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https://www.kidsacademy.mobi/printable-worksheets/online/easy/age-9/math/two-step-word-problems/ | 1,719,279,870,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00002.warc.gz | 740,377,793 | 54,698 | # Easy Two-step Word Problems Worksheets for 9-Year-Olds
Introducing our Easy Two-step Word Problems worksheets, expertly crafted for nine-year-olds. Perfect for homeschooling families or supplementing traditional classroom learning, these online printables are designed to challenge young minds while reinforcing essential math skills. Through engaging and relatable scenarios, children will practice problem-solving and critical thinking as they navigate through each two-step question. Our user-friendly worksheets are easily accessible and printable, making learning both fun and convenient. Equip your child with the tools to excel in mathematics by integrating our worksheets into their daily learning routine. Ideal for fostering academic growth and confidence in young learners!
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• Interactive
• Two-step Word Problems
• Easy
## Subtracting Socks Worksheet
Before beginning this exercise with your children, warm them up with a counting game. If math is not their favorite subject, use this worksheet. Help them read the two word problems, then use their fingers to count and subtract. Ask them to select the correct answer and check the box.
Subtracting Socks Worksheet
Worksheet
## Changing Leaves Worksheet
Fall is the perfect time for kids to help Sully the scientist observe the changing leaves! Download the free worksheet with a subtraction word problem and bolded info to find the matching equation and picture. It's a fun way for your tots to learn while they admire the reddening, yellowing, and orange leaves.
Changing Leaves Worksheet
Worksheet
## Counting Seedlings Worksheet
Understanding math word problems is key. Multiple steps can prove challenging - this free worksheet provides one-to-one picture representation to help kids solve multi-step addition word problems. Strengthen addition skills by choosing the matching picture to the answer.
Counting Seedlings Worksheet
Worksheet
## Using Number Sentences to Solve Problems Worksheet
Solving math problems requires an efficient strategy. Picture representation can help young mathematicians with basic addition problems, especially when it involves repeated addends. As they gain confidence, they'll be able to apply their number sentence solving abilities to more complex equations.
Using Number Sentences to Solve Problems Worksheet
Worksheet
Take your kids to Fairytale Land! They'll meet witches, dragons, fairies, elves, knights, and princesses. This free worksheet adds up the fun, letting kids use three addends to solve addition equations and find the right answers. With friends like these, math won't even seem like math!
Worksheet
Learning Skills
## The Benefits of Easy Worksheets on Two-Step Word Problems for Nine-Year-Olds
When it comes to building a strong mathematical foundation in children, Easy worksheets on Two-step Word Problems are incredibly beneficial, especially for nine-year-olds. These worksheets not only help in sharpening kids' problem-solving skills but also prepare them for more advanced math concepts. For parents who opt for homeschooling, incorporating "Homeschool interactive worksheets" into the curriculum can greatly enhance the learning experience, making it engaging and effective.
### Why Focus on Two-Step Word Problems?
At the age of nine, children are typically in third or fourth grade, a crucial time when they start to encounter more complex math operations that involve multiple steps to reach a solution. Two-step word problems require students to engage in critical thinking as they must determine which operations to use and in what order. This not only tests their understanding of basic math operations—addition, subtraction, multiplication, and division—but also enhances their ability to think logically and methodically.
### The Interactivity of Homeschool Worksheets
Homeschool interactive worksheets specifically designed for two-step word problems offer an immersive learning experience. These worksheets are not just about solving problems; they are about understanding scenarios in real-world contexts. Interactive elements such as drag-and-drop, fill-in-the-blanks, or multiple-choice questions make the learning process dynamic and keep young learners engaged. This interactive aspect is crucial as it caters to different learning styles—whether auditory, visual, or kinesthetic—ensuring that each child can grasp the concepts effectively.
### Building Confidence and Competence
One of the primary advantages of Easy worksheets on Two-step Word Problems is the boost in confidence they provide to children. With each problem solved, children feel a sense of achievement, which is pivotal at this developmental stage. Confidence in their mathematical abilities encourages them to tackle more challenging problems, reducing math anxiety and fostering a positive attitude towards the subject.
### Preparation for Future Academic Success
Mastery of two-step word problems lays a solid foundation for future academic success. As children progress to higher grades, they will encounter increasingly complex problems that require multiple steps to solve. Early exposure through these worksheets ensures that students are well-prepared and can approach these challenges with confidence.
### Flexibility in Homeschooling
For homeschooling parents, the flexibility that these worksheets offer is invaluable. They can be integrated into the daily homeschool schedule easily and can be revisited as many times as needed to reinforce learning. This flexibility allows parents to tailor the pace according to their child’s individual learning needs and abilities. If a child struggles with a particular concept, more time can be spent on that area without the pressure of a rigid classroom schedule. Conversely, if a child grasps concepts quickly, they can move ahead at a faster pace, keeping them continually challenged and engaged.
### Enhanced Parental Involvement
Using Homeschool interactive worksheets for two-step word problems also enhances parental involvement in a child’s education. These resources allow parents to see exactly where their child might be struggling and provide immediate help. By working through the worksheets together, parents can explain concepts, model problem-solving strategies, and encourage logical thinking, which deepens the child’s understanding and retention.
### Development of Essential Life Skills
Beyond just math, solving two-step word problems helps children develop essential life skills. These include patience, perseverance, and attention to detail. The process of reading a problem, identifying the necessary information, deciding on the operational steps, and calculating the solution teaches children how to approach complex tasks systematically and patiently. These are skills that will benefit them outside of their academic environment and well into adulthood.
### Supporting Diverse Learning Needs
Easy worksheets on Two-step Word Problems can be particularly beneficial for children with diverse learning needs. For example, children who have dyscalculia or other math-related challenges can benefit from the clear, step-by-step approach that these worksheets provide. Additionally, interactive elements can be tailored to suit different sensory needs, such as using larger texts, contrasting colors for better visibility, or auditory feedback to enhance understanding.
### Conclusion
In conclusion, Easy worksheets on Two-step Word Problems are an invaluable tool for homeschooling parents of nine-year-olds. They not only help in making math a more enjoyable and less daunting subject but also play a crucial role in developing a child's analytical, logical, and problem-solving skills. By incorporating Homeschool interactive worksheets into their daily learning routine, parents can provide a robust educational foundation that prepares their children for future academic challenges and life beyond the classroom. | 1,412 | 7,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-26 | latest | en | 0.892878 |
http://mathhelpforum.com/advanced-algebra/70554-linearly-independent-dependent.html | 1,481,388,599,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543316.16/warc/CC-MAIN-20161202170903-00089-ip-10-31-129-80.ec2.internal.warc.gz | 176,613,016 | 10,070 | 1. ## Linearly independent/dependent
Determine if each of the following sets of vectors is linearly independent or linearly dependent.
(a)
v1 = [3 1 0] , v2 = [1 2 1] , v3 = [4 3 1].
(b)
v1 = [1
0
0]
v2 = [0
2
1]
v3 = [1
0
1]
Not sure how you determine that ... thanks!
2. Originally Posted by jlt1209
Determine if each of the following sets of vectors is linearly independent or linearly dependent.
(a)
v1 = [3 1 0] , v2 = [1 2 1] , v3 = [4 3 1].
(b)
v1 = [1
0
0]
v2 = [0
2
1]
v3 = [1
0
1]
Not sure how you determine that ... thanks!
For (a) v1 + v2 = v3
(b) Show that if $av_1 + bv_2 + cv_3 = 0$, then $a=b=c=0$ by solving a system of simple linear equations. | 259 | 673 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.3125 | 3 | CC-MAIN-2016-50 | longest | en | 0.778563 |
https://kidsworksheetfun.com/maths-worksheet-for-class-2-multiplication/ | 1,725,928,205,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00870.warc.gz | 310,061,420 | 27,408 | # Maths Worksheet For Class 2 Multiplication
### These word problem worksheets place 2nd grade math concepts in a context that grade 2 students can relate to.
Maths worksheet for class 2 multiplication. No login is needed. Free grade 2 math worksheets organized by grade and topic. We provide math word problems for addition subtraction multiplication time money and fractions.
Multiplication facts to 81 facts 2 to 9. Worksheets math grade 2 word problems. 100 per page 934 views this week multiplying by three 3 with factors 1 to 12 100 questions 717 views this week multiplying by six 6 with factors 1 to 12 100 questions 564 views this week five minute multiplying frenzy factor range 2 to 12 489 views this week multiplication facts tables in gray 1 to 12 482 views this week.
Our grade 4 math worksheets help build mastery in computations with the 4 basic operations delve deeper into the use of fractions and decimals and introduce the concept of factors. Multiplication of 2 digit number by 1 digit number. This page contains all our printable worksheets in section multiplication and division of second grade math as you scroll down you will see many worksheets for multiplication as repeated addition multiplication and addition sentences skip count equal groups model with arrays multiply in any order multiply with 1 and 0 times table to 12 multiplication sentences model division.
Free math worksheets from k5 learning. Our grade 3 multiplication worksheets emphasize the meaning of multiplication basic multiplication and the multiplication tables. Worksheets math grade 2 multiplication.
Skip counting addition subtraction place value multiplication division fractions rounding telling. Math word problem worksheets for grade 2. We emphasize mental multiplication exercises to improve numeracy skills.
Follow the below mentioned rules for this kind of multiplication where there is no carry over. Free math worksheets for grade 2 this is a comprehensive collection of free printable math worksheets for grade 2 organized by topics such as addition subtraction mental math regrouping place value clock money geometry and multiplication. Grade 4 math worksheets from k5 learning.
Grade 2 multiplication worksheets. Exercises also include multiplying by whole tens and whole hundreds as well as some column form multiplication missing factor questions are also included. Multiplication worksheets for grades 2 to 6.
In class i we have learnt multiplication table up to 5. Worksheets math grade 4. Try to memorize all the below mentioned tables for quicker problem resolution.
The worksheets are organized by grade and are free. These grade 2 multiplication worksheets emphasize early multiplication skills. All worksheets are printable pdf files.
Here we will learn till 15. In particular recall of the 2 5 and 10 times tables multiplying by whole tens and solving missing factor problems. Choose your grade 4 topic.
Our multiplication worksheets start with the basic multiplication facts and progress to multiplying large numbers in columns.
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Multiplication Add Multiply Acorns Worksheet Education Com 2nd Grade Math Worksheets Math Worksheets Kids Math Worksheets | 1,057 | 5,631 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2024-38 | latest | en | 0.880999 |
https://rdrr.io/cran/probout/f/vignettes/probout.Rmd | 1,643,222,564,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304959.80/warc/CC-MAIN-20220126162115-20220126192115-00617.warc.gz | 560,881,287 | 8,707 | # Introduction
Outliers are observations within a dataset that seem not to belong with the rest of the data. They could be caused, for example, by spurious entries that need to be eliminated before further analysis, or by hard-to-detect signals of interest in their own right.
The probout package provides unsupervised estimates of the probability of outlyingness for observations, based largely on separation in terms of distance. It is intended for multivariate numeric data with large numbers of sequentially accessible observations. The dimensionality of the data should not be too large, so that distances between individual observations can be computed efficiently.
The method relies on leader clustering (Hartigan,1975) to reduce the size of the data in an initial phase. Leader clustering partitions the data into groups that are within a user-specified radius $\rho$ of leader observations. The leader observations are those that are not within $\rho$ of an existing leader as the data is processed sequentially. The leader observations, and hence the associated groups, will typically vary with the order of the data. By default, the data is normalized through min-max scaling, in which each variable is mapped to the unit interval.
After leader clustering, an outlier probability is determined for each group, based on the group centroids and data simulated from a mixture model defined by the group proportions, centroids, and variances, accumulated as the data is processed sequentially. The centroids are included to ensure representation of any groups with proportions so small that it would be unlikely that a simulated observation would be drawn from those groups.
# Criteria for Outlyingness
probout estimates outlier probabilities by fitting an exponential distribution to a nonparametric outlier statistic from robust statistics (Stahel 1981, Donoho 1982). This statistic is essentially a robust $z$-score: for each observation, the median is subtracted and the absolute value of the result is divided by the median absolute deviation (MAD). For multivariate data, the univariate statistic is repeatedly computed for many random projections of the data, and the maximum value is retained as the value of the multvariate statistic. Outliers correspond to unusually large values of the outlier statistic.
# Example
## Example data
We use the 100, 400, 1500 meter timings from the Decathlon dataset from CRAN package GDAdata.
require(GDAdata)
data(Decathlon)
x <- Decathlon[,c("m100","m400","m1500")]
A projection of the data onto the first and third coordinates can be produced as follows:
plot(x[,1], x[,3], xlab = "100 meter timings", ylab = "1500 meter timings",
main = "", pch = 16, cex = .5)
To obtain outlier probabilities, first apply leader clustering:
require(probout)
require(FNN)
lead <- leader(x)
The leader function produces a list of leader clusterings for each radius supplied as a argument. The default is to compute the leader clustering for a single radius, which corresponds to the default radius $0.1 ~ / ~ log(n)^{(1/p)}$ from Wilkinson (2016) --- the same as in the HDoutliers package (Fraley 2016). A plot of the leaders can be produced as follows:
plot(x[,1], x[,3], xlab = "100 meter timings", ylab = "1500 meter timings",
main = "leader observations (blue)", pch = 16, cex = .5)
leads <- lead[[1]]$leaders points(x[leads,1],x[leads,3],pch="+",cex=1.5,col="dodgerblue") Probability of outlyingness for the leader partitions can be estimated through an exponential fit to an outlier statistic to data simulated from the partition proportions, centroids, and variances (function \verb|simData|). The process is repeated for a number of simulations: require(mclust) require(MASS) ntimes <- 100 P <- matrix( NA, length(leads), ntimes) for (i in 1:ntimes) { P[,i] <- partProb( simData(lead[[1]]), method = "distance") } pprobs <- apply( P, 1, median) Here we've used the "distance" option for the outlier method; other options are available. To show the leaders with estimated outlier probability greater than$.95\$:
thresh <- .95
plot(x[,1], x[,3], xlab = "100 meter timings", ylab = "1500 meter timings",
main = "leaders: outlier prob > .95 red else blue", pch = 16, cex = .5)
points(x[out,1],x[out,3],pch="*",cex=1.5,col="red")
We now obtain outlier probabilities for all of the data:
probs <- allProb(lead[[1]],pprobs)
plot(x[,1], x[,3], xlab = "100 meter timings", ylab = "1500 meter timings",
main = "outlier prob > .95 (red)", pch = 16, cex = .5)
out <- (1:nrow(x))[probs > thresh]
points(x[out,1],x[out,3],pch=1,col="red")
# Parameter Tuning
While probout is written as a general tool for estimating outlier probabilities, we expect it to be most useful when tuned to specific applications. Probability estimates can vary with parameter settings, and parameters can often be effectively tuned via profiling. Here is an example involving adjusting the number of simulations until a value is reached after which the estimates remain stable. For this example, at least 8,000 simulations should be used to achieve this, given the other parameter settings.
