun but not predicting new trends.
Its reliance on historical data means it’s excellent for confirming ongoing trends but less effective for spotting the start or end of these trends.
In contrast, leading indicators like the Relative Strength Index (RSI) aim to predict future price movements by identifying early signs of overbought or oversold conditions. These indicators can
offer early warnings about potential market reversals, although they are susceptible to false signals and generally need to be used with other analysis tools for best results.
RSI (Relative Strength Index)
The Relative Strength Index (RSI) is an oscillator in technical analysis that measures the speed and change of price movements of a stock or asset. It operates on a scale from 0 to 100 and helps
identify overbought or oversold conditions, suggesting potential price reversals. The RSI is calculated by comparing the average gains and losses over a specific period, typically 14 days.
How Does RSI Work?
• Calculation: To determine the RSI, start by calculating the average price increase (gains) and the average price decrease (losses) over the period. The ratio of these averages is known as the
“relative strength” (RS).
• Formula: The RSI is then calculated using the formula: RSI = 100 – (100 / (1 + RS)). This formula helps normalize the index to fit within the 0 to 100 range.
Interpreting RSI Values
Overbought Conditions: When the RSI value exceeds 70, it often suggests that the stock might be overbought. This condition implies that recent gains are disproportionately high compared to any losses
during the same period. Essentially, the stock has experienced strong upward momentum without significant pullbacks, making it potentially overvalued or primed for a price correction.
Here’s a breakdown of the mechanics:
• The RSI compares the magnitude of recent gains to recent losses.
• If the average gains over the RSI’s calculation period (commonly 14 periods) are larger than the average losses, the RSI increases.
• An RSI above 70 indicates that the stock has been gaining significantly, often due to heightened buying activity, which might not be sustainable. This can lead investors to anticipate a pullback
as some traders might start taking profits, expecting that the rapid price increases are likely to slow down or reverse.
This high RSI value signals that the upward price momentum is possibly overextended, making it a critical indicator for traders looking to predict potential reversals from recent trends.
Oversold Conditions: When the RSI value drops below 30, it typically indicates that the stock may be oversold. This condition suggests that recent losses are disproportionately high compared to any
gains during the same period, potentially signaling that the stock is undervalued or due for a price rebound.
• An RSI below 30 signifies that the stock has been experiencing significant selling pressure, often leading to a downward price push that might be excessive relative to its true value.
This low RSI value signals that the downward price momentum may be overextended. It acts as a critical alert for traders who might see this as a buying opportunity, anticipating a potential price
increase as the market corrects itself from this oversold state.
Why Use RSI?
RSI is particularly useful in volatile markets where prices oscillate within a consistent range. It provides clear signals that can help traders:
• Identify potential entry and exit points based on perceived overbought or oversold conditions.
• Detect divergences where the price movement differs from the RSI trend, often indicating a potential price reversal.
RSI and Divergence
Bearish divergence is a situation where the stock price hits a new high while the RSI fails to reach a new high suggests weakening momentum and a possible price decline. This interpretation suggests
that while prices are reaching higher highs, the momentum behind those price increases is weakening. The RSI, which measures the magnitude and velocity of directional price movements, isn’t
confirming the new highs with higher readings of its own. This lack of confirmation can be a warning signal that the upward price movement might not be sustained, possibly leading to a future
The RSI is like the thermometer of the stock market. It measures the speed and change of price movements, helping you identify if a security is overbought or oversold.
MACD (Moving Average Convergence Divergence)
The MACD, or Moving Average Convergence Divergence, is a powerful tool used by traders to understand market trends. At its core, the MACD is about comparing two moving averages of stock prices — one
that covers a shorter period and one that covers a longer period. This comparison helps traders spot changes in the direction and strength of stock prices over time. By observing how these two
averages converge (come together) or diverge (move apart), the MACD provides clear signals that can guide traders on when to buy or sell a stock. It’s like having a personal assistant that helps you
decide when to get into or out of the market, making it easier to manage your trades effectively.
How Does MACD Work?
• MACD Line: The MACD Line is the foundation of the indicator and is calculated by subtracting the 26-day exponential moving average (EMA) of the stock’s price from its 12-day EMA.
• Signal Line: This is the 9-day EMA of the MACD Line, acting as a trigger for buy and sell signals.
• Histogram: The histogram simplifies the relationship between the MACD Line and the Signal Line. It’s a visual tool on a stock chart, showing the difference between the MACD Line and the Signal
Line through bars. If the bars are above zero (MACD Line is above the Signal Line), it suggests a potential increase in stock price, and bars below zero (MACD Line is below the Signal Line)
indicate a possible decline. The height of these bars helps traders quickly gauge the strength of a trend and spot changes in momentum.
• Why these numbers (12 and 26)? These specific numbers are standard in the industry and were originally chosen based on the traditional working days in a month, roughly corresponding to two weeks
and one month of trading data. The 12-day EMA is a faster, more responsive average because it considers fewer days, making it sensitive to recent price movements. The 26-day EMA is slower,
smoothing out volatility and providing a longer-term trend line.
• What does this show? The difference between these two EMAs gives an immediate visual representation of the trend’s momentum. If the MACD Line is above zero, it suggests that the short-term
momentum is higher than the long-term momentum, indicating bullish conditions. If the MACD Line is below zero, it suggests bearish conditions.
Signal Line
This is essentially a smoother or “slower” version of the MACD Line, being the 9-day EMA of the MACD itself. It’s an average of an average so it smooths out the fluctuations more extensively.
• Role of the Signal Line: It acts as a trigger for trading signals. When the MACD Line crosses above the Signal Line, it’s typically seen as a bullish signal (good time to buy), and when it
crosses below, it’s seen as bearish (good time to sell). The slower nature of the Signal Line makes it useful as a trigger for trading signals. When the faster MACD Line crosses the slower Signal
Line, it indicates a potential change in momentum and trend strength, which traders might use to make buy or sell decisions.
Histogram: The Histogram is a graphical representation of the difference between the MACD Line and the Signal Line.
• What does the Histogram show? It provides a clear visual of how far apart the MACD Line and the Signal Line are, indicating the strength of the trend. Positive bars suggest that the MACD Line is
above the Signal Line (bullish), and negative bars indicate the MACD Line is below the Signal Line (bearish). Increasing bar length signals growing momentum, whereas decreasing length suggests
fading momentum.
Why ‘Convergence’ and ‘Divergence’?
• Convergence occurs when the two moving averages move towards each other. In the context of MACD, it happens when the MACD Line and the Signal Line draw closer together. This suggests that the
difference between the fast and slow EMAs of price is decreasing, often indicating a weakening momentum in the trend. Convergence can signal a potential end to the current trend or a weakening in
• Divergence occurs when the two moving averages move away from each other. For MACD, divergence is observed when the MACD Line moves away from the Signal Line. This indicates that the difference
between the fast and slow EMAs is increasing. Divergence can be a sign of strengthening momentum and often precedes a potential reversal in the trend.
A Good Summary: What Happens When the Lines Cross?
Here’s a detailed breakdown of what happens when the MACD Line crosses above the Signal Line:
• MACD Line Dynamics: The MACD Line represents the difference between two exponential moving averages (EMAs) of the stock’s prices – specifically, the 12-day EMA and the 26-day EMA. When the MACD
Line crosses above the Signal Line, it indicates that the 12-day EMA (which responds more quickly to recent price changes) has moved above the 26-day EMA. This shows that recent prices are rising
faster than they were over the longer period, suggesting an increase in short-term momentum relative to the longer term.
• Signal Line Dynamics: The Signal Line, being a 9-day EMA of the MACD Line, represents a smoothed version of these movements. It moves slower because it accumulates this data over a short period
but in a more averaged way. It trails behind the more immediate reactions seen in the MACD Line.
• Interpretation of the Crossover: When the MACD Line crosses above the Signal Line, it suggests that recent price momentum is strengthening and moving faster than the average of the recent past,
as reflected in the Signal Line. This is typically interpreted as a bullish signal, indicating that the prices might continue to rise and that it could be a good time to consider buying.
Essentially, this crossover points to a shift where short-term prices are gaining strength and outpacing the more moderated view of recent price trends, which can indicate a potential upward movement
in the market.
Why Does MACD Work?
The MACD is a helpful tool for traders because it puts together a couple of important trading concepts into one easy-to-read indicator. Essentially, it measures both the direction a stock price is
heading and how fast it’s moving in that direction. It does this by comparing two averages of the stock’s prices over different time periods: a shorter one for recent trends and a longer one for
overall direction. The difference between these averages forms the MACD Line, which shows if the stock’s momentum is increasing or decreasing.
Additionally, the MACD uses a Signal Line, which is a smoothed version of the MACD Line, and a Histogram, which shows the distance between the MACD Line and the Signal Line. These features help
traders see when the stock might be ready for a buy or sell by showing entry and exit signals clearly. This combination of features makes the MACD very valuable for traders looking to catch and
follow trends in fast-moving markets.
MACD is like the traffic light of the stock market. It shows the relationship between two moving averages and helps you decide when to enter or exit a trade. | {"url":"https://stokestrades.com/technical-indicators-and-oscillators/","timestamp":"2024-11-02T08:24:13Z","content_type":"text/html","content_length":"95866","record_id":"<urn:uuid:7ce406e7-5d1b-4ace-a1d1-d7fd855c7338>","cc-path":"CC-MAIN-2024-46/segments/1730477027709.8/warc/CC-MAIN-20241102071948-20241102101948-00145.warc.gz"} |
Algebraic proof - Higher
How do you represent even and odd numbers in proof?
An even number can be represented by 2n and an odd number can be represented by 2n+1.
What is disproof by counter example?
It is possible to disprove a statement by finding an example to show that a mathematical statement is false. | {"url":"https://evulpo.com/en/uk/dashboard/lesson/uk-m-ks4-02algebra-24proof","timestamp":"2024-11-10T12:59:53Z","content_type":"text/html","content_length":"955872","record_id":"<urn:uuid:3ef6920a-e57e-4fc4-9c4e-e8f52398f6fa>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00067.warc.gz"} |
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A seismic slope stability probabilistic model based on
Bishop's method using analytical approach", Scientia
Iranica, 22(3), pp. 728-741 (2015). | {"url":"https://scientiairanica.sharif.edu/article_4202.html","timestamp":"2024-11-09T13:42:53Z","content_type":"text/html","content_length":"51373","record_id":"<urn:uuid:306e8293-36e7-4205-b2fb-a090ad9cc5fa>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00827.warc.gz"} |
Binary Calculator | Advanced & Simple Calculations
Binary Calculator: Mastering Binary Operations and Conversions
Introduction to Binary Calculator
Understanding the Binary System
The binary system, often referred to as the base-2 numeral system, is fundamental in the world of computing and digital technology. A binary number is composed exclusively of two numbers represented
zeros and ones. For instance, 10001110101010 is an example of a binary number. Contrary to the decimal system where we count from 0 to 9 before adding another digit, the binary number system quickly
moves from 1 to 10, as it only uses two digits.
Binary Calculator: A Comprehensive Tool for Binary Operations
A binary calculator is an indispensable calculator tool designed to handle various operations in the binary system. It's essentially a base-2 calculator, equipped to perform binary calculation,
addition, subtraction, multiplication, and division, as well as binary to decimal conversions and vice versa.
How to Use the Binary Calculator
Performing Binary Calculations
Binary Addition, Subtraction, Multiplication, and Division
To execute binary calculations like binary addition calculator of, subtraction, division, or multiplication:
1. Enter two binary numbers.
2. Select the desired operation (+, -, ×, ÷).
3. Press “Calculate” to get the result in both binary and decimal formats.
Binary to Decimal Conversion
Converting Binary Value to Decimal
For converting a binary value to a decimal:
1. Enter the binary number in the designated section.
2. Press “Calculate” to obtain the decimal equivalent.
Decimal to Binary Conversion
Converting Decimal Value to Binary
To convert decimal to binary:
1. Input the decimal number.
2. Click “Calculate” to get the binary equivalent.
In-Depth Understanding of Binary Numbers
Binary Number Formation
In the binary system, numbers are formed similarly to the decimal system, but reaching the number 10 happens much sooner due to the use of only two digits (0 and 1 convert binary). For example, 2 in
decimal equals 10 in binary.
Binary and Decimal Equivalents
Here are some examples of decimal and binary equivalents:
• Decimal: 0 | Binary: 0
• Decimal: 1 | Binary: 1
• Decimal: 2 | Binary: 10
• Decimal: 3 | Binary: 11
• Decimal: 4 | Binary: 100
Note: In both systems, adding zeros in front values does not change the value (e.g., 06 in decimal or 0110 in binary for the number 6).
Comprehensive Guide to Binary Conversions
Converting Decimal to Binary
To convert a decimal number to binary, repeatedly divide or multiply the decimal number by 2 and record the remainders in equal whole. Writing down these remainders in reverse order gives the binary
Converting Binary to Decimal
For converting binary to decimal:
1. Start with the left-most digit, multiplying the previous number by 2, then add the current digit.
2. Repeat this process for each digit to get the decimal value.
Binary Operations Explained
Binary Addition Rules
Binary addition follows similar rules to decimal addition, but carrying over occurs when the sum reaches 2.
Binary Subtraction Process
Binary subtraction is akin to decimal subtraction, with borrowing rules slightly adjusted for the binary system.
Understanding Binary Multiplication
Binary multiplication follows straightforward rules, similar to basic arithmetic multiplication.
Binary Division Methodology
Binary division mirrors the long division process used in decimal numbers, adhering to specific binary division rules.
Historical Perspective of Binary Numbers
The Origin and Evolution
Binary numbers date back to the 17th century, conceptualized by Gottfried Wilhelm Leibniz. Significant contributions were later made by George Boole in the 19th century, forming the basis of Boolean
algebra. The real breakthrough came with the advent of electronic computing in the 20th century, establishing binary numbers as a cornerstone of digital technology.
Real-world Applications of Binary Numbers
Binary Numbers in Everyday Technology
Binary numbers find extensive use in various fields, from computer memory and digital imaging to telecommunications and automated machinery. They are integral to the functioning of modern cars,
medical equipment, and digital devices, showcasing the versatility, power and ubiquity of the binary system in our daily lives.
FAQ on Binary Calculators and Conversions
What is a Binary Calculator and How Does it Work?
A binary calculator is a tool designed for performing arithmetic operations using the binary number system. This system uses only two digits, 0 and 1, and is commonly used in computing and digital
electronics. The binary calculator performs basic operations like addition, subtraction, multiplication, and division, using binary values as input.
How Do You Convert Binary to Decimal?
Converting binary to decimal involves a step process where each digit of the binary number is multiplied by the power of 2 based on its position. The sum of these values gives the decimal equivalent
to convert binary to. For example, to convert 1010 from binary to decimal, you multiply each digit by the corresponding power of 2 and add them up.
Can Binary Calculators Handle Negative Numbers?
Yes, binary calculators can handle negative numbers using a method called two's complement. In this system, the first bit represents the sign of the number (0 for positive, 1 for negative), and the
remaining bits represent the value.
What is the Decimal System and How is it Different from Binary?
The decimal system is a base-10 system using ten digits (0 through 9) and is the most commonly used system for representing numbers. In contrast, the binary system is a base-2 system, using only two
digits (0 and 1). The decimal system is used in everyday counting, while the binary system is fundamental in computing.
How Do You Perform Binary Addition?
Binary addition follows similar rules to decimal addition but with two digits only. When you add 1 and 1 in binary, the sum is 10, where 0 is written in the sum's position in third column, and 1 is
carried over to the next column.
What Does it Mean to Multiply in Binary?
Multiplying in binary follows the same concept as in the decimal system but is simpler since the digits involved are only 0 and 1. When you multiply any binary number with 0 or 1, the product either
becomes 0 or the number itself.
How Do You Subtract in Binary?
Subtraction in binary requires borrowing, similar to the decimal system. However, since there are only two digits, for example, when you subtract 1 from 0, you need to borrow from the next
higher-order bit, turning the 0 into 2 (in binary form) before performing the subtraction.
How Does Binary Division Work?
Binary division is akin to long division in the decimal system. The dividend is divided by the divisor, and the quotient is written above the dividend. The remainder, if any, is represented in binary
What are the Steps Involved in Converting Decimal to Binary?
To convert a decimal number to binary, repeatedly divide the decimal number by 2 and keep track of the remainders. Write these remainders in reverse order to form the binary equivalent. This step
process is straightforward but requires attention to detail to ensure accuracy.
Why are Binary Numbers So Important in Computers?
Binary numbers are crucial in computers because they can easily represent the two states of electronic components: on and off. Computers use binary numbers to perform calculations and store data.
Each binary digit (bit) represents a power of 2, and combined, these bits can represent complex data and carry out instructions efficiently.
These FAQs cover the essential concepts of computer binary calculators, binary and decimal systems, and the process of converting between these two number systems, highlighting the importance of
binary numbers in computing and digital technology. | {"url":"https://www.size.ly/calculator/binary-calculator","timestamp":"2024-11-05T00:01:11Z","content_type":"text/html","content_length":"63580","record_id":"<urn:uuid:f1c8f1b2-1516-41cb-9661-f23c87fc80ff>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00637.warc.gz"} |
FY2022 Annual Report
Quantum Gravity Unit
Assistant Professor Yasha Neiman
(From left to right) Dr. Slava Lysov, Dr. Sebastian Murk, Prof. Yasha Neiman, Dr. Aritra Banerjee, Julian Lang, David O'Connell
This year, we worked on various projects. Yasha and Slava Lysov worked on a reformulation of higher-spin gravity using its BPS solution. Yasha also worked on the quartic locality of higher-spin
gravity, and on self-dual General Relativity in de Sitter space, while Slava worked on the tropical-geometry version of mirror symmetry. Mirian Tsulaia and Dorin Weissman worked on higher-spin
interaction vertices. Mirian also worked on supersymmetric quantum mechanics, while Dorin worked on holographic computations of QCD scattering amplitudes. Aritra Banerjee worked on fundamental and
condensed-matter applications of Carrollian symmetry. Sebastian Murk worked on consistency conditions for astrophysical black-hole-like compact objects. Ardak Kussainova worked on solid-state
physics. David O’Connell (Ph.D. student) worked on the topology and geometry of non-Hausdorff manifolds. Julian Lang (Ph.D. student) worked on the twistor space picture of type-B higher-spin
1. Staff
• Dr. Mirian Tsulaia, Staff Scientist
• Dr. Vyacheslav Lysov, Postdoctoral Scholar
• Dr. Dorin Weissman, Postdoctoral Scholar
• Dr. Aritra Banerjee, Postdoctoral Scholar
• Dr. Sebastian Murk, Postdoctoral Scholar
• Dr. Ardak Kussainova, Postdoctoral Scholar
• David O'Connell, Graduate Student
• Julian Lang, Graduate Student
• Lena Hashimoto, Research Unit Administrator
2. Collaborations
2.1 Supersymmetric cubic interactions from higher-spin approach
• Description: Continued collaboration from previous FY
• Type of collaboration: Joint research
• Researchers:
□ Dr. Mirian Tsulaia (OIST)
□ Dr. Dorin Weissman (OIST)
□ Prof. Joseph Buchbinder (Tomsk Pedagogical Institute and Tomsk State University)
□ Prof. Vladimir Krykhtin (Tomsk Pedagogical Institute)
2.2 Supersymmetric approach to discrete-time Schrodinger equation
• Description: Inter-unit collaboration within OIST (continued from last FY)
• Type of collaboration: Joint research
• Researchers:
□ Dr. Mirian Tsulaia (OIST Quantum Gravity Unit)
□ Dr. Jonas Sonnenchein (OIST Theory of Quantum Matter Unit)
2.3 Mirror symmetry from tropical geometry
• Description: Continued collaboration from previous FY
• Type of collaboration: Joint research
• Researchers:
□ Dr. Vyacheslav Lysov (OIST)
□ Prof. Andrei Losev (National Research University Higher School of Economics, Moscow)
2.4 Holographic description of hadron scattering
• Description: Continued collaboration from previous FY
• Type of collaboration: Joint research
• Researchers:
□ Dr. Dorin Weissman (OIST)
□ Prof. Jacob Sonnenschein (Tel Aviv University)
□ Prof. Massimo Bianchi (Rome University Tor Vergata)
□ Dr. Maurizio Ferrotta (Rome University Tor Vergata)
2.5 String and field theory in the Carrollian limit
• Description: Continued & expanded collaboration from previous FY
• Type of collaboration: Joint research
• Researchers:
□ Dr. Aritra Banerjee (OIST)
□ Prof. Shankhadeep Chakrabortty (IIT Ropar)
□ Prof. Arpan Bhattacharyya (IIT Gandhinagar)
□ Prof. Arjun Bagchi (IIT Kanpur)
□ Dr. Kedar S. Kolekar (IIT Kanpur)
□ Sudipta Dutta (IIT Kanpur)
□ Punit Sharma (IIT Kanpur)
□ Ritankar Chatterjee (IIT Kanpur)
□ Saikat Mondal (IIT Kanpur)
□ Dr. Hisayoshi Muraki (POSTECH, Pohang)
□ Dr. Aditya Mehra (BITS Pilani and University of Edinburgh)
2.6 Carrollian symmetry in a Graphene system
• Type of collaboration: Joint research
• Researchers:
□ Dr. Aritra Banerjee (OIST)
□ Prof. Rudranil Basu (BITS Pilani)
□ Prof. Arjun Bagchi (IIT Kanpur)
□ Saikat Mondal (IIT Kanpur)
□ Minhajul Islam (IIT Kanpur)
2.7 Bootstrap method for quantum mechanics with periodic potential
• Type of collaboration: Joint research
• Researchers:
□ Dr. Aritra Banerjee (OIST)
□ Matthew Blacker (Cambridge University and OIST)
□ Prof. Arpan Bhattacharyya (IIT Gandhinagar)
2.8 Consistency conditions for alternative black hole models
• Type of collaboration: Joint research
• Researchers:
□ Dr. Sebastian Murk (OIST)
□ Prof. Robert Mann (University of Waterloo)
□ Prof. Daniel Terno (Macquarie University)
□ Ioannis Soranidis (Macquarie University)
3. Activities and Findings
3.1 New formulation of higher-spin gravity
Yasha Neiman and Slava Lysov have developed a new formulation of higher-spin gravity. The formulation uses novel diagrammatic rules that interpolate between those of field and string theory. The main
new ingredient is the Didenko-Vasiliev “BPS black hole”, which we understood to be the holographic dual of a bilocal operator in the boundary theory. The formulation uses only cubic vertices, some of
which are still unknown, but have been proved to be local. The infinite tower of unknown and potentially non-local vertices at quartic and higher orders is completely avoided.
3.2 Quartic locality of higher-spin gravity
Yasha Neiman has re-examined the argument that the higher-spin gravity (in the standard, field-theoretic formulation) must be non-local at the quartic order. I found that the argument is valid for
Lorentzian Anti-de Sitter space, but can be falsified by direct computation in Euclidean Anti-de Sitter or Lorentzian de Sitter space. Thus, higher-spin gravity provides a first example of a theory
whose non-locality depends on the spacetime signature and the sign of the cosmological constant. Moreover, while my interest in this theory was motivated by its compatibility with de Sitter space, we
now see that it prefers de Sitter space from the locality point of view.
3.3 Self-dual gravity in de Sitter space
Yasha Neiman has applied the novel Krasnov formulation of General Relativity to the problem of (non-linear) self-dual perturbations over de Sitter space. I found a lightcone ansatz for the solutions
in Poincare coordinates, and showed how they can be applied to compute scattering in an observer’s static patch. Since the geometry is fluctuating, the static-patch scattering problem must be posed
with great care. I showed how this can be done for the self-dual sector.
3.4 Topology and geometry of non-Hausdorff manifolds
David O’Connell continued developing the geometry of non-Hausdorff manifolds, with the aim of describing spacetimes with splitting and re-joining causal structures. In a preliminary result, he
managed to demonstrate the Gauss-Bonnet theorem, which plays a key role in 2-dimensional gravity theories, in the perturbative expansion of string theory, and in black hole thermodynamics.
3.5 Holographic QCD scattering computations
Dorin Weissman and collaborators have pursued the study of scattering amplitudes in AdS/CFT, in which the boundary theory is a model of Quantum Chromodynamics, while the bulk is described by
scattering strings. In particular, they characterized the signatures of chaos in these processes.
3.6 Application of Carrollian symmetry to a Graphene system
Aritra Banerjee and collaborators have continued their study of field theories with Carrollian symmetry and their applications. While the initial motivation for this work was the BMS symmetry of
scattering amplitudes, they recently discovered a surprising and completely different arena for Carrollian dynamics, in a 2-layer Graphene system. The Carrollian symmetry predicts a special
electronic band structure for this novel material.
3.7 Modern methods for old-fashioned Quantum Mechanical problems
Two recent projects involved the application of a technique from modern fundamental theory to problems in ordinary Quantum Mechanics. One was by Mirian Tsulaia and Jonas Sonnenschein (Shannon Unit at
OIST), who applied supersymmetry to solve the time-discretized Schordinger equation for a variety of potentials. Another was by Aritra Banerjee and Matthew Blacker (intern), who applied the bootstrap
technique from Confromal Field Theory to solve the Schrodinger equation with a periodic potential.
4. Publications
4.1 Journals
1. Bagchi, A., Banerjee, A., Basu, R., Islam, M., Mondal, S. “Magic fermions: Carroll and flat bands”. Journal of High Energy Physics, doi:10.1007/JHEP03(2023)227 (2023).
2. Tsulaia, M., Weissman, D. “Supersymmetric chiral quantum higher spin gravity”. Journal of High Energy Physics, doi:10.1007/JHEP12(2022)002 (2022).
3. Blacker, MJ., Bhattacharyya, A., Banerjee, A. “Bootstrapping the Kronig-Penney model”. Physical Review D, doi:10.1103/PhysRevD.106.085005 (2022).
4. Lysov, V., Neiman, Y. “Bulk locality and gauge invariance for boundary-bilocal cubic correlators in higher-spin gravity”. Journal of High Energy Physics, doi:10.1007/JHEP12(2022)142 (2022).
5. Neiman, Y. “Five questions about higher-spin holography in de Sitter space”. Proceedings of Science, doi:10.22323/1.406.0355 (2022).
6. Bianchi, M., Firrotta, M., Sonnenschein, J., Weissman, D. “Measure for chaotic scattering amplitudes”. Physical Review Letters, doi:10.1103/PhysRevLett.129.261601 (2022).
7. Lysov, V., Neiman, Y. “Higher-spin gravity’s ‘string’: new gauge and proof of holographic duality for the linearized Didenko-Vasiliev solution”. Journal of High Energy Physics, doi:10.1007/JHEP10
(2022)054 (2022).
8. Banerjee, A., Mehra, A. “Maximally symmetric nonlinear extension of electrodynamics with Galilean conformal symmetries”. Physical Review D, doi:10.1103/PhysRevD.106.085005 (2022).
9. Banerjee, A., Bhattacharyya, A., Drashni, P., Pawar, S. “From CFTs to theories with Bondi-Metzner-Sachs symmetries: Complexity and out-of-time-ordered correlators”. Physical Review D, doi:10.1103
/PhysRevD.106.126022 (2022).
10. Tsulaia, M., Sonnenschein, J. “A note on shape invariant potentials for discretized Hamiltonians”. Modern Physics Letters A, doi:10.1142/S021773232250153X (2022).
11. Bagchi, A., Banerjee, A., Muraki, H. “Boosting to BMS”. Journal of High Energy Physics, doi:10.1007/JHEP09(2022)251 (2022).
12. Bagchi, A., Banerjee, A., Dutta, S., Kolekar, KS., Sharma, P. “Carroll covariant scalar fields in two dimensions”. Journal of High Energy Physics, doi:10.1007/JHEP01(2023)072 (2023).
13. Buchbinder, IL., Krykhtin, VA., Tsulaia, M., Weissman, D. “Supersymmetric cubic interactions for lower spins from ‘higher spin’ approach”. Proceedings of Science, doi:10.22323/1.412.0035 (2022).
14. Bianchi, M., Firrotta, M., Sonnenschein, J., Weissman, D. “Partonic behavior of string scattering amplitudes from holographic QCD models”. Journal of High Energy Physics, doi:10.1007/JHEP05(2022)
058 (2022).
15. Bagchi, A., Banerjee, A., Chakrabortty, S., Chatterjee, R. “A Rindler road to Carrollian worldsheets”. Journal of High Energy Physics, doi:10.1007/JHEP04(2022)082 (2022).
16. O’Connell, D., “Non-Hausdorff manifolds via adjunction spaces”. Topology and its Applications, doi:10.1016/j.topol.2022.108388 (2023).
4.2 Books and other one-time publications
1. Murk, S. “Constraining modified gravity theories with physical black holes”. The 16^th Marcel Grossmann meeting, doi:10.1142/9789811269776_0109 (2023).
4.3 Oral and Poster Presentations
1. Bagchi, A., Banerjee, A., Basu, R., Islam, M., Mondal, S. “Magic fermions: Carroll and flat bands”. Carroll fermions, flat bands, Graphene and all that, seminar at National Sun-Yat-Sen
University, Sizihwan, Taiwan, Mar 24 (2023).
2. Losev, A., Lysov, V. “Tropical mirror symmetry: correlation functions”. Tropical geometry and mirror symmetry, OIST workshop “Women at the intersection of mathematics and theoretical physics meet
in Okinawa”, Onna-son, Japan, Mar 20-24 (2023).
3. Bagchi, A., Banerjee, A., Basu, R., Islam, M., Mondal, S. “Magic fermions: Carroll and flat bands”. Carroll fermions, flat bands, Graphene and all that, seminar at Kyoto University, Kyoto, Japan,
Mar 7 (2023).
4. Neiman, Y. “Self-dual gravity in de Sitter space: lightcone ansatz and static-patch scattering”. Perturbation theory for self-dual gravity in de Sitter space, seminar at Nottingham University,
Nottingham, UK, Feb 24 (2023).
5. Neiman, Y. “Quartic locality of higher-spin gravity in de Sitter and Euclidean Anti-de Sitter space”. Locality of higher-spin gravity in de Sitter vs. Anti-de Sitter space, seminar at King’s
College London, London, UK, Feb 20 (2023).
6. Losev, A., Lysov, V. “Tropical mirror symmetry: correlation functions”. Tropical mirror symmetry, RIKKYO MathPhys 2023 conference, Tokyo, Japan, Jan 7-8 (2023).
7. Mann, R., Murk, S., Terno, DR. “Black holes and their horizons in semiclassical and modified theories of gravity”. Physical black holes in modified theories of gravity, 24^th Australian Institute
of Physics Congress, Adelaide, Australia, Dec 11-16 (2022).
8. Mann, R., Murk, S., Terno, DR. “Surface gravity and the information loss problem”. Surface gravity and information loss, 24^th Australian Institute of Physics Congress, Adelaide, Australia, Dec
11-16 (2022).
9. Neiman, Y. “New diagrammatic framework for higher-spin gravity”. On the quartic locality problem in higher-spin gravity, Joint Canada-Asia Pacific Conference on General Relativity and
Relativistic Astrophysics, Pohang, Korea, Nov 28-Dec 2 (2023).
10. Neiman, Y. “New diagrammatic framework for higher-spin gravity”. A workaround and a direct objection to the quartic non-locality result, APCTP workshop “Higher Spin Gravity and its Applications”,
Pohang, Korea, Oct 12-17 (2023).
11. Neiman, Y. “New diagrammatic framework for higher-spin gravity”. Local reformulation of higher-spin gravity, seminar at Weizmann Institute, Rehovot, Israel, Sep 19 (2022).
12. Bagchi, A., Banerjee, A., Muraki, H. “Boosting to BMS”. The curious case of Carrollian theories, seminar at BITS Pilani, Pilani, India, Sep 17 (2022).
13. Buchbinder, IL., Krykhtin, VA., Tsulaia, M., Weissman, D. “Cubic vertices for N=1 supersymmetric massless higher spin fields in various dimensions”. Strings, supersymmetry and higher spins:
foundations and modern developments, Open Physics Seminar Series at V.N. Karazin Kharkiv National University, Kharkiv, Ukraine, Aug 15 (2022).
14. Bagchi, A., Banerjee, A., Muraki, H. “Boosting to BMS”. The curious case of Carrollian theories, seminar at Le Center de Physique Theorique, Paris, France, Aug 3 (2022).
15. Skvortsov, E., Tran, T., Tsulaia, M. “A stringy theory in three dimensions and massive higher spins”. Models with interacting massless and massive higher spin fields, seminar at Yukawa Institute
for Theoretical Physics, Kyoto, Japan, Jun 9 (2022).
16. Bagchi, A., Banerjee, A., Dutta, S., Kolekar, KS., Sharma, P. “Carroll covariant scalar fields in two dimensions”. The curious case of Carrollian theories, seminars at IIT Kharagpur, Kharagpur,
India, Apr 23-25 (2022).