nsim <- (1:15)*1000
ntimes <- 100
P <- array( NA, c(nleads, length(nsim), ntimes))
dimnames(P) <- list(NULL, nsim, NULL)
for (i in 1:ntimes) {
for (j in seq(along = nsim)) {
P[,j,i] <- partProb( simData(lead[[1]], nsim[j]), method = "distance")
}
}
probs <- apply( P, c(1,2), median)
plot( nsim, sample(range(probs),size=length(nsim),replace=T), ylim = c(.90,1),
type="n", xlab = "# simulated observations", ylab = "leader group probability")
dummy <- apply( probs, 1, function(p) lines(nsim,p))
abline( v = 8000, lty = 2)
# Bibiliography
J. Hartigan (1975), Clustering Algorithms, Wiley.
W. A. Stahel (1981), Breakdown of Covariance Estimators, doctoral thesis, Fachgruppe fur Statistik, Eidgenossische Technische Hochschule (ETH)
D. L. Donoho (1982), Breakdown Properties of Multivariate Location Estimators, doctoral thesis, Department of Statistics, Harvard University
L. Wilkinson (2016), Visualizing Outliers, Department of Computer Science, University of Illinois at Chicago (2016)
https://www.cs.uic.edu/~wilkinson/Publications/outliers.pdf
C. Fraley (2016), HDoutliers: Leland Wilkinson's Algorithm for Detecting Multidimensional Outliers (R package available on CRAN) https://cran.r-project.org/package=HDoutliers
## Try the probout package in your browser
Any scripts or data that you put into this service are public.
probout documentation built on May 1, 2019, 6:58 p.m. | 1,610 | 6,533 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2022-05 | latest | en | 0.940482 |
https://answers.search.yahoo.com/search?p=SPE&ei=UTF-8&flt=ranking%3Adate | 1,590,519,622,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347391277.13/warc/CC-MAIN-20200526160400-20200526190400-00446.warc.gz | 258,624,308 | 27,179 | # Yahoo Web Search
related to SPE
1. Sort by
1. ### Physics Question?
...to the weight: z = x - 8m = 2.3 m KE = initial GPE - final GPE - SPE KE = m*g*(12.7+2.3) - ½*[m*g/2.3]*(2.3)² KE...
1 Answers · Science & Mathematics · 11/04/2020
2. ### Physics Question?
...and so the change is -0.813 J C) KE becomes SPE 0.902 J = ½kx² = ½ * 50N/m * x²...
1 Answers · Science & Mathematics · 27/02/2020
3. ### How much will the spring be compressed when the block momentarily comes to a rest?
.... For a spring compression of "x," initial KE + GPE = SPE means that ½mv² + mg(d+x)sinΘ + ½...
3 Answers · Science & Mathematics · 10/11/2019
4. ### One last physics question?
... = ½ * 2kg * v² solves to v = 3.3 m/s b) convert SPE to GPE ½kx² = mgh = mgLsinΘ ½...
2 Answers · Science & Mathematics · 02/07/2019
5. ### Physics Question?
GPE becomes SPE mgh = ½kh² k = 2mg / h = 2*96*9.81 / 192 = 9.81 N/m period...
3 Answers · Science & Mathematics · 14/04/2018
6. ### Why don’t we give more money to teachers so they don’t have to spe n so much of their own money on the classroom and kids?
Amen. Ours is a misguided culture that pays entertainers millions of dollars and that will not pay teachers a living wage. And we wonder why we're falling behind in technology. Really?
13 Answers · Politics & Government · 08/04/2018
7. ### What does Chemical Enhanced Oil Recovery (CEOR) screening mean?
1 Answers · Science & Mathematics · 15/02/2018
8. ### Physics homework?? I'm lost!!?
Initial GPE becomes SPE: mgh = ½kx² 2.200kg * 9.8m/s² * (ℓ + 0.17m) * ...
2 Answers · Science & Mathematics · 04/01/2018
... / 100N/m = 0.49 m After pulling down 0.0200 m, SPE = ½ * 100N/m * (0.51m)² = 13.005 J and after...
1 Answers · Science & Mathematics · 23/12/2017
10. ### Physics Help!?
GPE becomes SPE mgh = ½kx² so you need the mass m to proceed. My mass is...
1 Answers · Science & Mathematics · 21/12/2017 | 642 | 1,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2020-24 | latest | en | 0.690781 |
https://community.openenergymonitor.org/t/calculating-the-pence-per-kwh-stored-in-my-battery/17171 | 1,660,713,917,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572833.95/warc/CC-MAIN-20220817032054-20220817062054-00314.warc.gz | 179,972,358 | 5,709 | # Calculating the pence per kwh stored in my battery
I have just installed solar battery storage which is linked to octopus agile tariff and solar panels, so can charge from either. I’m trying to get the processing right to calculate the current p/kwh of the energy stored in the battery so I can determine if its better to charge or discharge based on grid prices.
Complicating the matter is that each 30mins the grid price changes and the battery also soaks up excess solar which has a cost of 0p.
Has anyone tackled this already, I’ve been tweaking for 3 days and my processing never seems to get it right. The p/kwh keeps going up too fast…
I can share my process flows if needed, but wondered if anyone already has the answer?!
Interesting question, it’s not something I’ve done but I can imagine it might be quite complicated to get right with input processing.
Is it a matter of calculating a cumulative cost of energy stored in the battery value?
``````solar_kwh_inc = (solar_power x timestep) / 3600000
agile_kwh_inc = (agile_charge_power x timestep) / 3600000
Value of electric in battery = Integration of:
((solar_kwh_inc x solar_price) + (agile_kwh_inc x agile_price))
unit cost of electric in battery = Value of electric in battery / state of charge in kwh
``````
If you are treating the cost of the solar power as zero then
``````agile_kwh_inc = (agile_charge_power x timestep) / 3600000
Value of electric in battery = Integration of:
(agile_kwh_inc x agile_price))
unit cost of electric in battery = Value of electric in battery / state of charge in kwh
``````
Next we would need to subtract the value of any discharge and perhaps account for battery wear?
Or perhaps it is easier to first try and calculate the number of kWh that are stored in the battery from solar vs agile and then just multiply the result of this with the unit costs… that would be a bit cleaner?
Then subtracting the amount discharged from the recorded stored solar and agile:
``````solar_discharge_kwh = battery_discharge_kwh * (stored_solar_kwh / stored_total_kwh)
agile_discharge_kwh = battery_discharge_kwh * (stored_agile_kwh / stored_total_kwh)
stored_solar_kwh -= solar_discharge_kwh
stored_agile_kwh -= agile_discharge_kwh
``````
I think this would need a number of new input processors to achieve… | 556 | 2,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2022-33 | latest | en | 0.937864 |
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# Larry, Michael, and Doug have five donuts to share. If any
Author Message
Intern
Joined: 10 Jan 2006
Posts: 25
Followers: 0
Kudos [?]: 4 [0], given: 0
Larry, Michael, and Doug have five donuts to share. If any [#permalink]
### Show Tags
19 May 2006, 22:22
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
In general this problem can take multiple forms
1. how many ways can u separate N numbered balls (1..N) into X buckets
2. How many ways can u separate N identical balls into X buckets.
3. How many ways can u add X numbers to get N as the sum where
a) 1 or more numbers can be 0
b) Each number is > 0.
Senior Manager
Joined: 09 Mar 2006
Posts: 445
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### Show Tags
20 May 2006, 03:52
The simplest aproach for this kind of problem that I am aware of would be to add (X -1) separators to the bunch of Y objects that you want to distribute among X people.
For instance, in your current question, the permutation that looks like OO|O|OO means that Larry got 2 , Michael got 1 and Doug got 2 donuts.
Hence the formula: (Y + X - 1 )! / ( Y! * (X-1)! )
You have to divide by ( Y! * (X-1)! ) since the separators and distributed objects are identical.
So the answer is 7!/(2!*5!) = 21
1) This problem is different, so its approach is different as well. Since each
ball can go to one of X baskets, total number of distributions will be:
X^N
2) This one is a general case of the problem explained in the beginning of this post.
3) I am not sure I understood this question. Please explain.
Intern
Joined: 10 Jan 2006
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### Show Tags
20 May 2006, 07:05
deowl wrote:
3) I am not sure I understood this question. Please explain.
Should have been obvious.
Anyway.
3a) How many ways can we get distinct non-negative integers (x1,x2,x3,x4...xr) such that x1+x2+x3..xr = N where N is another distinct non-negative number.
Solved the same way as separating N identical balls into R urns.
b) Here we just add the condition that x1,x2,x3 are all greater than 0.
We can again solve this as the urns+balls problem i.e the number of ways of placing N balls into R urns with each urn having at least one ball. Keep the N balls in a row. We get N-1 spots between the balls which is a demarcator. We'll now choose (R-1) spots from the N-1 spots.
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### Show Tags
20 May 2006, 09:28
1
KUDOS
saha wrote:
Larry, Michael, and Doug have five donuts to share. If any one of the men can be given any whole number of donuts from 0 to 5, in how many different ways can the donuts be distributed?
(A) 21
(B) 42
(C) 120
(D) 504
(E) 5040
(1) 5-0-0 in 3!/2! or 3 ways.
(2) 4-1-0 in 3! or 6 ways
(3) 3-1-1 in 3!/2 or 3 ways
(4) 3-2-0 in 3! or 6 ways
(5) 2-2-1 in 3!/2 or 3 ways
21 ways..
Re: Worth revising once [#permalink] 20 May 2006, 09:28
Display posts from previous: Sort by | 1,209 | 4,017 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-04 | latest | en | 0.890288 |
https://www.geteasysolution.com/derivative_of_sin(2x%5E2)%5E3 | 1,723,781,399,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00802.warc.gz | 598,601,721 | 5,403 | # Derivative of sin(2x^2)^3
## Derivative of sin(2x^2)^3. Simple step by step solution, to learn. Simple, and easy to understand, so dont hesitate to use it as a solution of your homework.
Below you can find the full step by step solution for you problem. We hope it will be very helpful for you and it will help you to understand the solving process.
If it's not what You are looking for type in the derivative calculator your own function and let us solve it.
## Derivative of sin(2x^2)^3:
((sin(2*x^2))^3)'3*(sin(2*x^2))^(3-1)*(sin(2*x^2))'3*(sin(2*x^2))^(3-1)*cos(2*x^2)*(2*x^2)'3*(sin(2*x^2))^(3-1)*cos(2*x^2)*((2)'*x^2+2*(x^2)')3*(sin(2*x^2))^(3-1)*cos(2*x^2)*(0*x^2+2*(x^2)')3*(sin(2*x^2))^(3-1)*cos(2*x^2)*(0*x^2+2*2*x^(2-1))3*(sin(2*x^2))^(3-1)*cos(2*x^2)*(0*x^2+2*2*x)3*(sin(2*x^2))^(3-1)*4*x*cos(2*x^2)12*x*(sin(2*x^2))^2*cos(2*x^2)`
The calculation above is a derivative of the function f (x) | 373 | 909 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-33 | latest | en | 0.834292 |
https://www.jiskha.com/questions/23078/how-do-you-write-equations-for-perpendicular-lines-i-will-assume-you-know-how-to-find-the | 1,586,090,325,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371604800.52/warc/CC-MAIN-20200405115129-20200405145629-00254.warc.gz | 961,152,050 | 6,379 | # Math
How do you write equations for perpendicular lines?
I will assume you know how to find the equation of a line if you know the slope and a point on that line.
The key thing to remember is:
Two perpendicular lines have slopes that are opposite reciprocals of each other.
e.g if one has slope -8/13 then its perpendicular has slope of 13/8.
Simply use that fact and you should have no problem.
I understand that much...
The problem that I'm having an issue with is this...
Find an equation of the line that contains (1,8) and is perpendicular to y = 3/4x + 1.
I always get the answer y = -4/3 + 8, but the answer is y = -4/3 + 9 1/3.
I just don't understand how they get the "9 1/3" part.
Ok
the slope of the old line is 3/4 so the slope of the new line must be -4/3
let it be y = (-4/3)x + b
but the point (1,8) is on it, so
8 = -4/3(1)+b
8+4/3=b
b=28/3
therefore y = -4/3 x + 28/3 or
y = -4/3 x + 9 1/3
You said "I always get the answer y = -4/3 + 8, but the answer is y = -4/3 + 9 1/3."
Your equation makes no sense, it has no x, and why did you put the 8 where the y-intercept should be????
I forgot to put the x's in, sorry. I MENT to say "y = (-4/3)x + 8" and "y = (-4/3) + 9 1-3"
I put the 8 where the y-intercept should go because that's where it said to put it in the packet that I have for over the summer Geometry homework. It showed step by step instructions and that's where it said to put it. Maybe it's teachng it wrong then.
Thanks, the way you did it made WAYY more sence than the way the page explained it. They made it way more complicated than it truley is. I'm definatly going to come here for any other homework help.
I am pretty sure that your packet says to put Y-intercept where you put the 8
But the 8 was NOT the y-intercept, the point was (1,8) so it could not have been on the y axis.
You should recognize the y-intercept if the point has the form (0,?)
If you are given any point, simply replace the x and y of your equation with the x and y values of the given point.
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More Similar Questions | 1,533 | 5,208 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2020-16 | longest | en | 0.964743 |
https://convertoctopus.com/24-5-ounces-to-kilograms | 1,637,971,810,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358074.14/warc/CC-MAIN-20211126224056-20211127014056-00170.warc.gz | 254,730,759 | 7,851 | ## Conversion formula
The conversion factor from ounces to kilograms is 0.028349523125, which means that 1 ounce is equal to 0.028349523125 kilograms:
1 oz = 0.028349523125 kg
To convert 24.5 ounces into kilograms we have to multiply 24.5 by the conversion factor in order to get the mass amount from ounces to kilograms. We can also form a simple proportion to calculate the result:
1 oz → 0.028349523125 kg
24.5 oz → M(kg)
Solve the above proportion to obtain the mass M in kilograms:
M(kg) = 24.5 oz × 0.028349523125 kg
M(kg) = 0.6945633165625 kg
The final result is:
24.5 oz → 0.6945633165625 kg
We conclude that 24.5 ounces is equivalent to 0.6945633165625 kilograms:
24.5 ounces = 0.6945633165625 kilograms
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 kilogram is equal to 1.4397535489625 × 24.5 ounces.