5. Intellectual Property Rights and Other Specific Achievements
Nothing to report
6. Meetings and Events
6.1 BMS field theories with U(1) symmetry
• Date: February 15, 2023
• Venue: OIST Campus Lab4
• Speaker: Dr. Max Riegler (University of Vienna)
6.2 Probing quantum gravity and non-locality through R^2-like inflation
• Date: May 11, 2022
• Venue: OIST Campus Lab4
• Speaker: Dr. Sravan Kumar (Tokyo Institute of Technology)
• Date: June 17, 2010
• Venue: OIST Campus Lab1
• Co-organizers: The Institute of All But Cats (IABC)
• Speakers:
□ Dr. First Last (Affiliation)
□ Dr. First Last (Affiliation)
□ Dr. First Last (Affiliation)
7. Other
Nothing to report. | {"url":"https://groups.oist.jp/ja/qgu/fy2022-annual-report","timestamp":"2024-11-11T14:39:32Z","content_type":"text/html","content_length":"65075","record_id":"<urn:uuid:6afc2e71-3cbf-40d5-9b8d-2c5679b6d3be>","cc-path":"CC-MAIN-2024-46/segments/1730477028230.68/warc/CC-MAIN-20241111123424-20241111153424-00626.warc.gz"} |
Local Shape Blending using Coherent Weighted Regions
We present a new local shape blending system that maps a sparse configuration of facial markers captured from an actor on to target meshes, by blending highly detailed target key meshes. The
traditional local blending computes weight vectors of each region on the mesh with respect to the key meshes. Here all the vertices in a region share the weight vector of the region. While respecting
the coherence or quazi-rigidness of control points within each region, this method decouples the natural correlation between regions. This problem has been partly addressed by a recent soft-region
method which computes the weight vectors of each control point independently of each other, and then computes the weight vector of each vertex by scattered data interpolation over the weight vectors
of the control points. But this method ignores the coherence that may exist among control points. The underlying issue here is how to compare the observed con guration of control points and their con
guration computed by blending key meshes. We solve the problem by comparing the observed configuration and the computed con guration region by region, by designing regions to be coherent weighted
regions. These regions are de ned by weight functions that use the generalized distance between control points called the coherency-based distance: two control points are more coherent or more closer
when they are closer to each other in geometric distance, and they have more similar blending weight vectors. To use coherent weighted regions systematically, we formulate local shape blending as a
problem of nding an optimal con guration of blending weight vectors, assuming that the con gurations of weight vectors obey a Markov Random Field. | {"url":"https://dcollection.sogang.ac.kr/dcollection/srch/srchDetail/000000046338","timestamp":"2024-11-02T06:15:28Z","content_type":"text/html","content_length":"27301","record_id":"<urn:uuid:5cbd3b58-c3a7-41e8-829f-384bb0d25417>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00313.warc.gz"} |
: Book IX, Proposition 20
The following is as given in Sir Thomas L. Heath's translation, which can be found in the book The Thirteen Books of The Elements, Vol. 2.
Proposition 20.
Prime numbers are more than any assigned multitude of prime numbers.
Let A, B, C be the assigned prime numbers;
I say that there are more prime numbers than A, B, C.
For let the least number measured by A, B, C be taken,
and let it be DE;
let the unit DF be added to DE.
Then, EF is either prime or not.
First, let it be prime;
then the prime numbers A, B, C, EF have been found which are more than A, B, C.
Next, let EF not be prime;
therefore it is measured by some prime number. [VII. 31]
Let it be measured by the prime number G.
I say that G is not the same with any of the numbers A, B, C.
For, if possible, let it be so.
Now A, B, C measure DE;
therefore G also will measure DE.
But it also measures EF.
Therefore G, being a number, will measure the remainder, the unit DF;
which is absurd.
Therefore G is not the same with an y one of the numbers A, B, C.
And by hypothesis it is prime.
Therefore the prime numbers A, B, C, G have been found which are more than the assigned multitude of A, B, C. | {"url":"http://mathlair.allfunandgames.ca/elements9-20.php","timestamp":"2024-11-12T15:16:28Z","content_type":"text/html","content_length":"5368","record_id":"<urn:uuid:1f2f3b30-a7b7-4ae6-bde6-e776660d37bc>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.63/warc/CC-MAIN-20241112145015-20241112175015-00601.warc.gz"} |
2010 Symposia
A Ubiquity symposium is an organized debate around a proposition or point of view. It is a means to explore a complex issue from multiple perspectives. An early example of a symposium on teaching
computer science appeared in Communications of the ACM (December 1989).
To organize a symposium, please read our guidelines.
Ubiquity Symposium: What is Computation
Table of Contents
1. What is Computation, Editor's Introduction, by Peter J. Denning
2. What is Computation, Opening Statement, by Peter J. Denning
3. The Evolution of Computation, by Peter Wegner
4. Computation is Symbol Manipulation, by John S. Conery
5. Computation is Process, by Dennis J. Frailey
6. Computing and Computation, by Paul S. Rosenbloom
7. Computation and Information, by Ruzena Bajcsy
8. Computation and Fundamental Physics, by Dave Bacon
9. The Enduring Legacy of the Turing Machine, by Lance Fortnow
10. Computation and Computational Thinking, by Alfred V. Aho
11. What is the Right Computational Model for Continuous Scientific Problems?, by Joseph Traub
12. Computation, Uncertainty and Risk, by Jeffrey P. Buzen
13. Natural Computation, by Erol Gelenbe
14. Biological Computation, by Melanie Mitchell
15. What is Information?: Beyond the jungle of information theories, by Paolo Rocchi
16. What Have We Said About Computation?: Closing statement, by Peter J. Denning
• Ubiquity symposium 'What is computation?': Computation and information
by Ruzena Bajcsy
December 2010
In this sixth article in the ACM Ubiquity symposium,What is Computation? Ruzena Bajcsy of the University of California-Berkeley explains that computation can be seen as a transformation or
function of information.
• Ubiquity symposium 'What is computation?': Computing and computation
by Paul S. Rosenbloom
December 2010
In this fifth article in the ACM Ubiquity symposium on What is computation? Paul S. Rosenbloom explains why he believes computing is the fourth great scientific domain, on par with the physical,
life, and social sciences.
• Ubiquity symposium 'What is computation?': Computation and Fundamental Physics
by Dave Bacon
December 2010
In this seventh article in the ACM Ubiquity symposium, What is Computation?, Dave Bacon of University of Washington explains why he thinks discussing the question is as important as thinking
about what it means to be self-aware. —Editor
• Ubiquity symposium 'What is computation?': Computation is process
by Dennis J. Frailey
November 2010
Various authors define forms of computation as specialized types of processes. As the scope of computation widens, the range of such specialties increases. Dennis J. Frailey posits that the
essence of computation can be found in any form of process, hence the title and the thesis of this paper in the Ubiquity symposium discussion what is computation. --Editor
• Ubiquity symposium 'What is Computation?': The evolution of computation
by Peter Wegner
November 2010
In this second article in the ACM Ubiquity symposium on 'What is computation?' Peter Wegner provides a history of the evolution of comptuation. --Editor
• Ubiquity symposium 'What is computation?': Computation is symbol manipulation
by John S. Conery
November 2010
In the second in the series of articles in the Ubiquity Symposium What is Computation?, Prof. John S. Conery of the University of Oregon explains why he believes computation can be seen as symbol
manipulation. For more articles in this series, see table of contents in the http://ubiquity.acm.org/article.cfm?id=1870596 Editors Introduction to the symposium. --Editor
• Ubiquity symposium 'What is computation?': Opening statement
by Peter J. Denning
November 2010
Most people understand a computation as a process evoked when a computational agent acts on its inputs under the control of an algorithm. The classical Turing machine model has long served as the
fundamental reference model because an appropriate Turing machine can simulate every other computational model known. The Turing model is a good abstraction for most digital computers because the
number of steps to execute a Turing machine algorithm is predictive of the running time of the computation on a digital computer. However, the Turing model is not as well matched for the natural,
interactive, and continuous information processes frequently encountered today. Other models whose structures more closely match the information processes involved give better predictions of
running time and space. Models based on transforming representations may be useful. | {"url":"https://ubiquity.acm.org/symposia2010.cfm?volume=2010","timestamp":"2024-11-07T19:37:04Z","content_type":"text/html","content_length":"22501","record_id":"<urn:uuid:7c065b36-dd88-4379-8dd2-ca341d40c659>","cc-path":"CC-MAIN-2024-46/segments/1730477028009.81/warc/CC-MAIN-20241107181317-20241107211317-00389.warc.gz"} |
Re: Bootload the MIMXRT 10xx devices in the field
11-09-2022 05:26 PM
1,328 Views
Hi All,
I need to add bootloader support to my project so the customer can update the code remotely. I noticed there is a Kinetis Flash Tool and an open-source project in python. I need the project files to
modify the bootloader for my own needs. Is there a Visual C++ or C# project I could get so I can modify it? I will be using MIMXRT1011 and MIMXRT1062 devices in my products and need to change how the
program is used and gray out functions I don't want the customer to mess with.
11-09-2022 08:33 PM
1,318 Views
11-09-2022 09:22 PM
1,317 Views
11-10-2022 11:23 PM
1,304 Views
11-11-2022 10:55 AM
1,299 Views
11-13-2022 07:42 PM
1,293 Views | {"url":"https://community.nxp.com/t5/i-MX-Processors/Bootload-the-MIMXRT-10xx-devices-in-the-field/m-p/1552696","timestamp":"2024-11-10T02:46:11Z","content_type":"text/html","content_length":"259286","record_id":"<urn:uuid:b70ef763-22c4-4948-8e51-d26ed08e0fca>","cc-path":"CC-MAIN-2024-46/segments/1730477028164.3/warc/CC-MAIN-20241110005602-20241110035602-00190.warc.gz"} |
Good question, since it's kind of a subtle thing in ITensor. For codes using quantum-number conserving tensors, such as when doing DMRG calculations, the quantum numbers present in any wavefunction
are not allowed to change. So the number is set by how you make your initial state.
So if you set your initial state to have 3 up and 1 down particles, and use QN conserving tensors, then the number will not change.
Please let me know if you still have a question of course!
Could you please clarify about QN conserving tensors and how to use it?
After defining the hamiltonian
auto H = IQMPO(ampo);
I write the state as follows
auto psi0 = InitState(sites, "Emp");
auto psi = IQMPS(psi0);
(3 up and 1 down particles).
I verified the ground state for small system with the exact value and noticed that the values are slightly different.
For example, -7.34478928869732605733
instead of the correct answer -7.46410162
I noticed that the discrepancy mentioned above is because of the PBC. When I consider the same model with OBC, I obtain almost exact result for small systems.
Yes, PBC calculations are much harder to converge. Did you turn on the “noise” parameter when setting your DMRG accuracy paramters (maxim, cutoff, etc)? This can help a lot with convergence. | {"url":"https://www.itensor.org/support/1280/number_of_particles?show=1282","timestamp":"2024-11-15T04:18:05Z","content_type":"text/html","content_length":"26579","record_id":"<urn:uuid:35c2eec1-c0a0-4680-a8a5-5427254fffdc>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00807.warc.gz"} |
seminars - Symplectic Torelli classes of positive entropy
By a Torelli class, we mean a symplectic mapping class acting trivially on homology. The study of symplectic Torelli groups has been a central topic of modern symplectic topology. In this talk, we
embark on a voyage of Torelli classes and entropies. It turns out that Floer theory can genuinely detect a positive topological entropy, namely that there is a Torelli class in a symplectic K3
surface having a positive topological entropy. This is joint work with Myeonggi Kwon. | {"url":"http://www.math.snu.ac.kr/board/index.php?mid=seminars&sort_index=speaker&order_type=asc&l=ko&page=85&document_srl=1126444","timestamp":"2024-11-09T19:12:13Z","content_type":"text/html","content_length":"45407","record_id":"<urn:uuid:e261cf25-9335-4b2f-84b2-d266fe76736f>","cc-path":"CC-MAIN-2024-46/segments/1730477028142.18/warc/CC-MAIN-20241109182954-20241109212954-00607.warc.gz"} |
Formation of multipartite entanglement using random quantum gates
The formation of multipartite quantum entanglement by repeated operation of one- and two-qubit gates is examined. The resulting entanglement is evaluated using two measures: the average bipartite
entanglement and the Groverian measure. A comparison is made between two geometries of the quantum register: a one-dimensional chain in which two-qubit gates apply only locally between nearest
neighbors and a nonlocal geometry in which such gates may apply between any pair of qubits. More specifically, we use a combination of random single-qubit rotations and a fixed two-qubit gate such as
the controlled-phase gate. It is found that in the nonlocal geometry the entanglement is generated at a higher rate. In both geometries, the Groverian measure converges to its asymptotic value more
slowly than the average bipartite entanglement. These results are expected to have implications on different proposed geometries of future quantum computers with local and nonlocal interactions
between the qubits.
Dive into the research topics of 'Formation of multipartite entanglement using random quantum gates'. Together they form a unique fingerprint. | {"url":"https://cris.huji.ac.il/en/publications/formation-of-multipartite-entanglement-using-random-quantum-gates","timestamp":"2024-11-06T18:17:54Z","content_type":"text/html","content_length":"47764","record_id":"<urn:uuid:9d4cc3e6-7770-4855-81a4-ffa67105da5c>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00856.warc.gz"} |
Review using column addition with regrouping in the ones and tens columns | Oak National Academy
Hello, my name's Mrs. Hopper and I'm really looking forward to working with you in our maths lesson today, from the unit "Reviewing Column Addition and Subtraction." Hopefully, there'll be things in
here that are familiar to you and you'll be able to revisit some ideas that you've learned about before.
So if you're ready, let's make a start.
So welcome to this lesson where we're going to be thinking about using column addition with regrouping in the ones and the tens.
So you may have been doing a lot of column addition recently, and you may have done some regrouping in the ones perhaps, but we're going to regroup in the ones and the tens today, so let's have a
look at what's going to be in our lesson.
Well, we've got two key words in our lesson, column addition and regroup.
So I'll take my turn to say them and then it'll be your turn.
My turn, column addition.
Your turn.
My turn, regroup.
Your turn.
Now you may well be very familiar with these words, but let's just remind ourselves what they mean 'cause they're going to be useful to us as we go through the lesson.
Column addition is a way of adding numbers by writing a number below another.
It's a way of representing our addition so that we can keep track of what we've added and what we've still got to add.
The process of unitizing and exchanging between place values is known as regrouping.
So for example, 10 tens can be regrouped for 100, and 100 can be regrouped for 10 tens.
And you can see in our image, 10 tens are the same as 100, and 100 is the same as 10 tens.
Sometimes we want to think about that number as 10 tens, and sometimes we want to think about it as 100, and thinking in both ways will be useful as we go through our lesson today.
So in the first part of our lesson, we're going to be regrouping ones and tens, and in the second part, we're going to be solving problems with regrouping.
So let's make a start.
And we've got Alex and Lucas helping us out today.
Alex is trying to remember how to use column addition with two digit numbers.
He says, "I remember working on this in Year 3, but I can't remember exactly how to do it." He says, "What happens when the sums of the ones and tens digits are both 10 or greater?" And he's got an
example here, 48 add 76.
8 ones add 6 ones have a sum greater than 10 ones, and 4 tens and 7 tens have a sum greater than 10 tens, so what would Alex do when he was adding these numbers together? Lucas says, "It's okay, we
have to use regrouping in the ones and the tens." So Alex wants to use column addition, regrouping the ones and tens to help him to add these numbers.
So he says, "What is the sum of 48 and 76?" And Lucas says, "Well, 8 ones add 6 ones is equal to 14 ones." And the 14 ones is regrouped into 1 ten and 4 ones, 'cause we know that's what 14 is, 1 ten
and 4 ones.
Then we can move the 10 into the tens column, so we can add it in with the tens when we get to the tens.
4 tens add 7 tens, and add the regrouped 1 ten is equal to 12 tens.
So 4 tens plus 7 tens is equal to 11 tens, plus the additional 10 is equal to 12 tens.
And 12 tens can be regrouped into 1 hundred and 2 tens.
The 100 is moved to the hundreds column, and the 2 is recorded in the tens column.
There are no other hundreds except the regrouped hundred, so 1 is written in the hundreds column because we've got that 1 hundred.
So 48 add 76 is equal to 124.
Alex wants to add three digit numbers using column addition with regrouping in the ones and tens, so Lucas sets him a challenge, "What's 257 add 186?" And Alex has set it out carefully with the
columns carefully lined up.
And he says, "7 ones add 6 ones is equal to 13 ones." 13 is regrouped into 1 ten and 3 ones.
The 10 is moved into the tens column.
So there we have it, the 10 ready to be added with the tens and the 3 of the 13 recorded in the ones column.
5 tens add 8 tens, and add the regrouped 10, is equal to 14 tens.
5 tens add 8 tens is 13 tens, add another one is 14 tens.
14 tens is regrouped into 1 hundred and 4 tens, and the 100 moved into the hundreds column.
So there we have it.
2 hundreds add 1 hundred, plus the regrouped 100, is equal to 4 hundreds.
So we record that in the hundreds column.
So 257 add 186 is equal to 443.
Sometimes, regrouping the ones means regrouping in the tens.
Have a look carefully at this one.
What do you notice? Alex says, "It only looks like I need to regroup in the ones because 4 tens add 5 tens is equal to 9 tens." Lucas says, "Watch what happens!" Oh, let's have a go at adding these
So starting with the ones, 7 ones add 8 ones is equal to 15 ones.
15 is regrouped into 1 ten and 5 ones, and the 10 is moved into the tens column.
So there we have it.
4 tens add 5 tens, add the regrouped 10, is equal to 10 tens.
Ah, Lucas was right to tell us to watch out.
So 4 tens plus 5 tens is equal to 9 tens.
Add one more 10 is equal to 10 tens.
10 tens is regrouped into 100 and is moved into the hundreds column.
But we've got no extra tens, so we put a zero in our tens column.
3 hundreds add 3 hundreds, add the regrouped 100, is equal to 7 hundreds.
So our answer, our sum, is 705.
You only needed to regroup in the tens because you'd had to regroup a 10 from the ones column.
So, time to check your understanding.
Use column addition to add together these numbers.
488 add 219.
And there they are, set out as a column addition.
So pause the video, have a go, and we'll come back for some feedback.
How did you get on? Alex says, "8 ones add 9 ones is equal to 17 ones." 17 is regrouped into 1 ten and 7 ones.
The 10 is moved into the tens column.
8 tens add 1 ten, add the regrouped 10, is 10 tens.
8 add 1 is 9, and one more is equal to 10.
10 tens is regrouped into 100, and the 100 is moved into the hundreds column, and we've got no additional tens to record.
4 hundreds plus 2 hundreds, plus the regrouped 100, is equal to 7 hundreds.
So the answer to our equation is 707.
488 add 219 is equal to 707.
Well done if you got all that regrouping correct.
Alex tries to find two numbers with a sum of 432, and he's got to choose two of these numbers on the cards.
Alex says, "The ones digits must add to a sum of 12.
I know that 5 ones add 7 ones is equal to 12 ones." How did he know that they must add up to a sum of 12? Well, let's have a look at the numbers we've got.
We've got numbers with 5, 6, and 7 in the ones, so we can't have anything that just adds to a 2.
So it's got to be a 12 to give us that 1 in the ones digit in the sum 432.
So he says that I know that 5 ones add 7 ones is equal to 12 ones.
He says, "I can use estimation too.
275 is close to 300, and 147 is close to 150." And he wants something that's close to 432, so that's not bad is it? And he says, "300 plus 150 is equal to 450, and 450 is close to the sum of 432, so
I'm going to add 275 and 147." So he set it out as a column addition.
Let's see if he's right.
5 ones add 7 ones is equal to 12 ones.
12 is regrouped into 1 ten and 2 ones, and the 10 is moved to the tens column.
7 tens add 4 tens, add the regrouped 10, is equal to 12 tens.
And 12 tens is regrouped into one hundred and 2 tens, and the 100 is moved into the hundreds column.
Can you see a problem already? What was the sum we were aiming for? 432.
I'm not sure he's quite there, is he? 2 hundreds add one hundred, add the regrouped 100, is equal to 4 hundreds.
So he's got a sum of 422.
And Lucas says, "Your answer is 422, Alex.
It's very close to 432, but it's not the same number." Time to check your understanding.
Which two numbers have a sum of 432? Now Alex has started you off on this, I wonder if you can use his ideas and help you to get the right answer.
Alex says, "Look carefully at the ones numbers and think about using estimation." Pause the video, have a go, and we'll come back for some feedback.
Alex said, "I'm gonna try 256 add 176." Ah, now the 5 plus 7 to equal the 12 didn't work, so he's looking at 6 plus 6 to equal the 12 to give him that 2 in the ones column.
I wonder how his estimation worked.
Let's have a go.
6 ones add 6 ones is equal to 12 ones.
12 is regrouped into 1 ten and 2 ones, and the regrouped 10 is moved into the tens column.
5 tens add 7 tens, add the regrouped 10, is equal to 13 tens.
13 tens is regrouped into 1 hundred and 3 tens, and the 100 is moved to the hundreds column.
It's looking good so far, isn't it? 2 hundreds add 1 hundred, add another 100, is equal to 4 hundreds.
So, yes, we've got our sum of 432.
So the two numbers were 256 and 176.
I wonder if you've got there as well.
Did you use some estimating to help you to work that out? Time to have a go for yourself now.
So you are going to calculate the sum of each pair of numbers.
So here they are and you're going to add them together, remembering to think about how you might need to regroup the ones, and possibly regroup the tens as well.
And for the second part of your task, you are going to add any two of these numbers to make each of the sums that you've been given.
And Lucas says, "Look carefully at the ones digits." And Alex says, "Use estimation.
Think about the size of your numbers and whether your answer is reasonable." So pause the video, have a go, and we'll get together for some feedback.
How did you get on? Here are the answers to the first question.
So you had to do regrouping in all of them, but did you spot in C, that we only had a two digit number that we were writing? And Alex spotted in C, that regrouping was needed in the tens because
there's regrouping in the ones.
4 tens plus 5 tens, add the regrouped 10, is equal to 1 hundred, so we had an extra 100 to add in there.
And for part two, you had to choose the pairs of numbers, the correct pairs of numbers to give those sums. Did you get those correctly? Though Lucas says here, 6 ones add 7 ones is equal to 13 ones,
so you can see that regrouped 10.
And for the final one, Alex said 550 plus 300 is equal to 850.
So the sum of 546 and 275 is quite close to 850.
So he used some estimation to help him.
I wonder what other things you thought about as you were choosing the numbers to add.
Did you think about, as Lucas did, thinking about adding those ones digits together to find the correct ones digit in your sum? I hope you had fun experimenting, and I hope you got the answers.
Let's move on to part two.
So we're going to be solving problems with regrouping.
Lucas is using these number cards.
He makes two 3-digit numbers.
He uses column addition to add them together.
"The sum is 403," he says, "which two add-ins did I make?" Hmm, I wonder.
Alex says, "I'm going to work out how Lucas arranged the cards." Good luck, Alex.
Let's see if we can help you out.
Alex tries again to make the sum of 403 using the cards.
He says, "I'm going to use 1 and 2 as the ones digits." Oh, 1 add 2 equals 3, doesn't it? 1 one add 2 ones.
So there we go, he's used 1 and 2 and he's crossed them out to keep track of what he's used.
So 1 one add 2 ones equals 3 ones, that's fine.
He says the tens digits must have a total of 10 because we've got that zero there, so we must have 10 tens, which will mean we've got an extra 100.
So he's going to use 7 and 3.
So 7 and 3 equals 10.
So 7 and 3 is equal to 10, and these tens are regrouped as 100, so he's got to remember that when he's finding the total for his hundreds column.
So he's got a 1 there to add in, can you see a problem here? "Oh no," he says, "there are 4 hundreds, but I've used the digit cards 1 and 2." He'd have to have 1 hundred plus 2 hundreds, plus the
extra 1 hundred to equal 4 hundreds, wouldn't he? He's not got the right cards left.
I don't think that was the right way to start, was it, Alex? Can you think of another way that he could get a three in the ones of his sum? And Lucas says, "Yeah, you can only use each card once in a
solution." So he can't just use the 1 and the 2 again, I'm afraid.
Alex tries to make the sum of 403 once more.
So the sum has a 4 as the hundreds digit.
He says, "I need to use the digit cards 1 and 2 as the hundreds." So he's actually gonna fix those ones there first.
How does he know he's got to have only the 1 and the 2? Why couldn't he have 1 and 3, do you think? What do you spot about that tens digit of 0? Yeah, we must have regrouped somewhere, mustn't we? So
there must be a hundred being regrouped from the tens column.
So he's gonna think about the 1 and the 2 for the hundreds column this time.
"The ones digits must add to 13 then," he says.
"I'm going to start with 9 and 4 as the ones digits." So he spotted that if we've got to use 1 and 2 in the hundreds, we can't use them in the ones, so therefore, the sum of our ones digits must be
13 and not 3.
So he's got 9 and 4, 9 and 4 equals 13, and 13 ones can be regrouped into 1 ten and 3 ones.
So he's got that extra 1 in there.
Oh, now what? Hmm, last time, we had the tens digits having a sum of 10, didn't we? But I don't think that's going to work this time, is it? He said the tens digit cards must have a total of 9 tens,
and then we can have the extra 1 ten to make the 10 tens.
So he says I can use 6 and 3.
6 tens add 3 tens, add the regrouped 10, is equal to 10 tens.
And these 10 tens are regrouped to 100.
And now he was right to save those cards, wasn't he? 1 hundred add 2 hundreds, add the regrouped 100, is equal to 4 hundreds, so he can use his 1 and 2 cards there.
And Lucas says, "Excellent work, Alex." And that was good work, wasn't it? I wonder if you were thinking a step ahead of Alex all the way.
But it was a good puzzle to solve, and lots of thinking about what our column addition means.
Time to check your understanding.
Can you use the digit cards to make two 3-digit numbers that sum to 625? And Lucas says, "Think about the size of the digits and whether you need to regroup in the ones and the tens." So pause the
video, have a go, and we'll come back to some feedback.
How did you get on? "So here's one possible answer," says Lucas.
So Alex says, "7 ones add 8 ones is equal to 15 ones.
15 ones can be regrouped into 1 ten and 5 ones." So 7 add 8 equals 15, and we've got that 1 ten extra there, so what does that mean about our tens? Well, they've got to total 12, haven't they? So 5
tens add 6 tens, plus the regrouped 10, is equal to 12 tens, and 10 tens are regrouped as 100.
So 6 plus 6 is equal to 11, plus another 1 is 12, and that regrouped 10 tens goes into our hundreds.
So now what do we spot? We've got an extra one there.
So our 1 hundreds digits of our two add-ins have got to some to 5.
So 3 hundreds add 2 hundreds, add the regrouped 100, is equal to 6 hundreds.
You might have gone for 1 and 4 there, which would've worked as well, but Alex went for 3 and 2 in the hundreds.
So 3 hundreds plus 2 hundreds, plus the extra 1, is equal to 6 hundreds, and so that is now correct.
Our sum is 625.
Our add-ins are 357 and 268.
You may have found a different solution.
Time for you to have some practise.
Can you use these number cards and make two 3-digit numbers, and then use column addition to add them? Can you make the sums below? And you can see, you've got some regroupings in there marked in.
So can you pick the correct cards to make two 3-digit numbers to make those sums? And because we've got those regrouping ones there, Alex is saying, "Can you find solutions with regrouping in the
ones and the tens?" And then can you use the number cards to make two 3-digit numbers and find three different ways of getting the sum of 843? So you may not have to use regrouping in all of these.
But Lucas says, "Can you find solutions with regrouping in the ones and the tens amongst your answers?" So pause the video, have a go, and we'll come back for some feedback.
How did you get on? There were different solutions for these, but here are some possible solutions.
So you could have had 198 add 326 to give you a sum of 524.
You could have had 769 add 148 to give you a sum of 917.
And you could have had 524 add 176 to give you a sum of 700.
And Lucas says to get 700, you had to regroup in the ones and the tens.
So here are some possible solutions with 843 as the sum.
I wonder if you found any other ones.
These ones all involve regrouping in the ones and the tens.
And Lucas says, "9 plus 4, 8 plus 5, and 7 plus 6 all have a sum of 13, which would give you that 3 in the ones." So there were different ways that you could start your calculation.
And we've come to the end of our lesson.
Thank you for all your hard work and your thinking, especially when we were finding those different ways of making a sum using those digit cards.
Lots of thinking going on there, lots of reasoning, so well done.
So what have we been learning about today? We've been learning that when using column addition, we start by adding the numbers with the smallest place value first.
Really important, especially when there's regrouping.
If the sum of the ones or tens digits is 10 or greater, then regrouping is needed when we're adding with three digits.
Any complete tens are regrouped into the tens column, and any complete hundreds are regrouped into the hundreds column.
You've worked really hard today, and I hope you've enjoyed it as much as I have.
See you soon. | {"url":"https://www.thenational.academy/pupils/programmes/maths-primary-year-4/units/review-of-column-addition-and-subtraction-roman-numerals/lessons/review-using-column-addition-with-regrouping-in-the-ones-and-tens-columns/video","timestamp":"2024-11-10T12:08:36Z","content_type":"text/html","content_length":"134920","record_id":"<urn:uuid:6a20b0ad-232b-4c77-bf11-086a9da76b83>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00595.warc.gz"} |
Collecting like terms
Our chosen students improved 1.19 of a grade on average - 0.45 more than those who didn't have the tutoring.
In order to access this I need to be confident with:
Here we will learn about simplifying algebraic expressions by collecting like terms.
There are also collecting like terms worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if youβ re still stuck.
Collecting like terms is a way of simplifying algebraic expressions. It is also known as combining like terms.To do this we identify the like terms in an algebraic expression and combine them by
adding or subtracting.
3a and +2a are like terms (they have the same letter).+4b and -2b are also like terms, but they are different to the terms with the letter a. The plus or minus sign in front of a term belongs to that
\[\begin{align*} 3a +4b + 2a – b &= 3a + 2a+ 4b – b\\ &=5a+2b\\ \end{align*}\]
This expression cannot be simplified any more as the 5a and the +2b are unlike terms.
In order to simplify algebraic expressions by collecting like terms:
Get your free Collecting like terms worksheet of 20+ questions and answers. Includes reasoning and applied questions.
All the terms involve an x. They are all the same type of term.
All the terms involve an a. They are all the same type of term.
The terms involving a are like terms (7a and +2a) The terms involving b are like terms (+4b and +3b)The plus (or minus) sign belongs to the term after it.
The terms involving x are like terms (5x and -3x)The terms involving y are like terms ( +4y and +5y)The plus or minus sign belongs to the term after it.
The terms involving c are like terms (2c and +3c) The terms involving d are like terms (-5d and +7d) The terms that are only numbers are known as constants (+4 and +2) The plus or minus sign belongs
to the term after it.
The terms involving x are like terms (-6x and +2x) The terms involving y are like terms (+5y and -3y) The terms that are only numbers are known as constants (+4 and -7) The plus or minus sign belongs
to the term after it.
The terms involving x2 are like terms (8x^{2} and -4x^{2}) The terms involving x are like terms (+5x and +7x) The plus or minus sign belongs to the term after it.
The terms involving x2y are like terms (6x^{2}y and +4x^{2}y) The terms involving x2 are like terms (-2x^{2} and -5x^{2})The plus or minus sign belongs to the term after it.
If there is no coefficient (number) seen in front of a term then the coefficient is 1, but we do not write the number 1.
It is possible for all the terms to be cancelled out and the answer is zero.
The terms involving x have cancelled out. We do not write 0x.
Terms involving y and y2 are unlike terms and cannot be collected together by adding or subtracting.
The order of the terms is not critical as long as the plus and minus signs are with the correct term.
3. The diagram shows a pentagon. It has one line of symmetry.
Write an expression for the perimeter in its simplest form.
CD = x+2 DE = 2x+3 for using symmetry to find one of the sides CD or DE
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors. | {"url":"https://thirdspacelearning.com/gcse-maths/algebra/collecting-like-terms/","timestamp":"2024-11-13T18:07:43Z","content_type":"text/html","content_length":"250526","record_id":"<urn:uuid:040c4374-4f3e-49d5-a3be-87d6a6379381>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00688.warc.gz"} |
Arithmetic Sequences 1
Arithmetic Sequences 1
Arithmetic sequences are sequences where the difference between terms in the sequence are the same each time. Finding an unknown term in the sequence is called finding the nth term.
Is there a pattern in this sequence of numbers? Answer yes or no.
How big is the difference between $5$ and $8$?
How big is the difference between $8$ and $11$?
This is an example of an arithmetic sequence
There is a constant pattern in the sequence of numbers. The difference between each number and the following number is $+3$
Is this an arithmetic sequence? Answer yes or no.
This is not an example of an arithmetic sequence
The numbers appear more or less random. There is no constant pattern in the sequence.
Is this an arithmetic sequence? Answer yes or no.
This is an example of an arithmetic sequence
There is a constant pattern. Each number is followed by a number that is $+12$ greater.
What is the next term in this sequence? $4 \space\space\space 13 \space\space\space 22 \space\space\space 31 \space\space\space 40 \space\space\space ...$
How many terms are listed in this sequence?
There are 5 terms in this sequence
You can label them term number $1$, $2$, $3$ , $4$ and $5$ and call that number label $n$
What would be the 6th term in this sequence?
It's probably not hard to find the 7th, 8th and 9th terms either
But what if you wanted to find the 100th term?!
You can work out a formula that helps you find the value of any nth term in this sequence.
To work out this formula, you have to go through a few steps
First you work out how much bigger the value of the term gets every time $n$ gets 1 bigger.
What is the value of the term when $n=1$ in this sequence?
What is the value of the term when $n=2$?
How much bigger does the value of the term get every time $n$ gets $1$ bigger?
Every time $n$ gets one bigger the value of the term gets $+3$ bigger
So the first part of your formula is $3n$
But $3n$ is not the finished formula! You can call the missing piece of the formula $x$
Now, what term value should the finished formula give you when $n=1$?
So you can substitute $1$ and $7$ into the formula like this $3 \times 1 + x = 7$. What is $x$?
So now you have the formula. Try testing it. What do you get if you let $n=5$?
What would be the value of the 50th term in this sequence?
What is the difference between the each of the numbers in this sequence?
So what do you multiply $n$ by in the formula for this sequence?
Now, let $n=1$ and work out what the $x$ in your formula should be.
To recap! You found the formula like this
You worked out the difference between each of the numbers in the sequence, which gave you $5n$
When $n=1$ the formula should give you $8$, so you worked out $x$ like this $5 \times 1 + x = 8$, which means $x=3$
What is the 100th term in this sequence?
What is the 20th term in this sequence? $7 \space\space\space 11\space\space\space15\space\space\space19...$
Summary! This is an arithmetic sequence
There is a constant pattern in the terms in the sequence.
Each term can be labelled by a number $n$
You use that $n$ to work out a formula for the sequence, so you can find the value of any nth term in the sequence.
To work out the formula, you first find the pattern between the terms
Here they get $+3$ bigger every time $n$ gets 1 bigger, so the first part of your formula is $3n$
Then you find out what $x$ is by letting $n=1$
When $n=1$, the formula should give $5$, so you can work out $x$ like this $3 \times 1+x=5$.
That means that $x=2$
You could now use this formula to work out any nth term
For example, the 10th term in this sequence would be $3 \times 10 +2 = 32$ | {"url":"https://albertteen.com/uk/gcse/mathematics/algebra/arithmetic-sequences-1","timestamp":"2024-11-03T10:35:19Z","content_type":"text/html","content_length":"194381","record_id":"<urn:uuid:3e135161-18f5-40e8-bc5c-5b0f73f102ec>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00053.warc.gz"} |
How big is 12 Gigabytes (GB)? Is It Enough For You?
ByAdamonDecember 16, 2023
Just how much storage is 12GB? For you nerds out there, 12GB is 12,000,000,000 bytes. But, if those numbers don't mean anything to you, you're not alone!
Even if you know your bits and bytes, numbers alone won't help you understand how much you'll get out of 12GB. Let's go over a few practical ways you can think of 12GB. So you'll know if 12GB is
enough for you.
How much content fits in 12GB?
Is 12GB a lot? Even if you know how many gigabytes you have, it won't make sense unless you have a good analogy for it. Think of 12GB as:
12GB is 12,000 high quality photos
Assuming photos are around 1MB each
12GB is 8.571 hours of video
Assuming each hour of video is around 1.4GB
12GB is 4,000 songs
Assuming each song is around 3 minutes long, and each minute is 1MB
12GB is 9.231 hours scrolling through tiktok
Assuming each video is around 13MB and you watch around 100 tiktoks per hour
12GB is 4,615 ebooks
Assuming each ebook is 2.6MB
12GB is 8.333 days of music
Assuming one minute of music is 1MB
12GB is 6.25 days playing video games
Assuming one hour of gaming uses 80MB
Is 12GB Enough for You?
If you're wondering if 12 Gigabytes is enough for you, you'll first have to answer: is this for storage (e.g. laptop, flash drive) or for data transfer (e.g. cell phone plan or internet service)?
If it is for storage, try to estimate the number of photos and videos you need to store and compare with the section above to see if 12GB is enough for you.
Otherwise, if it is for data transfer, try to estimate the number of hours of streaming and social media you'll use and compare with the section above to see if 12GB is enough for you.
What are other ways to say 12 Gigabytes?
Here are some useful conversions for 12GB:
Abbreviations of Gigabytes
You might hear people abbreviate Gigabytes as one of these:
Is 12 Gigabytes the same as 12 Gigabits? (12 Gigabytes vs 12 Gigabits)
No, 12 Gigabytes is not the same as 12 Gigabits!
When internet service providers talk about data usage, they refer to the amount in terms of bits rather than bytes. For example, a typical download speed from a internet provider is often advertised
as 200 megabits per second (abbreviated as "Mb" with a lower case "b").
When talking about storage on your phone or computer, people almost always refer to bytes rather than bits. You'll hear about the iPhone having 128 gigabytes (abbreviated as "GB" with a capital "B").
You'll almost never hear that the iPhone has 1024 gigabits of storage (which is the equivalent to 128GB in gigabits).
The difference between bytes and bits is important so you know exactly how much data or storage you're getting. 12 Gigabytes is eight times more than 12 Gigabits!
Looking for Some Other Size? Type in Any Size Below
Type in any size into the search bar below, and then click "Go". Here are some example sizes you could try out:
What are some common storage and data transfer sizes?
It might help to compare 12 Gigabytes to some common sizes that are used in devices and services you already know. Here are a few examples:
Storage Sizes for Phones and Computers
In 2023, iPhones came with 5 different storage options: 64GB, 128GB, 256GB, 512GB, and 1TB.
Storage Sizes for External Hard Drives, Flash Drives and SD cards
Flash drives and SD cards can have sizes ranging from very small 1.5GB or 10GB all the way up to 2TB and even 10TB. Flash drives storage sizes typically come in multiples of two like 2GB, 4GB, 8GB,
16GB and 32GB.
Storage Sizes for Cloud Storage Providers
Google Drive has cloud storage plans for these different sizes: 15GB (for free), 30GB ($6 per month), 2TB ($12 per month), and 5TB ($18 per month).
iCloud has personal cloud storage plans for these different sizes: 50GB ($0.99 per month), 200GB ($2.99 per month), 2TB ($9.99 per month), 6TB ($29.99 per month), and 12TB ($59.99 per month).
Foyer a secure client portal has two different sizes: 1GB (free during trial) and 100GB (per month per user).
Data Limits for Cell Phone Plans
As of December 2023, Verizon has these 4 prepaid data plans: 5GB ($40 per month), 25GB ($60 per month), 100GB ($80 per month), and 150GB ($100 per month).
Data Download Rates for Internet Service Providers (ISP)
Spectrum has 3 popular internet plans, with these 300 megabits per second downloads ($49.99 per month), 500 megabits per second downloads ($69.99 per month), and 1 gigabit per second downloads
($89.99 per month). | {"url":"https://usefoyer.com/how-big-is/12-gigabytes","timestamp":"2024-11-13T03:20:28Z","content_type":"text/html","content_length":"97740","record_id":"<urn:uuid:01291d6c-efbc-4796-a285-f8b7ee22c59a>","cc-path":"CC-MAIN-2024-46/segments/1730477028303.91/warc/CC-MAIN-20241113004258-20241113034258-00896.warc.gz"} |
SimplifyAsFeature Method—ArcGIS Pro
Simplifies the given geometry to make it topologically consistent according to the geometry type for storage in a database. For instance, it rectifies polygons that may be self-intersecting.
The geometry to be simplified.
When true, it forces the simplification code to be applied to the geometry even if the geometry comes from a trusted source or has already been simplified. When false, the method will do nothing
if called on the same geometry a second time.
Return Value
The simplified geometry.
The method tries to have the same behavior as IFeatureSimplify.SimplifyFeature on the Feature class. If the input geometry is a polyline, the method is equivalent to IPolyline::SimplifyNonPlanar().
If the input geometry is a polygon, the method is equivalent to IPolygon::SimplifyPreserveFromTo(). For all other geometry types, the method is equivalent to ITopologicalOperator::Simplify().
var g1 = PolygonBuilder.FromJson("{\"rings\": [ [ [0, 0], [10, 0], [10, 10], [0, 10] ] ] }");
var result = GeometryEngine.Instance.Area(g1); // result = -100.0 - negative due to wrong ring orientation
// simplify it
var result2 = GeometryEngine.Instance.Area(GeometryEngine.Instance.SimplifyAsFeature(g1, true));
// result2 = 100.0 - positive due to correct ring orientation (clockwise)
Target Platforms: Windows 10, Windows 8.1, Windows 7
See Also | {"url":"https://pro.arcgis.com/en/pro-app/2.6/sdk/api-reference/topic8287.html","timestamp":"2024-11-06T04:45:34Z","content_type":"application/xhtml+xml","content_length":"28197","record_id":"<urn:uuid:ce70157b-14cd-4a10-9db3-a1ae6192b7db>","cc-path":"CC-MAIN-2024-46/segments/1730477027909.44/warc/CC-MAIN-20241106034659-20241106064659-00006.warc.gz"} |
Maximum Matching and a Polyhedron With 0,1-Vertices
• Published in 1964
• Added on
A matching in a graph $G$ is a subset of edges in $G$ such that no two meet the same node in $G$. The convex polyhedron $C$ is characterised, where the extreme points of $C$ correspond to the
matchings in $G$. Where each edge of $G$ carries a real numerical weight, an efficient algorithm is described for finding a matching in $G$ with maximum weight-sum.
Other information
BibTeX entry
key = {item49},
type = {misc},
title = {Maximum Matching and a Polyhedron With 0,1-Vertices},
author = {Jack Edmonds},
abstract = {A matching in a graph {\$}G{\$} is a subset of edges in {\$}G{\$} such that no two meet the same node in {\$}G{\$}. The convex polyhedron {\$}C{\$} is characterised, where the extreme points of {\$}C{\$} correspond to the matchings in {\$}G{\$}. Where each edge of {\$}G{\$} carries a real numerical weight, an efficient algorithm is described for finding a matching in {\$}G{\$} with maximum weight-sum.},
comment = {},
date_added = {2015-03-07},
date_published = {1964-10-09},
urls = {http://nvlpubs.nist.gov/nistpubs/jres/69B/jresv69Bn1-2p125{\_}A1b.pdf},
collections = {},
url = {http://nvlpubs.nist.gov/nistpubs/jres/69B/jresv69Bn1-2p125{\_}A1b.pdf},
urldate = {2015-03-07},
year = 1964 | {"url":"https://read.somethingorotherwhatever.com/entry/item49","timestamp":"2024-11-08T16:54:56Z","content_type":"text/html","content_length":"5111","record_id":"<urn:uuid:93b48f36-5cc8-4446-a182-af0aedd7dc5b>","cc-path":"CC-MAIN-2024-46/segments/1730477028070.17/warc/CC-MAIN-20241108164844-20241108194844-00304.warc.gz"} |
The mean systolic blood pressure of adults is 120 millimeters of
mercury (mm Hg) with a...
The mean systolic blood pressure of adults is 120 millimeters of mercury (mm Hg) with a...
The mean systolic blood pressure of adults is 120 millimeters of mercury (mm Hg) with a standard deviation of 5.6. Assume the variable is normally distributed.
1) If an individual is randomly selected, what is the probability that the individual's systolic pressure will be between 120 and 121.8 mm Hg.
2) If a sample of 30 adults are randomly selected, what is the probability that the sample mean systolic pressure will be between 120 and 121.8 mm Hg.
-Central Limit Theorem -
please solve the following problem and explain how you approached each step (include how you solved the problem with the calculator):
For P(x1<X<x2), use TI-83 function normalcdf(x1, x2, mean, sd).
1) For individual, use
2) The Central limit theorem states that the sampling distribution of the sample mean is approximately normally distributed with mean μ and standard deviation σ/√n if either n is large or population
is normal.
For sample mean, use
No need to solve this. In calculator, you can enter this expression directly. | {"url":"https://justaaa.com/statistics-and-probability/108514-the-mean-systolic-blood-pressure-of-adults-is-120","timestamp":"2024-11-10T05:39:07Z","content_type":"text/html","content_length":"43490","record_id":"<urn:uuid:c9ebd715-582d-4eb1-b3e9-9e82cff99591>","cc-path":"CC-MAIN-2024-46/segments/1730477028166.65/warc/CC-MAIN-20241110040813-20241110070813-00341.warc.gz"} |
5. AC Circuit Analysis
• There are a number of techniques used for analysing non-DC circuits.
phasors: for single frequency, steady state systems
Laplace transforms: to find steady state as well as transient responses
5.1 Phasors
• Phasors are used for the analysis of sinusoidal, steady state conditions.
• Sinusoidal means that if we measure the voltage (or current) at any point ‘i’ in the circuit it will have the general form,
• Steady state means that the transients have all stopped. This can be crudely though of as the circuit has ‘charged-up’ or ‘warmed-up’.
• Steady state is another important concept, it means that we are not concerned with the initial effects when we start a circuit (these effects are known as the transients). The typical causes of
transient effects are inductors and capacitors.
• We typically deal with these problems using phasor analysis. In the example before we had a voltage represented in the time domain,
• Basically to do this type of analysis we represent all components voltages and currents in complex form, and then do calculations as normal.
• Consider the simple example below,
5.1.1 RMS Values
• When dealing with alternating currents we are faced with the problem of how we represent the signal magnitude. One easy way is to use the peak values for the wave.
• Another common method is to use the effective value. This is also known as the Root Mean Squared value.
5.1.2 LR Circuits
• One common combination of components is an inductor and resistor.
5.1.3 RC Circuits
• Capacitors are often teamed up with resistors to be used as filters,
5.1.4 LRC Circuits
• These circuits tend to weigh off capacitors and inductors to have a preferred frequency.
5.1.5 LC Circuits
• Inductor capacitor combinations can be useful when attempting to filter certain frequencies,
5.2 AC Power
• Consider the power system shown below,
• The generator converts some form of mechanical force into electrical power. This power is then distributed to consumers over wires (and through transformers). Finally at the point of application,
each load will draw a certain current, at the supply voltage: operating at a rated power. The voltages supplied this way are almost exclusively AC. Also in an ideal situation the load will be pure
resistance, but in reality it will be somewhat reactive.
• Another important example of power delivered is when impedence matching between audio amplifiers and audio speakers. Most consumer systems are 50ohm for maximum power transfer and minimum
5.2.1 Complex Power
• Consider the basic power equation,
5.2.1.1 - Real Power
• The relationship for real power is shown below where the current and resistance are in phase (although the values are rarely perfectly in phase).
• When the current and voltage are D.C. (not charging) the circuit contains pure resistance, and the power is constantly dissipated as heat or otherwise. Notice that the value of P will always be
positive, thus it never returns power to the circuit.
5.2.1.2 - Average Power
• An average power can be a good measure of real power consumption of a resistive component.
5.2.1.3 - Reactive Power
• When we have a circuit component that has current ±90° out of phase with the voltage it uses reactive power. In this case the net power consumption is zero, in actuality the power is stored in and
released from magnetic or electric fields.
• Consider the following calculations,
5.2.1.4 - Apparent Power
• In all circuits we have some combination of Real and Reactive power. We can combine these into one quantity called apparent power,
5.2.1.5 - Complex Power
• We can continue the examination of power by assuming each is as below,
5.2.1.6 - Power Factor
• The power factor (p.f.) is a good measure of how well a power source is being used.
• It is common to try to correct power factor values when in industrial settings. For example, if a large motor were connected to a power grid, it would introduce an inductive effect. Capacitors can
be added to compensate.
5.2.1.7 - Average Power Calculation
• If we want to find the average power, consider the following,
5.2.1.8 - Maximum Power Transfer
• Consider the Thevenin circuit below. We want to find the maximum power transfered from this circuit to the external resistance.
5.3 3-Phase Circuits
• 3-phase circuits are common in large scale power generators and delivery systems.
• These systems carry 3 phases of voltage, each 120 degrees out of phase, on three separate conductors. If these three wires are connected through a balanced load the sum of currents is zero. Most
systems provide a fourth wire as a neutral.
• As a result loads can be connected in a delta configuration with no neutral. | {"url":"https://engineeronadisk.com/V3/engineeronadisk-34.html","timestamp":"2024-11-11T01:14:25Z","content_type":"text/html","content_length":"13891","record_id":"<urn:uuid:16e0b94d-fb5e-4ed6-bd9b-88c0ee8fadf1>","cc-path":"CC-MAIN-2024-46/segments/1730477028202.29/warc/CC-MAIN-20241110233206-20241111023206-00589.warc.gz"} |
MOD: Google Sheets Function ☝️Explained (Definition, Syntax, How to Use It, Examples) - Spreadsheet Daddy
This guide covers everything you need to know about the Google Sheets MOD function, including its definition, syntax, use cases, and how to use it.
What is the MOD Function? How Does It Work?
The MOD function in Google Sheets returns the result of the modulo operation, which is the remainder left over after a division operation. Its purpose is to provide a way to perform calculations that
involve dividing one number by another and determining what remainder, if any, is left over.
The beauty of the MOD function lies in its simplicity and versatility. As long as you have two numbers – the dividend (the number to be divided) and the divisor (the number to divide by) – you can
use the MOD function to find out the remainder of their division operation.
One important thing to note is that the MOD function is not just for integers. It also works with non-integer numbers. However, when you use non-integer numbers, the MOD function may return an
approximate result due to the use of floating-point arithmetic. If you need a more precise result, you can use the ROUND function to round the output up or down to get the exact remainder.
The MOD function can be used in a variety of situations. For example, you can use it to calculate the remaining days of the week after a certain number of days have passed or to determine if a number
is even or odd (if the remainder of a number divided by 2 is 0, the number is even; if not, it’s odd). So, the MOD function is quite a handy function to have in your Google Sheets arsenal. It’s
simple, versatile, and useful in many different scenarios.
MOD Syntax
The syntax and arguments for the function are as follows:
MOD(dividend, divisor)
In this syntax:
• ‘dividend‘ refers to the number that will be divided to find the remainder.
• ‘divisor‘ is the number by which the dividend will be divided.
When using the MOD function, there are a few important notes to consider regarding its syntax and arguments:
• Both the ‘dividend’ and ‘divisor’ arguments need to be numeric values. The function will not work with text or other non-numeric data types.
• The ‘divisor’ argument cannot be zero, as division by zero is undefined in mathematics. If you try to use zero as the divisor, the function will return an error.
• While the MOD function can be used with non-integer numbers, this may result in an approximate result due to the use of floating-point arithmetic. This is because the function calculates the
remainder after division, and division of non-integers can lead to fractional remainders.
• If you’re using non-integer numbers and want a more accurate result, you can use the ROUND function to round the output of the MOD function up or down to the nearest integer.
• The order of the arguments matters in the MOD function. The ‘dividend’ argument should always come before the ‘divisor’ argument. If they are switched, the function will return a different
Examples of How to Use the MOD Function
Here are some practical examples of how you can use the MOD function in Google Sheets:
Example #1: Calculate the Remainder of a Division
If you want to find out the remainder of a division operation, you can use the MOD function. Let’s say you want to divide 10 by 3 and want to know the remainder. You would input the formula as =MOD
(10, 3) in a cell. The result would be 1, because when you divide 10 by 3, the remainder is 1.
Example #2: Determine if a Number is Even or Odd
You can use the MOD function to check if a number is even or odd. If you divide a number by 2 and the remainder is 0, then the number is even.
If the remainder is 1, then the number is odd. For example, if you want to check if the number 7 is even or odd, you would input the formula as =MOD(7, 2). The result would be 1, indicating that 7 is
an odd number.
Example #3: Create a Custom Recurring Schedule
Let’s say you’re managing a team, and you want to create a recurring schedule where tasks are assigned to team members in a round-robin fashion.
You have 5 team members and a list of tasks in consecutive rows. You can use the MOD function to assign each task to a team member. If you input the formula =MOD(row(), 5) + 1, it will return a
number between 1 and 5 for each row, which you can associate with a team member.
Example #4: Highlight Rows Based on a Condition
You can use the MOD function in conditional formatting to highlight alternate rows or columns. For instance, if you want to highlight every 3rd row, you would use the formula =MOD(row(), 3) = 0 in
the conditional formatting rule. This will highlight every row where the row number divided by 3 has a remainder of 0, i.e., every 3rd row.
Example #5: Calculate Age from Birth Date
If you have a column of birth dates and you want to calculate the age of each person, you can use the MOD function along with the TODAY function.
The formula would be =YEAR(TODAY()) – YEAR(A1) – IF(MOD(TODAY(),1) < MOD(A1,1), 1, 0) where A1 is the cell with the birth date. This formula calculates the number of years between the birth date and
today’s date and adjusts for whether or not the birthday has occurred this year.
Why MOD Is Not Working? Troubleshooting Common Errors
If you’re working with the MOD function in Google Sheets and you’re encountering issues, it’s important to understand the common errors that may arise, their causes, and how to troubleshoot them.
These issues can stem from incorrect syntax, invalid arguments, or other issues. Below are some common errors you may encounter:
#DIV/0! Error
Cause: The #DIV/0! error is caused when the second argument (the divisor) in the MOD function is zero. Division by zero is undefined in mathematics and therefore Google Sheets will display this
Solution: Always ensure the divisor in your MOD function is not zero. If you’re pulling the divisor from another cell, make sure that cell doesn’t contain a zero.
#NUM! Error
Cause: The #NUM! error is less common but can occur when the dividend is non-numeric and the divisor is zero. As mentioned earlier, both arguments need to be numbers, and the divisor cannot be zero.
Solution: Similar to the solutions above, always ensure that both arguments are numbers and that the divisor is not zero. If you’re pulling data from other cells, check that those cells contain valid
numbers and that the divisor is not zero.
#N/A Error
Cause: The #N/A error occurs when either or both arguments are missing in the MOD function. Google Sheets requires two arguments for this function, and if one or both are missing, it will return this
Solution: Always ensure you have provided both arguments to the MOD function. If you’re pulling data from other cells, make sure those cells contain valid data.
Using MOD With Other Google Sheets Functions
Combining the MOD function with other Google Sheets functions can enhance its utility and allow you to perform more complex calculations. Here are a few examples of how you can use the MOD function
with other Google Sheets functions:
With SUM
Usage: You can use the SUM function along with the MOD function to calculate the sum of the remainder of the division of a range of numbers by a specific divisor.
Example: Suppose you have a list of numbers in cells A1 to A5 and want to calculate the sum of the remainder of these numbers when divided by 2. You can use the following formula:
=SUM(ARRAYFORMULA(MOD(A1:A5, 2)))
This formula calculates the remainder of each number in the range A1:A5 when divided by 2 and then sums these remainders.
With IF
Usage: You can use the IF function along with the MOD function to perform calculations based on whether the remainder of a division operation is equal to a specific value.
Example: Suppose you have a list of numbers in cells A1 to A5 and want to determine if each number is even or odd. You can use the following formula:
=ARRAYFORMULA(IF(MOD(A1:A5, 2)=0, “Even”, “Odd”))
This formula uses the MOD function to calculate the remainder of each number in the range A1:A5 when divided by 2. If the remainder is 0, the number is even, and the formula returns “Even”. If the
remainder is not 0, the number is odd, and the formula returns “Odd”.
With COUNTIF function
Usage: You can use the COUNTIF function with the MOD function to count the number of cells within a range that have a remainder of a specific value when divided by a certain number.
Example: Suppose you have a list of numbers in cells A1 to A5, and you want to count how many numbers are even. You can use the following formula:
=COUNTIF(ARRAYFORMULA(MOD(A1:A5, 2)), 0)
This formula calculates the remainder of each number in the range A1:A5 when divided by 2, then counts how many of these remainders are equal to 0 (indicating an even number).
For more details on the MOD function, check out the official documentation at the Google Docs Editors Help Center. | {"url":"https://spreadsheetdaddy.com/google-sheets/functions/mod","timestamp":"2024-11-15T00:16:47Z","content_type":"text/html","content_length":"214837","record_id":"<urn:uuid:75c85f41-0f98-46f1-8d38-38a4ee978334>","cc-path":"CC-MAIN-2024-46/segments/1730477397531.96/warc/CC-MAIN-20241114225955-20241115015955-00027.warc.gz"} |
How to Introduce Decimals with Base Ten BlocksHow to Introduce Decimals with Base Ten Blocks
There's a reason math teachers start the year by introducing or reviewing place value concepts. Understanding place value is essential to developing a solid foundation of mathematical understanding.
Introducing Place Value
Whether you're introducing whole number concepts or decimal place value, it's important to start at the concrete level, and base 10 blocks work perfectly because they are sized according to their
Even 5th graders aren't too old for base 10 blocks. Primary teachers often use them to introduce whole numbers, but base 10 blocks are also effective with upper elementary students when exploring
decimals. Kids have to understand that each place to the left is 10 times the size of the place to the right, and base 10 blocks are the best way to explore that concept.
How to Introduce Decimals with Base 10 Blocks
When introducing decimals, you can use the cubes to represent tens, the flats to represent ones, the rods for tenths, and the units for hundredths. I found that it's really important to create a
place value mat (like the one shown above) for these lessons because it helps students remember what place is represented by each model. To create mats like the ones I used in my lesson, draw the
4-column chart above on a large sheet of heavy paper (18" x 24") and laminate it. Use a dry erase marker to draw a decimal point between the ones place and the tenths place. Or you can print the
patterns in my Build a Number freebie to create the place value mat on 2 sheets of 8.5" by 11" paper.
Seat your students in teams of four and give each team one set of Base 10 manipulatives. It's best if each student has a dry erase board and marker, too. Ask your students to divide up the materials
so that one person has the cubes, one has the flats, one has the rods, and one has the units. Introduce each piece and explain how it represents a particular decimal place. I explained that even
though they had been taught that the "flat" was equal to 100, I wanted them to think of it as one whole... maybe one whole cake for a family of mice! If they sliced the cake into 10 parts, each part
was 1/10 or 0.1. If they cut those 10 slices into 10 parts, each part was 1/100 or 0.01 of the whole.
Team Practice: Build a Decimal
After you introduce the value of each part of the model, lead them through the "Build a Decimal" lesson. Start by writing a decimal in standard form on the board and asking students work to with
their team to "build" that number on the team mat. You can make up your own numbers or use the Build a Decimal task cards provided in this freebie. After you check to make sure that they represented
the number properly on the mat, ask students to write the word name and expanded form.
Expanded form is particularly difficult when representing decimals, and using base 10 manipulatives seems to help illustrate the concept. For example, it's easy to see that the number on the mat can
be written as 20 + 4 + 0.6 + 0.09 because students can see each place represented with physical objects. The expanded form of this number could also be expressed as 20 + 4 + 6/10 + 9/100 or
completely broken down to 20 + 4 + 6 x (1/10) + 9 x (1/100). All of those ways to express decimals are much more easily viewed when looking at a mat such as the one below. Students can also see that
each place to the left is 10 times greater than the one on its right.
Place Value Games Combo
As with most math concepts, it's not enough to introduce them with hands-on materials and then move on to the next lesson. Kids have to practice the terminology and work with the concepts until they
are fluent with them. That's where math games come in handy. These four place value games are bundled together in my
Place Value Games Combo pack
, which is great for 4th and 5th grade. My students enjoyed all of these activities, and they perfect for math centers and small group instruction.
Place Value Partners Game
CCSS 2.NBT.A.3, 4.NBT.2, and 5.NBT.A.3
Place Value Partners
is my most popular game for reviewing whole number and/or decimal place value. It's similar to Battleship, but students use a game board with lines for placing numbers and number cards instead of
ships and a coordinate grid. See the image below to understand how the game is set up and played. The Sender calls out each digit, one at a time, and tells the Receiver where to place them. When all
the numbers are in place, they compare their game boards, check the final number and write its standard form, word name, and expanded form in their math journals or on a recording page. The packet
shown here includes four different variations of the game board to include everything from 4-digit whole numbers to decimals.
Bingo Showdown Decimal Review
CCSS 5.NBT.3 and 5.NBT.4.
Decimal Place Value Bingo Showdown
is a variation of the classic Bingo game that can be used for whole group instruction, small guided math groups, cooperative learning teams, or in learning centers. What more can I say? It's Bingo
and kids love it!
Place Value Spinner Games
CCSS 4.NBT.2 and 4.NBT.3
Another game my students enjoyed was the
Spin 4 Cash Place Value Review Game
. In this activity, students practice word names and expanded forms using task cards and a spinner. When they get a correct answer, they may spin for a certain amount of "math cash." I also created
an international version called Spin 2 Win that awards tokens instead of cash.
I'm the Greatest - Comparing Numbers Game
CCSS 2.NBT.4, 4.NBT.2, 5.NBT.3
The final game I want to share with you is
I'm the Greatest
, a simple activity for comparing numbers. The teacher can play against the class, or students can play against each other in cooperative learning teams. In this game, players attempt to create the
largest number by placing randomly-selected numbers on a game board. There's a bit of luck involved, but in order to win, students have to be able to read word names and compare numbers accurately.
All four games are available from my
TeachersPayTeachers store
, either individually or as a part of the
Place Value Game Combo
. Even if you have already introduced place value concepts, these games make great review and practice activities throughout the year. You've heard that practice makes perfect, and that definitely
applies to place value concepts. Plenty of place value practice is needed to make perfect!
2 comments:
1. Thank you for posting this about place value. Looking forward to using some of these in my class.
2. This comment has been removed by a blog administrator. | {"url":"https://corkboardconnections.blogspot.com/2013/09/placevalue.html?m=0","timestamp":"2024-11-01T19:54:20Z","content_type":"application/xhtml+xml","content_length":"117296","record_id":"<urn:uuid:fc41fabc-baee-4fa2-b408-8ffed6a995d4>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00315.warc.gz"} |
How Important is the Seventh Game of the Set?
Italian translation at settesei.it
Few nuggets of tennis’s conventional wisdom are more standard than the notion that the seventh game of each set is particularly crucial. While it’s often difficult to pin down such a well-worn
conceit, it seems to combine two separate beliefs:
1. If a set has reached 3-3, the pressure is starting to mount, and the server is less likely to hold serve.
2. The seventh game is somehow more important than its immediate effect on the score, perhaps because the winner gains momentum by taking such a pivotal game.
Let’s test both.
Holding at 3-3
Drawing on my database of over 11,000 ATP tour-level matches from the last few years, I found 11,421 sets that reached three-all. For each, I calculated the theoretical likelihood that the server
would hold (based on his rate of service points won throughout the match) and his percentage of service games won in the match. If the conventional wisdom is true, the percentage of games won by the
server at 3-3 should be noticeably lower.
It isn’t. Using the theoretical model, these servers should have held 80.5% of the time. Based on their success holding serve throughout these matches, they should have held 80.2% of the time. At
three-all, they held serve 79.5% of the time. That’s lower, but not enough lower that a human would ever notice. The difference between 80.2% and 79.5% is roughly one extra break at 3-3 per Grand
Slam. Not Grand Slam match–an entire tournament.
None of that 0.7% discrepancy can be explained by the effect of old balls [1]. Because new balls are introduced after the first seven games of each match, the server at three-all in the first set is
always using old balls, which should, according to another bit of conventional wisdom, work against him. However, the difference between actual holds and predicted holds at 3-3 is slightly greater
after the first set: 78.9% instead of the predicted 79.8%. Still, this difference is not enough to merit the weight we give to the seventh game.
The simple part of our work is done: Servers hold at three-all almost as often as they do at any other stage of a match.
Momentum from the seventh game
At 3-3, a set is close, and every game matters. This is especially true in men’s tennis, where breaks are hard to come by. Against many players, getting broken so late in the set is almost the same
as losing the set.
However, the focus on the seventh game is a bit odd. It’s important, but not as important as serving at 3-4, or 4-4, or 4-5, or … you get the idea. The closer a game to the end of the set, the more
important it is–theoretically, anyway. If 3-3 is really worth the hoopla, it must grant the winner some additional momentum.
To measure the effect of the seventh game, I took another look at that pool of 11,000-plus sets that reached three-all. For each set, I calculated the two probabilities–based on each player’s service
points won throughout the match–that the server would win the set:
1. the 3-3 server’s chance of winning the set before the 3-3 game
2. his chance of winning the set after winning or losing the 3-3 game
In this sample of matches, the average server at three-all had a 48.1% chance of winning the set before the seventh game. The servers went on to win 49.4% of the sets [2].
In over 9,000 of our 3-3 sets, the server held at 3-3. These players had, on average, a 51.3% chance of winning the set before serving at 3-3, which rose to an average of a 57.3% chance after
holding. In fact, they won the set 58.6% of the time.
In the other 2,300 of our sets, the server failed to hold. Before serving at three-all, these players had a 35.9% chance of winning the set, which fell to 12.6% after losing serve. These players went
on to win the set 13.7% of the time. In all of these cases, the model slightly underestimates the likelihood that the server at 3-3 goes on to win the set.
There’s no evidence here for momentum. Players who hold serve at three-all are slightly more likely to win the set than the model predicts, but the difference is no greater than that between the
model and reality before the 3-3 game. In any event, the difference is small, affecting barely one set in one hundred.
When a server is broken at three-all, the evidence directly contradicts the momentum hypothesis. Yes, the server is much less likely to win the set–but that’s because he just got broken! The same
would be true if we studied servers at 3-4, 4-4, 4-5, or 5-5. Once we factor in the mathematical implications of getting broken in the seventh game, servers are slightly more likely to win the set
than the model suggests. Certainly the break does not swing any momentum in the direction of the successful returner.
There you have it. Players hold serve about as often as usual at three-all (whether they’re serving with new balls or not), and winning or losing the seventh game doesn’t have any discernible
momentum effect on the rest of the set [3]. Be sure to tell your friendly neighborhood tennis pundits.
1. Using a more limited dataset, Magnus and Klaassen found that new balls did not result in more holds of serve.
2. It’s not entirely clear why these numbers aren’t 50%. My best guess is that underdogs are able to stay close early in sets, reaching 3-3 a bit more often than the model would predict. That’s a
project for another day.
3. I ran the same tests against WTA, women’s ITF, Challenger, and Futures matches to see if the results would be different by gender or level. The ITF numbers are the reverse of most of the other
groups, but overall, none of these subsets contradict anything I’ve generalized from the ATP numbers.
LEVEL WTA ITF CHALL FUT
Matches 11203 17143 18717 14052
Hold% 64.3% 54.9% 75.8% 69.9%
Hold at 3-3 63.4% 57.1% 74.6% 69.4%
Hold% (no 1st set) 63.9% 54.4% 75.4% 69.6%
Hold at 3-3 (no 1st) 64.0% 56.4% 73.6% 68.4%
Prob at 3-3 49.2% 49.1% 47.8% 48.2%
Server set% 50.0% 49.4% 48.0% 48.7%
WIN at 3-3:
Prob at 3-3 54.6% 56.6% 51.8% 53.2%
Prob at 4-3 65.0% 69.2% 58.8% 61.5%
Set won% 65.8% 68.7% 58.9% 61.2%
LOSE at 3-3:
Prob at 3-3 40.0% 39.1% 36.1% 36.8%
Prob at 3-4 21.5% 24.2% 14.9% 17.8%
Set won% 22.8% 23.8% 16.1% 20.3%
4 thoughts on “How Important is the Seventh Game of the Set?”