Another way is saying that 24.5 ounces is equal to 1 ÷ 1.4397535489625 kilograms.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that twenty-four point five ounces is approximately zero point six nine five kilograms:
24.5 oz ≅ 0.695 kg
An alternative is also that one kilogram is approximately one point four four times twenty-four point five ounces.
## Conversion table
### ounces to kilograms chart
For quick reference purposes, below is the conversion table you can use to convert from ounces to kilograms
ounces (oz) kilograms (kg)
25.5 ounces 0.723 kilograms
26.5 ounces 0.751 kilograms
27.5 ounces 0.78 kilograms
28.5 ounces 0.808 kilograms
29.5 ounces 0.836 kilograms
30.5 ounces 0.865 kilograms
31.5 ounces 0.893 kilograms
32.5 ounces 0.921 kilograms
33.5 ounces 0.95 kilograms
34.5 ounces 0.978 kilograms | 499 | 1,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-49 | latest | en | 0.753357 |
https://www.thestudentroom.co.uk/showthread.php?t=212740 | 1,527,487,623,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794871918.99/warc/CC-MAIN-20180528044215-20180528064215-00190.warc.gz | 841,869,943 | 40,059 | You are Here: Home >< Maths
# C2 Logs watch
1. Can anyone help:
Write as a single logarithm:
2log a + 3log b
log x-3 log y+
4 log z.
2. 2loga + 3logb = loga2 + logb3 = log(a2b3)
Do similarly for the second problem.
3. (Original post by z_2j96)
Can anyone help:
Write as a single logarithm:
2log a + 3log b
log x-3 log y+
4 log z.
It probably seems simple, but I've written out the method in a bit more detail for both questions here (see examples section):
http://wikinotes.cdpuk.net/wiki/Laws_of_logarithms
-Chris
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Top tips from students who have already aced their exams | 307 | 1,129 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2018-22 | latest | en | 0.893128 |
http://mathoverflow.net/questions/55572/is-there-a-conjunction-bias?answertab=active | 1,455,203,838,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701162035.80/warc/CC-MAIN-20160205193922-00149-ip-10-236-182-209.ec2.internal.warc.gz | 141,155,258 | 15,266 | # Is there a conjunction bias?
This is slightly related to question The unprecedented success of the “intersection” operator .
Apart from a set of maths books of null measure, most have the following property:
Objects definitions are presented as a conjunction of properties.
Most axiomatic are also clearly conjunctive in their presentation.
It is uncommon to have say "By definition a Zorglub is a red zorg or a white zorg".
Q1 : Do you agree with the bias (if not, give enough examples)?.
Q2: Is this bias mainly a discourse convention or does it lie deeper (where?) ?
-
We generally want to talk about very specific things; conjunctions increase specificity, and hence they're useful. Maybe I'm missing the point---are you asking for more than this sort of answer? – Scott Morrison Feb 16 '11 at 2:35
Conjunctivitis can be extremely contagious. – Tom Goodwillie Feb 16 '11 at 4:35
@Scott: Specificity is relative (level) . A 'human being' is a man of a woman ( Yet of course you could argue that the right name is 'social human being' and that is certainly less natural as a concept as it's longer name shows).Yes I ask for more but your idea is good as one of the reason for the bias, yet it look to me as not being the only one. – Jérôme JEAN-CHARLES Feb 16 '11 at 5:19
I don't think I agree with this bias- every time we say "by abuse of notation" we mean a disjunction. A common disjunction in mathematics would be define a zorg to be either a specific zorg or an equivalence class of zorgs, and similarly for maps between zorgs. For instance, a knot is a PL embedding of S^1 in S^3 or in R^3; or an ambient isotopy class thereof. – Daniel Moskovich Feb 16 '11 at 14:31
When I first saw this question, my impression was that it was too broad to get useful answers. But I wasn't sure, so I let it lie for a while to see what kind of answers would be given. Now, 14 hours later, the response I like best is Tom Goodwillie's conjunctivitis joke...so I have voted to close. – Pete L. Clark Feb 16 '11 at 16:26
A slightly tongue-in-cheek answer: definitions are the hypotheses of theorems. If the hypothesis of a theorem is a disjunction, you can always split the theorem up into two theorems with separate hypotheses, and doing so will often clarify the statement and proof anyway. So it's natural that single definitions are not usually disjunctive.
(Dually, if the conclusion of a theorem is a conjunction, then you can split it up into two theorems with separate conclusions. But I think we don't as often give names to the conclusions of theorems, unless they happen to coincide with a definition we gave for some other reason elsewhere.)
-
@Mike : OK very nice point! It covers definitions that are tailor maid for a theorem (to clarify exposition) and this does not preclude them to be meaningful, There also the case of specialization that fit the conjunction bias yet does not seem to enter your case ? Do you agree? – Jérôme JEAN-CHARLES Feb 18 '11 at 0:45
"Specialization" meaning what? – Mike Shulman Feb 18 '11 at 5:35
I was not precise : A green zorg is a specialization of a zorg, and this fit the bill, this was my idea. A vector space on the reals can also clearly be called a specialization (of vector space), it fit the bill of conjunction (may be not so well): One says V IS a vector space on a field F AND F = R. I use specialization as adding/picking 1 or more specifics. – Jérôme JEAN-CHARLES Feb 19 '11 at 1:06
But why do we specialize a definition in some way? Often, because there is some theorem we can prove in the specialized case but don't (yet) know how to prove in the general case. I didn't just mean definitions tailored for particular theorems; more generally we tend to make definitions so that we can then prove numerous theorems using those definitions as hypotheses. – Mike Shulman Feb 19 '11 at 21:21
Ok agreed. You sometimes may also specialize a true general result to illustrate (obtain mental pictures). I think I use specialization in the sense of particularization. – Jérôme JEAN-CHARLES Feb 20 '11 at 0:09 | 1,010 | 4,065 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2016-07 | longest | en | 0.947872 |
https://sv2.sekifusha.com/99706 | 1,670,078,653,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710933.89/warc/CC-MAIN-20221203143925-20221203173925-00143.warc.gz | 582,649,497 | 15,906 | # How To Play Blackjack For Beginners
紹介
See how casinos have made it tougher for system players. These are the blackjack terms you must know to determine the right play to make in each situation. Regardless of how good this strategy appears, you can give it a try but it doesn’t mean you always win.
However, this strategy is eliminated on Will there be A magical Gambling Formula? just the second round. The chance of making it through 500 rounds at 55% is practically impossible – although you would have earned \$67 billion by bet 27. The only system other than proportional betting to avoid losses was fixed betting , earning \$6,400 after 500 bets. This method has similar drawbacks to the Martingale system, but it reduces how quickly the bet increases if you’re on a losing streak,as well as reducing the rate at which you win.
## Blackjack Betting Strategies Debunked
If you lose, then you informative post end up \$1 less than when you started your winning streak. He must hit until he has 17 or higher, and then he must stand. Even if all the players at the table have 18, the dealer must stand if he ends up with a 17. Once the cards have been dealt, you are not allowed to touch the bet in the circle. If you need to know how much you have bet for doubling or splitting , the dealer will count down the chips for you. Ok this one’s an extremely rare variation which I doubt you will see in any casinos today but I thought I’d mention.
## Know The Rules
Understanding blackjack payout can also give you a general idea of risks and return rates when playing blackjack. I wrote a book about the true mathematics of blackjack, insofar as precise probability calculations are concerned. You might be shocked to hear, but the mathematical truth is that your knowledge of blackjack probabilities or odds is dead wrong. Everything you had known was based on guesswork, albeit it educated guesswork.
Play this hand out entirely before continuing to the second hand. This version of Blackjack is generally played with two decks of cards. Namely, there are some rules that are more benevolent toward the players. In this variation, the dealers don’t check for Blackjack until all the players have finally played their hand. For starters, the players have to choose how much their bets are gonna be.
## House Advantage With Multiple Number Of Decks
This way the losses can be cut down after the dealing of the cards. This way players can surrender when the cards they have been dealt are weak or the dealer has an ace or a card that worths 10 points. The exact moment when the player can surrender depends on the version of blackjack that you are playing. There is the option of “early surrender” which allows players to give up before the dealer checks his cards for a blackjack. Another variant that you may encounter while playing is the “late surrender” which can be done after the dealer checks whether he/she has 21 points.
Most casinos require an additional stake equal to half of your original bet to make this side bet. Blackjack has inspired a wide array of different betting systems, including options such as Martingale. Aside from betting systems, you can also learn game strategies like splitting. Some players swear by these systems, but of course none have been proven to be 100% effective. We’ll go into more details about how these systems can be used further down this page.
The dealer’s decisions, then, are automatic on all plays, whereas the player always has the option of taking one or more cards. The standard 52-card pack is used, but in most casinos several decks of cards are shuffled together. When four or more decks are used, they are dealt from a shoe . If you bet \$10 and get a blackjack in a traditional game (3-to-2 payoff on blackjack) you will win \$15. In a 6-to-5 game that same \$10 bet will net you only \$12. So you’re out \$3 for every blackjack hand that you get.
Even if you think you can afford it, when you finally win a hand you will be up a whopping \$25 for all your work and emotional stress. Maybe not in the first hour of play, but there is no question that at some point you will lose more than 8 hands in a row no matter how good of a player you are. A split is a move which can be made if your hand contains two cards of equal value.
Live blackjack is as close as you can get to a real casino. Experience an immersive environment with tables, felt, cards, and live streaming dealers. Enjoy playing with other peoplewithout having to leave your house. After you click on a live blackjack table, the first thing you see is the dealer behind the felt. They will have the shoe full of cards next to them and a few active hands in front of them. | 997 | 4,712 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-49 | latest | en | 0.966919 |
https://acedmypaper.com/1001547/ | 1,643,370,706,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00164.warc.gz | 145,647,585 | 11,190 | ## [answered] 1. You may use Excel (as written in my instructions) or man
1. You may use Excel (as written in my instructions) or manual calculations as shown in the Lane?s book
To test if the mileage of three different makes of cars is the same, randomly selected trips on those makes are recorded and the data is shown below for the miles per gallon of the trips.
Make 1 Make 2 Make 3 29.9 45.8 34.8 31.1 40.3 41.4 23.5 36.6 40 30.5 29 37.6 35.4 40.3 41.4 33 33 48.1 28.5 35.5 37.2 32.4 34.9 44.1 25.7 31 42 40.3 44.9
Use one way ANOVA to test if the means of the mileage of the all the cars of these make are the same. Assume that the data may be treated as simple random samples.
Show
a. Use Excel or other technology to compute the standard deviations of the three groups and verify that the largest standard deviation is no larger than the smallest standard deviation.
b. Refer to the box plot below.
Are there any extreme outliers in any of the groups?
c. State the F statistic and the P_value and your conclusion at a 5% level of significance.
2. Jamal suspects that his diastolic blood pressure is higher when he is at work than his diastolic blood pressure when he is not at work. He is going to test this statement based on blood pressure readings at randomly selected times in the above two situations.
Here are the data (units in mmHg)
at work
84 84 86 86 87 87 87 88 90 97 99
not at work
77 78 79 79 80 83 84 85 85 89
Refer to the following box plots
a) Why should we not use the two sample t-test in this case?
b) Use the Wilcoxon Rank Sum test with normal approximation to test if the BP readings at work are higher than the readings when not at work. Write the P_value and your conclusion at 5% level of significance.
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Tutor | 620 | 2,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-05 | latest | en | 0.917019 |
https://pty.vanderbilt.edu/for-students/savy/courses/savy-fall-2015/ | 1,726,392,260,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651622.79/warc/CC-MAIN-20240915084859-20240915114859-00896.warc.gz | 426,511,834 | 10,788 | # SAVY Fall 2015 Courses
### Kindergarten | 1st & 2nd Grades | 3rd & 4th Grades | 5th & 6th Grades
#### Kindergarten
The Unseen World of Microorganisms
Why do people get sick? Why does medicine make you feel better? Is there such a thing as good bacteria? In this hands-on and minds-on science course, you will learn the answer to these questions and many others as we investigate the fascinating world of microbiology. Microorganisms come in many shapes and sizes and include bacteria, viruses, fungi, protists and more! You might not be able to see them, but by the end of this course you will know where to find microorganisms, what your cells look like, and how microorganisms impact your life. Come ready to take on the role of a scientist as you investigate microbiology by conducting experiments and participating in lab work alongside a real microbiologist!
Robotic Engineering
The Math Behind Games
When you flip a coin, does it matter if you started with heads or tails? How important is it to stick to the rules? What is the difference between skill and luck when it comes to playing games? In this exciting course, you will use advanced mathematics to explore the laws of probability and how to determine if a game is fair. As you play a variety of games, you will discover the chances that each player can win, and you will use creative problem solving to change the rules that aren’t fair. With a combination of math skills and logic, we will analyze all types of games and even create our own games to test out in our probability lab!
Pattern Geometry: Tiles & Tessellations
Have you ever wondered how math connects to what you see in nature? Do you ever think about the relationship between art and math? If so, then you have already started your exploration of pattern geometry! In this course, you will continue this investigation as you explore the math associated with tessellations and develop a code for describing patterns. You will learn how patterns of polygons can describe the Fibonacci sequence and also how patterns in a geometric picture can be used to explain arithmetic arrangements or number sequences. As you take on the role of a mathematician you will create your own Escher-esque tessellation and you may even learn to use this creation to explore the relationship to fractions and irrational numbers. In this course you will realize that math may be the art of numbers but numbers can be used to think about some pretty cool art!
Writing and the Imagination
A good story or poem can make us feel like we’re somewhere else entirely—in another country, another world, or sometimes just inside another person’s head. In this class, we will explore the ways authors help us to imagine different lives and different lands. How do writers create interesting characters, strong stories, and vivid places? Why do certain sentences make us want to cry, while other sentences make us laugh? How does an author turn something imaginary into something that feels real? How can words alone convey smells, sounds, and other senses? While we think about these questions as a class, you will begin to write your own creative pieces. Using your imagination and the techniques we learn in class, you, too, will transport your readers to other places and other minds. | 675 | 3,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2024-38 | latest | en | 0.944244 |
https://gomathanswerkey.com/texas-go-math-grade-6-module-17-quiz-answer-key/ | 1,709,456,688,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00117.warc.gz | 288,807,404 | 54,379 | Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Module 17 Quiz Answer Key.
17.1 Measures of Center
Question 1.
Find the mean and median of the data set. ____.
Determine the mean and median.