1. Why you didn’t do it for 4-4?
In this case the returner can serve out the set if he breaks?
1. because commentators don’t breathlessly proclaim the importance of the 9th game.
1. Off topic sort of, but I would be curious about the % of winning/losing a set after a 4-4 game, solely only because if you lose that game the odds should drop more since one must win 2
games and a tiebreak or three games to win. But not sure what specifically we could gain from it.
2. If I have learned one thing from years of reading your blog it is:
Tennis commentators, like all humans, are very eager to divine patterns where there are none.
Thank you for thoroughly debunking them time after time. | {"url":"https://www.tennisabstract.com/blog/2015/09/24/how-important-is-the-seventh-game-of-the-set/","timestamp":"2024-11-03T22:15:38Z","content_type":"text/html","content_length":"69027","record_id":"<urn:uuid:33937bab-d412-4329-a4c2-392c068cd1f1>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00240.warc.gz"} |
6 Best Homeschool Math Curriculum
6 Best Homeschool Math Curriculum
Navigating the world of homeschooling can feel like charting a course through an ocean of resources, especially when it comes to a subject as vital as mathematics. In this post, I we’ve compiled a
list of the best homeschool math curriculums to share with you.
These comprehensive, engaging, and highly-rated programs cater to a variety of learning styles, ensuring that every child has the opportunity to shine. I also invite you to check the FAQs section at
the bottom of this post to explore answers to frequently asked questions about math curriculum.
Best Homeschool Math Curriculum
Here are our picks for the best homeschool math curriculum:
Khan Academy offers a comprehensive math curriculum that starts from the basics such as counting and early math and goes all the way through high school and college-level courses, including algebra,
geometry, trigonometry, calculus, statistics, and linear algebra.
The curriculum aligns with the U.S. Common Core standards, which can be useful for homeschooling families who want to ensure their curriculum matches what is taught in traditional schools.
Students can learn at their own pace through engaging video lessons, interactive activities and quizzes. They can pause and rewind videos, practice problems as much as they want, and master concepts
before moving on to the next.
When students practice problems on Khan Academy, they receive instant feedback. If they answer a question incorrectly, they can see the correct answer and an explanation right away.More importantly,
Khan Academy’s system can identify gaps in a student’s understanding and suggest targeted practice problems to help fill those gaps.
The platform includes tools for tracking student progress. Parents, teachers, or students can see at a glance how many skills a student has mastered, and where they might need more practice.
IXL Math is an online math platform that offers a comprehensive, curriculum-aligned math program from pre-K through high school. It’s part of the larger IXL Learning suite, which includes subjects
beyond math, such as language arts, science, and social studies.
IXL Math covers a wide range of math topics following the standards of common core. This includes everything from basic number sense, through fractions and decimals, all the way up to calculus and
IXL provides an abundance of interactive practice questions for each math topic. This allows students to gain proficiency in each skill through repetition and application. IXL provides immediate
feedback when a student answers a question. If the answer is incorrect, IXL will not just reveal the correct answer but will also provide a detailed explanation to help the student understand the
concept better.
The questions on IXL are adaptive, meaning they adjust to the learner’s level. As a student answers questions correctly, the questions gradually become more challenging to continually stretch the
student’s learning.
IXL provides detailed reporting on student progress. To keep students motivated, IXL uses virtual awards. Students can earn awards for reaching learning milestones, which can be motivating for many
Saxon Math is another helpful resource for K-12 homeschool math curriculum. Saxon Math introduces new mathematical concepts in small, manageable increments. Concepts covered include algebra,
geometry, and many more. After a new concept is introduced, students have the opportunity to practice this new concept along with older concepts they’ve already learned.
One of the hallmarks of Saxon Math is the continuous review of older material. Every lesson includes mixed practice problems that cover both the new material from the current lesson and review
material from previous lessons. This ensures that students retain what they’ve learned and that older concepts stay fresh in their minds.
Each Saxon Math lesson begins with a “warm-up” section that includes facts practice, mental math, and problem-solving exercises. This is followed by a new concept introduction and lesson practice,
where the new concept of the day is introduced and practiced.
As for assessments, Saxon Math includes regular assessments to monitor students’ understanding and retention of concepts. There are also “investigations,” which are activity-based lessons focused on
real-world applications of math.
Math-U-See is a multi-sensory, mastery-based homeschool math curriculum for grades K-12, designed to teach students math concepts in a logical and straightforward way. Math-U-See uses blocks and
other manipulatives to represent math concepts visually and tangibly. This hands-on learning approach is particularly useful for tactile or visual learners, helping them to understand abstract
concepts more concretely.
The curriculum is divided into levels, each represented by a Greek or Latin prefix (like Primer, Alpha, Beta, Gamma, etc.). Each level focuses on a specific set of skills or a particular area of math
(like addition and subtraction, multiplication, division, fractions, pre-algebra, etc.)
Additionally, each lesson begins with a video where Steve Demme (the creator of Math-U-See) teaches the concept. These videos are designed for the parent-teacher to watch and learn from, but they can
also be watched by students. Lessons also include plenty of practice problems so students can reinforce what they’ve learned. This also includes review problems from previous lessons to ensure
retention of those concepts.
As for assessment, Math-U-See offers periodic tests to assess comprehension and mastery of the material. These tests serve as a helpful tool for parents to track their child’s progress.
Time4Learning is an online curriculum that offers homeschooling programs for students from pre-K through high school. While it includes multiple subjects, its math curriculum is particularly
Time4Learning’s math curriculum covers all the main areas of math from early number skills to more advanced topics like algebra, geometry, and calculus, depending on grade level. The curriculum is
aligned to state standards.
Time4Learning’s lessons are presented in a multimedia format, combining text, audio, animations, interactive exercises, and more to keep learning engaging and varied. This can be particularly helpful
for students who struggle with traditional textbook formats.
Additionally, the program includes numerous interactive activities designed to reinforce the concepts taught in the lessons. This can help students solidify their understanding and apply their new
For additional reinforcement and practice, Time4Learning offers printable worksheets that accompany the online lessons. This allows students to practice offline and provides a balance between screen
time and traditional pencil-and-paper work.
As for assessments,Time4Learning offers regular quizzes and tests to assess student understanding and track progress over time. Parents receive detailed reporting on their child’s progress. To ease
the burden on homeschooling parents, Time4Learning automatically grades lessons and keeps detailed records of student progress.
ALEKS (Assessment and LEarning in Knowledge Spaces) is an online, artificially intelligent assessment and learning system that uses adaptive questioning to accurately determine a student’s knowledge
in a specific subject area. The system, developed by McGraw Hill, covers a wide range of subjects, including a comprehensive math curriculum for grades 3-12.
ALEKS uses artificial intelligence to adapt to each student’s knowledge level. When a student begins a new topic, ALEKS assesses what the student already knows and what they’re ready to learn next.
The system will then present the student with problems and lessons based on this information, ensuring the student is always working at the appropriate level.
Periodic knowledge checks are used to re-assess student learning and recalibrate the learning path as necessary. This helps ensure that the student has retained what they’ve learned and can apply it
ALEKS uses a unique “Learning Pie” to visually represent a student’s knowledge. Each slice of the pie represents a different topic, and the portion of the slice that’s filled in represents what the
student has mastered in that topic. This provides an intuitive way for students and parents to track progress.
Based on its assessments, ALEKS creates an individualized learning path for each student. The system focuses on teaching the concepts that the student is most ready to learn, while also ensuring they
maintain mastery of what they’ve already learned.
When a student answers a problem, ALEKS provides immediate feedback. If the student answers incorrectly, the system shows the correct solution and an explanation to help the student understand their
ALEKS provides detailed reports to track student progress. These reports can show what a student has learned, what they’re ready to learn next, and what they’re struggling with, helping parents or
teachers to identify where the student might need additional support.
I went ahead and created this section where I answered some of the frequently asked questions about math curriculum:
What is a math curriculum?
A math curriculum refers to the structure and content of a math education program, typically comprising a sequence of topics, learning objectives, teaching methods, resources, and assessments. It
sets out what students should know and be able to do in mathematics at different grade levels.
What’s the best homeschool math curriculum?
The “best” homeschool math curriculum can vary depending on a student’s individual learning style, strengths, weaknesses, and interests. However, some highly regarded options include Khan Academy,
Math-U-See, Saxon Math, Time4Learning, IXL Math, and ALEKS, among others. It’s often a good idea to try several and see which works best for your child.
What should I look for in a math curriculum?
Key elements to consider when choosing a math curriculum include:
Comprehensive coverage of math topics
Alignment with national or state standards
Clear, logical progression of concepts
Engaging and varied instructional methods
Availability of practice problems and assessments
Adaptability to a student’s pace and learning style
Quality of supporting materials and resources
Feedback and support for parents, if homeschooling
What is the best homeschool math for visual learners?
For visual learners, a math curriculum that offers visual aids, manipulatives, and graphical representations of concepts is often beneficial. Math-U-See is a popular choice because it heavily
incorporates manipulatives. Time4Learning also includes many visual and interactive components.
What are the 5 content areas in the math curriculum?
Although the specific categorization can vary, the math curriculum is often divided into five content areas:
Number and Operations
Data Analysis and Probability
What type of math is used in high school?
High school math typically includes the following subjects, though the order and specific curriculum can vary:
Algebra I
Algebra II
Calculus (often an AP or advanced course)
Statistics (also often an AP or advanced course)
What are the 6 principles of mathematics?
The National Council of Teachers of Mathematics (NCTM) identifies six principles for school mathematics:
Equity: High expectations and strong support for all students
Curriculum: A coherent, focused, and well-articulated curriculum
Teaching: Effective mathematics teaching requires understanding what students know and need to learn
Learning: Students must learn mathematics with understanding
Assessment: Assessment should support the learning of mathematics and guide instruction
Technology: Technology is essential in teaching and learning mathematics
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Numericals on Electrostatics
In this page we have Numericals on Electrostatics(Electric Charge,Field & Gauss Laws) for JEE advanced , Main & NEET . Hope you like them and do not forget to like , social share and comment at the
end of the page.
Question 1
Two point charges q
[1 ]
and q
are located with points having position vectors
1. Find the position vector r[3] where the third charge q[3] should be placed so that force acting on each of the three charges would be equal to zero.
2. Find the amount of charge q[3]
To solve the problem we choose Cartesian co-ordinates system where $\mathbf{r_1}$ , $\mathbf{r_2}$ and $\mathbf{r_3}$ are position vectors of charge $q_1$, $q_2$ and $q_3$ respectively with respect
to the origin O as shown below in the figure.
Question 2
Consider a thin wire ring of radius R and carrying uniform charge density λ per unit length.
1. Find the magnitude of electric field strength on the axis of the ring as a function of distance x from its centre.
2. What would be the form of electric field function for x>>R.
3. Find the magnitude of maximum strength of electric field.
Consider the figure given below
Thus, from symmetry considerations total field along Y-axis is zero. Hence,
$E_Y=0$ and $E=E_X$
Question 3
Two equally charged metal balls each of mass m Kg are suspended from the same point by two insulated threads of length l m long. At equilibrium, as a result of mutual separation between balls, balls
are separated by x m. Determine the charge on each ball.
Let Q be the charge on the each of the metal balls A and B suspended from a point O as shown below in the figure. Let r be the distance between two metal balls when they are at equilibrium.
Forces acting on the metal ball are
(1) Weight of the metal ball
W= mg
Acting in vertically downward direction.
(2) Force of electrostatic repulsion
$F= \frac {1}{4 \pi \epsilon _0} \frac {Q^2}{r^2}$
directed away from each other.
(3) Tension T in each of the insulated threads directed towards O. Since each of the metal ball is in equilibrium under these three forces, the tension must be equal and opposite to the resultant R
of F and W. Considering similar triangles AWR and AOC we have
$\frac {W}{OC} = \frac {F}{CA} = \frac {T}{AO}$
$F= W \frac {CA}{OC}$
CA=r/2 and $OC \approx OA=l$ m
$F= W \frac {r/2}{l}$
Putting the value of F and W we get
$\frac {1}{4 \pi \epsilon _0} \frac {Q^2}{r^2}= \frac {mgr}{2l}$
$Q = \sqrt {\frac {2 \pi \epsilon _0 mg r^3}{l}}$
Question 4
There are two identical particles each of mass m and carrying charge Q. Initially one of them is at rest and another charge moves with velocity v directly towards the particle at rest. Find the
distance of closest approach.
Given that v is the initial velocity of first charged particle. Due to the repulsion second particle also moves. The velocity of particle moving initially decreases and that of second particle
increases. At distance of closest approach velocity of both the particles become equal. From the law of conservation of momentum,
or, $v^{'}=\frac {v}{2}$
Let r be the distance between charged particles at the distance of closest approach then potential energy between them is
$U= \frac {1}{4 \pi \epsilon _0} \frac {Q^2}{r}$
From law of conservation of energy we have
$\frac {1}{2} mv^2 = \frac {1}{4 \pi \epsilon _0} \frac {Q^2}{r} + \frac {1}{2} mv_{'}^2 + \frac {1}{2} mv_{'}^2$
$\frac {1}{4} mv^2= \frac {1}{4 \pi \epsilon _0} \frac {Q^2}{r}$
$r = \frac {1}{4 \pi \epsilon _0} \frac {4Q^2}{mv^2}$
Question 5
Find the electric field at the centre of uniformly charged semi circular arc having linear charge density λ.
Consider the figure given below
Question 6
Two opposite corners of square carry charge –q and other tow opposite corners of same square carry charge +q as shown below in the figure
All the four charges are equal in magnitude. Find the magnitude and direction of force on the charge on the upper right corner by the other three charges.
The forces acting on the charge +q at point C are $F_A$, $F_B$, and $F_D$ by the charges at corners A, B, and D respectively. Magnitude of force $F_B$ and $F_D$ is
$F_B=F_D= \frac {1}{4 \pi \epsilon _0} \frac {q^2}{a^2}$
and that of $F_A$ is
$F_A= \frac {1}{4 \pi \epsilon _0} \frac {q^2}{(\sqrt{2} a)^2}=\frac {1}{4 \pi \epsilon _0} \frac {q^2}{2a^2}$
$F_B$ and $F_D$ can be regarded as the sum of vector along diagonal CA and one normal to it as shown below in the figure.
$F_D^{"} + F_B^{"}=0$
$F_B+ F_D = F_B^{'} + F_D^{'}=2 F_B \cos 45=\frac {\sqrt {2} q^2}{ 4 \pi \epsilon _0 a^2}$
Having direction from C to A. This force is larger then $F_A$. Hence resultant force on charge +q at C is directed from C to A and has magnitude
$F_{res}= ( \sqrt{2} - \frac {1}{\sqrt {2}}) \frac {q^2}{4 \pi \epsilon _0 a^2}$
Question 7
A rigid insulated wire frame in form of a right angled triangle ABC is set in vertical plane as shown below in the figure.
Two beads of equal masses m and each carrying charges q
and q
are connected by a chord of length l and can slide without friction on the wires. Considering the case when beads are stationary determine
1. the normal reaction on the beads
2. the angle α
3. tension in the chord
If the chord is now cut what are the value of charges for which beads continue to remain stationary.
To solve the problem, we would first have to consider the forces acting on the bead P. Figure below shows the diagrammatic representation of force acting on the bead P.
Thus, forces acting on bead P are
(i) W=mg, weight of bead acting in vertically downward direction.
(ii) Tension T in the string
(iii) $F=\frac {1}{4 \pi \epsilon _0} \frac {q_1 q_2}{l^2}$ electric force between two beads
(iv) Force of normal reaction.
(T-F) is the net force acting along the string. Bead P would be in equilibrium if net force acting on the bead is zero. We now resolve the components of mg and (T-F) for bead P as shown below in the
Because of equilibrium of charge $q_1$
$N_P=mg \sin {60}^0+\left(T-F \right) \sin {\alpha}$ -(1)
$(T-F) \cos {\alpha}=mg \cos {60}^0$ -(2)
We now resolve the components of mg and (T-F) for bead Q as shown below in the figure
Because of equilibrium of charge $q_2$
$N_Q=mg \sin {30}^0+ (T-F) \cos {\alpha}$ --(3)
$ (T-F) \sin{\alpha}=mgcos{30}^0$ --(4)
(A) From 1 and 3
$N_P=mg \sin{60}^0+ mg \cos{30}^0= \sqrt{3} mg$
from 2 and 4
$N_Q= mg \sin {30}^0+ mg \cos{60}^0=mg$
This is the required normal reaction on the bead.
(B) dividing equation 4 by 3 we get
$ \tan \alpha= \frac {\cos 30^0}{\cos 60^0}=\sqrt {3}= \tan 60^0$
thus $ \alpha=60^0$
(C) using $\alpha=60^0$ and $N_Q$ in equation 3
$mg=mg \sin {30}^0 + (T-F) \cos{60}^0$
$mg (1- \sin{30}^0)= (T-F)cos{60}^0$
solving it for T and using $F= \frac {1}{4 \pi \epsilon _0} \frac {q_1 q_2}{l^2}$ , we get
$ T=mg + \frac {1}{4 \pi \epsilon_0} \frac{q_1q_2}{l^2}$ --(5)
This is the tension on the chord.
When the chord is cut T=0 thus from equation 5 we have
$mg= -\frac{1}{4 \pi \epsilon_0}\frac{q_1q_2}{l^2}$
The right hand side of this equation should be positive which is possible if charges $q_1$ and $q_2$ have opposite signs. Thus for equilibrium beads must have unlike
charges. Magnitude of product of charges is
Question 8
Inside a ball charged uniformly with volume density ρ , there is a spherical cavity. The centre of cavity is displaced with respect to the centre of the ball by a distance a. Find the field strength
inside the cavity assuming the permittivity to be equal to unity.
Consider the figure given below
From Gauss's law we know that electric field strength inside a uniformly charged sphere is
$\mathbf{E_{sphere}}=\frac{\rho}{3 \epsilon_0}\mathbf{r}$
So electric field at any point P having position vector $\mathbf{OP}=\mathbf{r}_{+}$ is given by
$\mathbf{E_{sphere}}=\frac{\rho}{3 \epsilon_0}\mathbf{r_{+}}$
We now treat the cavity as the sphere of charge density $-\rho$. Then electric field due to cavity at point P acting along O'P is
Now net electric field at p is the vector sum of $\mathbf{E}_{sphere}$ and $\mathbf{E}_{cavity}$
$\mathbf{E}=\frac{\rho}{3\varepsilon_0}\mathbf{r}_+-\frac{\rho}{3\varepsilon_0}\mathbf{r}_-=\frac{\rho}{3\varepsilon_0}(\boldsymbol{r}_+ - \boldsymbol{r}_-)=\frac{\rho}{3\varepsilon_0}\mathbf{a}$
As seen from the expression, the electric field found is uniform and constant.
Question 9
An electric dipole is placed at a distance x from a infinitely long rod of linear charge density λ.
1. Find the net amount of force acting on the dipole.
2. Assuming that dipole is fixed at its centre find its time period of oscillations if the dipole is slightly rotated about its equilibrium position.
Question 10
An electric charge Q is uniformly distributed over the surface of a sphere of radius R. Show that the force on a small charge element dq is radially outwards and is given by
is the electric field at the surface of the sphere.
Surface charge density of the sphere is
$\sigma=\frac{Q}{4\pi R^2}$
Consider the figure given below
where A is any point inside the sphere close to an area element da. Let dq be the amount of charge on this area element da. This charge dq will produce an electric field at point A that is
approximately the one due to a uniformly charged infinitesimal charged plate. So,
$\mathbf{E}_{A'}=-\frac{\sigma}{2\epsilon_0}\mathbf{n }$
where n is the unit vector normal to da in the outward direction.
We know that the electric field is zero inside the sphere. So, total field at point A should also be equal to zero. If is the electric field at A due to all the charges on the spherical surface
except da then,
Since point A is close to da, $E^{"}$ may be considered as the field strength at da due to all charges on the spherical surface. Hence force acting on da is
Where E is the field strength on the spherical surface
Question 11
A thin fixed ring of radius R has a positive charge of +q C uniformly distributed over it. A particle of mass m and charge –q is placed on axis at a distance x from the centre of the ring. Show that
the motion of negatively charged particle is approximately simple harmonic.
Consider two small and equal elements of charge $\Delta q$ on the opposite sides of a diameter of a ring as shown below in the figure.
Let P is the point on the axis of the ring at a distance x from the centre. The force on charge -q placed at point P would be
$\Delta F = \frac {1}{4 \pi \epsilon _0} \frac {-q \Delta q}{AP^2}$
which is directed from P to A
This force $\Delta F$ can be resolved into two components $\Delta F \sin \theta$ along PO and $\Delta F \cos \theta$ along PQ. Similarly element of charge $\Delta q$ at B exerts a force on negative
charge -q at P equal in magnitude to $\Delta F$ but directed along PB. This force$\Delta F$ can also be resolved into two components $\Delta F \sin \theta$ along PO and $\Delta F \cos \theta$ along
PR which cancels out $\Delta F \cos \theta$ along PQ. Summing over all such elements over the ring , the net force along PO would be
$F=\sum\Delta F s i n\theta=-\frac{1}{4\pi\epsilon_0}\frac{q^2sin\theta}{{(AP)}^2}$
From triangle AOP
$\sin \theta=\frac {OP}{AP}$
Now OP=x and $AP=\sqrt {x^2 + R^2}$
$F=-\frac {q^2x}{4\pi\epsilon_0 {(x^2+R^2)}^{\frac{3}{2}}}$
For R>>x , $(x^2+R^2)^{3/2} \approx R^3$
$F=-\frac {q^2x}{4\pi\epsilon_0R^3}$
Since acceleration a=F/m comparing this wit above equation we get
$a=\frac{F}{m}=-\omega^2x $
Where $\omega=\left[\frac{q^2}{4\pi\varepsilon_0mR^3}\right]^{1/2}$
Since acceleration is proportional to negative of displacement x motion of charge would be simple harmonic motion
Question 12
Consider the figure given below
A positive charge +q is placed at corner of the cube. Find the electric flux through the right face BCGDB of the cube.
Think of this cube as one of 8 surrounding the charge. Each of the 24 squares which make up the surface of this larger cube gets the same flux as every other one, so $\int _{one face} E.da=\frac {1}
{24} \int _{whole cube} E.da$
From Gauss's law $ \int E.da=\frac {q}{\epsilon _0}$
Hence electric flux through face BCDGB is
$Flux = \frac {q}{24 \epsilon _0}$
Question 13
Consider a sphere of radius r having charge q C distributed uniformly over the sphere. This sphere is now covered with a hollow conducting sphere of radius R>r.
1. Find the electric field at point P away from the centre O of the sphere such that r<OP<R.
2. Find the surface charge density on the outer surface of the hollow sphere if charge q’ C is placed on the hollow sphere.
(a)Consider the figure given below
We need to find the field at the point P. First, we draw the concentric spherical Gaussian surface through point P. Field at all points on the Gaussian surface would be equal in magnitude and
direction of the field would be in radial outwards direction. Now flux through the surface is
$\phi=\oint{\mathrm{E.ds}=E\oint d s}=E(4\pi{r}^2)$
Where OP=r
From Gauss s Law
$\oint{\mathrm{E.ds}} = \frac {q_{enc}}{\epsilon _0}$
Since $q_{enc} = q$ the charge enclosed inside the Gaussian surface
$E \times 4 \pi r^2 = \frac {q}{\epsilon _0}$
$E=\frac{q}{4\pi\epsilon_0 r^2}$
(b) For finding the charge on the outer surface of the hollow sphere we draw Gaussian surface through the material of the hollow sphere as shown below in the figure
We know that electric field inside the conductor is zero so the flux through the Gaussian surface would also be zero. From Gauss's law we conclude that total charge enclosed in Gaussian surface is
zero. Hence charge on inner surface of hollow sphere is q C. Hence total charge on outer surface of hollow sphere is
$q_{outer}=(q^{'}+q)$ C
Question 14
1. Find the electric field inside the uniformly charged sphere of radius R and volume charge density ρ using Gauss’s law.
2. Use Gauss’s law to find the electric field outside, at a point on the surface and at any point inside a spherical shell of radius R, carrying a uniform surface charge density σ.
(a) We need to find the electric field inside the uniformly charged sphere of radius R and volume charge density $\rho$ using Gauss's law. In this case, Gaussian surface is a spherical surface, whose
centre is at O and radius equal to r where r
$E\cdot4\pi r^2=\frac{q'}{\epsilon_0}$
$q'=\frac{q}{4/3(\pi R^3)}\times\frac{4}{3}\pi r^3=\frac{qr^3}{R^3}$
From this
$\because q=\frac{4}{3}\pi R^3\rho$
$E=\frac{\rho r}{3\epsilon_0}$
Thus, as we get near to the centre, the intensity falls from maximum to zero at the centre. Hence intensity at point P inside a charged sphere, in which the charge is uniformly distributed, is
directly proportional to distance of point P from the centre of the sphere as shown below in the figure.
(b)Now we need to use Gauss's law to find the electric field outside, at a point on the surface and at any point inside a spherical shell of radius R, carrying a uniform surface charge density $\
sigma$. Now total charge on the shell is
$q=4 \pi R^2 \sigma$
For solving the problem consider the figure given below which shows the Gaussian surface (dotted surface) for points inside and outside the shell.
(1) For points outside the shell:-
Draw a sphere of radius a concentric with the shell to represent the Gaussian surface. Total electric flux over a spherical surface of radius a is
$\int_{s}{E\cdot d}A=|E|\int_{s} d A=|E|4\pi a^2$
From Gauss's law flux
$\int_{s}{E\cdot d}A=\frac{q}{\epsilon_0}$
$\Rightarrow|E|4\pi a^2=\frac{q}{\epsilon_0}$
For a>R. Thus above equation gives the electric field intensity at a point outside the charged spherical shell. In terms of surface charge density field intensity can be calculated by putting $q=4 \
pi R^2 \sigma$ in above equation and the result obtained is
$\mathbf{E}=\frac{\sigma R^2}{\varepsilon_0a^2}\hat{\mathbf{n}}$
(2) At a point on the surface of the shell
For point on the surface of the shell a=R
$E=\frac{\sigma a^2}{\epsilon_0a^2}=\frac{\sigma}{\epsilon_0}$
(3) At a point inside the shell
In this case r $\int_{s}{E\cdot d}A=\frac{q}{\epsilon_0}$
$|E|4\pi a^2=0$
inside the spherical shell having uniform surface charge density.
Question 15
(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
(a) We know that statement of Gauss's law in integral form is
$\oint E.dS= \frac {Q_{enc}}{\epsilon_0}$
Now we have to show that that the electric field always undergoes a discontinuity when surface charge $\sigma $ is crossed.
Let's draw a wafer-thin Gaussian pillbox, extending just over the edge in each direction. Apply Gauss's law to Gaussian pillbox S of cross-sectional area A whose two ends are parallel to the
interface. The ends of the box can be made arbitrarily close together. In this limit, the flux of the electric field out of the sides of the box is obviously negligible, and the only contribution to
the flux comes from the two ends. Thus
$\oint \mathbf{E}.d\boldsymbol{S}= ( \mathbf{E_1 -E_2}).\mathbf{n} A$
where $E_1$ is the component of electric field normal to the interface immediately above the surface and $E_2$ is the component of electric field normal to the interface immediately below the surface
The charge enclosed by the pill-box is simply $\sigma A$, where $\sigma$ is the sheet charge density on the interface . Thus, Gauss' law yields
$( \mathbf{E_1 -E_2}).\mathbf{n} =\frac {\sigma}{\epsilon _0}$
Where $\mathbf{n}$ a unit vector is normal to the surface at a point and $\sigma$ is the surface charge density at that point. (The direction of $\mathbf{n}$ is from side 1 to side 2.) Thus that the
normal component of electrostatic field has a discontinuity from one side of a charged surface to another i.e., the presence of a charge sheet on an interface causes a discontinuity in the
perpendicular component of the electric field.
Just outside a conductor, the electric field is $\frac {\sigma}{\epsilon _0}\mathbf{n}$
(b) Now we have to show that tangential component of E, by contrast, is always continuous. For this consider the figure shown below
In the above figure consider the thin rectangular loop of length l and width e -> 0
Now we consider one of the Maxwell's equation known Faraday's law in integral form which is
$\oint _c \mathbf{E}.d\mathbf{l}=- \frac {d}{dt}\int _s \mathbf{B}.\mathbf{n} da$
Here right-hand side leads to A(B)[normal] , but A -> 0 as sides of the loop are very short nearly approaching zero i.e., e -> 0 , hence contribution due to magnetic effects vanishes. We thus have,
$\oint _c \mathbf{E}.d\mathbf{l}=0$
Now if we apply this integral to thin rectangular loop as shown in figure 2 then the dominant contribution to the loop integral comes from the long sides because he length of the short sides is
assumed to be arbitrarily small which ends up giving nothing. Thus, we have
$E_{above} - E_{below}=0$
$E_{above} =E_{below}$
Where, $E_{above}$ and $E_{below}$ are tangential components of electric field thus there can be no discontinuity in the parallel component of the electric field across an interface.
Question 16
Consider a cylinder as given below in the figure
Volume between radius r
and r
contains uniform charge density ρ C/m
. Use Gauss’s law to find electric field in all regions.
Consider the figure given in question.
(a) For region $0 < r < r_1$ from Gauss s law we have
$\oint E.dS= \frac {Q_{enc}}{\epsilon_0}$
or $\frac {q_{enc}}{\epsilon_0}=E(2 \pi r L)$
where L is the length of cylindrical segment under consideration. Since charge enclosed in this region is
$q_{enc}=0$ therefore from gauss's law E=0 in this region.
(b) For region $r_1 \leq r \leq r_2$
from gauss's law for cylinders we have
$\frac {q_{enc}}{\epsilon_0}=E(2 \pi r L)$
Consider a Gaussian cylinder of radius r inside the cylindrical shell such that $r_1=r=r_2$. Volume of cylinder of radius r and length L is given by
$V= \pi r^2L$
Thus charge enclosed is
$q_{enc}=\rho \pi [r^2-(r_1^2]L$
using gauss's law
$\frac{\rho\pi\left(r^2-r_1^2\right)L}{\epsilon_0}=E(2\pi rL)$
(c) For $r>r_2$
Gaussian surface here is a cylinder of radius $r > r_2$. Now charge enclosed inside the Gaussian surface is
$q_{enc}= \rho \pi [(r_2)^2-(r_1)^2]$
Form Gauss's law ,
$\frac {q_{enc}}{\epsilon_0}=E(2 \pi r L)$
$\frac{\rho\pi\left({r_2}^2-r_1^2\right)L}{\epsilon_0}=E(2\pi rL)$ | {"url":"https://physicscatalyst.com/elec/electrostatics-numericals.php","timestamp":"2024-11-07T07:25:32Z","content_type":"text/html","content_length":"93010","record_id":"<urn:uuid:77f08104-d558-418e-9fdf-d990b5e5ffed>","cc-path":"CC-MAIN-2024-46/segments/1730477027957.23/warc/CC-MAIN-20241107052447-20241107082447-00179.warc.gz"} |
3 Divided By 63
63 PICTURES OF PEOPLE WHO ARE COMFORTABLE AND COMFORTABLE Page 45 of
3 Divided By 63. Subtract 126 126 from 133 133. The result of division of 763÷63 763 ÷ 63 is 12 12 with a remainder of 7 7.
63 PICTURES OF PEOPLE WHO ARE COMFORTABLE AND COMFORTABLE Page 45 of
Web the answer to 63 divided by 3 can also be written as a mixed fraction as follows: Note that the result of the division 3÷63 is not an exact value and is equivalent to recurring decimal
0.047619047619. Subtract 126 126 from 133 133. Web generally, a division problem has three main parts: Web division calculator with remainder (÷) home › calculators › math calculators › division
calculator division calculator online division calculator. The result of division of 763÷63 763 ÷ 63 is 12 12 with a remainder of 7 7. Multiply the newest quotient digit (2) ( 2) by the divisor 63
63. There are no more digits to move down from the dividend, which means we have completed the long division problem. 21 0/3 note that the numerator in the fraction above is the remainder and the
denominator is the divisor. Bring down the next number of the dividend.
Place this digit in the quotient on top of the division symbol. Web divide two numbers, a dividend and a divisor, and find the answer as a quotient with a remainder. Web so, what is the answer to 3
divided by 63? Web generally, a division problem has three main parts: 205 ÷ 2 = 102.5 more math calculators square root with division fraction gcd lcm The dividend, divisor, and quotient. Web
subtract 63 63 from 76 76. There are no more digits to move down from the dividend, which means we have completed the long division problem. Quotient (decimal) quotient (integer) remainder
multiplication calculator see also multiplication calculator remainder calculator addition calculator Bring down the next number of the dividend. A decimal number, say, 3 can be written as 3.0, 3.00
and so on. | {"url":"https://math.virtualuncp.edu.pe/3-divided-by-63.html","timestamp":"2024-11-10T08:27:30Z","content_type":"text/html","content_length":"20974","record_id":"<urn:uuid:06276122-988f-4343-93fe-0ac0e54b8072>","cc-path":"CC-MAIN-2024-46/segments/1730477028179.55/warc/CC-MAIN-20241110072033-20241110102033-00361.warc.gz"} |
Childcare Costs By State: 2023 Statistics | Self.inc
The cost of childcare can take up a large chunk of a family’s spending, with the average annual cost being 29.43% of a typical household’s income. [1] Household Spending - U.S. Census https://
data.census.gov/table?g=0500000US24001,24003 We’ve analyzed some of the latest data and reports on childcare costs in the U.S. to find out which states are paying the most.
Key Statistics
• A household with a national median income of $64,529 will spend 29.43% of its earnings on childcare.
• The average cost of daycare across the United States is $18,866 per child per year.
• Childcare costs the most in Connecticut, at an average of $27,125 per year for daycare, whereas it’s the cheapest in South Dakota at $14,813 per year.
• In all states, infant care (up to 18 months) is the most expensive form of daycare, with a U.S. average of $22,350 per child per year, compared to $17,800 for toddler care and $13,950 for
preschooler care.
• Across all states, the average annual cost to hire a full-time nanny is $35,432.
• 55% of Millennials were living in families with a spouse and/or children in 2019, compared to 85% of the Silent Generation in 1968.
Average childcare costs by state
The average cost of childcare across the United States is $18,886 per year. The state with the most expensive average childcare costs is Connecticut, at $27,125 per year. South Dakota has the lowest
costs for childcare, at $14,813 per year.
*Estimated figures based on inflation adjustment using a rate of 6.4%.
Disclaimer: Average annual childcare costs by state were taken from the average costs for infant, toddler, pre-school and family childcare, and combined to determine one average cost figure for all
Source [2] The True Cost of High Quality Childcare - American Progress https://www.americanprogress.org/article/true-cost-high-quality-child-care-across-united-states
Childcare by type in the U.S.
The cost of care for different types of daycare also varies, typically depending on the child’s age. The data below shows the breakdown of costs for infant, toddler, and preschooler care and family
childcare settings.
• Infant care - The most expensive state is Massachusetts with an average cost of $36,200 per child per year, while the cheapest is South Dakota at $16,450. The average annual cost for infant care
in the U.S. is $22,350.
• Toddler care - Connecticut has the most expensive cost for toddler care at $30,700, compared to Louisiana with the cheapest at $13,350. The average annual cost for toddler care in the U.S. is
• Preschooler care - The most expensive preschooler care is in Washington D.C. at $18,650 per child per year, but the state with the highest priced preschooler care is Rhode Island at $18,600 per
year. Louisiana has the cheapest costs for preschooler daycare at $10,750. The U.S. average is $13,950 for preschooler daycare.
• Family childcare - This type of care involves a caregiver looking after children in their own home rather than a daycare center. These programs are typically licensed for children between the
ages of 6 and 12. Connecticut has the most expensive family childcare costs at $24,525 per year, while Louisiana has the cheapest at $12,050. The average cost for family childcare in the U.S. is
$15,875 per child per year.
Average childcare costs by type
State Infant Toddler Preschooler Family Childcare
StateUnited States Infant$22,350 Toddler$17,800 Preschooler$13,950 Family Childcare$15,875
StateAlabama Infant$18,550 Toddler$14,650 Preschooler$12,000 Family Childcare$13,325
StateAlaska Infant$27,800 Toddler$22,500 Preschooler$17,250 Family Childcare$19,875
StateArizona Infant$20,300 Toddler$15,650 Preschooler$12,200 Family Childcare$13,925
StateArkansas Infant$18,600 Toddler$14,350 Preschooler$11,350 Family Childcare$12,850
StateCalifornia Infant$29,350 Toddler$22,700 Preschooler$16,850 Family Childcare$19,775
StateColorado Infant$23,200 Toddler$18,350 Preschooler$14,550 Family Childcare$16,450
StateConnecticut Infant$30,700 Toddler$30,700 Preschooler$18,350 Family Childcare$24,525
StateDelaware Infant$24,600 Toddler$18,500 Preschooler$15,200 Family Childcare$16,850
StateDistrict of Columbia Infant$27,850 Toddler$27,850 Preschooler$18,650 Family Childcare$23,250
StateFlorida Infant$21,650 Toddler$15,250 Preschooler$12,350 Family Childcare$13,800
StateGeorgia Infant$19,600 Toddler$15,150 Preschooler$12,000 Family Childcare$13,575
StateHawaii Infant$23,000 Toddler$17,900 Preschooler$13,350 Family Childcare$15,625
StateIdaho Infant$18,000 Toddler$14,600 Preschooler$11,650 Family Childcare$13,125
StateIllinois Infant$23,900 Toddler$19,050 Preschooler$15,400 Family Childcare$17,225
StateIndiana Infant$20,300 Toddler$15,450 Preschooler$12,400 Family Childcare$13,925
StateIowa Infant$21,450 Toddler$16,850 Preschooler$14,600 Family Childcare$15,725
StateKansas Infant$23,750 Toddler$15,950 Preschooler$12,200 Family Childcare$14,075
StateKentucky Infant$21,050 Toddler$15,500 Preschooler$12,450 Family Childcare$13,975
StateLouisiana Infant$16,900 Toddler$13,350 Preschooler$10,750 Family Childcare$12,050
StateMaine Infant$24,800 Toddler$21,400 Preschooler$14,500 Family Childcare$17,950
StateMaryland Infant$31,800 Toddler$20,000 Preschooler$15,300 Family Childcare$17,650
StateMassachusetts Infant$32,600 Toddler$27,450 Preschooler$17,550 Family Childcare$22,500
StateMichigan Infant$20,400 Toddler$16,350 Preschooler$13,300 Family Childcare$14,825
StateMinnesota Infant$24,950 Toddler$18,700 Preschooler$14,800 Family Childcare$16,750
StateMississippi Infant$17,850 Toddler$14,250 Preschooler$11,450 Family Childcare$12,850
StateMissouri Infant$22,250 Toddler$16,350 Preschooler$13,400 Family Childcare$14,875
StateMontana Infant$19,400 Toddler$15,600 Preschooler$12,400 Family Childcare$14,000
StateNebraska Infant$20,900 Toddler$17,750 Preschooler$13,600 Family Childcare$15,675
StateNevada Infant$22,150 Toddler$17,500 Preschooler$14,150 Family Childcare$15,825
StateNew Hampshire Infant$22,650 Toddler$17,850 Preschooler$14,450 Family Childcare$16,150
StateNew Jersey Infant$26,250 Toddler$22,350 Preschooler$17,350 Family Childcare$19,850
StateNew Mexico Infant$19,300 Toddler$14,900 Preschooler$12,200 Family Childcare$13,550
StateNew York Infant$28,950 Toddler$22,600 Preschooler$17,550 Family Childcare$20,075
StateNorth Carolina Infant$19,450 Toddler$15,950 Preschooler$11,850 Family Childcare$13,900
StateNorth Dakota Infant$21,700 Toddler$16,950 Preschooler$12,850 Family Childcare$14,900
StateOhio Infant$20,350 Toddler$16,700 Preschooler$12,950 Family Childcare$14,825
StateOklahoma Infant$19,550 Toddler$14,350 Preschooler$11,400 Family Childcare$12,875
StateOregon Infant$27,600 Toddler$23,850 Preschooler$16,450 Family Childcare$20,150
StatePennsylvania Infant$24,200 Toddler$19,000 Preschooler$14,800 Family Childcare$16,900
StateRhode Island Infant$29,550 Toddler$23,000 Preschooler$18,600 Family Childcare$20,800
StateSouth Carolina Infant$18,250 Toddler$14,250 Preschooler$11,400 Family Childcare$12,825
StateSouth Dakota Infant$16,450 Toddler$13,900 Preschooler$11,850 Family Childcare$12,875
StateTennessee Infant$20,800 Toddler$16,200 Preschooler$13,150 Family Childcare$14,675
StateTexas Infant$19,950 Toddler$14,550 Preschooler$11,700 Family Childcare$13,125
StateUtah Infant$22,100 Toddler$16,750 Preschooler$13,100 Family Childcare$14,925
StateVermont Infant$25,350 Toddler$21,900 Preschooler$14,950 Family Childcare$18,425
StateVirginia Infant$24,850 Toddler$18,250 Preschooler$14,900 Family Childcare$16,575
StateWashington Infant$27,800 Toddler$20,900 Preschooler$16,500 Family Childcare$18,700
StateWest Virginia Infant$21,950 Toddler$16,000 Preschooler$12,800 Family Childcare$14,400
StateWisconsin Infant$22,600 Toddler$17,750 Preschooler$13,800 Family Childcare$15,775
StateWyoming Infant$22,150 Toddler$20,200 Preschooler$13,650 Family Childcare$16,925
Source [2] The True Cost of High Quality Childcare - American Progress https://www.americanprogress.org/article/true-cost-high-quality-child-care-across-united-states
As we can see from the figures, infant care is the most expensive type of daycare, and the cost typically increases as children get older, with preschool care being the cheapest form of childcare in
every state.
The cost of childcare compared to median income
The average U.S. household that pays for childcare will spend 29.43% of their income on care for one child, based on a median household income of $64,529.
The highest percentage of income spend on childcare is in West Virginia, where childcare costs 37.18% of the median income. Hawaii has the lowest income-to-childcare ratio, with the cost being 22% of
the median household income.
State Median household income Average childcare cost per child per year Percentage of income spent on childcare
StateWest Virginia Median household income$48,037 Average childcare cost per child per year$17,863 Percentage of income spent on childcare37.18%
StateOregon Median household income$65,667 Average childcare cost per child per year$23,250 Percentage of income spent on childcare35.41%
StateMaine Median household income$59,489 Average childcare cost per child per year$20,513 Percentage of income spent on childcare34.48%
StateRhode Island Median household income$70,305 Average childcare cost per child per year$24,150 Percentage of income spent on childcare34.35%
StateConnecticut Median household income$79,855 Average childcare cost per child per year$27,125 Percentage of income spent on childcare33.97%
StateKentucky Median household income$52,238 Average childcare cost per child per year$17,613 Percentage of income spent on childcare33.72%
StateNew York Median household income$71,117 Average childcare cost per child per year$23,850 Percentage of income spent on childcare33.54%
StateVermont Median household income$63,477 Average childcare cost per child per year$20,938 Percentage of income spent on childcare32.98%
StateNew Mexico Median household income$51,243 Average childcare cost per child per year$16,788 Percentage of income spent on childcare32.76%
StateMississippi Median household income$46,511 Average childcare cost per child per year$14,963 Percentage of income spent on childcare32.17%
StateArkansas Median household income$49,475 Average childcare cost per child per year$15,713 Percentage of income spent on childcare31.76%
StatePennsylvania Median household income$63,627 Average childcare cost per child per year$20,000 Percentage of income spent on childcare31.43%
StateMassachusetts Median household income$84,385 Average childcare cost per child per year$26,238 Percentage of income spent on childcare31.09%
StateTennessee Median household income$54,833 Average childcare cost per child per year$17,025 Percentage of income spent on childcare31.05%
StateMissouri Median household income$57,290 Average childcare cost per child per year$17,488 Percentage of income spent on childcare30.52%
StateCalifornia Median household income$78,672 Average childcare cost per child per year$23,863 Percentage of income spent on childcare30.33%
StateNevada Median household income$62,043 Average childcare cost per child per year$18,763 Percentage of income spent on childcare30.24%
StateIllinois Median household income$68,428 Average childcare cost per child per year$20,638 Percentage of income spent on childcare30.16%
StateOhio Median household income$58,116 Average childcare cost per child per year$17,463 Percentage of income spent on childcare30.05%
StateAlabama Median household income$52,035 Average childcare cost per child per year$15,550 Percentage of income spent on childcare29.88%
StateFlorida Median household income$57,703 Average childcare cost per child per year$17,125 Percentage of income spent on childcare29.68%
StateKansas Median household income$61,091 Average childcare cost per child per year$18,100 Percentage of income spent on childcare29.63%
StateMichigan Median household income$59,234 Average childcare cost per child per year$17,538 Percentage of income spent on childcare29.61%
StateU.S. Average Median household income$64,529 Average childcare cost per child per year$18,866 Percentage of income spent on childcare29.43%
StateNorth Carolina Median household income$56,642 Average childcare cost per child per year$16,663 Percentage of income spent on childcare29.42%
StateAlaska Median household income$77,790 Average childcare cost per child per year$22,875 Percentage of income spent on childcare29.41%
StateLouisiana Median household income$50,800 Average childcare cost per child per year$14,925 Percentage of income spent on childcare29.38%
StateOklahoma Median household income$53,840 Average childcare cost per child per year$15,800 Percentage of income spent on childcare29.35%
StateMontana Median household income$56,539 Average childcare cost per child per year$16,563 Percentage of income spent on childcare29.29%
StateDelaware Median household income$69,110 Average childcare cost per child per year$20,038 Percentage of income spent on childcare28.99%
StateWashington Median household income$77,006 Average childcare cost per child per year$22,238 Percentage of income spent on childcare28.88%
StateIowa Median household income$61,836 Average childcare cost per child per year$17,850 Percentage of income spent on childcare28.87%
StateNebraska Median household income$63,015 Average childcare cost per child per year$18,013 Percentage of income spent on childcare28.58%
StateWisconsin Median household income$63,293 Average childcare cost per child per year$17,950 Percentage of income spent on childcare28.36%
StateIndiana Median household income$58,235 Average childcare cost per child per year$16,500 Percentage of income spent on childcare28.33%
StateWyoming Median household income$65,304 Average childcare cost per child per year$18,488 Percentage of income spent on childcare28.31%
StateGeorgia Median household income$61,224 Average childcare cost per child per year$17,063 Percentage of income spent on childcare27.87%
StateSouth Carolina Median household income$54,864 Average childcare cost per child per year$15,288 Percentage of income spent on childcare27.86%
StateMinnesota Median household income$73,382 Average childcare cost per child per year$20,088 Percentage of income spent on childcare27.37%
StateArizona Median household income$61,529 Average childcare cost per child per year$16,675 Percentage of income spent on childcare27.10%
StateNew Jersey Median household income$85,245 Average childcare cost per child per year$23,100 Percentage of income spent on childcare27.10%
StateNorth Dakota Median household income$65,315 Average childcare cost per child per year$17,500 Percentage of income spent on childcare26.79%
StateTexas Median household income$63,826 Average childcare cost per child per year$16,725 Percentage of income spent on childcare26.20%
StateVirginia Median household income$76,398 Average childcare cost per child per year$19,863 Percentage of income spent on childcare26.00%
StateIdaho Median household income$58,915 Average childcare cost per child per year$15,238 Percentage of income spent on childcare25.86%
StateColorado Median household income$75,231 Average childcare cost per child per year$19,288 Percentage of income spent on childcare25.64%
StateMaryland Median household income$87,063 Average childcare cost per child per year$22,288 Percentage of income spent on childcare25.60%
StateSouth Dakota Median household income$59,896 Average childcare cost per child per year$14,813 Percentage of income spent on childcare24.73%
StateNew Hampshire Median household income$77,923 Average childcare cost per child per year$19,075 Percentage of income spent on childcare24.48%
StateUtah Median household income$74,197 Average childcare cost per child per year$17,538 Percentage of income spent on childcare23.64%
StateHawaii Median household income$83,173 Average childcare cost per child per year$18,300 Percentage of income spent on childcare22.00%
Sources [2] The True Cost of High Quality Childcare - American Progress https://www.americanprogress.org/article/true-cost-high-quality-child-care-across-united-states
Change in the cost of childcare over time
The price of childcare hasn’t always been the same. Let’s take a look at how the average costs have changed over time in the U.S.
In 2004, the average cost of childcare in the U.S. was $13,332 per child which made up 31% of the average household income. Since then, the average childcare cost has increased but the percentage of
household income has declined. This suggests that the average household income across all households has risen faster than the cost of childcare. However, this does not mean that households who rely
on daycare have necessarily seen a rapid rise in their annual income.
Net childcare costs for parents using childcare facilities
Year USD % of net household income
2004 $13,332 31%
2008 $15,591 31%
2012 $16,844 30%
2015 $15,446 27%
2018 $18,486 29%
2019 $18,949 29%
2020 $18,183 27%
2021 $12,226 14%
2030 (predicted)* $18,323 19%
2035 (predicted)* $19,035 16%
2040 (predicted)* $19,748 13%
2050 (predicted)* $21,173 8%
*Figures predicted using TREND
Source [3] Net Childcare Costs - OECD https://stats.oecd.org/Index.aspx?DataSetCode=NCC
Cost differences between daycare and nanny care
The table below shows the difference in costs between the annual cost of daycare compared to the annual cost of hiring a nanny to care for a child. The state with the highest difference in price is
Arizona where a nanny costs 76.08% more at $37,152 compared to $16,675 for daycare.
The state with the lowest difference is Connecticut where a full-time nanny only costs 39.5% more at $40,476, compared to $27,125 for daycare. As we mentioned earlier, Connecticut’s annual daycare
costs are also the highest of all the states.
Across all states, the average annual cost to hire a nanny is $35,432. The most expensive state to hire a nanny in is Massachusetts with an average annual cost of $43,656.
State Daycare cost (Annual) Nanny cost (Annual) % Difference in price
StateArizona Daycare cost (Annual)$16,675 Nanny cost (Annual)$37,152 % Difference in price76.08%
StateNew Mexico Daycare cost (Annual)$16,788 Nanny cost (Annual)$37,212 % Difference in price75.65%
StateSouth Dakota Daycare cost (Annual)$14,813 Nanny cost (Annual)$32,640 % Difference in price75.14%
StateHawaii Daycare cost (Annual)$18,300 Nanny cost (Annual)$39,828 % Difference in price74.07%
StateLouisiana Daycare cost (Annual)$14,925 Nanny cost (Annual)$32,100 % Difference in price73.05%
StateIdaho Daycare cost (Annual)$15,238 Nanny cost (Annual)$32,760 % Difference in price73.01%
StateSouth Carolina Daycare cost (Annual)$15,288 Nanny cost (Annual)$32,796 % Difference in price72.83%
StateMontana Daycare cost (Annual)$16,563 Nanny cost (Annual)$35,448 % Difference in price72.62%
StateColorado Daycare cost (Annual)$19,288 Nanny cost (Annual)$40,848 % Difference in price71.71%
StateArkansas Daycare cost (Annual)$15,713 Nanny cost (Annual)$32,844 % Difference in price70.56%
StateMississippi Daycare cost (Annual)$14,963 Nanny cost (Annual)$30,852 % Difference in price69.36%
StateFlorida Daycare cost (Annual)$17,125 Nanny cost (Annual)$35,148 % Difference in price68.96%
StateTexas Daycare cost (Annual)$16,725 Nanny cost (Annual)$34,296 % Difference in price68.88%
StateNorth Carolina Daycare cost (Annual)$16,663 Nanny cost (Annual)$33,984 % Difference in price68.40%
StateOklahoma Daycare cost (Annual)$15,800 Nanny cost (Annual)$31,920 % Difference in price67.56%
StateAlabama Daycare cost (Annual)$15,550 Nanny cost (Annual)$30,996 % Difference in price66.37%
StateTennessee Daycare cost (Annual)$17,025 Nanny cost (Annual)$33,900 % Difference in price66.27%
StateGeorgia Daycare cost (Annual)$17,063 Nanny cost (Annual)$33,972 % Difference in price66.27%
StateNew Hampshire Daycare cost (Annual)$19,075 Nanny cost (Annual)$37,236 % Difference in price64.50%
StateIndiana Daycare cost (Annual)$16,500 Nanny cost (Annual)$32,112 % Difference in price64.23%
StateWashington Daycare cost (Annual)$22,238 Nanny cost (Annual)$43,200 % Difference in price64.07%
StateMichigan Daycare cost (Annual)$17,538 Nanny cost (Annual)$33,924 % Difference in price63.68%
StateMissouri Daycare cost (Annual)$17,488 Nanny cost (Annual)$33,360 % Difference in price62.43%
StateUtah Daycare cost (Annual)$17,538 Nanny cost (Annual)$33,324 % Difference in price62.08%
StateMaine Daycare cost (Annual)$20,513 Nanny cost (Annual)$38,832 % Difference in price61.74%
StateNevada Daycare cost (Annual)$18,763 Nanny cost (Annual)$35,280 % Difference in price61.13%
StateOhio Daycare cost (Annual)$17,463 Nanny cost (Annual)$32,640 % Difference in price60.59%
StateVermont Daycare cost (Annual)$20,938 Nanny cost (Annual)$39,000 % Difference in price60.27%
StateVirginia Daycare cost (Annual)$19,863 Nanny cost (Annual)$36,336 % Difference in price58.63%
StateCalifornia Daycare cost (Annual)$23,863 Nanny cost (Annual)$43,596 % Difference in price58.51%
StateIllinois Daycare cost (Annual)$20,638 Nanny cost (Annual)$37,584 % Difference in price58.21%
StateWisconsin Daycare cost (Annual)$17,950 Nanny cost (Annual)$32,676 % Difference in price58.18%
StateMinnesota Daycare cost (Annual)$20,088 Nanny cost (Annual)$36,312 % Difference in price57.53%
StateKentucky Daycare cost (Annual)$17,613 Nanny cost (Annual)$31,572 % Difference in price56.76%
StateWyoming Daycare cost (Annual)$18,488 Nanny cost (Annual)$33,036 % Difference in price56.47%
StateKansas Daycare cost (Annual)$18,100 Nanny cost (Annual)$32,340 % Difference in price56.46%
StateNorth Dakota Daycare cost (Annual)$17,500 Nanny cost (Annual)$31,224 % Difference in price56.33%
StateNebraska Daycare cost (Annual)$18,013 Nanny cost (Annual)$32,100 % Difference in price56.22%
StateIowa Daycare cost (Annual)$17,850 Nanny cost (Annual)$31,344 % Difference in price54.86%
StatePennsylvania Daycare cost (Annual)$20,000 Nanny cost (Annual)$35,088 % Difference in price54.78%
StateMaryland Daycare cost (Annual)$22,288 Nanny cost (Annual)$38,808 % Difference in price54.08%
StateDelaware Daycare cost (Annual)$20,038 Nanny cost (Annual)$34,668 % Difference in price53.49%
StateWest Virginia Daycare cost (Annual)$17,863 Nanny cost (Annual)$30,768 % Difference in price53.08%
StateNew Jersey Daycare cost (Annual)$23,100 Nanny cost (Annual)$39,600 % Difference in price52.63%
StateNew York Daycare cost (Annual)$23,850 Nanny cost (Annual)$40,680 % Difference in price52.16%
StateOregon Daycare cost (Annual)$23,250 Nanny cost (Annual)$39,600 % Difference in price52.03%
StateMassachusetts Daycare cost (Annual)$26,238 Nanny cost (Annual)$43,656 % Difference in price49.84%
StateRhode Island Daycare cost (Annual)$24,150 Nanny cost (Annual)$37,668 % Difference in price43.73%
StateAlaska Daycare cost (Annual)$22,875 Nanny cost (Annual)$34,860 % Difference in price41.52%
StateConnecticut Daycare cost (Annual)$27,125 Nanny cost (Annual)$40,476 % Difference in price39.50%
Nanny costs based on a 40-hour work week.
Sources [4] Average Nanny Salary - Care.com https://www.care.com/c/average-nanny-salary-by-state/
Additional costs for caring for children
Childcare is, of course, not the only cost to consider when it comes to looking after children. The cost of raising a child from birth includes a wide variety of expenses from food and medical care
to toys and activities.
Below are some of the most common baby items and the average annual amount spent on these per baby. Some of these items will be purchased once or twice, like a car seat or crib, while others, like
diapers and clothing, will be purchased regularly over the first few years of a child’s life.
Item Avg. Price (annual)
Diapers $720
Formula $1,350
Clothing $600
Crib $200
Crib mattress $100
Car seats $240
Bottles $70
Sources [5] How Much Does It Cost to Have a Baby? - What to Expect https://www.whattoexpect.com/getting-pregnant/preparing-for-baby/work-and-finance/what-babies-really-cost.aspx [6] How Much Does it
Cost to Raise a Baby? - Healthline https://www.healthline.com/health/parenting/how-much-does-it-cost-to-raise-a-baby-and-what-you-can-do-to-prepare#clothing
How many millennials have children?
Data from Pew Research Center found that the percentage of adults having children has reduced significantly over the past 60 years. The research defines people aged 23 to 38 as being in a family if
they live with a spouse, their own children (either biological, adopted, or step children), or both.
In 2019, 55% of Millennials were living in a family of this type, compared to 66% of Gen X in 2003, 69% of Baby Boomers in 1987, and 85% of the Silent Generation in 1968. This change over time could
show that fewer people are choosing to settle down with families and have children than in previous generations.
Generation Living with a spouse and children Not living with a spouse and children
Millennials (2019) 55% 45%
Gen X (2003) 66% 34%
Baby Boomer (1987) 69% 31%
Silent Generation (1968) 85% 15%
Source [7] Millennials Family Life - Pew Research https://www.pewresearch.org/social-trends/2020/05/27/as-millennials-near-40-theyre-approaching-family-life-differently-than-previous-generations/
Getting help with childcare costs
If you are struggling to cover your childcare costs, there are a few ways you can get help. [8] Childcare Financial Assistance - U.S. Gov. https://childcare.gov/consumer-education/
Government programs
• Childcare financial assistance - States receive funding from the federal government for childcare financial assistance for low-income families through vouchers, subsidies or certificates.
Eligibility requirements vary by state so check your local childcare assistance program for more details.
• State-funded prekindergarten - For children aged 3-5, some states offer prekindergarten services for free or at low costs for eligible families. These aim to prepare children for starting
• Head Start and Early Head Start - These programs help children prepare for school from birth to age 5 with early learning in development support. They are available at no cost for families on low
incomes who are eligible.
• Military childcare financial assistance - A number of programs help military families pay for childcare where they are stationed. See local guidance for more details in your state.
Local assistance and discounts
• Sibling discounts - Some local childcare providers will offer a discount for families that enroll more than one child into their care. This might include a percentage or dollar amount reduction
in the cost, or a waiver of other fees.
• Local scholarships and assistance - In some areas, local nonprofit organizations will offer scholarships or financial assistance to eligible families to cover childcare costs.
• Military discounts - Some local childcare providers offer discounts for children of military families. Check with your chosen provider if this is available.
Tax-related support
• Earned-income tax credit - Families on low to moderate incomes can get a tax break if they qualify. This may reduce the taxes you pay or possibly give you a tax refund, depending on if you
• Child and dependent care tax credit - This is available to people who need to pay for childcare while they work and look for work and is applicable to children under the age of 13. [8] Childcare
Financial Assistance - U.S. Gov. https://childcare.gov/consumer-education/get-help-paying-for-child-care
To calculate the average annual daycare costs for infants, toddlers, and preschoolers we took an average from the typical costs for base quality and high-quality daycare in each state. | {"url":"https://www.self.inc/info/childcare-costs-by-state/","timestamp":"2024-11-13T12:19:21Z","content_type":"text/html","content_length":"119465","record_id":"<urn:uuid:680aba96-c34c-4643-8be2-6caeb4d4715d>","cc-path":"CC-MAIN-2024-46/segments/1730477028347.28/warc/CC-MAIN-20241113103539-20241113133539-00572.warc.gz"} |
A 20 mW 360 nm laser is incident on a metal anode. The metal in
For the external photoelectric effect we have
Photon energy = Work function + Electron kinetic energy
h*F = W + Ek
h*F = W +eU
where U is the stopping potential that need to be applied to completely stop all electrons
U =[(h*c/lambda)-W)/ e = 6.62*10^-34*3*10^8/360*10^-9/1.6*10^-19 - 1.8 =1.648 V
If N/t is number of photons in time unit then
Power = Energy/time = (N/t) h*F = (N/t)*(h*c/lambda)
N/t = P*lambda/(h*c) = 0.02*360*10^-9/6.62*10^-34/3*10^8 =3.62*10^16 photons/sec
If 1 photon = 1 electron emitted then current is
I = Q/t = (N/t)*e =3.62*10^16*1.6*10^-19 =5.8*10^-3 A =5.8 mA
These are unrealistic assumption because of two main reasons:
- efficiency of photoelectric effect is ussualy 5-20%.
- electrons emitted in their way to cathode scatter on wir molecules that remain in tube and do not reach the cathode. The vacuum in the tube need to be less than 10^-9 Torr. (Ultra high vacuum)
De Broglie wavelength is
lambda = h/p = h/sqrt(2*m*Ek) = h/sqrt(2*m*e*U) =6.62*10^-34/sqrt(2*9.1*10^-31*1.6*10^-19*1.648) =9.56*10^-10 m | {"url":"https://wizedu.com/questions/1391048/a-20-mw-360-nm-laser-is-incident-on-a-metal-anode","timestamp":"2024-11-06T11:50:20Z","content_type":"text/html","content_length":"35663","record_id":"<urn:uuid:0cd50456-bd56-464c-a9d7-f27e8bca8e30>","cc-path":"CC-MAIN-2024-46/segments/1730477027928.77/warc/CC-MAIN-20241106100950-20241106130950-00294.warc.gz"} |
Bone Age Height Calculator
Home » Simplify your calculations with ease. » Biological Calculators »
Bone Age Height Calculator
The Bone Age Height Calculator is a medical tool used to estimate the future adult height of children based on their bone age, current height, and chronological age. This tool is especially valuable
for pediatricians, endocrinologists, and parents who want to understand the growth potential of a child. Bone age is typically measured through X-rays, which provide insight into the maturity of the
bones compared to average growth rates at a certain chronological age.
This calculator is used for assessing whether a child’s growth is on track, whether there are any delays or advancements in growth, and what the likely adult height might be. It’s particularly useful
in cases where growth hormone therapy or other medical interventions are being considered. By using the calculator, users can obtain a more informed view of how a child’s growth may progress, helping
them to make decisions regarding potential treatments or lifestyle changes to promote healthy development.
Formula of Bone Age Height Calculator
To estimate adult height using the Bone Age Height Calculator, follow these steps:
1. Current Height (H): The height of the individual, typically measured in centimeters or inches.
2. Bone Age (BA): The bone age of the child, determined through a radiograph, compared to the average bone development for their chronological age. Bone age is measured in years.
3. Chronological Age (CA): The actual age of the child, measured in years.
4. Growth Remaining Factor (GRF): This factor estimates the potential for further growth based on the bone age. It is usually obtained from standardized growth prediction charts or tables.
5. Adult Height Prediction (AHP): The predicted adult height, calculated using the current height and the growth remaining factor.
The formula used to calculate adult height is:
Adult Height Prediction (AHP) = Current Height (H) + (Remaining Growth Factor (GRF) × (Bone Age (BA) - Chronological Age (CA)))
• Current Height (H): The current height of the child.
• Bone Age (BA): The estimated bone age based on a radiograph.
• Chronological Age (CA): The child's current age in years.
• Growth Remaining Factor (GRF): A factor derived from medical literature, which estimates how much more growth is possible.
• Adult Height Prediction (AHP): The estimated final adult height.
Key Points:
• Bone age is crucial because it gives a more precise understanding of physical development than chronological age alone.
• The formula includes a Growth Remaining Factor (GRF), which takes into account how much more growth the child is likely to experience.
• Accurate results depend on using up-to-date growth charts and tables.
• It is important to note that this prediction provides an estimate, and individual outcomes may vary based on health, nutrition, and genetics.
General Terms and Conversion Table
To help users quickly reference common data, here’s a table with estimated adult height predictions based on some average inputs:
Current Height (H) Bone Age (BA) Chronological Age (CA) Growth Remaining Factor (GRF) Predicted Adult Height (AHP)
140 cm 12 years 10 years 0.25 145 cm
150 cm 13 years 11 years 0.20 155 cm
160 cm 14 years 12 years 0.15 165 cm
170 cm 15 years 13 years 0.10 173 cm
180 cm 16 years 14 years 0.05 182 cm
This table shows potential adult height predictions for different ages and growth factors. While this provides a general idea, it’s essential to use an accurate Bone Age Height Calculator that
factors in individual characteristics.
Example of Bone Age Height Calculator
Let’s work through an example. Suppose you have a 12-year-old child whose current height is 150 cm. After an X-ray, the bone age is determined to be 13 years, and based on growth charts, the Growth
Remaining Factor (GRF) is estimated to be 0.20.
Here’s how we calculate the predicted adult height:
1. Current Height (H) = 150 cm
2. Bone Age (BA) = 13 years
3. Chronological Age (CA) = 12 years
4. Growth Remaining Factor (GRF) = 0.20
Using the formula:
Adult Height Prediction (AHP) = Current Height (H) + (Remaining Growth Factor (GRF) × (Bone Age (BA) - Chronological Age (CA)))
Adult Height Prediction (AHP) = 150 cm + (0.20 × (13 - 12))
AHP = 150 cm + (0.20 × 1) = 150 cm + 0.20 = 150.2 cm
So, the predicted adult height for the child is 150.2 cm. This estimate can be refine using more precise medical data, but this formula provides a reliable approximation.
Most Common FAQs
1. What is bone age and how is it measure?
Bone age refers to the maturity of a child’s bones compared to standard growth charts. It is typically measure through an X-ray of the hand and wrist, where doctors assess the development of bones
and growth plates.
2. How accurate is the Bone Age Height Calculator?
The Bone Age Height Calculator provides an estimate based on current height, bone age, and growth remaining factors. While it is reasonably accurate, actual growth may vary due to genetics, health,
and environmental factors. Medical professionals can offer more precise assessments.
3. What is the significance of the Growth Remaining Factor (GRF)?
The Growth Remaining Factor (GRF) is a value that helps predict how much more a child can grow. It is based on medical studies and growth charts, offering a scientific estimate of future growth based
on bone age and current development.
Leave a Comment | {"url":"https://calculatorshub.net/biological-calculators/bone-age-height-calculator/","timestamp":"2024-11-10T14:54:46Z","content_type":"text/html","content_length":"121771","record_id":"<urn:uuid:67cbce7e-3a0e-459d-9eea-b89a92e9d242>","cc-path":"CC-MAIN-2024-46/segments/1730477028187.60/warc/CC-MAIN-20241110134821-20241110164821-00188.warc.gz"} |
Holding Period Return - What Is It, Formula, How To Calculate?