Mean = 10.5
Median = 10
17.2 Box Plots
Make a box plot for the data set.
Box plot of the given data
Using the given data, identify the least value, first quartile, median, third quartile, and the greatest value to create a box plot of the data.
17.3 Dot Plots and Data Distribution
A baseball team scored the following number of runs over a 10-game period:
6, 6, 8, 5, 4, 6, 4, 3, 8, 4.
Question 3.
Make a dot plot for the data.
Dot plot of the given data
Use the frequency of the data to create a dot plot graph.
Question 4.
Find the mean, median, and range.
Determine the mean, median, and range.
Mean = 5.4
Median = 5.5
Range = 3
17.4 Stem-and-Leaf Plots and Histograms
Texas Go Math Grade 6 Pdf Module 17 Quiz Answer Key Question 5.
Make a histogram for the data set.
Histogram of the given data
Using the frequency of the data and intervals to create the histogram.
17.5 Categorical Data
Question 6.
On one day, a pet store sells 2 birds, 6 gerbils, 4 puppies, 3 fish, and 3 hamsters. Identify the mode of the data and find its relative frequency expressed as a percent.
Graph of the data
Based from the data set and graph, it shows that the mode of the given data indicates that the most sold out pet is the gerbil.
Texas Go Math Grade 6 Module 17 Mixed Review Texas Test Prep Answer Key
Selected Response
Question 1.
Over 6 days, Dan jogged 7.5 miles, 6 miles, 3 miles, 3 miles, 5.5 miles, and 5 miles. What is the mean distance that Dan jogged each day?
(A) 3 miles
(B) 5 miles
(C) 5.25 miles
(D) 7.5 miles
(B) 5 miles
Explanation:
Determine the mean of the data
mean = $$\frac{7.5+6+3+3+5.5+5}{6}$$ add all the values then divide by 6
= $$\frac{30}{6}$$ simplify
= 5 miles mean of the dat
The mean of the data is B. 5 miles.
What is the interquartile range of the data represented by the box plot shown below?
(A) 15
(B) 20
(C) 35
(D) 40
(A) 15
Explanation:
The interquartile range is the difference between the first quartile and the third quartile.
IQR = 35 – 20 subtract the first quartile from the third quartile
= 15 value of the interquartile range
The interquartile range of the data is A. 15.
Question 3.
What is the median of the data represented by the dot plot ?
(A) 21
(B) 21.5
(C) 22
(D) 25
(C) 22
Explanation:
First count the number of dots above the line, then get the middle value of the data. The middle dot is located at the 7th dot with a value of 22. Therefore, the median of the data is 22.
The median of the data is C. 22.
Question 4.
What is the range of the data represented by the stem-and-leaf plot shown below?
Key: 3|7 means 37
(A) 25
(B) 26
(C) 28
(D) 65
(C) 28
Explanation:
Range is the difference between the least and greatest data value.
range = 65 – 37 subtract the least value from the greatest value
= 28 range of the data
The range of the data is C. 28.
The percent bar graph below shows what day of the week students have music class.
What is the mode of the data?
(A) 20%
(B) 35%
(C) Tuesday
(D) Wednesday
(C) Tuesday
Explanation:
Mode indicates how frequent a value occurs or how often a value was chosen. In the given graph, the bar with the tallest height is Tuesday. It was the most frequent day of the week for students to have music class.
The mode of the data is C. Tuesday.
Question 6.
The graph displays data collected from 200 students. How many more students have music class on Tuesday than on Friday?
(A) 1o
(B) 20
(C) 50
(D) 70
Determine the number of students for each day.
There are C. 50 more students on Tuesday than on Friday.
Gridded Response
$$\frac{5 x}{5}$$ = $$\frac{195.5}{5}$$ divide both sides of the equation by 5 | 1,139 | 3,932 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-10 | latest | en | 0.807389 |
https://webwork.libretexts.org/webwork2/html2xml?answersSubmitted=0&sourceFilePath=Library/AlfredUniv/diffeq/series/ordinarypoints/OrdinaryPoint2.pg&problemSeed=123567&courseID=anonymous&userID=anonymous&course_password=anonymous&showSummary=1&displayMode=MathJax&problemIdentifierPrefix=102&language=en&outputformat=sticky | 1,591,410,722,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348509264.96/warc/CC-MAIN-20200606000537-20200606030537-00071.warc.gz | 588,661,833 | 3,525 | In this problem you will solve the differential equation
(1) By analyzing the singular points of the differential equation, we know that a series solution of the form $y = \sum_{k=0}^{\infty}c_k\ x^k$ for the differential equation will converge at least on the interval .
(2) Substituting $y = \sum_{k=0}^{\infty}c_k\ x^k$ into $(x^2 + 11) y'' + 9 x y' - y =0$, you get that
$\Big($ c $-$ c $\Big)+\Big($ c $+$ c $\Big)x+$ $\hskip 3pt\infty$$\displaystyle\sum$$n$$=$$2$ $\Bigg\lbrack$ c $+$ c $\Bigg\rbrack x^n = 0$
The subscripts on the $c$'s should be increasing and numbers or in terms of $n$.
(3) In this step we will use the equation above to solve for some of the terms in the series and find the recurrence relation.
(a) From the constant term in the series above, we know that
c $=$ c
(b) From the coefficient of $x$ in the series above, we know that
c $=$ c
(c) From the series above, we find that the recurrence relation is
c $=$ c for $\geq$
(4) The general solution to $(x^2 + 11) y'' + 9 x y' - y=0$ converges at least on and is
$y = c_0\Big($ $+$ $x^2+$ $x^4+$ $x^6+\cdots\Big) + c_1\Big($ $x+$ $x^3+$ $x^5+$ $x^7+\cdots\Big)$
Your overall score for this problem is | 411 | 1,192 | {"found_math": true, "script_math_tex": 33, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-24 | latest | en | 0.824051 |
https://fsharpforfunandprofit.com/posts/monoids-without-tears/ | 1,722,873,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00214.warc.gz | 220,899,424 | 18,242 | Part of the "Understanding monoids" series (link)
# Monoids without tears
A mostly mathless discussion of a common functional pattern
If you are coming from an OO background, one of the more challenging aspects of learning functional programming is the lack of obvious design patterns. There are plenty of idioms such as partial application, and error handling techniques, but no apparent patterns in the GoF sense.
In this post, we’ll look at a very common “pattern” known as a monoid. Monoids are not really a design pattern; more an approach to working with many different types of values in a common way. In fact, once you understand monoids, you will start seeing them everywhere!
Unfortunately the term “monoid” itself is a bit off-putting. It originally comes from mathematics but the concept as applied to programming is easy to grasp without any math at all, as I hope to demonstrate. In fact, if we were to name the concept today in a programming context, we might call it something like `ICombinable` instead, which is not nearly as scary.
Finally, you might be wondering if a “monoid” has any connection with a “monad”. Yes, there is a mathematical connection between them, but in programming terms, they are very different things, despite having similar names.
## Uh-oh… some equations
On this site, I generally don’t use any math, but in this case I’m going to break my self-imposed rule and show you some equations.
``````1 + 2 = 3
``````
Could you handle that? How about another one?
``````1 + (2 + 3) = (1 + 2) + 3
``````
And finally one more…
``````1 + 0 = 1 and 0 + 1 = 1
``````
Ok! We’re done! If you can understand these equations, then you have all the math you need to understand monoids.
## Thinking like a mathematician
“A mathematician, like a painter or poet, is a maker of patterns. If his patterns are more permanent than theirs, it is because they are made with ideas” – G H Hardy
Most people imagine that mathematicians work with numbers, doing complicated arithmetic and calculus.
This is a misconception. For example, if you look at typical high-level math discussions, you will see lots of strange words, and lots of letter and symbols, but not a lot of arithmetic.
One of the things that mathematicians do do though, is try to find patterns in things. “What do these things have in common?” and “How can we generalize these concepts?” are typical mathematical questions.
So let’s look at these three equations through a mathematician’s eyes.
### Generalizing the first equation
A mathematician would look at `1 + 2 = 3` and think something like:
• We’ve got a bunch of things (integers in this case)
• We’ve got some way of combining two of them (addition in this case)
• And the result is another one of these things (that is, we get another integer)
And then a mathematician might try to see if this pattern could be generalized to other kinds of things and operations.
Let’s start by staying with integers as the “things”. What other ways are there of combining integers? And do they fit the pattern?
Let’s try multiplication, does that fit this pattern?
The answer is yes, multiplication does fit this pattern because multiplying any two integers results in another integer.
What about division? Does that fit the pattern? The answer is no, because in most cases, dividing two integers results in a fraction, which is not an integer (I’m ignoring integer division).
What about the `max` function? Does that fit the pattern? It combines two integers and returns one of them, so the answer is yes.
What about the `equals` function? It combines two integers but returns a boolean, not an integer, so the answer is no.
Enough of integers! What other kinds of things can we think of?
Floats are similar to integers, but unlike integers, using division with floats does result in another float, so the division operation fits the pattern.
How about booleans? They can be combined using operators such as AND, OR and so on. Does `aBool AND aBool` result in another bool? Yes! And `OR` too fits the pattern.
Strings next. How can they be combined? One way is string concatenation, which returns another string, which is what we want. But something like the equality operation doesn’t fit, because it returns a boolean.
Finally, let’s consider lists. As for strings, the obvious way to combine them is with list concatenation, which returns another list and fits the pattern.
We can continue on like this for all sorts of objects and combining operations, but you should see how it works now.
You might ask: why is it so important that the operation return another thing of the same type? The answer is that you can chain together multiple objects using the operation.
For example, because `1 + 2` is another integer, you can add 3 to it. And then because `1 + 2 + 3` is an integer as well, you can keep going and add say, 4, to the result. In other words, it is only because integer addition fits the pattern that you can write a sequence of additions like this: `1 + 2 + 3 + 4`. You couldn’t write `1 = 2 = 3 = 4` in the same way, because integer equality doesn’t fit the pattern.
And of course, the chain of combined items can be as long as we like. In other words, this pattern allows us to extend a pairwise operation into an operation that works on lists.
Mathematicians call the requirement that “the result is another one of these things” the closure requirement.
### Generalizing the second equation
Ok, what about the next equation, `1 + (2 + 3) = (1 + 2) + 3`? Why is that important?
Well, if you think about the first pattern, it says we can build up a chain of operations such as `1 + 2 + 3`. But we have only got a pairwise operation. So what order should we do the combining in? Should we combine 1 and 2 first, then combine the result with 3? Or should we combine 2 and 3 first and then combine 1 with that result? Does it make a difference?
That’s where this second equation is useful. It says that, for addition, the order of combination doesn’t matter. You get the same result either way.
So for a chain of four items like this: `1 + 2 + 3 + 4`, we could start working from the left hand side: `((1+2) + 3) + 4` or from the right hand side: `1 + (2 + (3+4))` or even do it in two parts and then combine them like this: `(1+2) + (3+4)`.
Let’s see if this pattern applies to the examples we’ve already looked at.
We’ll start with multiplication again. Does `1 * (2 * 3)` give the same result as `(1 * 2) * 3`? Yes. Just as with addition, the order doesn’t matter.
Now let’s try subtraction. Does `1 - (2 - 3)` give the same result as `(1 - 2) - 3`? No. For subtraction, the order does matter.
What about division? Does `12 / (2 / 3)` give the same result as `(12 / 2) / 3`? No. For division also, the order matters.
But the `max` function does work. `max( max(12,2), 3)` gives the same result as `max(12, max(2,3))`.
What about strings and lists? Does concatenation meet the requirement? What do you think?
Here’s a question… Can we come up with an operation for strings that is order dependent?
Well, how about a function like “subtractChars” which removes all characters in the right string from the left string. So `subtractChars("abc","ab")` is just `"c"`. `subtractChars` is indeed order dependent, as you can see with a simple example: `subtractChars("abc", subtractChars("abc","abc"))` is not the same string as `subtractChars(subtractChars("abc","abc"),"abc")`.
Mathematicians call the requirement that “the order doesn’t matter” the associativity requirement.
Important Note: When I say the “order of combining”, I am talking about the order in which you do the pairwise combining steps – combining one pair, and then combining the result with the next item.
But it is critical that the overall sequence of the items be left unchanged. This is because for certain operations, if you change the sequencing of the items, then you get a completely different result! `1 - 2` does not mean the same as `2 - 1` and `2 / 3` does not mean the same as `3 / 2`.
Of course, in many common cases, the sequence order doesn’t matter. After all, `1+2` is the same as `2+1`. In this case, the operation is said to be commutative.
### The third equation
Now let’s look at the third equation, `1 + 0 = 1`.
A mathematician would say something like: that’s interesting – there is a special kind of thing (“zero”) that, when you combine it with something, just gives you back the original something, as if nothing had happened.
So once more, let’s revisit our examples and see if we can extend this “zero” concept to other operations and other things.
Again, let’s start with multiplication. Is there some value, such that when you multiply a number with it, you get back the original number?
Yes, of course! The number one. So for multiplication, the number `1` is the “zero”.
What about `max`? Is there a “zero” for that? For 32 bit ints, yes. Combining `System.Int32.MinValue` with any other 32 bit integer using `max` will return the other integer. That fits the definition of “zero” perfectly.
What about booleans combined using AND? Is there a zero for that? Yes. It is the value `True`. Why? Because `True AND False` is `False`, and `True AND True` is `True`. In both cases the other value is returned untouched.
What about booleans combined using OR? Is there a zero for that as well? I’ll let you decide.
Moving on, what about string concatenation? Is there a “zero” for this? Yes, indeed – it is just the empty string.
``````"" + "hello" = "hello"
"hello" + "" = "hello"
``````
Finally, for list concatenation, the “zero” is just the empty list.
``````[] @ [1;2;3] = [1;2;3]
[1;2;3] @ [] = [1;2;3]
``````
You can see that the “zero” value depends very much on the operation, not just on the set of things. The zero for integer addition is different from the “zero” for integer multiplication, which is different again from the from “zero” for `Max`.
Mathematicians call the “zero” the identity element.
### The equations revisited
So now let’s revisit the equations with our new generalizations in mind.
``````1 + 2 = 3
1 + (2 + 3) = (1 + 2) + 3
1 + 0 = 1 and 0 + 1 = 1
``````
But now we have something much more abstract, a set of generalized requirements that can apply to all sorts of things:
• You start with a bunch of things, and some way of combining them two at a time.
• Rule 1 (Closure): The result of combining two things is always another one of the things.
• Rule 2 (Associativity): When combining more than two things, which pairwise combination you do first doesn’t matter.