Holding Period Return
Last Updated :
21 Aug, 2024
What Is Holding Period Return (HPR)?
Holding Period Return refers to total returns over the period for which an investment was held, usually expressed in percentage of initial investment, and is widely used for comparing returns
from various investments held for different periods of time. It also captures any additional income from the investment apart from helping calculate the growth or decline in value over multiple
An investor can easily compare between its different set of assets and portfolios based on the holding period return received for each of them as and when the period for which they are held gets
over. Additionally, these yields also help in determining tax implications based on the period for which the asset or portfolio is held.
Holding Period Return Explained
Holding period return, as the name suggests, is the return or yield received from the assets or portfolios held for a specific period of time. The period for which investors keep a security or asset
with them is the holding period. It starts from the dafter after the asset or portfolio is purchased and ends on the day it is sold or disposed. The return received for this holding period is
expressed as a percentage in most cases.
If the asset or portfolio in question is held for less than a year, it makes investor receive short-term gain or loss as return, while if the holding period is of more than a year, investors realize
for long-term returns, be it a gain or loss.
The holding period return becomes an important metric for the businesses or individual investors as it gives them all an opportunity to compare between the sets of securities they own and hold for
different periods.
For the calculation of HPR, there are two components that hold relevance – capital appreciation and income (dividend or interest). Capital appreciation occurs when the sales price of an asset is more
than the purchase price of it. This means, the capital appreciation value is determined based on the gains that are recorded for a period.
On the other hand, the income can be in the form of interests or dividends. If the investment is made in the shares of the company, the income is in the form of dividends earned. On the contrary, if
the investments is made in the debt securities, the interest payment received becomes the income.
Let us have a look at the equation below that can help to calculate the asset or bond holding period return:
Holding Period Return Formula = Income + (End of Period Value - Initial Value)/Initial Value
An alternative version of the formula can be used for calculating return over multiple periods from an investment. It is useful for calculating returns over regular intervals, which could include
annualized or quarterly returns. Here, t = number of years
Annualized HPR = * 1/t[-1]
Alternatively, returns for regular time intervals can be calculated thus:
(1 + HPR) = (1+r[1]) x (1+r[2]) x (1+r[3]) x (1+r[4])
Here, r[1], r[2], r[3], r[4] are periodic returns.
It can also be represented thus:
HPR = - 1
Here, r = return per period
n = number of periods
How To Calculate?
The calculation of HPR involves a series of steps. Let us have a look at those steps for clarity:
• The first step is to calculate the capital appraisal. For example, one buys an asset worth $10 and the value increases to $15 after during the holding period. Here, the capital appreciation value
will be $15-$10 = $5.
• The next step is to calculate the income earned from investments. Now, the gain in the form of capital appreciation will be added to the dividend or income investors receive for their investments
in shares or debt securities, respectively. This accounts for the total income.
• The final step is to calculate the holding period income. To get this figure, the total income derived in the second step is divided by the initial price of the asset, i.e., $10. The value
derived is represented in percentage form.
Let us consider the following instances to understand the calculation of the total holding period return:
Example 1
Suppose if an individual bought a stock which paid dividends of $50 and its price reached $170 from the initial price of $140 at which it was bought a year ago.
Now, we can calculate the HPR as follows:
Now, we would try to calculate the annualized returns for the same stock over a period of 3 years. Let us suppose the stock paid dividends worth $50 each year, and returns varied with 21% growth for
the first year, followed by 30% returns for the second year and -15% returns for the third year.
Now, we would calculate the annualized HPR as below:
• HPR = - 1
• = -1 = 33.70%
• The result would be HPR of 33.71 for all 3 years.
The advantage of using this method is that it would help take into account the effect of compounding over the years, which would lead to a realistic outcome.
Example 2
A report released on “Global Private Equity Report 2018” stated large private equity firms launched buyout funds with longer holding period and earned double than those assured by traditional buyout
funds. These firms included major names, like CVC Capital Partners, Blackstone, and The Carlyle Group, etc.
Among these were two first-time funds that made over one billion as returns for holding them for almost 15 years.
Holding Period Return Formula in Excel (with excel template)
Let us now do the same example above in Excel. This is very simple. You need to provide the three inputs of Income, end of the period value, and initial value.
You can easily calculate the Holding Period in the template provided.
Now, we can calculate the HPR as follows:
Now, we would calculate the annualized HPR as below:
HPR can be used to calculate total returns for an investment for a single or multiple periods, including various forms of returns, which might be added improperly otherwise when calculating total
returns. For instance, if someone holds a stock for a certain amount of time, and it pays dividends periodically, these dividends also need to be taken into account along with changes in stock
prices. It would also require keeping in mind that a rise in the value of the investment during multiple periods of return leads to a compounding effect, which might be left out in simpler
For instance, if an investment grew by 10% annually, it would be erroneous to assume that in two years, the growth on initial value would be 20%. It has to be calculated, taking into account 10%
growth for the first year and then calculating 10% growth over ‘this’ amount for the 2nd year, leading to a total return of 21.1% in two years, instead of 20%.
Relevance and Use
The HPR is used as an important metric to compare the returns of assets held for different holding periods. Another key application of the holding period return formula is inaccurately calculating
the effect of compounding while estimating total returns on investment for multiple periods. Apart from that, it has great utility in comparing various investments held for different time intervals
in terms of their total returns over these periods.
You can use the following Calculator.
Recommended Articles
This has been a guide to what is a Holding Period Return. Here, we explain the concept with its formula, how to calculate it, examples, and components. You can refer the following articles as well -
• Abnormal Return
• Accounting Period Definition
• Rate of Return Formula
• Cost-Benefit Analysis Formula | {"url":"https://www.wallstreetmojo.com/holding-period-return-formula-hpr/","timestamp":"2024-11-04T14:17:50Z","content_type":"text/html","content_length":"328936","record_id":"<urn:uuid:bddea969-b01d-4378-ab25-dac944c9e0d0>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00820.warc.gz"} |
What is Brandywine Total Current Assets from 2010 to 2024 | Stocks: BDN - Macroaxis
BDN Stock USD 5.18 0.10 1.97%
Brandywine Realty Total Current Assets yearly trend continues to be very stable with very little volatility. Total Current Assets are likely to drop to about 252.9
. Total Current Assets is the total value of all assets that are expected to be converted into cash within one year or during the normal operating cycle.
View All Fundamentals
First Reported Previous Quarter Current Value Quarterly Volatility
Total Current Assets
1989-12-31 239.2 M 233 M 1.4 B
Oil Shock Dot-com Bubble Housing Crash Credit Downgrade Yuan Drop Covid
Check Brandywine Realty
financial statements
over time to gain insight into future company performance. You can evaluate financial statements to find patterns among Brandywine Realty's main balance sheet or income statement drivers, such as
Interest Expense of 62.3 M
, Total Revenue of 366.6
Gross Profit of 96 M
, as well as many indicators such as
Price To Sales Ratio of 1.84
, Dividend Yield of 0.0957 or
PTB Ratio of 0.76
. Brandywine financial statements analysis is a perfect complement when working with
Brandywine Realty Valuation
Brandywine Total Current Assets
Check out the analysis of
Brandywine Realty Correlation
against competitors.
Latest Brandywine Realty's Total Current Assets Growth Pattern
Below is the plot of the Total Current Assets of Brandywine Realty Trust over the last few years. It is the total value of all assets that are expected to be converted into cash within one year or
during the normal operating cycle. Brandywine Realty's Total Current Assets historical data analysis aims to capture in quantitative terms the overall pattern of either growth or decline in
Brandywine Realty's overall financial position and show how it may be relating to other accounts over time.
Total Current Assets 10 Years Trend
Brandywine Total Current Assets Regression Statistics
Arithmetic Mean 297,431,537
Geometric Mean 226,075,173
Coefficient Of Variation 63.76
Mean Deviation 137,948,976
Median 252,908,050
Standard Deviation 189,649,560
Sample Variance 35967T
Range 792.5M
R-Value (0)
Mean Square Error 38733T
R-Squared 0.000016
Significance 0.99
Slope (166,988)
Total Sum of Squares 503537.4T
Brandywine Total Current Assets History
About Brandywine Realty Financial Statements
Brandywine Realty investors utilize fundamental indicators, such as Total Current Assets, to predict how Brandywine Stock might perform in the future. Analyzing these trends over time helps investors
make informed
market timing
decisions. For further insights, please visit our
fundamental analysis
Last Reported Projected for Next Year
Total Current Assets 266.2 M 252.9 M
Pair Trading with Brandywine Realty
One of the main advantages of trading using pair correlations is that every trade hedges away some risk. Because there are two separate transactions required, even if Brandywine Realty position
performs unexpectedly, the other equity can make up some of the losses. Pair trading also minimizes risk from directional movements in the market. For example, if an entire industry or sector drops
because of unexpected headlines, the short position in Brandywine Realty will appreciate offsetting losses from the drop in the long position's value.
0.82 BXP Boston Properties PairCorr
0.75 CUZ Cousins Properties PairCorr
0.58 EQC Equity Commonwealth PairCorr
0.57 UK Ucommune International PairCorr
0.51 RC Ready Capital Corp PairCorr
0.43 AHT-PF Ashford Hospitality Trust PairCorr
0.4 AHT-PH Ashford Hospitality Trust PairCorr
The ability to find closely correlated positions to Brandywine Realty could be a great tool in your tax-loss harvesting strategies, allowing investors a quick way to find a similar-enough asset to
replace Brandywine Realty when you sell it. If you don't do this, your portfolio allocation will be skewed against your target asset allocation. So, investors can't just sell and buy back Brandywine
Realty - that would be a violation of the tax code under the "wash sale" rule, and this is why you need to find a similar enough asset and use the proceeds from selling Brandywine Realty Trust to buy
The correlation of Brandywine Realty is a statistical measure of how it moves in relation to other instruments. This measure is expressed in what is known as the correlation coefficient, which ranges
between -1 and +1. A perfect positive correlation (i.e., a correlation coefficient of +1) implies that as Brandywine Realty moves, either up or down, the other security will move in the same
direction. Alternatively, perfect negative correlation means that if Brandywine Realty Trust moves in either direction, the perfectly negatively correlated security will move in the opposite
direction. If the correlation is 0, the equities are not correlated; they are entirely random. A correlation greater than 0.8 is generally described as strong, whereas a correlation less than 0.5 is
generally considered weak.
Correlation analysis
and pair trading evaluation for Brandywine Realty can also be used as hedging techniques within a particular sector or industry or even over random equities to generate a better risk-adjusted return
on your portfolios.
Pair CorrelationCorrelation Matching
When determining whether Brandywine Realty Trust
offers a strong return on investment
in its stock, a comprehensive analysis is essential. The process typically begins with a thorough review of Brandywine Realty's
financial statements
, including income statements, balance sheets, and cash flow statements, to assess its
financial health
. Key financial ratios are used to gauge profitability, efficiency, and growth potential of Brandywine Realty Trust Stock.
Outlined below are crucial reports that will aid in making a well-informed decision on Brandywine Realty Trust Stock:
Check out the analysis of
Brandywine Realty Correlation
against competitors. You can also try the
Equity Forecasting
module to use basic forecasting models to generate price predictions and determine price momentum.
Is Diversified REITs space expected to grow? Or is there an opportunity to expand the business' product line in the future? Factors like these will boost
the valuation of Brandywine Realty
. If investors know Brandywine will grow in the future, the company's valuation will be higher. The financial industry is built on trying to define current growth potential and future valuation
accurately. All the valuation information about Brandywine Realty listed above have to be considered, but the key to understanding future value is determining which factors weigh more heavily than
Quarterly Earnings Growth Dividend Share Earnings Share Revenue Per Share Quarterly Revenue Growth
5.506 0.6 (1.84) 1.784 (0.95)
The market value of Brandywine Realty Trust
is measured differently than its book value, which is the value of Brandywine that is recorded on the company's balance sheet. Investors also form their own opinion of Brandywine Realty's value that
differs from its market value or its book value, called intrinsic value, which is Brandywine Realty's true underlying value. Investors use various methods to calculate intrinsic value and buy a stock
when its market value falls below its intrinsic value. Because Brandywine Realty's market value can be influenced by many factors that don't directly affect Brandywine Realty's underlying business
(such as a pandemic or basic market pessimism), market value can vary widely from intrinsic value.
AltmanZ ScoreDetails PiotroskiF ScoreDetails BeneishM ScoreDetails FinancialAnalysisDetails Buy or SellAdviceDetails
Please note, there is a significant difference between Brandywine Realty's value and its price as these two are different measures arrived at by different means. Investors typically determine
if Brandywine Realty is a good investment
by looking at such factors as earnings, sales, fundamental and technical indicators, competition as well as analyst projections. However, Brandywine Realty's price is the amount at which it trades on
the open market and represents the number that a seller and buyer find agreeable to each party. | {"url":"https://widgets.macroaxis.com/financial-statements/BDN/Total-Current-Assets","timestamp":"2024-11-05T06:31:40Z","content_type":"text/html","content_length":"342274","record_id":"<urn:uuid:9d2f87b6-f913-4f01-8ee4-3ebb5c47cc08>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00580.warc.gz"} |
Potential Visibility for Domains & TLDs using Branded Links
A TLD (Top Level Domain) has the potential to be viewed close to 8,000,000 times in total per each domain which uses it. This figure was calculated as follows:
Sample Size
Based on a sample size of 10,000 domains chosen at random from the Rebrandly's database spanning across 173 nations and 350 TLDs. The average number of branded links created during April 2017 was 189
links per domain.
Distribution of Links
NLC x NLS x NCV x NMD = Average number of TLD impressions per domain
189 x 35 x 10 x 120 = 7,938,000*
• 189 = average number of links created per month. (NLC)
• 35 = average number of times a link is shared per month. (NLS)
• 10 = average number of times content is viewed per month. (NCV)
• 120 = average number of months a domain is used. (NMD)
*Please note NLS, NCV, NMD are based on estimations as these numbers cannot be calculated for definite. We estimate that on average: a link is shared 35 times per month, each link is viewed 10 times
per month and that a domain is actively used for 10 years (120 months).
What Does This Mean?
This means that each of the domains you register and the associated TLDs have the potential to be viewed close to 8 million times. Essentially, every time you share a branded link you are conceivably
sharing your brand name a further 8 million times!
This Article is About:
• Top Level Domain Impressions
• Number of impressions a TLD can receive
• Branded Domains
See Also:
0 comments
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What is an Inception Score (IS)? Definition from TechTarget
Prev Next multimodal AI reinforcement learning from human feedback (RLHF)
What is the inception score (IS)?
The inception score (IS) is a mathematical algorithm used to measure or determine the quality of images created by generative AI through a generative adversarial network (GAN). The word "inception"
refers to the spark of creativity or initial beginning of a thought or action traditionally experienced by humans.
Without an inception score, humans are left to observe a generative image and make a visual evaluation of the image -- but such visual evaluations are highly subjective and can vary widely based on
the preferences and biases of the human viewer. The inception score, and other metrics such as Fréchet inception distance (FID), offer objective and consistent measures of generated images; and by
extension, the quality and capability of the underlying generative model.
The score produced by the IS algorithm can range from zero (worst) to infinity (best). The inception score algorithm measures two factors:
• Quality. How good the generated image is. Generated images should be believable or realistic as if a real person painted a picture or took a photograph. For example, if the AI produces images of
cats, each image should include a clearly identifiable cat. If the object is not clearly identifiable as a cat, the corresponding IS will be low.
• Diversity. How diverse the generated image is. Generated images should have high randomness (entropy), meaning that the generative AI should produce highly varied images. For example, if the AI
produces images of cats, each image should be a different cat breed and perhaps a different cat pose. If the AI is producing images of the same cat breed in the same pose, the diversity and
corresponding IS will be low.
Generative AI developers use the inception score as a measure of image quality. The IS may be employed as a training mechanism by feeding the IS back to the AI model. This kind of training can
provide more objective and explainable feedback than solely allowing human viewers to subjectively "score" generative images.
How does the inception score work?
The inception score, first defined in a 2016 technical paper, is based on Google's "Inception" image classification network.
Calculating an inception score starts by using the image classification network to ingest a generated image and return a probability distribution for the image. The image classification network is
fundamentally a pre-trained Inception v3 model, which can predict class probabilities -- what something might be -- for each computer-generated image. A probability distribution is simply a numbered
list of what the image classification network "thinks" the image might be -- each with a fractional score that adds up to 1.0.
For example, the image classification network might see the generated image of a cat and return a series of potential results such as the following:
Cat 0.5
Flower 0.2
Car 0.2
House 0.1
Total 1.0
The probability distribution helps to determine whether the generated image contains one well-defined thing, or a series of things that are harder (if not impossible) for the image classification
network to identify. This is the foundation of the quality factor -- does the generated image look like something specific and identifiable?
Next, the inception score process compares the probability distribution for all the generated images. There may be as many as 50,000 generated images in a sample. This creates a second factor called
marginal distribution, which indicates the amount of variety present in the generative AI's images.
For the cat example, the labels utilized in probability distribution are summed to show the focused distribution (the number of same images such as cats), and the uniform distribution (the number of
flowers, cars, houses, and so on). These factors illustrate the variety in the generative AI's output. This is the foundation of the diversity factor -- can the AI produce varied items and scenes?
The last step is to combine probability distribution and marginal distribution into a single score, which can represent both the distinctiveness of the object as well as the diversity of the output.
The more those two distributions differ, the higher the inception score. The actual score is calculated using a statistical method called the Kullback-Leibler divergence, or KL divergence.
When there is high KL divergence, there is a strong probability distribution and an even (flat) marginal distribution -- each image has a distinct label (such as a cat), but the overall set of images
has many different labels. This yields the highest inception score.
Finally, the IS algorithm takes the exponent of the KL divergence and produces an average of the final number for every image in the sample set.
What are the limitations of the inception score?
Although the inception score algorithm provides an objective means of measuring the quality and diversity of AI-generated images, the IS poses three principal limitations for AI developers:
• Small image sizes. The IS algorithm only works on small, square image sizes -- roughly 300 x 300 pixels.
• Limited samples. The IS measures image diversity, so a limited sample size -- such as only one seascape image or the same image produced many times -- will produce an artificially high inception
score because there just are not enough images of that type or class to adequately judge diversity.
• Unusual images. The inception score is calculated against a pre-trained data set within the image classification network that represents about 1000 image types or classes. The IS will produce an
artificially low inception score if the AI generates an image that is not within those 1000 classes. This is because there is no similar pre-trained data to compare the new image against. Any
generative work with labels that are not in the image classification network -- such as different fish or varieties of trees -- may score lower.
Inception score vs. Fréchet inception distance
Another metric used to evaluate the quality of AI-generated images is the Fréchet inception distance. FID was introduced in 2017 and has generally superseded inception score as the preferred measure
of generative image model performance.
The principal difference between IS and FID is the comparative use and evaluation of real images, referred to as "ground truth." This allows FID to analyze real images alongside computer-generated
images in a bid to better simulate human perception. By comparison, IS only evaluates computer-generated images.
Although FID has generally edged out IS as the preferred quality metric for GANs, FID has also been shown to demonstrate some statistical bias, and does not always accurately reflect human
For more information on generative AI-related terms, read the following articles:
How to calculate the inception score
The actual formula to calculate inception score requires the use of calculus and is beyond the scope of this definition. For a more complete explanation, however, an abbreviated mathematical
expression for inception score can be shown as the following:
IS(G) = exp (E[x][∼][pg] D[KL] (p(y|x) || p(y) ) )
The major components of the formula are as follows:
• IS is the final inception score.
• D[KL] is the KL divergence.
• p(y|x) is the conditional probability distribution.
• p(y) is the marginal probability distribution.
• E[x~pg] is the sum and average of all results.
The common process for resolving this expression and determining a final inception score involves five basic steps:
1. Process the AI-generated images through the image classification network to obtain the conditional probability distribution or p(y|x).
2. Calculate the marginal probability distribution or p(y).
3. Calculate the KL divergence (between p(y) and p(y|x)).
4. Calculate the sum for classes and calculate an average score for all images (basically repeat the previous steps for all images in the computer-generated set or sample).
5. Calculate the average value of all results (E[x~pg]) and take its exponent (exp).
This final result is the inception score for the given set of computer-generated images.
How to implement the inception score
Although the mathematical formula for inception score can be resolved manually, the process of repeating advanced multi-step calculations across thousands of images can be a daunting and error-prone
human challenge.
Instead of manual calculations, AI developers working with generative image models will typically implement a metric such as inception score using a mathematical software package. Common math
processing alternatives include the following:
• Keras. An open source software library and Python interface for artificial neural networks supporting the TensorFlow library. Keras can interoperate with the Inception v3 model directly.
• NumPy. A Python library for scientific computing which supports multidimensional array objects, derived objects and various routines for fast operations on arrays, along with statistical
Implementing IS in a math package will require some amount of coding to derive probability distributions (or access to data where distributions are stored) and perform other required calculations.
Coding may be performed by AI scientists already working on generative AI systems or supporting development staff.
This was last updated in May 2024
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Electromagnetic Simulation Solvers
CST Studio Suite Simulation Solvers for Electromagnetic Systems and Devices
A Powerful Portfolio of Electromagnetic Field Solvers
CST Studio Suite® gives customers access to multiple electromagnetic (EM) simulation solvers. The range includes methods such as the finite element method (FEM) the finite integration technique (FIT)
, and the transmission line matrix method (TLM). The named three methods represent the most powerful general-purpose solvers for electromagnetic simulation tasks. CST Studio Suite offers solution
methods in time domain and in frequency domain. The applicability range of the CST Studio Suite solution spans from statics to optical frequencies.
Electromagnetic Solvers by Frequency Range and Application
High-Frequency Electromagnetic Simulation Solvers
With the FIT, FEM and TLM methods CST Studio Suite provides solvers that are ideally suited for high-frequency simulations. FIT and TLM as classical time domain methods can play their advantages for
broadband, antenna, and complex and detail-rich applications. These solvers can also analyze the electromagnetic compatibility (EMC) of devices and signal and power integrity. Additional solvers for
specialist high-frequency applications such as electrically large or highly resonant structures complement the general-purpose solvers.
Low-Frequency Electromagnetic Simulation Solvers
CST Studio Suite includes FEM solvers dedicated to static and low-frequency applications such as electromechanical devices, motors, generators, transformers, and sensors. Opera technology complements
this solver set for a comprehensive and highly accurate solution.
Solvers for Charged Particle Dynamics
The simulation of particles in electromagnetic fields is a particular strength of CST Studio Suite. There is a wide range of applications, from electron guns to microwave tubes, from magnetron
sputtering to particle accelerator components. We can deliver suitable solvers for the efficient simulation of particle-based devices.
Multiphysics with CST Studio Suite
The presence of electromagnetic fields causes effects from other physics. The losses in materials lead to increases in temperature. The increased temperature can cause deformations to the components
that compromise their performance. CST Studio Suite offers multiphysics simulation to analyze these effects. The range of applications includes electronics cooling as well as bio-heat consideration
for medical devices. The 3DEXPERIENCE platform enables a far wider range of multiphysics applications.
Electromagnetic Analysis with the Best Solver for Your Application
The seamless integration of the solvers into one user interface in CST Studio Suite enables the easy selection of the most appropriate simulation method for a given problem class. Being able to
choose between simulation approaches leads to improved simulation performance and unprecedented simulation reliability through cross-verification.
• High Frequency Solvers
• Low Frequency Solvers
• Multiphysics Solvers
• Particle Solvers
• EMC and EDA Solvers
Time Domain Solver
The Time Domain Solver is a powerful and versatile multi-purpose transient 3D full-wave solver, with both finite integration technique (FIT) and transmission line matrix (TLM) implementations
included in a single package. The Time Domain Solver can perform broadband simulations in a single run. Support for hardware acceleration and MPI cluster computing also makes the solver suitable for
extremely large, complex and detail-rich simulations.
Applications of the Time Domain Solver:
• General high-frequency applications using medium-to-large models
• Transient effects
• 3D electronics
Frequency Domain Solver
The Frequency Domain Solver is a powerful multi-purpose 3D full-wave solver, based on the finite element method (FEM), that offers excellent simulation performance for many types of component.
Because the Frequency Domain Solver can calculate all ports at the same time, it is also a very efficient way to simulate multi-port systems such as connectors and arrays. The Frequency Domain Solver
includes a model-order reduction (MOR) feature which can accelerate the simulation of resonant structures such as filters.
Applications of the Frequency Domain Solver:
• General high-frequency applications using small-to-medium-sized models
• Resonant structures
• Multi-port systems
• 3D electronics
Asymptotic Solver
The Asymptotic Solver is a ray tracing solver which is efficient for extremely large structures where a full-wave solver is unnecessary. The Asymptotic Solver is based on the Shooting Bouncing Ray
(SBR) method. SBR is an extension to physical optics, and capable of tackling simulations with an electric size of many thousands of wavelengths.
Applications of the Asymptotic Solver:
• Electrically very large structures
• Installed performance of antennas
• Scattering analysis
Eigenmode Solver
The Eigenmode Solver is a 3D solver for simulating resonant structures, incorporating the Advanced Krylov Subspace method (AKS), and the Jacobi-Davidson method (JDM). Common applications of the
Eigenmode Solver are highly resonant filter structures, high-Q particle accelerator cavities, and slow wave structures such as traveling wave tubes. The Eigenmode Solver supports sensitivity
analysis, allowing the direct calculation of the detuning effect of structural deformation.
Applications of the Eigenmode Solver:
• Filters
• Cavities
• Metamaterials and periodic structures
Filter Designer 3D
A synthesis tool for designing bandpass and diplexer filters, producing a range of coupling matrix topologies for the application in arbitrary coupled-resonator based technology. It also offers a
choice of building blocks to realize 3D filters through Assembly Modeling. From the Component Library, the user can choose between combline/interdigital coaxial cavities and rectangular waveguides.
Alternatively, the user can define customized building blocks of any type of single-mode technology (for example SIW or dielectric pucks).
The functionality provided includes the coupling matrix extraction. It can directly be used as a goal for optimization of a simulation model or for assistance in tuning complex hardware by real-time
measurements using a network analyzer.
Applications of Filter Designer3D:
• Cross-coupled filters for different electromagnetic technologies (for example cavities, microstrips, dielectrics)
• Assistive tuning for filter hardware (with vector network analyzer link)
Integral Equation Solver
The Integral Equation Solver is a 3D full-wave solver, based on the method of moments (MOM) technique with multilevel fast multipole method (MLFMM). The Integral Equation Solver uses a surface
integral technique, which makes it much more efficient than full volume methods when simulating large models with lots of empty space. The Integral Equation Solver includes a characteristic mode
analysis (CMA) feature which calculates the modes supported by a structure.
Applications of the Integral Equation Solver:
• High-frequency applications using electrically large models
• Installed performance
• Characteristic mode analysis
Multilayer Solver
The Multilayer Solver is a 3D full-wave solver, based on the method of moments (MOM) technique. The Multilayer Solver uses a surface integral technique and is optimized for simulating planar
microwave structures. The Multilayer Solver includes a characteristic mode analysis (CMA) feature which calculates the modes supported by a structure.
Applications of the Multilayer Solver:
• MMIC
• Feeding networks
• Planar antennas
Hybrid Solver Task
The Hybrid Solver Task allows the Time Domain, Frequency Domain, Integral Equation and Asymptotic Solvers to be linked for hybrid simulation. For simulation projects that involve very wide frequency
bands or electrically large structures with very fine details, calculations can be made much more efficient by using different solvers on different parts. Simulated fields are transferred between
solvers through field sources, with a bidirectional link between the solvers for more accurate simulation.
Applications of the Hybrid Solver Task:
• Small antennas on very large structures
• EMC simulation
• Human body simulation in complex environments
Electrostatic Solver
The Electrostatic Solver is a 3D solver for simulating static electric fields. This solver is especially suitable for applications such as sensors where electric charge or capacitance is important.
The speed of the solver also means that it is very useful for optimizing applications such as electrodes and insulators.
Applications of the Electrostatic Solver:
• Sensors and touchscreens
• Power equipment
• Charged particle devices and X-ray tubes
Magnetostatic Solver
The Magnetostatic Solver is a 3D solver for simulating static magnetic fields. This solver is most useful for simulating magnets, sensors, and for simulating electrical machines such as motors and
generators in cases where transient effects and eddy currents are not critical.
Applications of the Magnetostatic Solver:
• Sensors
• Electrical machines
• Particle beam focusing magnets
Low Frequency – Frequency Domain Solver
The Low-Frequency Frequency Domain (LF-FD) Solver is a 3D solver for simulating the time-harmonic behavior in low frequency systems, and includes magneto-quasistatic (MQS), electro-quasistatic (EQS)
and full-wave implementations. This solver is most useful for simulations that involve frequency-domain effects and where the sources are coils.
Applications of the Low Frequency Frequency Domain Solver:
• Sensors and non-destructive testing (NDT)
• RFID and wireless power transfer
• Power engineering – bus bar systems
Low Frequency – Time Domain Solver
The Low-Frequency Time Domain (LF-FD) Solver is a 3D solver for simulating the transient behavior in low-frequency systems, and includes both magneto-quasistatic (MQS) and electro-quasistatic (EQS)
implementations. The MQS solver is suitable for problems involving eddy currents, non-linear effects, and transient effects such as motion or inrush. The EQS solver is suitable for
resistive-capacitive problems and HV-DC applications.
Applications of the Low-Frequency Time Domain Solver:
• Electrical machines and transformers
• Electromechanical - motors, generators
• Power engineering – insulation, bus bar systems, switchgear
Stationary Current Solver
The Stationary Current Field Solver is a 3D solver for simulating the flow of DC currents through a device, especially with lossy components. This solver can be used to characterize the electrical
properties of a component that is DC or in which eddy currents and transient effects are irrelevant.
Applications of the Stationary Current Solver:
• High-power equipment
• Electrical machines
• PCB power distribution network
Conjugate Heat Transfer Solver
The Conjugate Heat Transfer (CHT) Solver uses CFD technique to predict fluid flow and temperature distribution in a system. The CHT solver includes the thermal effects from all heat transfer modes -
conduction, convection and radiation, and can include heat sources from electromagnetic losses just as the Steady State and Transient Thermal solvers do. Devices such as fans, perforated screens, and
thermal interface materials can be directly modeled. Compact thermal models (CTM), such as two-resistor CTM, can also be considered.
Applications of the Conjugate Heat Transfer Solver:
• Electronics cooling: natural and forced convection of high-power electronics components and devices, such as
□ PCBs
□ filters
□ antennas
□ chassis
• with installed cooling devices such as
Thermal Transient Solver
The Thermal Transient Solver can predict the time-varying temperature response of a system. Heat sources can include losses generated by electric and magnetic fields, currents, particle collisions,
human bio-heat, and other user-defined sources. Tightly linked to our electromagnetic solvers, the Thermal Transient Solver enables transient temperature prediction of devices and resulting impact on
their electromagnetic performance.
Applications of the Thermal Transient Solver:
• High-power electronics components and devices, such as PCBs, filters, antennas…
• Medical devices and human bio-heating
Thermal Steady State Solver
The Thermal Steady State Solver can predict the temperature distribution of a steady-state system. Heat sources can include losses generated by electric and magnetic fields, currents, particle
collisions, human bio-heat, and other user-defined sources. Seamlessly linked to our electromagnetic solvers, the Thermal Steady State Solver enables temperature prediction of devices and resulting
impact on their electromagnetic performance.
Applications of the Thermal Steady State Solver:
• High-power electronics components and devices, such as printed circuit boards (PCBs), filters, antennas etc.
• Medical devices and human bio-heating
Mechanical Solver
The Mechanical Solver can predict mechanical stress of structures and deformation caused by electromagnetic forces and thermal expansion. It is used with the EM and thermal solvers to assess the
possible performance impact of the force and heating to the device.
Applications of the Mechanical Solver:
• Filter detuning
• PCB deformation
• Lorentz forces on particle accelerators
Particle-in-Cell Solver
The Particle-in-Cell (PIC) Solver is a versatile, self-consistent simulation method for particle tracking. It calculates both particle trajectories and electromagnetic fields in the time-domain,
considering space charge effects and mutual coupling between particles and fields. The PIC solver can simulate a vast variety of devices where the interaction between particles and high-frequency
fields is important. Another application area is high-power devices where electron multipacting is a risk.
Applications of the Particle-In-Cell Solver:
• Accelerator components
• Slow-wave devices
• Multipaction
Electrostatic Particle-In-Cell Solver
The Electrostatic Particle-In-Cell (Es-PIC) solver technology computes the space charge dynamics in a transient approach captures the time domain behavior neglected by tracking analysis. Es-PIC
calculates space charge versus time, considering the electrostatic effects only. Compared to a pure Particle-In-Cell (PIC) approach, there is no current and H-Field induced but it is very well suited
for structures with large time scales.
Applications of the Electrostatic Particle-In-Cell Solver:
• Plasma Ion Source
• Electron Gun with Ionization
• Low-pressure Breakdown analysis
Particle Tracking Solver
The Particle Tracking Solver is a 3D solver for simulating particle trajectories through electromagnetic fields. It can consider the space charge effect on the electric field through the Gun
Iteration option. Several emission models including fixed, space charge limited, thermionic and field emission are available, and secondary electron emissions can be simulated.
Applications of the Particle Tracking Solver:
• Particle sources
• Focusing and beam steering magnets
• Accelerator components
Wakefield Solver
The Wakefield Solver calculates the fields around a particle beam, represented by a line current, and the wake fields produced through interactions with discontinuities in the surrounding structure.
Applications of the Wakefield Solver:
• Cavities
• Collimators
• Beam position monitors
PCB Solvers
The PCBs and Packages Module of CST Studio Suite is a tool for signal integrity (SI), power integrity (PI), and electromagnetic compatibility (EMC) analysis on printed circuit boards (PCB). It
integrates into the EDA design flow by providing powerful import filters for popular layout tools from Cadence, Zuken, and Altium. Effects like resonances, reflections, crosstalk, power/ground bounce
and simultaneous switching noise (SSN) can be simulated at any stage of product development, from pre-layout to post-layout phase.
CST Studio Suite includes three different solver types:
• 2D Transmission Line method
• 3D Partial Element Equivalent Circuit (PEEC) method
• 3D Finite-Element Frequency-Domain (FEFD) method
and pre-defined workflows for IR drop, PI and SI analysis
Rule Check
Rule Check is an EMC, SI and PI design rule checking (DRC) tool that reads popular board files from Cadence, Mentor Graphics, and Zuken as well as ODB++ (for example Altium). It checks the PCB design
against a suite of EMC or SI design rules. The kernel used by Rule Check is the well-known software tool EMSAT.
The user can designate various nets and components that are critical for EMC, such as I/O nets, power/ground nets, and decoupling capacitors. Rule Check examines each critical net, in turn, to check
that it does not violate any of the selected EMC or SI design rules. After the completion of the rule checking, it displays the EMC rule violations graphically or as an HTML document.
Applications of the Rule Check:
• Electromagnetic compatibility (EMC) PCB design rule checking
• Signal integrity and power integrity (SI/PI) PCB design rule checking
Cable Harness Solver
The Cable Harness Solver analyzes signal integrity (SI), conducted emission (CE), radiated emission (RE), and electromagnetic susceptibility (EMS) of complex cable structures in electrically large
systems in three-dimensions. It incorporates a fast and accurate transmission line modeling technique for cable harness configurations in 3D metallic or dielectric environment. Hybrid simulation with
the Cable Harness Solver and other high-frequency solvers allows structures containing complex cable harnesses to be simulated in 3D efficiently.
Applications of the Cable Harness Solver:
• General SI and EMC simulation of cables
• Cable harness layout in vehicles and aircraft
• Hybrid cables in consumer electronics
Fullwave em-simulators solve Maxwell's equations without approximations based on a problem's special physical nature. They typically provide solutions to high-frequency electromagnetic applications
such as antennas or components. SIMULIA CST Studio Suite provides time-domain and frequency-domain em-simulators.
An electromagnetic solver is an implementation of a numerical method that solves Maxwell's equations. It has to cover all the relevant physics and consider the material properties and geometrical
structures of the system being analyzed.
The best electromagnetic simulation software is the one, that gets your jobs done accurately and quickly. One fundamental requirement to accomplish this challenge within one software package is the
availability of a range of numerical simulation methods within this software because not one simulation method can solve all the simulation challenges. The SIMULIA electromagnetic simulation
portfolio offers a wide range of em-simulators for frequency ranges from DC to light.
Electromagnetic or EM simulation describes approaches to solving Maxwell's equations in space and time. Methods based on volume discretization are, for example, the finite element method (FEM), the
finite integration technique (FIT), the finite difference time domain method (FDTD) and the transmission-line matrix method TLM). These methods are very general and can be used to simulate all
classes of problems. However, there are methods that are far more efficient for specific types of electromagnetic analysis, such as the method of moments (MoM), boundary element method (BEM) mode
matching, physical optics,...
Also Discover
Automatic Optimization
CST Studio Suite Offers Automatic Optimization Routines for Electromagnetic Systems and Devices
Workflow Integration
CST Studio Suite Import and Data Exchange Options to Streamline Electromagnetic Design
Learn What SIMULIA Can Do for You
Speak with a SIMULIA expert to learn how our solutions enable seamless collaboration and sustainable innovation at organizations of every size.
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Courses and classes are available for students, academia, professionals and companies. Find the right SIMULIA training for you.
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Find information on software & hardware certification, software downloads, user documentation, support contact and services offering | {"url":"https://www.3ds.com/products/simulia/cst-studio-suite/electromagnetic-simulation-solvers","timestamp":"2024-11-07T15:35:43Z","content_type":"application/xhtml+xml","content_length":"552304","record_id":"<urn:uuid:9bd6e1d4-0162-4b16-92ce-71e1bbfde7a4>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00249.warc.gz"} |
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Machine Learning Logistic Regression: Python, Trading and more
Imagine a world where you can predict market movements with uncanny accuracy, where gut feelings give way to data-driven insights, and where every trade is a calculated step towards profit. This, my
friend, is the alluring promise of machine learning in trading.
Among the many algorithms vying for dominance in this arena, logistic regression stands out as a versatile and beginner-friendly tool. But how exactly does it work in the world of trading?
Think of Machine learning logistic regression as a binary classifier. It analyses mountains of historical data – prices, volumes, indicators – and learns to distinguish between two distinct outcomes:
up or down. Delve into the intricacies of logistic regression in machine learning for trading as we harness its capabilities to forecast stock price movements using Python.
Integrating artificial intelligence in trading allows us to enhance the logistic regression model further, enabling it to adapt and optimize its predictions based on evolving market conditions,
ultimately leading to more informed trading decisions.
This blog covers:
Machine learning tasks generally bifurcate into two realms:
1. The expected outcome is defined
2. The expected outcome is not defined
The former characterised by input data paired with labelled outputs, is termed supervised learning. Conversely, when input data lacks labelled responses, that is, in the latter case, it's known as
unsupervised learning. Explore the 'unsupervised learning course' from Quantra.
Additionally, there's reinforcement learning, which refines models through iterative feedback to enhance performance. Now, we explore Machine Learning Logistic Regression.
What is logistic regression?
Logistic regression falls under the category of supervised learning; it measures the relationship between the categorical dependent variable and one or more independent variables by estimating
probabilities using a logistic/sigmoid function.
It is primarily used for binary classification problems where the outcome can take on only two possible categorical values, often denoted as 0 and 1. Some examples are, "success" or "failure", "spam"
or "not spam", etc.
Despite the name ‘logistic regression’, this is not used for machine learning regression problems where the task is to predict the real-valued output. It is a classification problem which is used to
predict a binary outcome (1/0, -1/1, True/False) given a set of independent variables.
Linear regression and logistic regression
Logistic regression is a bit similar to linear regression, or we can say it is a generalised linear model. In linear regression, we predict a real-valued output 'y' based on a weighted sum of input
$$y=c + x_1*w_1 + x_2*w_2+ x_3*w_3 +........+ x_n*w_n$$
The aim of linear regression is to estimate values for the model coefficients c, w1, w2, w3 ….wn and fit the training data with minimal squared error and predict the output y.
Machine learning Logistic regression does the same thing but with one addition. The logistic regression model computes a weighted sum of the input variables similar to the linear regression, but it
runs the result through a special non-linear function, the logistic function or sigmoid function to produce the output y. Here, the output is binary or in the form of 0/1 or -1/1.
$$y = logistic(c + x_1*w_1 + x_2*w_2+ x_3*w_3 +........+ x_n*w_n)$$
$$y = 1/1 + e[-(c + x_1*w_1 + x_2*w_2+ x_3*w_3 +........+ x_n*w_n)]$$
The sigmoid/logistic function is given by the following equation.
y = 1 / 1+ e-x
As you can see in the graph, it is an S-shaped curve that gets closer to 1 as the value of the input variable increases above 0 and gets closer to 0 as the input variable decreases below 0.
In the context of Machine learning logistic regression, the decision boundary is commonly set at 0.5, meaning that if the predicted probability is greater than 0.5, the outcome is classified as 1
(positive), and if it is less than 0.5, the outcome is classified as 0 (negative).
Now, let us consider the task of predicting the stock price movement. If tomorrow’s closing price is higher than today’s closing price, then we will buy the stock (1), else we will sell it (-1). If
the output is 0.7, then we can say that there is a 70% chance that tomorrow’s closing price is higher than today’s closing price and classify it as 1.
Further, you can see this video below for learning about machine learning regression models.
Example of logistic regression in trading
Logistic regression can be used in trading to predict binary outcomes (stock price will “increase” or decrease”) or classify data based on predictor variables (technical indicators). Here's an
example of how Machine learning logistic regression might be applied in a trading context:
Example: Predicting Stock Price Movement
Suppose a trader wants to predict whether a stock price will increase (1) or decrease (0) based on certain predictor variables or indicators. The trader collects historical data and selects the
following predictor variables:
• Moving Average Crossover: A binary variable indicating whether there was a recent crossover of the short-term moving average (e.g., 50-day MA) above the long-term moving average (e.g., 200-day
MA) (1 = crossover occurred, 0 = no crossover).
• Relative Strength Index (RSI): A continuous variable representing the RSI value, which measures the momentum of the stock (values range from 0 to 100).
• Trading Volume: A continuous variable representing the trading volume of the stock, which may indicate the level of interest or activity in the stock.
The trader builds a logistic regression model using historical data, where the outcome variable is the binary indicator of whether the stock price increased (1) or decreased (0) on the next trading
After training the logistic regression model, the trader can use it to make predictions on new data. For example, if the model predicts a high probability of a stock price increase (p > 0.7) based on
past or current data, the trader may decide to buy the stock.
Types of logistic regression
Logistic regression is a versatile statistical method that can be adapted to various types of classification problems. Depending on the nature of the outcome variable and the specific requirements of
the analysis, different types of logistic regression models can be employed.
Here are some common types of logistic regression:
• Binary Logistic Regression: This is the most basic form of logistic regression, where the outcome variable is binary and can take on two categorical values, such as "yes" or "no," "success" or
"failure". These values are read as "1" or "0" by the machine learning model.
Hence, binary logistic regression is used to model the relationship between the predictor variables (an indicator such as RSI, MACD etc.) and the probability of the outcome being in a particular
category (“increase” or “decrease” in stock price).
• Multinomial Logistic Regression: In multinomial logistic regression, the outcome variable is categorical and can have more than two unordered categories. This type of logistic regression is
suitable for modelling nominal outcome variables with three or more categories that do not have a natural ordering.
For example, classifying stocks into multiple categories such as "buy," "hold," or "sell" based on a set of predictor variables such as fundamental metrics, technical indicators, and market
• Ordinal Logistic Regression: Ordinal logistic regression is used when the outcome variable is ordinal, meaning that it has a natural ordering but the intervals between the categories are not
necessarily equal. Examples of ordinal variables include Likert scale ratings (e.g., "strongly disagree," "disagree," "neutral," "agree," "strongly agree"). Ordinal logistic regression models the
cumulative probabilities of the outcome variable categories.
For example, analysing the ordinal outcome of trader sentiment or confidence levels (e.g., "low," "medium," "high") based on predictor variables such as market volatility, economic indicators, and
news sentiment.
• Multilevel Logistic Regression (or Hierarchical Logistic Regression): Multilevel logistic regression is used when the data has a hierarchical or nested structure, such as individuals nested
within groups or clusters. This type of logistic regression accounts for the dependence or correlation among observations within the same cluster and allows for the estimation of both
within-group and between-group effects.
For example, Modelling the binary outcome of stock price movements within different industry sectors (e.g., technology, healthcare, finance) while accounting for the hierarchical structure of the
data (stocks nested within sectors).
• Mixed-effects Logistic Regression: Mixed-effects logistic regression combines fixed effects (predictor variables that are the same for all observations) and random effects (predictor variables
that vary across groups or clusters) in the model. This type of logistic regression is useful for analysing data with both individual-level and group-level predictors and accounting for the
variability within and between groups.
For example, examining the binary outcome of stock price movements based on both individual-level predictors (such as company-specific factors, technical indicators) and group-level predictors (such
as industry sector, market index etc.).
• Regularised Logistic Regression: Regularised logistic regression, such as Lasso (L1 regularisation) or Ridge (L2 regularisation) logistic regression, incorporates regularisation techniques to
prevent overfitting and improve the generalisability of the model. Regularisation adds a penalty term to the logistic regression model, which shrinks the coefficients of the predictor variables
and selects the most important predictors.
For example, building a binary classification model to predict whether a stock is likely to outperform the market based on a large number of predictor variables while preventing overfitting and
selecting the most important features.
Each type of logistic regression has its assumptions, advantages, and limitations, and the choice of the appropriate model depends on the nature of the data, the type of outcome variable, and the
specific research or analytical objectives.
Difference between logistic regression and linear regression
Now, let us see the difference between logistic regression and linear regression.
Feature/ Linear Regression Logistic Regression
Outcome Type Continuous where the variable can take any value within a given range (e.g., daily Binary or Categorical (e.g., Price is “Up” or “Down”)
stock price)
Prediction Value prediction. For example, stock price Probability prediction For example, likelihood of an event
Relationship Linear, that is, the dependent variable (such as predictive outcome) can be found out Log-Linear. For example, consider a situation where a quantity grows exponentially over time. A
Assumption with the help of independent variables (such as past values). For example, on the log-linear model would describe the relationship between the logarithm of the quantity and time as
basis of historical data, a trader can predict future prices of stock. linear, implying that the quantity grows or decays at a constant rate on a logarithmic scale.
Model Output Change in outcome per unit change in predictor Change in log odds per unit change in predictor
Applications Predicting in terms of amounts Classifying in categories
In essence:
• Linear Regression predicts a continuous outcome based on predictors.
• Logistic Regression estimates the probability of a categorical outcome based on predictors.
Key assumptions while using logistic regression
Logistic regression, like other statistical methods, relies on several key assumptions to ensure the validity and reliability of the results. Here are some of the key assumptions underlying logistic
• Binary Outcome: Logistic regression is specifically designed for binary outcome variables, meaning that the outcome variable should have only two categorical outcomes (e.g., 0/1, Yes/No).
• Linearity of Log Odds: The relationship between the predictor variables and the log odds of the outcome should be linear. This assumption means that the log odds of the outcome should change
linearly with the predictor variables.
• Independence of Observations: Each observation in the dataset should be independent of the other observations. This assumption ensures that the observations are not correlated or dependent on
each other, which could bias the estimates and inflate the Type I error rate.
• No Multicollinearity: The predictor variables in the model should not be highly correlated with each other, as multicollinearity can make it difficult to estimate the individual effects of the
predictor variables on the outcome.
• Large Sample Size: Logistic regression models perform better and provide more reliable estimates with a larger sample size (data values). While there is no strict rule for the minimum sample
size, having a sufficiently large sample size ensures that the estimates are stable and the model has enough power to detect significant effects.
• Correct Specification of Model: The logistic regression model should be correctly specified, meaning that all relevant predictor variables should be included in the model, and the functional form
of the model should accurately reflect the underlying relationship between the predictor variables and the outcome.
• Absence of Outliers: The presence of outliers in the dataset can influence the estimates and distort the results of the logistic regression model. It is essential to identify and handle outliers
during data cleaning to ensure the robustness and validity of the model.
In summary, while logistic regression is a powerful and widely used method for modelling binary outcomes, it is crucial to ensure that the key assumptions of the model are met to obtain valid and
reliable results. Violation of these assumptions can lead to biassed estimates, inaccurate predictions, and misleading conclusions, emphasising the importance of careful data preparation, model
checking, and interpretation in logistic regression analysis.
Steps to use logistic regression in trading
Below are the steps that are used in logistic regression for trading.
• Step 1 - Define the Problem: Identify what you want to predict or classify in trading, such as predicting whether a stock will go up or down based on certain factors.
• Step 2 - Collect Data: Gather historical data on stocks, including predictor variables (e.g., trading volume, market index, volatility) and the binary outcome (e.g., stock went up = 1, stock went
down = 0).
• Step 3 - Preprocess the Data: Clean the data, handle missing values, and transform variables if needed (e.g., normalise trading volume).
• Step 4 - Split the Data: Divide the dataset into training and test sets to train the model on one set and evaluate its performance on another.
• Step 5 - Select Variables: Choose the predictor variables (independent variables) that you believe will help predict the outcome (dependent variable).
• Step 6 - Build the Model: Use software or programming tools to build a logistic regression model with the selected variables and the binary outcome.
• Step 7 - Train the Model: Train the logistic regression model on the training dataset, adjusting the model's parameters to minimise errors and fit the data.
• Step 8 - Evaluate the Model: Test the trained model on the test dataset to evaluate its performance, using metrics such as accuracy, precision, recall, or the area under the ROC curve.
• Step 9 - Interpret the Results: Interpret the coefficients and odds ratios in the logistic regression model to understand the relationships between the predictor variables and the probability of
the outcome.
• Step 10 - Make Predictions: Use the trained logistic regression model to make predictions on new data or real-time data in trading, such as predicting the likelihood of a stock going up or down
based on current market conditions.
• Step 11 - Monitor and Update: Continuously monitor the performance of the logistic regression model in real trading scenarios and update the model as needed with new data and insights.
How to use logistic regression in Python for trading?
Now that we know the basics behind logistic regression and the sigmoid function, let us go ahead. Now, we will learn how to implement logistic regression in Python and predict the stock price
movement using the above condition.
This is how the Python code is used:
Step 1: Import Libraries
We will start by importing the necessary libraries such as TA-Lib.
Step 2: Import Data
We will import the AAPL data from 01-Jan-2005 to 30-Dec-2023. The data is imported from yahoo finance using ‘pandas_datareader’.
[*********************100%%**********************] 1 of 1 completed
Open High Low Close Adj Close Volume
2023-12-22 195.179993 195.410004 192.970001 193.600006 193.600006 37122800
2023-12-26 193.610001 193.889999 192.830002 193.050003 193.050003 28919300
2023-12-27 192.490005 193.500000 191.089996 193.149994 193.149994 48087700
2023-12-28 194.139999 194.660004 193.169998 193.580002 193.580002 34049900
2023-12-29 193.899994 194.399994 191.729996 192.529999 192.529999 42628800
Let us print the top five rows of column ‘Open’, ‘High’, ‘Low’, ‘Close’.
Step 3: Define Predictor/Independent Variables
We will use 10-days moving average, correlation, relative strength index (RSI), the difference between the open price of yesterday and today, the difference between the close price of yesterday and
the open price of today. Also, open, high, low, and close prices will be used as indicators to make the prediction.
You can print and check all the predictor variables used to make a stock price prediction.
Step 4: Define Target/Dependent Variable
The dependent variable is the same as discussed in the above example. If tomorrow’s closing price is higher than today’s closing price, then we will buy the stock (1), else we will sell it (-1).
Step 5: Split The Dataset
We will split the dataset into a training dataset and test dataset. We will use 70% of our data to train and the rest 30% to test. To do this, we will create a split variable which will divide the
data frame in a 70-30 ratio. ‘Xtrain’ and ‘Ytrain’ are train dataset. ‘Xtest’ and ‘Ytest’ are the test dataset.
Step 6: Instantiate The Logistic Regression in Python
We will instantiate the logistic regression in Python using the ‘LogisticRegression’ function and fit the model on the training dataset using the ‘fit’ function.
Step 7: Examine The Coefficients
0 Open [5.681225012480715e-18]
1 High [5.686127781664772e-18]
2 Low [5.6201381013603385e-18]
3 Close [5.5831060987233e-18]
4 Adj Close [5.0129246504381945e-18]
5 Volume [9.71316227735615e-11]
6 S_10 [5.471752301907141e-18]
7 Corr [3.1490350717776683e-19]
8 RSI [3.646275382070163e-17]
Step 8: Calculate Class Probabilities
We will calculate the probabilities of the class for the test dataset using the ‘predict_proba’ function.
[[0.49653675 0.50346325]
[0.49587905 0.50412095]
[0.49479691 0.50520309]
[0.49883229 0.50116771]
[0.49917317 0.50082683]
[0.49896485 0.50103515]]
Step 9: Predict Class Labels
Next, we will predict the class labels using the predict function for the test dataset.
Now, let us see what the prediction shows here.
[1 1 1 ... 1 1 1]
If you print the ‘predicted’ variable, you will observe that the classifier is predicting 1, when the probability in the second column of variable ‘probability’ is greater than 0.5. When the
probability in the second column is less than 0.5, then the classifier will be predicting -1.
In the output above, the signal shows 1, which is a buy signal. But, for which dates did it predict 1?
Let us find out below.
Date(s) with Buy Signal(s):
[1 1 1 ... 1 1 1]
Step 10: Evaluate The Model
Confusion Matrix
The Confusion matrix is used to describe the performance of the classification model on a set of test dataset for which the true values are known. We will calculate the confusion matrix using the
‘confusion_matrix’ function.
[[ 0 665]
[ 0 764]]
Classification Report
This is another method to examine the performance of the classification model.
print(metrics.classification_report(y_test, predicted))
precision recall f1-score support
-1 0.00 0.00 0.00 665
1 0.53 1.00 0.70 764
accuracy 0.53 1429
macro avg 0.27 0.50 0.35 1429
weighted avg 0.29 0.53 0.37 1429
The f1-score tells you the accuracy of the classifier in classifying the data points in that particular class compared to all other classes. It is calculated by taking the harmonic mean of precision
and recall. The support is the number of samples of the true response that lies in that class.
The accuracy of the model is at 0.53 or 53%.
Step 11: Create Trading Strategy Using The Model
We will predict the signal to buy (1) or sell (-1) and calculate the cumulative Nifty 50 returns for the test dataset. Next, we will calculate the cumulative strategy return based on the signal
predicted by the model in the test dataset. We will also plot the cumulative returns.
Challenges of logistic regression
Now, let us find out below which challenges can be faced while using logistic regression.
• Model Complexity: Financial markets are complex and influenced by numerous factors, including economic conditions, geopolitical events, investor sentiment, and market dynamics. Logistic
regression may not capture all the nonlinear relationships and interactions among variables, limiting its ability to model and predict market movements accurately.
• Overfitting and Underfitting: Overfitting occurs when the logistic regression model is too complex and fits the training data too closely, capturing noise and random fluctuations rather than the
underlying patterns. Underfitting, on the other hand, occurs when the model is too simple and fails to capture the relationships and variations in the data, leading to poor performance on both
training and test datasets.
• Imbalanced Data: In trading, the distribution of the outcome variable (e.g., stock price movements) may be imbalanced, with a disproportionate number of observations in one class (e.g., more
instances of stock price increases than decreases). Imbalanced data can lead to biassed models that prioritise the majority class and perform poorly in predicting the minority class.
• Dynamic Nature of Markets: Financial markets are dynamic and constantly evolving, with changing trends, volatility, and investor behaviour. Logistic regression models, which are trained on
historical data, may not adapt quickly to new market conditions and may require frequent updates and recalibrations to maintain their predictive accuracy.
• External Factors and Black Swan Events: Logistic regression models may not account for unexpected or rare events, such as black swan events, geopolitical crises, or sudden market shocks, which
can have a significant impact on market movements and cannot be fully captured by historical data alone.
Overcoming the challenges of logistic regression
Here’s how you can overcome the challenges that arise while using logistic regression:
• Model Complexity: Evaluate and refine model complexity using regularisation and feature selection.
• Overfitting and Underfitting: Mitigate overfitting and underfitting through cross-validation and ensemble methods.
• Data Quality and Availability: Invest in high-quality data and thorough preprocessing techniques.
• Imbalanced Data: Address class imbalance with oversampling, undersampling, or alternative metrics.
• Model Interpretability: Utilise model-agnostic tools for transparent and actionable insights.
• Dynamic Nature of Markets: Continuously monitor and update models with evolving market data.
• External Factors and Black Swan Events: Implement risk management strategies to mitigate unexpected events.
Also, you can check out this video below by Dr Thomas Starke (CEO, AAAQuants) in order to learn to use logistic regression more effectively in trading.
Logistic regression in trading offers a powerful tool for predicting binary outcomes, such as stock price movements, by leveraging historical data and key predictor variables. While logistic
regression shares similarities with linear regression, it utilises a sigmoid function to estimate probabilities and classify outcomes into discrete categories.
However, traders must be mindful of its assumptions, potential challenges, and the dynamic nature of financial markets. By adhering to best practices, continuous monitoring, and incorporating risk
management strategies, logistic regression can enhance decision-making processes and contribute to more informed and effective trading strategies.
Ready to revolutionise your algorithmic trading? Immerse yourself in Quantra's comprehensive learning track, "Machine Learning & Deep Learning in Trading - I."
With this learning track, you will find the uses of AI-powered market domination with the help of five expertly-crafted courses covering Introduction to ML for Trading, Data & Feature Engineering,
Decision Trees, ML Classification and SVM, and Python Libraries for Algorithmic Trading.
Moreover, you will get an interactive learning environment with video lectures, quizzes, assignments, and hands-on coding exercises. That is not it. With each course, you will learn at your own pace
from industry veterans, solidifying core ML concepts for application in financial markets for a successful trading journey!
File in the download
• Machine Learning Logistic Regression in trading - Python notebook
Author: Chainika Thakar (Originally written By Vibhu Singh)
Note: The original post has been revamped on 15th February 2024, for the accuracy and recentness.
Disclaimer: All data and information provided in this article are for informational purposes only. QuantInsti^® makes no representations as to accuracy, completeness, currentness, suitability, or
validity of any information in this article and will not be liable for any errors, omissions, or delays in this information or any losses, injuries, or damages arising from its display or use. All
information is provided on an as-is basis. | {"url":"https://blog.quantinsti.com/machine-learning-logistic-regression-python/","timestamp":"2024-11-13T09:17:59Z","content_type":"text/html","content_length":"218678","record_id":"<urn:uuid:233ef671-dfe0-4087-964b-a3fb9670d2eb>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00463.warc.gz"} |
What is Pi?
distance learning
Happy Pi Day
On March 14, let's take a look at the wonder of Pi.
Since March 14 is also written 3/14, it's become a day to celebrate the number Pi, which starts off 3.14... but what is Pi?
Image: Circle with lines showing circumference and diameter. Text: Pi is the ratio of a circle's circumference to its diameter.
Try this with any circle!
1) Measure the circumference, or distance all around the edge of the circle.
2) Measure the diameter, or distance across the circle (being sure to pass through the center point).
3) Divide the circumference by the diameter.
You should get Pi, or 3.14159... Using long division, you will find that the number keeps going and going and going!
Image: Various circles of different sizes with circumference and diameter lines marked. Text: The number Pi is called a constant because no matter the size of the circle, the ratio is always the
same. Pi equals circumference divided by diameter.
No matter what size circle you measure, the answer to circumference divided by diameter will always be Pi.
Some people have tried to see if the number ever ends. But it has been carried out to 31.4 trillion decimal places and still continues!
Text: Pi is also an irrational number. Its digits continue forever without ever repeating in a pattern. (Pi digits shown). Pi is useful in science and engineering whenever circles are involved. On Pi
Day 2019, Google calculated 31.4 trillion decimal places of Pi. Wow!
Have you ever tried to memorize the numbers of Pi? How far did you get?
It may take a little while to load, but you can see 1,000,000 digits of Pi on one page on the website https://www.piday.org/million/.
Here are the first 100 decimal places (shown in groups of 10):
3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679
Scientists use Pi in their calculations whenever circles are involved. Many things are circular, so this number comes in handy. Imagine going to outerspace. Orbits, planets and all kinds of things
have circles in them.
Check out places to find Pi in the universe with NASA here: https://nasa.tumblr.com/post/612574998435135488/cosmic-piece-of-pi
For fun challenges at home, try the NASA Pi Day challenge, which gives real-life problems scientists solve using Pi as they explore the universe.
Student Slideshow: The NASA Pi Day Challenge | NASA/JPL Edu
Can you use π (pi) to solve these stellar math problems faced by NASA scientists and engineers?
Happy Pi Day! | {"url":"https://blog.davincisv.org/happy-pi-day/","timestamp":"2024-11-12T03:35:07Z","content_type":"text/html","content_length":"23760","record_id":"<urn:uuid:3beefc7f-54d7-456f-8ce4-f1a9edc7c315>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00264.warc.gz"} |
ram air pressure calculator
Ram Air Pressure Calculator
In aerodynamics, understanding the impact of Ram air pressure is crucial. It plays a significant role in various applications, including aviation, automotive engineering, and even sports. To simplify
calculations related to Ram air pressure, we’ve developed a user-friendly calculator. This article will guide you through its usage, formula, an example solve, frequently asked questions, and a
conclusive overview.
How to Use
1. Enter the required parameters into their respective fields.
2. Click the “Calculate” button to obtain the result.
3. Review the calculated Ram air pressure.
The formula for calculating Ram air pressure is as follows:
• Pram = Ram air pressure
• Ptotal = Total pressure
• Pstatic = Static pressure
Example Solve
Let’s consider an example where:
• Total pressure (Ptotal) = 101.3 kPa
• Static pressure (Pstatic) = 100.2 kPa
Using the provided formula:
Pram=101.3−100.2=1.1 kPa
Thus, the Ram air pressure is calculated to be 1.1 kPa.
Q: What are the practical applications of Ram air pressure?
A: Ram air pressure is utilized in various fields such as aviation for airspeed indicators, automotive engineering for turbocharged engines, and sports for wind speed measurements.
Q: How does Ram air pressure affect aircraft performance?
A: Ram air pressure impacts aircraft performance by affecting the accuracy of airspeed indicators and providing additional thrust in turbocharged engines.
Q: Is Ram air pressure affected by altitude?
A: Yes, Ram air pressure decreases with increasing altitude due to the decrease in total pressure.
Understanding Ram air pressure and its calculation is essential for engineers, pilots, and enthusiasts alike. With the provided calculator and accompanying guide, you can now effortlessly determine
Ram air pressure for various applications. | {"url":"https://calculatordoc.com/ram-air-pressure-calculator/","timestamp":"2024-11-12T07:08:45Z","content_type":"text/html","content_length":"84282","record_id":"<urn:uuid:4f139194-c6cd-4a79-8b7e-3e8a6807b49f>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.58/warc/CC-MAIN-20241112045844-20241112075844-00681.warc.gz"} |
Voltage Drop Study and Analysis Services in Dubai - UAE | Carelabz.com
In an electric circuit the existing energy of avoltagesupply decreases as electric current travels through the components that does not supplyvoltage i.e. passive elements of an electrical
circuit. This is called Voltage Drop. Each point in a circuit can be assigned a voltage that’s proportional to its electrical elevation, so to speak. Voltage drop is simply the arithmetical
difference between a higher voltage and a lower one. The National Electrical Code states that a voltage drop of 5% at the furthest receptacle in a branch wiring circuit is acceptable for normal
Why Perform Voltage Drop Study and Analysis?
Following are the harm caused by extensive voltage drop:
• Low voltage to the equipment being powered, causing improper, erratic, or no operation – and damage to the equipment.
• Poor efficiency and wasted energy.
• Heating at a high resistance connection/splice may result in a fire at high ampere loads.
One cannot determine exactly at what point voltage drop can cause fire, because it varies with the amount of current flowing through the high resistance connection, the resistance of that connection
etc. Number of factors must be considered regarding at what point ignition will occur, example:
• Is the high resistance connection in connection with a flammable matter?
• Is there air flow to dissipate the heat?
• Is the area around the connection insulated, so that heat cannot escape?
Voltage drop study and analysis must be taken into consideration so that it may hardly affect the instruments. Like some household appliances that have a fixed capacity power engines are strongly
influenced by voltage drop.
The main ways to reduce voltage drop are to:
• Increase power factor (add capacitors)
• Replace the conductor with a larger size.
What is Done During Voltage Drop Study and Analysis?
The reduction of any rate in the voltage leads to a rise in the current at the same rate and the problem is that this increase is often not huge. If we assume that this increase in the current was by
11%, it means that circuit breaker will not feel this increase and it will not operate while the device suffers from this increase in the current and thus leading to the increase in temperature
gradually with time until it reaches the stage of the combustion. Another example the voltage drop by 1% lead to decrease in illuminance of tungsten lamp by 3%.
The amount of power delivered to a component in a circuit is equal to the voltage drop across that component’s terminals multiplied by the current flow through the component:
P = V*I
V = voltage drop in volts
I = current flow in amperes and
P =power in watts.
Obviously, if either V or I is zero, no power or energy is delivered to that component, so it can’t fulfil any useful purpose. So voltage drop is a vital feature of all electric circuits and is
planned and controlled very carefully by the engineers that design those circuits.
So the transformers and the distribution boxes must be situated in a position such that maximum voltage drop from transformer to the furthest point doesn’t exceed 5% of the nominal voltage. lf it
hard to find a place for the transformer so that maximum voltage drop is less than 5%, then we must use cables with larger cross-section area but It will increase the cost.
How is Voltage Drop Study and Analysis Done?
There are two methods by which voltage drop study and analysis can be done:
Manual Voltage Drop Study and Analysis
Ohm’s Law Method – Single-Phase Only
Voltage drop of the circuit conductors is calculated by multiplying the complete resistance of the circuit conductors by the current through the circuit:
VD = I x R
I = Load in amperes
R = Resistance of the conductor
This method cannot be used for 3 phase circuits.
Voltage Drop Using the Formula Method
The voltage drop of circuit with already installed conductors can be calculated by:
VD = 2 * K * Q * I * D/CM – Single Phase
VD = 1.732 * K * Q * I * D/CM – Three Phase
VD = Volts Dropped
K = Direct Current Constant
Q = Alternating Current Adjustment Factor’
I = Amperes
D = Distance
CM = Circular-Mils
Electrical Transient Analyzer Programming (ETAP):
Since 1970s, ETAP has been a significant designing podium utilised for designing, analysing, and optimising electrical power systems. ETAP is a completely assimilated Electrical software solutions
comprising of load flow, arc flash, short circuit and more. Its flexible performance makes it apt for all companies in any shape or form.
Operation Technology, Inc. is the developer of ETAP, the most wide-ranging analysis software for the simulation, design, operation, control, monitoring and automation of power systems. ETAP is the
industry leader used worldwide in all types and sizes of power systems such as manufacturing, oil and gas, steel, mining, cement, and more.