• Rule 3 (Identity element): There is a special thing called “zero” such that when you combine any thing with “zero” you get the original thing back.
With these rules in place, we can come back to the definition of a monoid. A “monoid” is just a system that obeys all three rules. Simple!
As I said at the beginning, don’t let the mathematical background put you off. If programmers had named this pattern, it probably would been called something like “the combinable pattern” rather than “monoid”. But that’s life. The terminology is already well-established, so we have to use it.
Note there are two parts to the definition of a monoid – the things plus the associated operation. A monoid is not just “a bunch of things”, but “a bunch of things” and “some way of combining them”. So, for example, “the integers” is not a monoid, but “the integers under addition” is a monoid.
### Semigroups
In certain cases, you have a system that only follows the first two rules, and there is no candidate for a “zero” value.
For example, if your domain consists only of strictly positive numbers, then under addition they are closed and associative, but there is no positive number that can be “zero”.
Another example might be the intersection of finite lists. It is closed and associative, but there is no (finite) list that when intersected with any other finite list, leaves it untouched.
This kind of system is still quite useful, and is called a “semigroup” by mathematicians, rather than a monoid. Luckily, there is a trick that can convert any semigroup into a monoid (which I’ll describe later).
### A table of classifications
Let’s put all our examples into a table, so you can see them all together.
Things Operation Closed? Associative? Identity? Classification
Int32 Addition Yes Yes 0 Monoid
Int32 Multiplication Yes Yes 1 Monoid
Int32 Subtraction Yes No 0 Other
Int32 Max Yes Yes Int32.MinValue Monoid
Int32 Equality No Other
Int32 Less than No Other
Float Multiplication Yes No (See note 1) 1 Other
Float Division Yes (See note 2) No 1 Other
Positive Numbers Addition Yes Yes No identity Semigroup
Positive Numbers Multiplication Yes Yes 1 Monoid
Boolean AND Yes Yes true Monoid
Boolean OR Yes Yes false Monoid
String Concatenation Yes Yes Empty string "" Monoid
String Equality No Other
String "subtractChars" Yes No Empty string "" Other
List Concatenation Yes Yes Empty list [] Monoid
List Intersection Yes Yes No identity Semigroup
There are many other kinds of things you can add to this list; polynomials, matrices, probability distributions, and so on. This post won’t discuss them, but once you get the idea of monoids, you will see that the concept can be applied to all sorts of things.
[Note 1] As Doug points out in the comments, floats are not associative. Replace ‘float’ with ‘real number’ to get associativity.
[Note 2] Mathematical real numbers are not closed under division, because you cannot divide by zero and get another real number. However, with IEEE floating point numbers you can divide by zero and get a valid value. So floats are indeed closed under division! Here’s a demonstration:
``````let x = 1.0/0.0 // infinity
let y = x * 2.0 // two times infinity
let z = 2.0 / x // two divided by infinity
``````
## What use are monoids to a programmer?
So far, we have described some abstract concepts, but what good are they for real-world programming problems?
### The benefit of closure
As we’ve seen, the closure rule has the benefit that you can convert pairwise operations into operations that work on lists or sequences.
In other words, if we can define a pairwise operation, we can extend it to list operations “for free”.
The function that does this is typically called “reduce”. Here are some examples:
Explicit Using reduce
`1 + 2 + 3 + 4` `[ 1; 2; 3; 4 ] |> List.reduce (+)`
`1 * 2 * 3 * 4` `[ 1; 2; 3; 4 ] |> List.reduce (*)`
`"a" + "b" + "c" + "d"` `[ "a"; "b"; "c"; "d" ] |> List.reduce (+)`
`[1] @ [2] @ [3] @ [4]` `[ [1]; [2]; [3]; [4] ] |> List.reduce (@)`
You can see that `reduce` can be thought of as inserting the specified operation between each element of the list.
Note that in the last example, the input to `reduce` is a list of lists, and the output is a single list. Make sure you understand why this is.
### The benefit of associativity
If the pairwise combinations can be done in any order, that opens up some interesting implementation techniques, such as:
• Divide and conquer algorithms
• Parallelization
• Incrementalism
These are deep topics, but let’s have a quick look!
Divide and conquer algorithms
Consider the task of summing the first 8 integers; how could we implement this?
One way would be a crude step-by-step sum, as follows:
``````let sumUpTo2 = 1 + 2
let sumUpTo3 = sumUpTo2 + 3
let sumUpTo4 = sumUpTo3 + 4
// etc
let result = sumUpTo7 + 8
``````
But because the sums can be done in any order, we could also implement the requirement by splitting the sum into two halves, like this
``````let sum1To4 = 1 + 2 + 3 + 4
let sum5To8 = 5 + 6 + 7 + 8
let result = sum1To4 + sum5To8
``````
and then we can recursively split the sums into sub-sums in the same way until we get down to the basic pairwise operation:
``````let sum1To2 = 1 + 2
let sum3To4 = 3 + 4
let sum1To4 = sum1To2 + sum3To4
``````
This “divide and conquer” approach may seem like overkill for something like a simple sum, but we’ll see in a future post that, in conjunction with a `map`, it is the basis for some well known aggregation algorithms.
Parallelization
Once we have a divide and conquer strategy, it can be easily converted into a parallel algorithm.
For example, to sum the first 8 integers on a four-core CPU, we might do something like this:
Core 1 Core 2 Core 3 Core 4
Step 1 `sum12 = 1 + 2` `sum34 = 3 + 4` `sum56 = 5 + 6` `sum78 = 7 + 8`
Step 2 `sum1234 = sum12 + sum34` `sum5678 = sum56 + sum78` (idle) (idle)
Step 3 `sum1234 + sum5678` (idle) (idle) (idle)
There are still seven calculations that need to be done, but because we are doing it parallel, we can do them all in three steps.
Again, this might seem like a trivial example, but big data systems such as Hadoop are all about aggregating large amounts of data, and if the aggregation operation is a monoid, then you can, in theory, easily scale these aggregations by using multiple machines*.
* In practice, of course, the devil is in the details, and real-world systems don’t work exactly this way.
Incrementalism
Even if you do not need parallelism, a nice property of monoids is that they support incremental calculations.
For example, let’s say you have asked me to calculate the sum of one to five. Then of course I give you back the answer fifteen.
But now you say that you have changed your mind, and you want the sum of one to six instead. Do I have to add up all the numbers again, starting from scratch? No, I can use the previous sum, and just add six to it incrementally. This is possible because integer addition is a monoid.
That is, when faced with a sum like `1 + 2 + 3 + 4 + 5 + 6`, I can group the numbers any way I like. In particular, I can make an incremental sum like this: `(1 + 2 + 3 + 4 + 5) + 6`, which then reduces to `15 + 6`.
In this case, recalculating the entire sum from scratch might not be a big deal, but consider a real-world example like web analytics, counting the number of visitors over the last 30 days, say. A naive implementation might be to calculate the numbers by parsing the logs of the last 30 days data. A more efficient approach would be to recognize that the previous 29 days have not changed, and to only process the incremental changes for one day. As a result, the parsing effort is greatly reduced.
Similarly, if you had a word count of a 100 page book, and you added another page, you shouldn’t need to parse all 101 pages again. You just need to count the words on the last page and add that to the previous total.*
* Technically, these are scary sounding monoid homomorphisms. I will explain what this is in the next post.
### The benefit of identity
Having an identity element is not always required. Having a closed, associative operation (i.e. a semigroup) is sufficient to do many useful things.
But in some cases, it is not enough. For example, here are some cases that might crop up:
• How can I use `reduce` on an empty list?
• If I am designing a divide and conquer algorithm, what should I do if one of the “divide” steps has nothing in it?
• When using an incremental algorithm, what value should I start with when I have no data?
In all cases we need a “zero” value. This allows us to say, for example, that the sum of an empty list is `0`.
Regarding the first point above, if we are concerned that the list might be empty, then we must replace `reduce` with `fold`, which allows an initial value to be passed in. (Of course, `fold` can be used for more things than just monoid operations.)
Here are `reduce` and `fold` in action:
``````// ok
[1..10] |> List.reduce (+)
// error
[] |> List.reduce (+)
// ok with explicit zero
[1..10] |> List.fold (+) 0
// ok with explicit zero
[] |> List.fold (+) 0
``````
Using a “zero” can result in counter-intuitive results sometimes. For example, what is the product of an empty list of integers?
The answer is `1`, not `0` as you might expect! Here’s the code to prove it:
``````[1..4] |> List.fold (*) 1 // result is 24
[] |> List.fold (*) 1 // result is 1
``````
### Summary of the benefits
To sum up, a monoid is basically a way to describe an aggregation pattern – we have a list of things, we have some way of combining them, and we get a single aggregated object back at the end.
Or in F# terms:
``````Monoid Aggregation : 'T list -> 'T
``````
So when you are designing code, and you start using terms like “sum”, “product”, “composition”, or “concatenation”, these are clues that you are dealing with a monoid.
## Next steps
Now that we understand what a monoid is, let’s see how they can be used in practice.
In the next post in this series, we’ll look at how you might write real code that implements the monoid “pattern”. | 5,212 | 20,743 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-33 | latest | en | 0.963699 |
http://larswiki.larseighner.com/index.php/Main/CalculusSingleVariablesDerivativeOfVersedsine | 1,556,013,067,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578596571.63/warc/CC-MAIN-20190423094921-20190423120921-00383.warc.gz | 101,810,839 | 6,985 | # Lars Eighner's Homepage
## LarsWiki
### The Derivative of Versed Sine
Versine
##### Contents
The versed sine ({$\operatorname{ver}$}) is equal to {$1 - \cos$} as can be seen by inspection of the unit circle diagram. It is the rest of the unit circle radius after cosine leaves off. As with many things in trigonometry there are a variety of ways to express the versed sine including: {$$\operatorname{ver}\theta = 2\sin^2 \left( {\theta \over 2} \right)$$} which seems to be the preferred form in spite of not being particularly intuitive.
{$y = \operatorname{arcver}(x)$} and {$y=\operatorname{ver}(x)$}
{\begin{align} \operatorname{ver}\theta &= 1 - \cos\theta \cr {d \over {d\theta}}\operatorname{ver}\theta &= {d \over {d\theta}} \left( 1 - \cos\theta \right) \cr &= {d \over {d\theta}} 1 - {d \over {d\theta}}\cos\theta \cr &= 0 - (-\sin\theta) \cr \therefore \quad {d \over {d\theta}}\operatorname{ver}\theta &= \sin\theta \end{align}}
Verify this, as this value is not given in many tables.
Sources:
1. FooPlot: Online graphing calculator and function plotter
Recommended:
Category: Math Calculus Trigonometry
This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.
Figures are often enhanced by hand editing; the same results may not be achieved with source sites and source apps.
### December 25, 2018
• HomePage
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Help | 447 | 1,493 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2019-18 | latest | en | 0.82107 |
https://www.hellenicaworld.com/Science/Mathematics/en/Irreducibleideal.html | 1,716,555,755,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058719.70/warc/CC-MAIN-20240524121828-20240524151828-00006.warc.gz | 707,773,177 | 3,599 | In mathematics, a proper ideal of a commutative ring is said to be irreducible if it cannot be written as the intersection of two strictly larger ideals.[1]
Examples
Every prime ideal is irreducible.[2] Let two ideals $${\displaystyle J,K}$$ be contained in some commutative ring R. If the intersection $${\displaystyle J\cap K}$$ is a non-trivial ideal, then there exists some elements $${\displaystyle a\in J}$$ and $${\displaystyle b\in K}$$ , where neither is in the intersection but the product is, which means a reducible ideal is not prime. A concrete example of this are the ideals $${\displaystyle 2\mathbb {Z} }$$ and $${\displaystyle 3\mathbb {Z} }$$ contained in $$\mathbb {Z}$$ . The intersection is $${\displaystyle 6\mathbb {Z} }$$ , and $${\displaystyle 6\mathbb {Z} }$$ is not a prime ideal.
Every irreducible ideal of a Noetherian ring is a primary ideal,[1] and consequently for Noetherian rings an irreducible decomposition is a primary decomposition.[3]
Every primary ideal of a principal ideal domain is an irreducible ideal.
Every irreducible ideal is primal.[4]
Properties
An element of an integral domain is prime if and only if the ideal generated by it is a nonzero prime ideal. This is not true for irreducible ideals; an irreducible ideal may be generated by an element that is not an irreducible element, as is the case in $$\mathbb {Z}$$ for the ideal $$4{\mathbb Z}$$ since it is not the intersection of two strictly greater ideals.
An ideal I of a ring R can be irreducible only if the algebraic set it defines is irreducible (that is, any open subset is dense) for the Zariski topology, or equivalently if the closed space of spec R consisting of prime ideals containing I is irreducible for the spectral topology. The converse does not hold; for example the ideal of polynomials in two variables with vanishing terms of first and second order is not irreducible.
If k is an algebraically closed field, choosing the radical of an irreducible ideal of a polynomial ring over k is exactly the same as choosing an embedding of the affine variety of its Nullstelle in the affine space.
See also
Irreducible module
Irreducible space
Laskerian ring
References
Miyanishi, Masayoshi (1998), Algebraic Geometry, Translations of mathematical monographs, 136, American Mathematical Society, p. 13, ISBN 9780821887707.
Knapp, Anthony W. (2007), Advanced Algebra, Cornerstones, Springer, p. 446, ISBN 9780817645229.
Dummit, David S.; Foote, Richard M. (2004). Abstract Algebra (Third ed.). Hoboken, NJ: John Wiley & Sons, Inc. pp. 683–685. ISBN 0-471-43334-9.
Fuchs, Ladislas (1950), "On primal ideals", Proceedings of the American Mathematical Society, 1: 1–6, doi:10.2307/2032421, MR 0032584. Theorem 1, p. 3.