We can approximate the voltage drop along a circuit as:
Vdrop= |Vs| – |Vr| ≈ IR·R + IX·X
Where, Vdrop= voltage drop along the feeder
R= line resistance
X= line reactance
IR= line current due to real power flow (in phase with the voltage)
IX= line current due to reactive power flow (90°out of phase with the voltage)
The biggest fault happens under leading power factor and heavy current. The approximation has an error less than 1% for an angle between the sending and receiving end voltages.
This estimation brings two crucial factors about voltage drop into the light:
Resistive load: At large power factors, the voltage drop is related directly to the resistance of the conductors. The resistance plays a crucial role, even though the resistance is generally less
than the reactance.
Reactive load: At medium or small power factors, the voltage drop relates mostly to the reactance of the conductors. Because the reactance is usually larger than the resistance, the reactive load
causes most of the voltage drop. Poor power factor significantly increases voltage drop.
Carelabs is authorized provider of Electrical Installation’s Study, Analysis, Inspection, and Certification services in UAE, and offer voltage drop study and analysis services. | {"url":"https://carelabz.com/voltage-drop-study-analysis/","timestamp":"2024-11-04T14:59:00Z","content_type":"text/html","content_length":"116464","record_id":"<urn:uuid:ddf81c68-9a17-4730-8cd4-abc082cbefe8>","cc-path":"CC-MAIN-2024-46/segments/1730477027829.31/warc/CC-MAIN-20241104131715-20241104161715-00526.warc.gz"} |
Entropy of Extremal, Non-SUSY Blackholes
What with all the “fun” we’ve been having discussing LQG, I never did get around to posting about the recent paper of Emparan and Horowitz.
Better late than never…
One of the fundamental breakthroughs in quantum gravity was Strominger and Vafa’s tour de force calculation of the entropy of supersymmetric charged 5D blackhole. At weak coupling, the microstates
can be completely accounted-for as the BPS states of the collection of D-branes with those charges.
In 4 dimensions, to get a nonzero horizon area, a supersymmetric blackhole must carry at least 4 charges^1. And there, too, one finds perfect agreement between the microscopic D-brane and the
macroscopic Bekenstein-Hawking entropy.
There have been many important refinements of the supersymmetric story (some of which we’ve discussed here before). And there are positive results in the nearly-supersymmetric case.
But it’s hard to escape the feeling that perhaps it is supersymmetry (in particular, the fact that the counting of BPS states is independent of the string coupling) that is behind the success of
these calculations. Surely a weak-coupling D-brane description ought to be hopelessly poor in accounting for the micro-states of far-from-supersymmetric blackholes.
Emparan and Horowitz have a fascinating little paper, in which they study a far-from-supersymmetric (but still extremal) 2-charged blackhole in 4 dimensions.
Consider Type IIB string theory compactified on $T^6$. Write $T^6=T^2\times T^2\times T^2$, and let $a_\alpha,b_\alpha,\, \alpha=1,2,3$ be a basis of 1-cycles on each of the $T^2$s. There are many,
U-duality equivalent, ways to get a 4-charge, 1/8-BPS blackhole. One is to take 4 stacks of D3-branes wrapped, respectively around the 3-cycles
\begin{aligned} C_1&=a_1\times a_2\times a_3\\ C_2&=a_1\times b_2\times b_3\\ C_3&=b_1\times a_2\times b_3\\ C_4&=b_1\times b_2\times a_3 \end{aligned}
Note that
• Each pair of cycles intersects along a curve.
• All 4 intersect at a point.
• The full configuration preserves 1/8 of the original 32 supercharges.
If, for simplicity, we take the tori to be square, $\tau_\alpha=i$, then the ADM mass of this configuration is $M = \frac{V_3 \sum_{i=1}^4 N_i}{g l_s^4} = \frac{\sum_{i=1}^4 N_i}{\sqrt{G_4}}$ where
the volumes of the $C_i$ are $(2\pi)^3 V_3$, $G_4$ is the 4D Newton’s constant, and we’ve wrapped $N_i$ D3-branes around $C_i$.
For large $N_i$, the asymptotic degeneracy of states reproduces the Bekenstein-Hawking entropy $S_{\text{BH}}= \lim_{N_i\gg1} d(N_i) = 2\pi\sqrt{N_1 N_2 N_3 N_4}$
Emparan and Horowitz consider a closely related system. Let $k,l$ be a pair of positive coprime integers and let $c^\pm_\alpha = k a_\alpha \pm l b_\alpha$ be the circle which wraps $k$ times around
the $a$-cycle and $\pm l$ times around the $b$-cycle of the $\alpha$^th $T^2$. Instead of (1), consider the 3-cycles
\begin{aligned} C_1&=c^+_1\times c^+_2\times c^+_3\\ C_2&=c^+_1\times c^-_2\times c^-_3\\ C_3&=c^-_1\times c^+_2\times c^-_3\\ C_4&=c^-_1\times c^-_2\times c^+_3 \end{aligned}
Wrapping $N$ D3-branes around each of the $C_i$ breaks all supersymmetries. And there are only two net D3-brane charges for this configuration: $N_0=4N k^3$ times the charge corresponding to wrapping
$a_1\times a_2\times a_3$ and $N_6=4N l^3$ times the charge corresponding to wrapping $b_1\times b_2\times b_3$^2.
The lengths of the curves, $c^\pm_\alpha$ are a factor $(k^2+l^2)^{1/2}$ times longer than the previous case. So the volumes of the $C_i$ are $(2\pi)^3 (k^2+l^2)^{3/2} V_3$ and the ADM mass of this
configuration is $M = \frac{4 N(k^2+l^2)^{3/2} V_3}{g l_s^4}= \frac{4 N (k^2+l^2)^{3/2}}{\sqrt{G_4}}$
A weak-coupling calculation of the asymptotic degeneracy of states for this system gives $S = 16 \pi k^3 l^3 N^2$
Relative to the supersymmetric case (with $N_i=N$), the entropy is a factor $(2k l)^3$ larger, reflecting the fact that the $C_i$ in (2) intersect in $( 2 k l)^3$ points instead of 1 point.
These formulæ precisely reproduce the mass and entropy of the extremal dyonic blackholes. Indeed, lifted to M-theory, the Type-IIA description becomes a dyonic Kaluza-Klein blackhole ($\times T^6$).
(For a recent discussion, see here.) The mass is $M = \frac{\left(Q^{2/3} + P^{2/3}\right)^{3/2}}{2 G_4}$ where the electric and magnetic charges are, respectively $Q= \frac{2 G_4 N_0}{R},\quad P= \
frac{N_6 R}{4}$$R$ is the asymptotic radius of the Kaluza-Klein circle, $N_6$ is the Euler class of the $S^1$ bundle over the 2-sphere at spatial infinity, and $N_0$ is the (quantized) momentum
around the circle. The entropy $S = \frac{A}{4 G_4} = \frac{2\pi P Q}{G_4} = \pi N_0 N_6$ is independent of $R$. The $R\to \infty$ limit is the 5 dimensional extremal Myers-Perry blackhole. In the $R
\to 0$ limit, one obtains (a T-dual version of) the weakly-coupled D-brane description above.
The independence of the entropy of the radius, $R$, translates into independence of the string coupling, $g$, which, presumably, is the reason for the success of the weakly-coupled D-brane
calculation despite the absence of any BPS property to protect the counting of microstates.
^1 OK, there are the small blackholes which carry only two charges. Classically, they have zero horizon area, but quantum effects give them a finite horizon.
^2 The notation is suggested by the fact that, in a T-dual description, these are the numbers of D0- and D6-branes.
Posted by distler at July 10, 2006 4:36 PM
Re: Entropy of Extremal, Non-SUSY Blackholes
I cannot understand a word of what you are saying on this post. Aside from accents this usually isnt a problem for me with people from Texas.
Accordingly I am going to shamelessly hijack it on to the topic of intelligent design/creationism and given that you take an interest in this do you know the context of the following quote attributed
to Arno Penzias which suspiciously can only be found on ID/creationist websites:
“Astronomy leads us to a unique event, a universe which was created out of nothing and delicately balanced to provide exactly the conditions required to support life. In the absence of an
absurdly-improbable accident, the observations of modern science seem to suggest an underlying, one might say, supernatural plan.”
Posted by: Atiyah on July 10, 2006 7:42 PM | Permalink | Reply to this | {"url":"https://golem.ph.utexas.edu/~distler/blog/archives/000869.html","timestamp":"2024-11-03T09:52:51Z","content_type":"application/xhtml+xml","content_length":"35765","record_id":"<urn:uuid:df049d25-84fe-46fa-8afd-187bc99618d9>","cc-path":"CC-MAIN-2024-46/segments/1730477027774.6/warc/CC-MAIN-20241103083929-20241103113929-00502.warc.gz"} |
The American Mathematics Competitions (AMCs) are the first of a series of competitions in secondary school mathematics sponsored by the Mathematical Association of America that determine the United
States of America's team for the International Mathematical Olympiad (IMO). The selection process takes place over the course of roughly five stages. At the last stage, the US selects six members to
form the IMO team.
There are three AMC competitions held each year:
• the AMC 8, for students under the age of 14.5 and in grades 8 and below^[1]
• the AMC 10, for students under the age of 17.5 and in grades 10 and below
• the AMC 12, for students under the age of 19.5 and in grades 12 and below^[2]
The AMC 8 tests mathematics through the eighth grade curriculum.^[1] Similarly, the AMC 10 tests math through the tenth grade math curriculum, and the AMC 12 tests math through the twelfth grade
Before the 1999-2000 academic year, the AMC 8 was known as the AJHSME (American Junior High School Mathematics Examination), and the AMC 12 was known as the AHSME (American High School Mathematics
Examination). There was no AMC 10 prior to the 1999-2000 academic year.^[3]
Students who perform well on the AMC 10 or AMC 12 competitions are invited to participate in the American Invitational Mathematics Examination (ACsIME). Students who perform exceptionally well on the
AMC 12 and AIME are invited to the United States of America Mathematical Olympiad (USAMO), while students who perform exceptionally well on the AMC 10 and AIME are invited to United States of America
Junior Mathematical Olympiad (USAJMO). Students who do exceptionally well on the USAMO (typically around 45 students based on score and grade level) and USAJMO (typically around the top 15 students)
are invited to attend the Mathematical Olympiad Program (MOP).
The AMC contest series includes the American Mathematics Contest 8 (AMC 8) (formerly the American Junior High School Mathematics Examination) for students in grades 8 and below, begun in 1985; the
American Mathematics Contest 10 (AMC 10), for students in grades 9 and 10, begun in 2000; the American Mathematics Contest 12 (AMC 12) (formerly the American High School Mathematics Examination) for
students in grades 11 and 12, begun in 1950; the American Invitational Mathematics Examination (AIME), begun in 1983; and the USA Mathematical Olympiad (USAMO), begun in 1972.^[4]
Years Name No. of Comments
1950–1951 50 New York state only
1952–1959 Annual High School Contest Nationwide
1960–1967 40 -10 Questions
1968–1972 35 -5 Questions
1973 Annual High School 35
1974–1982 Mathematics Examination 30 -5 Questions
AIME introduced in 1983, now is a middle step between AHSME and USAMO
1983–1999 American High School 30
Mathematics Examination AJHSME, now AMC 8, introduced in 1985
-5 Questions
American Mathematics
2000–present Competition 25 AHSME split into AMC10 and AMC12
A&B versions introduced in 2002. USAMO split into USAJMO and USAMO in 2010. AMC 10 participants who pass AIME can qualify for and participate in
USAJMO, provided they don't also qualify for USAMO. USAJMO is meant to be easier than USAMO.
Rules and scoring
AMC 8
The AMC 8 is a 25 multiple-choice question, 40-minute competition designed for middle schoolers.^[4] No problems require the use of a calculator, and their use has been banned since 2008. The
competition was previously held on a Thursday in November. However, after 2022, the competition has been held in January. The AMC 8 is a standalone competition; students cannot qualify for the AIME
via their AMC 8 score alone.
The AMC 8 is scored based on the number of questions answered correctly only. There is no penalty for getting a question wrong, and each question has equal value. Thus, a student who answers 23
questions correctly and 2 questions incorrectly receives a score of 23.
Rankings and awards
Based on questions correct:
• Distinguished Honor Roll: Top 1% (has ranged from 19–25)
• Honor Roll: Top 5% (has ranged from 19-23)
• A Certificate of Distinction is given to all students who receive a perfect score.
• An AMC 8 Winner Pin is given to the student(s) in each school with the highest score.
• The top three students for each school section will receive respectively a gold, silver, or bronze Certificate for Outstanding Achievement.
• An AMC 8 Honor Roll Certificate is given to all high scoring students.
• An AMC 8 Merit Certificate is given to high scoring students who are in 6th grade or below.
AMC 10 and AMC 12
The AMC 10 and AMC 12 are 25 question, 75-minute multiple choice competitions in secondary school mathematics containing problems which can be understood and solved with precalculus concepts.
Calculators have not been allowed on the AMC 10/12 since 2008.^[6]
High scores on the AMC 10 or 12 can qualify the participant for the American Invitational Mathematics Examination (AIME).^[7]
The competitions are scored based on the number of questions answered correctly and the number of questions left blank. A student receives 6 points for each question answered correctly, 1.5 points
for each question left blank, and 0 points for incorrect answers. Thus, a student who answers 24 correctly, leaves 1 blank, and misses 0 gets ${\displaystyle 24\times 6+1.5\times 1=145.5}$ points.
The maximum possible score is ${\displaystyle 25\times 6=150}$ points; in 2020, the AMC 12 had a total of 18 perfect scores between its two administrations, and the AMC 10 also had 18.
From 1974 until 1999, the competition (then known as the American High School Math Examination, or AHSME) had 30 questions and was 90 minutes long, scoring 5 points for correct answers. Originally
during this time, 1 point was awarded for leaving an answer blank, however, it was changed in the late 1980s to 2 points. When the competition was shortened as part of the 2000 rebranding from AHSME
to AMC, the value of a correct answer was increased to 6 points and the number of questions reduced to 25 (keeping 150 as a perfect score). In 2001, the score of a blank was increased to 2.5 to
penalize guessing. The 2007 competitions were the first with only 1.5 points awarded for a blank, to discourage students from leaving a large number of questions blank in order to assure
qualification for the AIME. For example, prior to this change, on the AMC 12, a student could advance with only 11 correct answers, presuming the remaining questions were left blank. After the
change, a student must answer 14 questions correctly to reach 100 points.
The competitions have historically overlapped to an extent, with the medium-hard AMC 10 questions usually being the same as the medium-easy ones on the AMC 12. Problem 18 on the 2022 AMC 10A was the
same as problem 18 on the 2022 AMC 12A.^[3] Since 2002, two administrations have been scheduled, so as to avoid conflicts with school breaks. Students are eligible to compete in an A competition and
a B competition, and may even take the AMC 10-A and the AMC 12-B, though they may not take both the AMC 10 and AMC 12 from the same date.^[2] If a student participates in both competitions, they may
use either score towards qualification to the AIME or USAMO/USAJMO. In 2021, the competition format was changed to occur in the Fall instead of the Spring.^[8]
See also
External links | {"url":"https://www.knowpia.com/knowpedia/American_Mathematics_Competitions","timestamp":"2024-11-02T01:14:52Z","content_type":"text/html","content_length":"92795","record_id":"<urn:uuid:316f2e23-eeec-4ddb-a038-fbc9d152d10d>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00123.warc.gz"} |
Graphs of Direct Variation
Question Video: Graphs of Direct Variation Mathematics • Third Year of Preparatory School
Which of the given graphs represents the direct variation between π ₯ and π ¦? [A] Graph a [B] Graph b [C] Graph c [D] Graph d
Video Transcript
Which of the given graphs represents the direct variation between π ₯ and π ¦?
And then we have four different graphs to choose from. So, in order to be able to identify the relevant graph, we need to remind ourselves what we mean by direct variation or direct proportion. This
symbol that looks a little bit like the Greek letter π Ό but isnβ t represents direct proportion. This statement means π ¦ is directly proportional to π ₯. If π ¦ is directly proportional to π
₯, then this means that their ratio is constant. π ¦ divided by π ₯ is always equal to some value π . We can alternatively represent this as π ¦ is equal to π times π ₯. π is called the
constant of proportionality, or the constant of variation.
What this also in turn means is that one of the criteria for two variables to be in direct proportion to one another is that when one is equal to zero, the other is also equal to zero. And this makes
a lot of sense. If we compare the equation π ¦ equals π π ₯ to the equation for a straight line given in slopeβ intercept form, the equation in this form is π ¦ equals π π ₯ plus π . π ,
of course, is the value of the π ¦-intercept. But if when π ₯ is equal to zero, π ¦ is equal to zero, the π ¦-intercept, in fact, is zero. So, we get that equation in the form π ¦ equals π π
But letβ s think about what the value of π means in the equation π ¦ equals π π ₯ plus π . This is the slope. It tells us how steep the line itself is. So, thatβ s what the value of π
means when we think about this graphically. In fact, we donβ t really even need this information to answer the question. We need to find a graph that passes through the origin, zero, zero.
We notice graph (a) does not pass through the origin, graph (b) does, graph (c) does not, nor does graph (d). And so, graph (b) represents direct variation between π ₯ and π ¦. Specifically, since
the graph slopes downwards, it represents a situation where the constant of variation is negative; π is less than zero. The answer is (b). | {"url":"https://www.nagwa.com/en/videos/946106284505/","timestamp":"2024-11-06T17:49:14Z","content_type":"text/html","content_length":"249877","record_id":"<urn:uuid:c31e6182-ff38-4c45-a87a-3b1fee046f9b>","cc-path":"CC-MAIN-2024-46/segments/1730477027933.5/warc/CC-MAIN-20241106163535-20241106193535-00137.warc.gz"} |
equilateral polygons
A small polygon is a polygon of unit diameter. The maximal width of an equilateral small polygon with $n=2^s$ vertices is not known when $s \ge 3$. This paper solves the first open case and finds the
optimal equilateral small octagon. Its width is approximately $3.24\%$ larger than the width of the regular octagon: $\cos(\pi/8)$. … Read more
Tight bounds on the maximal perimeter of convex equilateral small polygons
A small polygon is a polygon of unit diameter. The maximal perimeter of a convex equilateral small polygon with $n=2^s$ vertices is not known when $s \ge 4$. In this paper, we construct a family of
convex equilateral small $n$-gons, $n=2^s$ and $s \ge 4$, and show that their perimeters are within $\pi^4/n^4 + O(1/n^5)$ … Read more | {"url":"https://optimization-online.org/tag/equilateral-polygons/","timestamp":"2024-11-03T16:53:42Z","content_type":"text/html","content_length":"85672","record_id":"<urn:uuid:de577e54-a0dc-4fff-9172-77d8216728f7>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00173.warc.gz"} |
Physics for everyone
The simplest kind of geometry, taught in schools, is the so called Euclidean geometry - named after an ancient Greek mathematician, Euclid, who described its basics in the 4th century BC in his
"Elements". It is based on the notions of points, straight lines and planes and it seems to correspond perfectly to our everyday experiences with various shapes. However, we can notice problems for
which Euclidean geometry is insufficient even in our immediate surroundings.
Let's imagine, for example, that we are airline pilots and our task is to fly as quickly as possible from Warsaw, Poland to San Francisco. We take a world map and knowing from Euclidean geometry that
a straight line is the shortest path between two points, we draw such a line from Warsaw to San Francisco. We're getting ready to depart and fly along the course we plotted... but fortunately, our
navigator friend tells us that we fell into a trap.
The trap is that the surface of the Earth isn't flat! The map we used to plot our straight line course is just a projection of a surface that is close to spherical in reality. Because of that, the
red line on the map below is not the shortest path - the purple line is:
Red line - a straight line between Warsaw and San Francisco on the map. The purple line is the actual shortest path.
Lorentz transformations, light cones
• What are events and spacetime?
• What are world lines?
• Simple spacetime diagrams
• How does the inseparability of space and time influence their perception by observers?
Most of the illustrations in the last article used rotations, but it turned out eventually that rotations aren't the correct transformations that would let us look at the spacetime from the point of
view of different observers. Now we will take a look at transformations that actually describe reality - the Lorentz transformations.
Events and space-time
The first entry in the series will be quite basic, but I think that some problems will nevertheless be quite interesting. We'll be talking about what is the space-time, events, and we will show where
the theory of relativity comes from. So, let's go :)
The notion of space-time is briefly mentioned at school, but usually the profound consequences of combining space and time into a single entity aren't explained too much. To understand this, one must
first go a bit deeper into the details of this idea. | {"url":"https://ebvalaim.net/en/category/articles/physics-for-everyone/","timestamp":"2024-11-01T19:11:33Z","content_type":"text/html","content_length":"47032","record_id":"<urn:uuid:02d2d7ae-c34b-4d40-a619-40affbf5fae7>","cc-path":"CC-MAIN-2024-46/segments/1730477027552.27/warc/CC-MAIN-20241101184224-20241101214224-00234.warc.gz"} |
Field Topology | Pagefind XKCD Demo
← Back
May 27, 2022
[A row of four signs, each held up by two posts, followed by a row of four rounded lozenge shapes, one for each sign. The signs and lozenge shapes are shaded as if three-dimensional objects, all
being flattish with a small third dimension; the four lozenge shapes each have one pair of sides horizontal and the other pair at a slight angle from vertical, denoting a horizontal plane
perpendicular to the signs extending “out” towards the viewer, which places each shape “in front” of its sign. All but the first lozenge shape have various numbers of ellipses within the shape -
ovoids shaded to denote holes piercing through the objects.]
[Leftmost sign:]
[The shape below this sign contains no ellipses.]
[Second sign from left:]
High jump
[This shape has one large ellipse in the center.]
[Third sign:]
Parallel bars
[This shape has two large ellipses - one in the top half and one in the bottom half.]
[Fourth and rightmost sign:]
Olympic swimming
[This shape has nine small ellipses - eight arranged symmetrically towards the edges of the shape and one in the center.]
[Caption underneath the signs and shapes:]
No one ever wants to use the topology department’s athletic fields. | {"url":"https://xkcd.pagefind.app/comics/2022-5-27-field-topology/","timestamp":"2024-11-04T21:03:19Z","content_type":"text/html","content_length":"5911","record_id":"<urn:uuid:48acd96c-54b6-42c4-9124-6173979723ef>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00339.warc.gz"} |
Important Change!
Just a quick post here. I wanted to announce as of May 1st there will be a 2 doll limit per household per month. I think most of you will laugh at this – seeing as so many of you have been trying for
a while to get just one doll – but it is something that needs to go into affect. The 2 doll per month also includes ‘Angel-ing’ (which I think most of you are aware of I’m not that keen on) – so if
you use your account/IP address to help someone out , that counts towards your 2 doll limit.
Thanks so much! I’ll be back for more of a ‘visit’ soon once I get over this head cold. Yuck!
*In case you are wondering what ‘angel-ing’ is – it’s the practice where someone will help a friend to get a doll using another computer.*
44 responses to “Important Change!”
I think this is a great new change, Christina! Thank you.
I think this change will make a big difference in how many new mamas are able to get dolls π
And not just the new mama’s but the ones who have siblings to buy for as well!
Great decision
Thrilled about this new change…I think it will make a big difference when May uploads begin……Thank you!!!
I am amazed that people are still able to score 2+ dolls per month- They need to teach classes in Bambo Acquisition!
People get more than 2 per month? Wow! They should spend their energy buying lotto tickets with that luck!
Thanks for the update. I am sure there will be a lot of people happy with this change (mommas of triplets being the exception lol).
I think the 2 per household limit is fine! I do however feel bad about the angel-ing thing because for instance on our board there is one person that will help someone and there have been many
mommas that have their first doll because of it!! So I respect what you are saying C, but just think the angel-ing is a good thing, sorry…
But they can still ‘angel’ but would use up their one spot for someone else. the vast vast majority of my customers have no idea about angel-ing and this sort of gives some people a real ‘leg up’
over everyone else. i know the intention is good! what i’m trying to do is make it more fair here π
thumbs up!
I think this is a great thing as it will hopefully level the playing field a bit for everyone. π
Love this rule! My wife actually gave up trying for a doll because of the stress and heartache of never being quick enough. I’m still trying for Mother’s Day, we have 4 girls she could give it to
but I’m wondering if she would just keep it! π
I just wanted to say that Luke’s comment is one of the sweetest ever! I hope so badly he can get a Bambo for his wife for Mother’s Day.
Great idea, spread the LOVE!
and REALLY people score more than 2 per month?!!! I thought I was getting good, and still haven’t been able to snag one, after trying for 7 months! It was easier to get my daughter than the
As someone who’s been trying for 6 months to purchase one of your gorgeous creations…
I thank you SO much, Christina.
I can’t imagine how difficult it must be for you to traverse these waters…
to balance such a high demand with your continued commitment to such beautiful craftmanship.
I think of how hard you and the other mommas are working every time I don’t get a doll…
trusting my time will come exactly when it’s meant to.
Love and blessings,
I’m truly amazed that anyone can get more than 2 dolls a month.That is still going to be a lot of dolls, if they only have to wait 30 days.
Must be some extremely fast internet over there.
Anyway, that is a fabulous change. Good luck to all those who are still trying for their first doll.
Thanks Christina for trying so hard to make it all fair. What a great job you do.
I kind of have to agree with Melissa in a sense. I understand why you are trying to control things, but on another hand it’s almost strange to me that you are such a caring, giving woman, but
that you don’t want other women to help others out. Does that make sense? Not trying to cause any drama here or anything, just a curious thought I had. π And fyi… I am not a gal that has
gotten more than 2 a month, more like 2 period! π I just know it will affect me helping my friends out because I don’t want to miss out on a possible future doll that might be perfect for us.
But I guess that’s the intent of the whole thing. π As a “giver” myself, it just makes me sad to be unable to help others who are having difficulty. Just my 2 cents! π π π Hugs!
Thanks for doing that! I’ve been trying for a boy doll for ds who will nearly be 7yrs and if I don’t get one soon it will be too late for him. Time is running out for us!
Thank you. That is HOPE to many of us.
Hmmm, that almost makes me sad. I’m pretty sure I would have never gotten an upload doll without an angel. Even though I tried every.single.upload for a few months and got an empty email 3 times.
My angel may not have wanted to give up her two dolls limit for me since she has more than one child to buy for, herself.
The fact is that if you don’t have a fast computer and fast internet you are at a disadvantage. Sure, some people luck out and get a doll regardless but living rurally does not help my odds when
our internet is slow.
I think it’s a great idea! Gives hope to those that don’t have access to angels.
I support this 100% C, people do not need more then 2 a month nor do the same angels have to keep getting dolls. If they sit out there is another doll for someone else to get π
I don’t understand angeling, but it seems that there have been kids waiting months and months for their dolls, while some people are buying more than 2 at a time, and not just for themselves.
Sounds more like an Angle than an Angel. I think a 2 per month limit is unbelievably generous to people who want several dolls.
If this frees things up a bit, then won’t those people who have lots of kids will also have a better go at buying them a doll? So where is the problem?
I ditto Jan’s comments and think this is wonderful! I think it is also much more enjoyable to purchase your own doll instead of someone getting it for you (even though I *do* like to give them as
gifts occasionally! :)). I truly believe people who have not been able to get a doll will find it ten times easier now without the “angels with an edge” competing for dolls and there will be less
traffic on the site overall during upload times. Good call, Christina! π
Yeah!!! Maybe I will start to try for one again! I was lucky enough to buy my Bambo before all of this craziness started and now I would like to get my daughter another for her tea party group –
but I had completely given up. I have to wonder if some of the people buying more than two are actually selling them somewhere?? I always keep an eye on this sight to see if there is any glimmer
of hope of getting one but haven’t been hopeful lately – maybe now there will be a chance??? Some of us would just like one doll a year!!! Maybe that could be the new limit? : )) Thanks
Angels are still going to angel but they will use up there slots on angeling. It’s great these ladies who get lots of dolls are giving them up to other mummas and this doesn’t take that away, it
just means they have to do it a little less often if they are looking to get dolls for themselves. And it opens up the chances for dolls to more people who don’t have an angel with “magic
fingers”. Some of you angels are wickedly fast and that is great but this allows for people who don’t have access to the boards you are on, a better chance at a doll.
This is a great idea. It is impossible to get a doll now. I hope it will give people who don’t have a doll, the opportunity to actually get one. I am hoping to get a doll for my daughter. She is
getting to the age where she will be interested in dolls.
I think it is a great idea too.
I just think that whatever Christina and crew decide is ultimately best for this amazing company. We won’t find another more heartfelt, dedicated, and down-to-earth group of magic-makers
anywhere, so I say that the sensible ground rules layed out by C lately are only going to make Bamboletta stronger. A stronger Bamboletta means more Bambos for us! Thanks for all you do!!
By the way, I loved your status update the other night that said something about being in full-production mode ~~ YAY!!
As a frustrated angel-less newbie, THANK YOU for these changes to help level the playing field!!! Also, I think the 2 per month limit is way more than generous, and wouldn’t have batted an eye if
it had been 1 per month instead.
I’m glad of the changes and hope it helps out.
Love the idea! I truly hope that this will get more Bambos out to their new homes! π The more kids that are able to get a Bamboletta doll, the better! π
As far as angel-ing goes, each angel will get to do it 24 times a year, which seems like plenty of opportunity, yes? Heck–this limit would even work for the Duggar’s! π
I agree – great policy! There are so many angel-less Bambo lovers out there that will benefit so much from this decision π
Great policy! Thanks for always updating your policies and the attention that you give your customers. I believe that it makes the dolls even more special to know that they are a product of
someone who cares so much. Thanks for all of your work!!!
Christina…your sense and confidence as a WAHM is wonderful. I wish you so much continued success!!
I will probably get a lot of flack for saying this but I almost think 2 dolls every six months is fair. I also have more than one little one to buy for (and I only have one doll so far) so I know
the frustration… but seriously think of all the other little ones out there waiting too.
I cannot imagine how many emails Christina gets from mommas feeling hopeless, frustrated and probably somewhat pissed off when all they really want to do is give their child a precious gift. I
cannot begin to know what it feels like for Christina to get those emails day in and day out and want to help so many but unsure how to best to this. For every angel trying to help one of their
friends, think about all those silent moms who are also vying for a doll and want an equal chance to be a bamboletta momma. I think balancing the playing field somewhat is a good thing.
I think we only need one angel at Bamboletta – and we have one – Christina – and her angelic team. This angel continues to share her talents, create dolls, and welcome us into her family of
bamboletta adoptees. You rock Christina.
Susan I think you ROCK with that comment I also agree that Christina may be all the angel we need. Although no Bambo’s for me yet I know the time will come π
Jan no it is not an ANGLE it is an ANGEL, I know many women who would not have gotten a doll if it weren’t for others helping them! I am sorry that everyone doesn’t have anyone who wants to help
them, but to say it’s an angle I think is really rude!! Why would anyone here who wants a doll so badly think that it is wrong for another momma to help a fellow momma get a doll for their child?
Hmmmmm does that really make sense?
I’m sure you have already thought of this but possibly you could charge a little more to cover your selling fees and sell some of the dolls on etsy again?? It seems like it would be easier to get
one on there for some reason, maybe not. I am just still trying to get our first doll.
Thanks for the rule change, maybe it will get easier to get one now. Fingers crossed!
“I know many women who would not have gotten a doll if it werenβ t for others helping them!” And this is exactly why, to doll-less me, it makes TOTAL sense to cap the angel/angle thing.
IMHO, I think someone helping another mom is ok if it’s a one-off type of thing (i.e. helping a real-life friend) that stops when they get a doll, but to provide a helping hand to various mommas
month after month provides an advantage, a connection, to those “in the loop” if you will. (p.s. I don’t even know if there’s truly an ongoing angel thing happening with an inner-circle of doll
winners… coz I’m out of the loop, lol)!
Anyways, to me it feels like waiting outside of a busy night club where there’s 2 diff lines, the long riff-raff line, and then the shorter VIP line. What you think of this rule change depends
greatly on which line you’ve been part of.
I think this is a great policy change Christina! I have been trying for 6 months unsuccessfully, and to know that you really want to see Newbies get dolls too is encouraging! I hope this helps π
Wow! I had no idea that this “angeling” thing exists. It seems that maybe the speed of people’s Internet collection wouldn’t be such an issue if it weren’t for the high-speed-possesing “angels”
swooping in for the dolls in the first place. In other words, it seems like these “angels” and their activity are a big part of why people need “angels” in the first place and why those without
access to “angels” cannot get dolls. | {"url":"https://blog.bamboletta.com/blog/important-change/","timestamp":"2024-11-02T12:35:02Z","content_type":"text/html","content_length":"159084","record_id":"<urn:uuid:1fc07886-47cf-492d-bb5e-171f4eb31568>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00822.warc.gz"} |
Find four numbers in AP whose sum is 28 and the sum of whose squares is 216. - Ask TrueMaths!Find four numbers in AP whose sum is 28 and the sum of whose squares is 216.
This is the Important question based on Arithmetic progression Chapter of R.S Aggarwal book for ICSE & CBSE Board.
Here you have to find the numbers of AP with the help of their given sum and the sum of their square.
Question Number 8 of Exercise 11 B of RS Aggarwal Solution | {"url":"https://ask.truemaths.com/question/find-four-numbers-in-ap-whose-sum-is-28-and-the-sum-of-whose-squares-is-216/","timestamp":"2024-11-13T07:57:33Z","content_type":"text/html","content_length":"123702","record_id":"<urn:uuid:ea5a95fd-9120-4fff-a4de-de677f88c89c>","cc-path":"CC-MAIN-2024-46/segments/1730477028342.51/warc/CC-MAIN-20241113071746-20241113101746-00557.warc.gz"} |