Undergraduate Texts in Mathematics
Graduate Texts in Mathematics
Graduate Studies in Mathematics
Mathematics Encyclopedia
World
Index
Retrieved from "http://en.wikipedia.org/"
All text is available under the terms of the GNU Free Documentation License | 770 | 3,000 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-22 | latest | en | 0.890991 |
https://www.cs.tufts.edu/~nr/toolkit/base-specs/mips-synth.spec | 1,542,421,727,000,000,000 | text/plain | crawl-data/CC-MAIN-2018-47/segments/1542039743248.7/warc/CC-MAIN-20181117020225-20181117041353-00032.warc.gz | 850,004,338 | 1,000 | #line 374 "mips.nw" constructors nop is sll (r0, r0, 0) mov rd, rs is addu(rd, rs, r0) b reloc is beq (r0, r0, reloc) #line 385 "mips.nw" constructors bge rs, rt, reloc is slt (r1, rs, rt); beq(r1, r0, reloc) bgeu rs, rt, reloc is sltu(r1, rs, rt); beq(r1, r0, reloc) blt rs, rt, reloc is slt (r1, rs, rt); bne(r1, r0, reloc) bltu rs, rt, reloc is sltu(r1, rs, rt); bne(r1, r0, reloc) bleu rs, rt, reloc is sltu(r1, rt, rs); beq(r1, r0, reloc) ble rs, rt, reloc is slt (r1, rt, rs); beq(r1, r0, reloc) bgt rs, rt, reloc is slt (r1, rt, rs); bne(r1, r0, reloc) bgtu rs, rt, reloc is sltu(r1, rt, rs); bne(r1, r0, reloc) #line 397 "mips.nw" constructors mul rd, rs, rt is multu(rs, rt); mflo(rd) #line 418 "mips.nw" constructors li rt, imm when { imm@[16:31]! = imm@[15]! } is addiu(rt, r0, imm@[0:15]!) when { imm@[0:15] = 0 } is lui (rt, imm@[16:31]) otherwise is lui(rt, imm@[16:31] + imm@[15]); addiu(rt, rt, imm@[0:15]!) #line 552 "mips.nw" constructors l.d ft,offset!(base) # { ft = 2 * _ } is lwc1(ft, offset!, base); lwc1(ft+1, offset!+4, base) s.d ft,offset!(base) # { ft = 2 * _ } is swc1(ft, offset!, base); swc1(ft+1, offset!+4, base) mtc1.d rt, fs # { rt = 2 * _, fs = 2 * _ } is mtc1(rt, fs); mtc1(rt+1, fs+1) mfc1.d rt, fs # { rt = 2 * _, fs = 2 * _ } is mfc1(rt, fs); mfc1(rt+1, fs+1) | 586 | 1,298 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2018-47 | latest | en | 0.519158 |
https://wiki.haskell.org/index.php?title=Haskell_Quiz/Maximum_Sub-Array/Solution_Kristof&oldid=15013 | 1,448,699,131,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398451744.67/warc/CC-MAIN-20151124205411-00260-ip-10-71-132-137.ec2.internal.warc.gz | 872,544,287 | 5,967 | ##### Views
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
This solution uses a left fold to find the maximum sum. It keeps track of the maximum sublist found so far (the list
max
and it's sum
sum
), and the maximum sublist that contains the current position (
r
and
rsum
).
```max_sublist l = reverse \$ fst \$ foldl find_max ([],(0,[],0)) l where
find_max (max, (sum, r, rsum)) x = (max', (sum', r', rsum'))
where
(r', rsum') = if (rsum > 0)
then (x:r, rsum+x)
else ([x], x)
(max', sum') = if (rsum' > sum)
then (r', rsum')
else (max, sum)``` | 186 | 574 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2015-48 | longest | en | 0.658691 |
https://www.jiskha.com/users?name=Mariam | 1,571,125,570,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986657586.16/warc/CC-MAIN-20191015055525-20191015083025-00029.warc.gz | 954,579,855 | 16,455 | # Mariam
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A=2i+3j-k, b=4i=2j-2k, find the vector x parallel to A but has magnitude of B | 5,266 | 19,875 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-43 | longest | en | 0.90322 |
https://origin.geeksforgeeks.org/arrange-consonants-vowels-nodes-linked-list/?ref=lbp | 1,656,805,181,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104205534.63/warc/CC-MAIN-20220702222819-20220703012819-00085.warc.gz | 483,566,956 | 55,202 | Arrange consonants and vowels nodes in a linked list
• Difficulty Level : Medium
• Last Updated : 25 Apr, 2022
Given a singly linked list, we need to arrange the consonants and vowel nodes of it in such a way that all the vowel nodes come before the consonants while maintaining the order of their arrival.
Examples:
```Input : a -> b -> c -> e -> d ->
o -> x -> i
Output : a -> e -> o -> i -> b ->
c -> d -> x```
Solution :
The idea is to keep a marker of the latest vowel we found while traversing the list. If we find another vowel, we take it out of the chain and put it after the existing latest vowel. Example: For linked list:
`a -> b -> c -> e -> d -> o -> x -> i`
say our latestVowel reference references the ‘a’ node, and that we currently reached the ‘e’ node. We do:
`a -> e -> b -> c -> d -> o -> x -> i`
So what was after the ‘a’ node is now after the ‘e’ node after deleting it, and linking ‘a’ directly to ‘e’.
To properly remove and add links, it’s best to use the node before the one you are checking. So if you have a curr, you will check curr->next node to see if it’s a vowel or not. If it is, we need to add it after the latestVowel node, and then it’s easy to remove it from the chain by assigning its next to curr’s next. Also, if a list only contains consonants, we simply return head.
C
`/* C program to arrange consonants and ` `vowels nodes in a linked list */` `#include ` `#include ` `#include ` `/* A linked list node */` `struct` `Node ` `{ ` ` ``char` `data; ` ` ``struct` `Node *next; ` `}; ` ` ` `/* Function to add new node to the List */` `struct` `Node *newNode(``char` `key) ` `{ ` ` ``struct` `Node *temp = (``struct` `Node*)``malloc``(``sizeof``(``struct` `Node)); ` ` ``temp->data = key; ` ` ``temp->next = NULL; ` ` ``return` `temp; ` `} ` ` ` `// utility function to print linked list ` `void` `printlist(``struct` `Node *head) ` `{ ` ` ``if` `(! head) ` ` ``{ ` ` ``printf``(``"Empty list \n"``); ` ` ``return``; ` ` ``} ` ` ``while` `(head != NULL) ` ` ``{ ` ` ``printf``(``"%c"``,head->data); ` ` ``if` `(head->next) ` ` ``printf``(``"->"``); ` ` ``head = head->next; ` ` ``} ` ` ``printf``(``"\n"``); ` `} ` ` ` `// utility function for checking vowel ` `bool` `isVowel(``char` `x) ` `{ ` ` ``return` `(x == ``'a'` `|| x == ``'e'` `|| x == ``'i'` `|| ` ` ``x == ``'o'` `|| x == ``'u'``); ` `} ` ` ` `/* function to arrange consonants and ` `vowels nodes */` `struct` `Node *arrange(``struct` `Node *head) ` `{ ` ` ``struct` `Node *newHead = head; ` ` ` ` ``// for keep track of vowel ` `struct` `Node *latestVowel; ` ` ` ` ``struct` `Node *curr = head; ` ` ` ` ``// list is empty ` ` ``if` `(head == NULL) ` ` ``return` `NULL; ` ` ` ` ``// We need to discover the first vowel ` ` ``// in the list. It is going to be the ` ` ``// returned head, and also the initial ` ` ``// latestVowel. ` ` ``if` `(isVowel(head->data)) ` ` ` ` ``// first element is a vowel. It will ` ` ``// also be the new head and the initial ` ` ``// latestVowel; ` ` ``latestVowel = head; ` ` ` ` ``else` ` ``{ ` ` ` ` ``// First element is not a vowel. Iterate ` ` ``// through the list until we find a vowel. ` ` ``// Note that curr points to the element ` ` ``// *before* the element with the vowel. ` ` ``while` `(curr->next != NULL && ` ` ``!isVowel(curr->next->data)) ` ` ``curr = curr->next; ` ` ` ` ` ` ``// This is an edge case where there are ` ` ``// only consonants in the list. ` ` ``if` `(curr->next == NULL) ` ` ``return` `head; ` ` ` ` ``// Set the initial latestVowel and the ` ` ``// new head to the vowel item that we found. ` ` ``// Relink the chain of consonants after ` ` ``// that vowel item: ` ` ``// old_head_consonant->consonant1->consonant2-> ` ` ``// vowel->rest_of_list becomes ` ` ``// vowel->old_head_consonant->consonant1-> ` ` ``// consonant2->rest_of_list ` ` ``latestVowel = newHead = curr->next; ` ` ``curr->next = curr->next->next; ` ` ``latestVowel->next = head; ` ` ``} ` ` ` ` ``// Now traverse the list. Curr is always the item ` ` ``// *before* the one we are checking, so that we ` ` ``// can use it to re-link. ` ` ``while` `(curr != NULL && curr->next != NULL) ` ` ``{ ` ` ``if` `(isVowel(curr->next->data)) ` ` ``{ ` ` ``// The next discovered item is a vowel ` ` ``if` `(curr == latestVowel) ` ` ``{ ` ` ``// If it comes directly after the ` ` ``// previous vowel, we don't need to ` ` ``// move items around, just mark the ` ` ``// new latestVowel and advance curr. ` ` ``latestVowel = curr = curr->next; ` ` ``} ` ` ``else` ` ``{ ` ` ` ` ``// But if it comes after an intervening ` ` ``// chain of consonants, we need to chain ` ` ``// the newly discovered vowel right after ` ` ``// the old vowel. Curr is not changed as ` ` ``// after the re-linking it will have a ` ` ``// new next, that has not been checked yet, ` ` ``// and we always keep curr at one before ` ` ``// the next to check. ` ` ``struct` `Node *temp = latestVowel->next; ` ` ` ` ``// Chain in new vowel ` ` ``latestVowel->next = curr->next; ` ` ` ` ``// Advance latestVowel ` ` ``latestVowel = latestVowel->next; ` ` ` ` ``// Remove found vowel from previous place ` ` ``curr->next = curr->next->next; ` ` ` ` ``// Re-link chain of consonants after latestVowel ` ` ``latestVowel->next = temp; ` ` ``} ` ` ``} ` ` ``else` ` ``{ ` ` ` ` ``// No vowel in the next element, advance curr. ` ` ``curr = curr->next; ` ` ``} ` ` ``} ` ` ``return` `newHead; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``struct` `Node *head = newNode(``'a'``); ` ` ``head->next = newNode(``'b'``); ` ` ``head->next->next = newNode(``'c'``); ` ` ``head->next->next->next = newNode(``'e'``); ` ` ``head->next->next->next->next = newNode(``'d'``); ` ` ``head->next->next->next->next->next = newNode(``'o'``); ` ` ``head->next->next->next->next->next->next = newNode(``'x'``); ` ` ``head->next->next->next->next->next->next->next = newNode(``'i'``); ` ` ` ` ``printf``(``"Linked list before :\n"``); ` ` ``printlist(head); ` ` ` ` ``head = arrange(head); ` ` ` ` ``printf``(``"Linked list after :\n"``); ` ` ``printlist(head); ` ` ` ` ``return` `0; ` `} `
C++
`/* C++ program to arrange consonants and ` ` ``vowels nodes in a linked list */` `#include ` `using` `namespace` `std; ` ` ` `/* A linked list node */` `struct` `Node ` `{ ` ` ``char` `data; ` ` ``struct` `Node *next; ` `}; ` ` ` `/* Function to add new node to the List */` `Node *newNode(``char` `key) ` `{ ` ` ``Node *temp = ``new` `Node; ` ` ``temp->data = key; ` ` ``temp->next = NULL; ` ` ``return` `temp; ` `} ` ` ` `// utility function to print linked list ` `void` `printlist(Node *head) ` `{ ` ` ``if` `(! head) ` ` ``{ ` ` ``cout << ``"Empty List\n"``; ` ` ``return``; ` ` ``} ` ` ``while` `(head != NULL) ` ` ``{ ` ` ``cout << head->data << ``" "``; ` ` ``if` `(head->next) ` ` ``cout << ``"-> "``; ` ` ``head = head->next; ` ` ``} ` ` ``cout << endl; ` `} ` ` ` `// utility function for checking vowel ` `bool` `isVowel(``char` `x) ` `{ ` ` ``return` `(x == ``'a'` `|| x == ``'e'` `|| x == ``'i'` `|| ` ` ``x == ``'o'` `|| x == ``'u'``); ` `} ` ` ` `/* function to arrange consonants and ` ` ``vowels nodes */` `Node *arrange(Node *head) ` `{ ` ` ``Node *newHead = head; ` ` ` ` ``// for keep track of vowel ` ` ``Node *latestVowel; ` ` ` ` ``Node *curr = head; ` ` ` ` ``// list is empty ` ` ``if` `(head == NULL) ` ` ``return` `NULL; ` ` ` ` ``// We need to discover the first vowel ` ` ``// in the list. It is going to be the ` ` ``// returned head, and also the initial ` ` ``// latestVowel. ` ` ``if` `(isVowel(head->data)) ` ` ` ` ``// first element is a vowel. It will ` ` ``// also be the new head and the initial ` ` ``// latestVowel; ` ` ``latestVowel = head; ` ` ` ` ``else` ` ``{ ` ` ` ` ``// First element is not a vowel. Iterate ` ` ``// through the list until we find a vowel. ` ` ``// Note that curr points to the element ` ` ``// *before* the element with the vowel. ` ` ``while` `(curr->next != NULL && ` ` ``!isVowel(curr->next->data)) ` ` ``curr = curr->next; ` ` ` ` ` ` ``// This is an edge case where there are ` ` ``// only consonants in the list. ` ` ``if` `(curr->next == NULL) ` ` ``return` `head; ` ` ` ` ``// Set the initial latestVowel and the ` ` ``// new head to the vowel item that we found. ` ` ``// Relink the chain of consonants after ` ` ``// that vowel item: ` ` ``// old_head_consonant->consonant1->consonant2-> ` ` ``// vowel->rest_of_list becomes ` ` ``// vowel->old_head_consonant->consonant1-> ` ` ``// consonant2->rest_of_list ` ` ``latestVowel = newHead = curr->next; ` ` ``curr->next = curr->next->next; ` ` ``latestVowel->next = head; ` ` ``} ` ` ` ` ``// Now traverse the list. Curr is always the item ` ` ``// *before* the one we are checking, so that we ` ` ``// can use it to re-link. ` ` ``while` `(curr != NULL && curr->next != NULL) ` ` ``{ ` ` ``if` `(isVowel(curr->next->data)) ` ` ``{ ` ` ``// The next discovered item is a vowel ` ` ``if` `(curr == latestVowel) ` ` ``{ ` ` ``// If it comes directly after the ` ` ``// previous vowel, we don't need to ` ` ``// move items around, just mark the ` ` ``// new latestVowel and advance curr. ` ` ``latestVowel = curr = curr->next; ` ` ``} ` ` ``else` ` ``{ ` ` ` ` ``// But if it comes after an intervening ` ` ``// chain of consonants, we need to chain ` ` ``// the newly discovered vowel right after ` ` ``// the old vowel. Curr is not changed as ` ` ``// after the re-linking it will have a ` ` ``// new next, that has not been checked yet, ` ` ``// and we always keep curr at one before ` ` ``// the next to check. ` ` ``Node *temp = latestVowel->next; ` ` ` ` ``// Chain in new vowel ` ` ``latestVowel->next = curr->next; ` ` ` ` ``// Advance latestVowel ` ` ``latestVowel = latestVowel->next; ` ` ` ` ``// Remove found vowel from previous place ` ` ``curr->next = curr->next->next; ` ` ` ` ``// Re-link chain of consonants after latestVowel ` ` ``latestVowel->next = temp; ` ` ``} ` ` ``} ` ` ``else` ` ``{ ` ` ` ` ``// No vowel in the next element, advance curr. ` ` ``curr = curr->next; ` ` ``} ` ` ``} ` ` ``return` `newHead; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``Node *head = newNode(``'a'``); ` ` ``head->next = newNode(``'b'``); ` ` ``head->next->next = newNode(``'c'``); ` ` ``head->next->next->next = newNode(``'e'``); ` ` ``head->next->next->next->next = newNode(``'d'``); ` ` ``head->next->next->next->next->next = newNode(``'o'``); ` ` ``head->next->next->next->next->next->next = newNode(``'x'``); ` ` ``head->next->next->next->next->next->next->next = newNode(``'i'``); ` ` ` ` ``printf``(``"Linked list before :\n"``); ` ` ``printlist(head); ` ` ` ` ``head = arrange(head); ` ` ` ` ``printf``(``"Linked list after :\n"``); ` ` ``printlist(head); ` ` ` ` ``return` `0; ` `} `
Java
`/* Java program to arrange consonants and ` `vowels nodes in a linked list */` `class` `GfG ` `{ ` ` ` `/* A linked list node */` `static` `class` `Node ` `{ ` ` ``char` `data; ` ` ``Node next; ` `} ` ` ` `/* Function to add new node to the List */` `static` `Node newNode(``char` `key) ` `{ ` ` ``Node temp = ``new` `Node(); ` ` ``temp.data = key; ` ` ``temp.next = ``null``; ` ` ``return` `temp; ` `} ` ` ` `// utility function to print linked list ` `static` `void` `printlist(Node head) ` `{ ` ` ``if` `(head == ``null``) ` ` ``{ ` ` ``System.out.println(``"Empty List"``); ` ` ``return``; ` ` ``} ` ` ``while` `(head != ``null``) ` ` ``{ ` ` ``System.out.print(head.data +``" "``); ` ` ``if` `(head.next != ``null``) ` ` ``System.out.print(``"-> "``); ` ` ``head = head.next; ` ` ``} ` ` ``System.out.println(); ` `} ` ` ` `// utility function for checking vowel ` `static` `boolean` `isVowel(``char` `x) ` `{ ` ` ``return` `(x == ``'a'` `|| x == ``'e'` `|| x == ``'i'` `|| ` ` ``x == ``'o'` `|| x == ``'u'``); ` `} ` ` ` `/* function to arrange consonants and ` `vowels nodes */` `static` `Node arrange(Node head) ` `{ ` ` ``Node newHead = head; ` ` ` ` ``// for keep track of vowel ` ` ``Node latestVowel; ` ` ` ` ``Node curr = head; ` ` ` ` ``// list is empty ` ` ``if` `(head == ``null``) ` ` ``return` `null``; ` ` ` ` ``// We need to discover the first vowel ` ` ``// in the list. It is going to be the ` ` ``// returned head, and also the initial ` ` ``// latestVowel. ` ` ``if` `(isVowel(head.data) == ``true``) ` ` ` ` ``// first element is a vowel. It will ` ` ``// also be the new head and the initial ` ` ``// latestVowel; ` ` ``latestVowel = head; ` ` ` ` ``else` ` ``{ ` ` ` ` ``// First element is not a vowel. Iterate ` ` ``// through the list until we find a vowel. ` ` ``// Note that curr points to the element ` ` ``// *before* the element with the vowel. ` ` ``while` `(curr.next != ``null` `&& ` ` ``!isVowel(curr.next.data)) ` ` ``curr = curr.next; ` ` ` ` ` ` ``// This is an edge case where there are ` ` ``// only consonants in the list. ` ` ``if` `(curr.next == ``null``) ` ` ``return` `head; ` ` ` ` ``// Set the initial latestVowel and the ` ` ``// new head to the vowel item that we found. ` ` ``// Relink the chain of consonants after ` ` ``// that vowel item: ` ` ``// old_head_consonant->consonant1->consonant2-> ` ` ``// vowel->rest_of_list becomes ` ` ``// vowel->old_head_consonant->consonant1-> ` ` ``// consonant2->rest_of_list ` ` ``latestVowel = newHead = curr.next; ` ` ``curr.next = curr.next.next; ` ` ``latestVowel.next = head; ` ` ``} ` ` ` ` ``// Now traverse the list. Curr is always the item ` ` ``// *before* the one we are checking, so that we ` ` ``// can use it to re-link. ` ` ``while` `(curr != ``null` `&& curr.next != ``null``) ` ` ``{ ` ` ``if` `(isVowel(curr.next.data) == ``true``) ` ` ``{ ` ` ``// The next discovered item is a vowel ` ` ``if` `(curr == latestVowel) ` ` ``{ ` ` ``// If it comes directly after the ` ` ``// previous vowel, we don't need to ` ` ``// move items around, just mark the ` ` ``// new latestVowel and advance curr. ` ` ``latestVowel = curr = curr.next; ` ` ``} ` ` ``else` ` ``{ ` ` ` ` ``// But if it comes after an intervening ` ` ``// chain of consonants, we need to chain ` ` ``// the newly discovered vowel right after ` ` ``// the old vowel. Curr is not changed as ` ` ``// after the re-linking it will have a ` ` ``// new next, that has not been checked yet, ` ` ``// and we always keep curr at one before ` ` ``// the next to check. ` ` ``Node temp = latestVowel.next; ` ` ` ` ``// Chain in new vowel ` ` ``latestVowel.next = curr.next; ` ` ` ` ``// Advance latestVowel ` ` ``latestVowel = latestVowel.next; ` ` ` ` ``// Remove found vowel from previous place ` ` ``curr.next = curr.next.next; ` ` ` ` ``// Re-link chain of consonants after latestVowel ` ` ``latestVowel.next = temp; ` ` ``} ` ` ``} ` ` ``else` ` ``{ ` ` ` ` ``// No vowel in the next element, advance curr. ` ` ``curr = curr.next; ` ` ``} ` ` ``} ` ` ``return` `newHead; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ``Node head = newNode(``'a'``); ` ` ``head.next = newNode(``'b'``); ` ` ``head.next.next = newNode(``'c'``); ` ` ``head.next.next.next = newNode(``'e'``); ` ` ``head.next.next.next.next = newNode(``'d'``); ` ` ``head.next.next.next.next.next = newNode(``'o'``); ` ` ``head.next.next.next.next.next.next = newNode(``'x'``); ` ` ``head.next.next.next.next.next.next.next = newNode(``'i'``); ` ` ` ` ``System.out.println(``"Linked list before : "``); ` ` ``printlist(head); ` ` ` ` ``head = arrange(head); ` ` ` ` ``System.out.println(``"Linked list after :"``); ` ` ``printlist(head); ` `} ` `} ` ` ` `// This code is contributed by Prerna Saini. `
C#
`/* C# program to arrange consonants and ` `vowels nodes in a linked list */` `using` `System; ` ` ` `class` `GfG ` `{ ` ` ` ` ``/* A linked list node */` ` ``public` `class` `Node ` ` ``{ ` ` ``public` `char` `data; ` ` ``public` `Node next; ` ` ``} ` ` ` ` ``/* Function to add new node to the List */` ` ``static` `Node newNode(``char` `key) ` ` ``{ ` ` ``Node temp = ``new` `Node(); ` ` ``temp.data = key; ` ` ``temp.next = ``null``; ` ` ``return` `temp; ` ` ``} ` ` ` ` ``// utility function to print linked list ` ` ``static` `void` `printlist(Node head) ` ` ``{ ` ` ``if` `(head == ``null``) ` ` ``{ ` ` ``Console.WriteLine(``"Empty List"``); ` ` ``return``; ` ` ``} ` ` ``while` `(head != ``null``) ` ` ``{ ` ` ``Console.Write(head.data +``" "``); ` ` ``if` `(head.next != ``null``) ` ` ``Console.Write(``"-> "``); ` ` ``head = head.next; ` ` ``} ` ` ``Console.WriteLine(); ` ` ``} ` ` ` ` ``// utility function for checking vowel ` ` ``static` `bool` `isVowel(``char` `x) ` ` ``{ ` ` ``return` `(x == ``'a'` `|| x == ``'e'` `|| x == ``'i'` `|| ` ` ``x == ``'o'` `|| x == ``'u'``); ` ` ``} ` ` ` ` ``/* function to arrange consonants and ` ` ``vowels nodes */` ` ``static` `Node arrange(Node head) ` ` ``{ ` ` ``Node newHead = head; ` ` ` ` ``// for keep track of vowel ` ` ``Node latestVowel; ` ` ` ` ``Node curr = head; ` ` ` ` ``// list is empty ` ` ``if` `(head == ``null``) ` ` ``return` `null``; ` ` ` ` ``// We need to discover the first vowel ` ` ``// in the list. It is going to be the ` ` ``// returned head, and also the initial ` ` ``// latestVowel. ` ` ``if` `(isVowel(head.data) == ``true``) ` ` ` ` ``// first element is a vowel. It will ` ` ``// also be the new head and the initial ` ` ``// latestVowel; ` ` ``latestVowel = head; ` ` ` ` ``else` ` ``{ ` ` ` ` ``// First element is not a vowel. Iterate ` ` ``// through the list until we find a vowel. ` ` ``// Note that curr points to the element ` ` ``// *before* the element with the vowel. ` ` ``while` `(curr.next != ``null` `&& ` ` ``!isVowel(curr.next.data)) ` ` ``curr = curr.next; ` ` ` ` ` ` ``// This is an edge case where there are ` ` ``// only consonants in the list. ` ` ``if` `(curr.next == ``null``) ` ` ``return` `head; ` ` ` ` ``// Set the initial latestVowel and the ` ` ``// new head to the vowel item that we found. ` ` ``// Relink the chain of consonants after ` ` ``// that vowel item: ` ` ``// old_head_consonant->consonant1->consonant2-> ` ` ``// vowel->rest_of_list becomes ` ` ``// vowel->old_head_consonant->consonant1-> ` ` ``// consonant2->rest_of_list ` ` ``latestVowel = newHead = curr.next; ` ` ``curr.next = curr.next.next; ` ` ``latestVowel.next = head; ` ` ``} ` ` ` ` ``// Now traverse the list. Curr is always the item ` ` ``// *before* the one we are checking, so that we ` ` ``// can use it to re-link. ` ` ``while` `(curr != ``null` `&& curr.next != ``null``) ` ` ``{ ` ` ``if` `(isVowel(curr.next.data) == ``true``) ` ` ``{ ` ` ``// The next discovered item is a vowel ` ` ``if` `(curr == latestVowel) ` ` ``{ ` ` ``// If it comes directly after the ` ` ``// previous vowel, we don't need to ` ` ``// move items around, just mark the ` ` ``// new latestVowel and advance curr. ` ` ``latestVowel = curr = curr.next; ` ` ``} ` ` ``else` ` ``{ ` ` ` ` ``// But if it comes after an intervening ` ` ``// chain of consonants, we need to chain ` ` ``// the newly discovered vowel right after ` ` ``// the old vowel. Curr is not changed as ` ` ``// after the re-linking it will have a ` ` ``// new next, that has not been checked yet, ` ` ``// and we always keep curr at one before ` ` ``// the next to check. ` ` ``Node temp = latestVowel.next; ` ` ` ` ``// Chain in new vowel ` ` ``latestVowel.next = curr.next; ` ` ` ` ``// Advance latestVowel ` ` ``latestVowel = latestVowel.next; ` ` ` ` ``// Remove found vowel from previous place ` ` ``curr.next = curr.next.next; ` ` ` ` ``// Re-link chain of consonants after latestVowel ` ` ``latestVowel.next = temp; ` ` ``} ` ` ``} ` ` ``else` ` ``{ ` ` ` ` ``// No vowel in the next element, advance curr. ` ` ``curr = curr.next; ` ` ``} ` ` ``} ` ` ``return` `newHead; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main(String[] args) ` ` ``{ ` ` ``Node head = newNode(``'a'``); ` ` ``head.next = newNode(``'b'``); ` ` ``head.next.next = newNode(``'c'``); ` ` ``head.next.next.next = newNode(``'e'``); ` ` ``head.next.next.next.next = newNode(``'d'``); ` ` ``head.next.next.next.next.next = newNode(``'o'``); ` ` ``head.next.next.next.next.next.next = newNode(``'x'``); ` ` ``head.next.next.next.next.next.next.next = newNode(``'i'``); ` ` ` ` ``Console.WriteLine(``"Linked list before : "``); ` ` ``printlist(head); ` ` ` ` ``head = arrange(head); ` ` ` ` ``Console.WriteLine(``"Linked list after :"``); ` ` ``printlist(head); ` ` ``} ` `} ` ` ` `// This code is contributed by PrinciRaj1992 `
Python3
`# Python3 program to remove vowels ` `# Nodes in a linked list ` ` ` `# A linked list node ` `class` `Node: ` ` ``def` `__init__(``self``, x): ` ` ``self``.data ``=` `x ` ` ``self``.``next` `=` `None` ` ` `# Utility function to print the ` `# linked list ` `def` `printlist(head): ` ` ``if` `(``not` `head): ` ` ``print``(``"Empty List"``) ` ` ``return` ` ` ` ``while` `(head !``=` `None``): ` ` ``print``(head.data, end ``=` `" "``) ` ` ``if` `(head.``next``): ` ` ``print``(end ``=` `"-> "``) ` ` ``head ``=` `head.``next` ` ``print``() ` ` ` `# Utility function for checking vowel ` `def` `isVowel(x): ` ` ``return` `(x ``=``=` `'a'` `or` `x ``=``=` `'e'` `or` `x ``=``=` `'i'` ` ``or` `x ``=``=` `'o'` `or` `x ``=``=` `'u'` `or` `x ``=``=` `'A'` ` ``or` `x ``=``=` `'E'` `or` `x ``=``=` `'I'` `or` `x ``=``=` `'O'` ` ``or` `x ``=``=` `'U'``) ` ` ` `#/* function to arrange consonants and ` `# vowels nodes */ ` `def` `arrange(head): ` ` ``newHead ``=` `head ` ` ` ` ``# for keep track of vowel ` ` ``latestVowel ``=` `None` ` ` ` ``curr ``=` `head ` ` ` ` ``# list is empty ` ` ``if` `(head ``=``=` `None``): ` ` ``return` `None` ` ` ` ``# We need to discover the first vowel ` ` ``# in the list. It is going to be the ` ` ``# returned head, and also the initial ` ` ``# latestVowel. ` ` ``if` `(isVowel(head.data)): ` ` ` ` ``# first element is a vowel. It will ` ` ``# also be the new head and the initial ` ` ``# latestVowel ` ` ``latestVowel ``=` `head ` ` ` ` ``else``: ` ` ` ` ``# First element is not a vowel. Iterate ` ` ``# through the list until we find a vowel. ` ` ``# Note that curr points to the element ` ` ``# *before* the element with the vowel. ` ` ``while` `(curr.``next` `!``=` `None` `and` ` ``not` `isVowel(curr.``next``.data)): ` ` ``curr ``=` `curr.``next` ` ` ` ` ` ``# This is an edge case where there are ` ` ``# only consonants in the list. ` ` ``if` `(curr.``next` `=``=` `None``): ` ` ``return` `head ` ` ` ` ``# Set the initial latestVowel and the ` ` ``# new head to the vowel item that we found. ` ` ``# Relink the chain of consonants after ` ` ``# that vowel item: ` ` ``# old_head_consonant.consonant1.consonant2. ` ` ``# vowel.rest_of_list becomes ` ` ``# vowel.old_head_consonant.consonant1. ` ` ``# consonant2.rest_of_list ` ` ``latestVowel ``=` `newHead ``=` `curr.``next` ` ``curr.``next` `=` `curr.``next``.``next` ` ``latestVowel.``next` `=` `head ` ` ` ` ``# Now traverse the list. Curr is always the item ` ` ``# *before* the one we are checking, so that we ` ` ``# can use it to re-link. ` ` ``while` `(curr !``=` `None` `and` `curr.``next` `!``=` `None``): ` ` ``if` `(isVowel(curr.``next``.data)): ` ` ` ` ``# The next discovered item is a vowel ` ` ``if` `(curr ``=``=` `latestVowel): ` ` ``# If it comes directly after the ` ` ``# previous vowel, we don't need to ` ` ``# move items around, just mark the ` ` ``# new latestVowel and advance curr. ` ` ``latestVowel ``=` `curr ``=` `curr.``next` ` ``else``: ` ` ` ` ``# But if it comes after an intervening ` ` ``# chain of consonants, we need to chain ` ` ``# the newly discovered vowel right after ` ` ``# the old vowel. Curr is not changed as ` ` ``# after the re-linking it will have a ` ` ``# new next, that has not been checked yet, ` ` ``# and we always keep curr at one before ` ` ``# the next to check. ` ` ``temp ``=` `latestVowel.``next` ` ` ` ``# Chain in new vowel ` ` ``latestVowel.``next` `=` `curr.``next` ` ` ` ``# Advance latestVowel ` ` ``latestVowel ``=` `latestVowel.``next` ` ` ` ``# Remove found vowel from previous place ` ` ``curr.``next` `=` `curr.``next``.``next` ` ` ` ``# Re-link chain of consonants after latestVowel ` ` ``latestVowel.``next` `=` `temp ` ` ` ` ``else``: ` ` ` ` ``# No vowel in the next element, advance curr. ` ` ``curr ``=` `curr.``next` ` ` ` ``return` `newHead ` ` ` `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` ` ` ``# Initialise the Linked List ` ` ``head ``=` `Node(``'a'``) ` ` ``head.``next` `=` `Node(``'b'``) ` ` ``head.``next``.``next` `=` `Node(``'c'``) ` ` ``head.``next``.``next``.``next` `=` `Node(``'e'``) ` ` ``head.``next``.``next``.``next``.``next` `=` `Node(``'d'``) ` ` ``head.``next``.``next``.``next``.``next``.``next` `=` `Node(``'o'``) ` ` ``head.``next``.``next``.``next``.``next``.``next``.``next` `=` `Node(``'x'``) ` ` ``head.``next``.``next``.``next``.``next``.``next``.``next``.``next` `=` `Node(``'i'``) ` ` ` ` ``# Print the given Linked List ` ` ``print``(``"Linked list before :"``) ` ` ``printlist(head) ` ` ` ` ``head ``=` `arrange(head) ` ` ` ` ``# Print the Linked List after ` ` ``# removing vowels ` ` ``print``(``"Linked list after :"``) ` ` ``printlist(head) ` ` ` `# This code is contributed by mohit kumar 29`
Javascript
` `
Output:
```Linked list before :
a -> b -> c -> e -> d -> o -> x -> i
a -> e -> o -> i -> b -> c -> d -> x```
TIME COMPLEXITY:- O(n)
SPACE COMPLEXITY:- O(1)
References:
Stackoverflow
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Another approach:-
Another approach to solve the above problem is to create two separate Linked List one containing only vowels and the other one containing only consonant.
Remember the below approach is simple to understand BUT IT REQUIRES A SPACE COMPLEXITY OF O(N)
While traversing the given input Linked List if the node data is vowel then add it to the vowel Linked List and if the node data is consonant then add it to the consonant Linked List.
After traversal you have to just link the last vowel node to the first consonant node of the respective Linked List.
C++
`/* C++ program to arrange consonants and ` `vowels nodes in a linked list */` `#include ` `using` `namespace` `std; ` ` ` `/* A linked list node */` `struct` `Node { ` ` ``char` `data; ` ` ``struct` `Node* next; ` ` ``Node(``int` `x) ` ` ``{ ` ` ``data = x; ` ` ``next = NULL; ` ` ``} ` `}; ` ` ` `/* Function to add new node to the List */` `void` `append(``struct` `Node** headRef, ``char` `data) ` `{ ` ` ``struct` `Node* new_node = ``new` `Node(data); ` ` ``struct` `Node* last = *headRef; ` ` ``if` `(*headRef == NULL) { ` ` ``*headRef = new_node; ` ` ``return``; ` ` ``} ` ` ``while` `(last->next != NULL) ` ` ``last = last->next; ` ` ``last->next = new_node; ` ` ``return``; ` `} ` ` ` `// utility function to print linked list ` `void` `printlist(Node* head) ` `{ ` ` ``if` `(!head) { ` ` ``cout << ``"Empty List\n"``; ` ` ``return``; ` ` ``} ` ` ``while` `(head != NULL) { ` ` ``cout << head->data << ``" "``; ` ` ``if` `(head->next) ` ` ``cout << ``"-> "``; ` ` ``head = head->next; ` ` ``} ` ` ``cout << endl; ` `} ` ` ` `/* function to arrange consonants and ` `vowels nodes */` ` ` `struct` `Node* arrange(Node* head) ` `{ ` ` ``Node *vowel = NULL, *consonant = NULL, *start = NULL, ` ` ``*end = NULL; ` ` ``while` `(head != NULL) { ` ` ``char` `x = head->data; ` ` ``// Checking the current node data is vowel or ` ` ``// not ` ` ``if` `(x == ``'a'` `|| x == ``'e'` `|| x == ``'i'` `|| x == ``'o'` ` ``|| x == ``'u'``) { ` ` ``if` `(!vowel) { ` ` ``vowel = ``new` `Node(x); ` ` ``start = vowel; ` ` ``} ` ` ``else` `{ ` ` ``vowel->next = ``new` `Node(x); ` ` ``vowel = vowel->next; ` ` ``} ` ` ``} ` ` ``else` `{ ` ` ``if` `(!consonant) { ` ` ``consonant = ``new` `Node(x); ` ` ``end = consonant; ` ` ``} ` ` ``else` `{ ` ` ``consonant->next = ``new` `Node(x); ` ` ``consonant = consonant->next; ` ` ``} ` ` ``} ` ` ``head = head->next; ` ` ``} ` ` ``// In case when there is no vowel in the incoming LL ` ` ``// then we have to return the head of the consonant LL ` ` ``if` `(start == NULL) ` ` ``return` `end; ` ` ``// Connecting the vowel and consonant LL ` ` ``vowel->next = end; ` ` ``return` `start; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``struct` `Node* head = NULL; ` ` ``append(&head, ``'a'``); ` ` ``append(&head, ``'b'``); ` ` ``append(&head, ``'c'``); ` ` ``append(&head, ``'e'``); ` ` ``append(&head, ``'d'``); ` ` ``append(&head, ``'o'``); ` ` ``append(&head, ``'x'``); ` ` ``append(&head, ``'i'``); ` ` ` ` ``printf``(``"Linked list before :\n"``); ` ` ``printlist(head); ` ` ` ` ``head = arrange(head); ` ` ` ` ``printf``(``"Linked list after :\n"``); ` ` ``printlist(head); ` ` ` ` ``return` `0; ` `} ` ` ` `// This code is contributed by Aditya Kumar`
Java
`/* Java program to arrange consonants and ` `vowels nodes in a linked list */` `import` `java.util.*; ` ` ` `class` `GFG{ ` ` ` `/* A linked list node */` `static` `class` `Node { ` ` ``char` `data; ` ` ``Node next; ` ` ``Node(``char` `x) ` ` ``{ ` ` ``this``.data = x; ` ` ``this``.next = ``null``; ` ` ``} ` `}; ` ` ` `/* Function to add new node to the List */` `static` `Node append(Node headRef, ``char` `data) ` `{ ` ` ``Node new_node = ``new` `Node(data); ` ` ``Node last = headRef; ` ` ``if` `(headRef == ``null``) { ` ` ``headRef = new_node; ` ` ``return` `headRef; ` ` ``} ` ` ``while` `(last.next != ``null``) ` ` ``last = last.next; ` ` ``last.next = new_node; ` ` ``return` `headRef; ` `} ` ` ` `// utility function to print linked list ` `static` `void` `printlist(Node head) ` `{ ` ` ``if` `(head == ``null``) { ` ` ``System.out.print(``"Empty List\n"``); ` ` ``return``; ` ` ``} ` ` ``while` `(head != ``null``) { ` ` ``System.out.print(head.data+ ``" "``); ` ` ``if` `(head.next!=``null``) ` ` ``System.out.print(``"-> "``); ` ` ``head = head.next; ` ` ``} ` ` ``System.out.println(); ` `} ` ` ` `/* function to arrange consonants and ` `vowels nodes */` ` ` `@SuppressWarnings``(``"null"``) ` `static` `Node arrange(Node head) ` `{ ` ` ``Node vowel = ``null``, consonant = ``null``, start = ``null``, ` ` ``end = ``null``; ` ` ``while` `(head != ``null``) { ` ` ``char` `x = head.data; ` ` ` ` ``// Checking the current node data is vowel or ` ` ``// not ` ` ``if` `(x == ``'a'` `|| x == ``'e'` `|| x == ``'i'` `|| x == ``'o'` ` ``|| x == ``'u'``) { ` ` ``if` `(vowel==``null``) { ` ` ``vowel = ``new` `Node(x); ` ` ``start = vowel; ` ` ``} ` ` ``else` `{ ` ` ``vowel.next = ``new` `Node(x); ` ` ``vowel = vowel.next; ` ` ``} ` ` ``} ` ` ``else` `{ ` ` ``if` `(consonant == ``null``) { ` ` ``consonant = ``new` `Node(x); ` ` ``end = consonant; ` ` ``} ` ` ``else` `{ ` ` ``consonant.next = ``new` `Node(x); ` ` ``consonant = consonant.next; ` ` ``} ` ` ``} ` ` ``head = head.next; ` ` ``} ` ` ` ` ``// In case when there is no vowel in the incoming LL ` ` ``// then we have to return the head of the consonant LL ` ` ``if` `(start == ``null``) ` ` ``return` `end; ` ` ` ` ``// Connecting the vowel and consonant LL ` ` ``vowel.next = end; ` ` ``return` `start; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ``Node head = ``null``; ` ` ``head = append(head, ``'a'``); ` ` ``head = append(head, ``'b'``); ` ` ``head = append(head, ``'c'``); ` ` ``head = append(head, ``'e'``); ` ` ``head = append(head, ``'d'``); ` ` ``head = append(head, ``'o'``); ` ` ``head = append(head, ``'x'``); ` ` ``head = append(head, ``'i'``); ` ` ` ` ``System.out.printf(``"Linked list before :\n"``); ` ` ``printlist(head); ` ` ` ` ``head = arrange(head); ` ` ` ` ``System.out.printf(``"Linked list after :\n"``); ` ` ``printlist(head); ` ` ` `} ` `} ` ` ` `// This code is contributed by Rajput-Ji `
Python3
`''' Python program to arrange consonants and ` `vowels Nodes in a linked list '''` ` ` `''' A linked list Node '''` `class` `Node: ` ` ``def` `__init__(``self``, data): ` ` ``self``.data ``=` `data; ` ` ``self``.``next` `=` `None``; ` ` ` `''' Function to add new Node to the List '''` `def` `append(headRef, data): ` ` ``new_Node ``=` `Node(data); ` ` ``last ``=` `headRef; ` ` ``if` `(headRef ``=``=` `None``): ` ` ``headRef ``=` `new_Node; ` ` ``return` `headRef; ` ` ` ` ``while` `(last.``next` `!``=` `None``): ` ` ``last ``=` `last.``next``; ` ` ``last.``next` `=` `new_Node; ` ` ``return` `headRef; ` ` ` `# utility function to print linked list ` `def` `printlist(head): ` ` ``if` `(head ``=``=` `None``): ` ` ``print``(``"Empty List"``); ` ` ``return``; ` ` ` ` ``while` `(head !``=` `None``): ` ` ``print``(head.data, end``=``""); ` ` ``if` `(head.``next` `!``=` `None``): ` ` ``print``(``"-> "``,end``=``""); ` ` ``head ``=` `head.``next``; ` ` ` ` ``print``(); ` ` ` ` ` `''' ` ` ``* function to arrange consonants and vowels Nodes ` ` ``'''` `def` `arrange(head): ` ` ``vowel ``=` `None``; ` ` ``consonant ``=` `None``; ` ` ``start ``=` `None``; ` ` ``end ``=` `None``; ` ` ``while` `(head !``=` `None``): ` ` ``x ``=` `head.data; ` ` ` ` ``# Checking the current Node data is vowel or ` ` ``# not ` ` ``if` `(x ``=``=` `'a'` `or` `x ``=``=` `'e'` `or` `x ``=``=` `'i'` `or` `x ``=``=` `'o'` `or` `x ``=``=` `'u'``): ` ` ``if` `(vowel ``=``=` `None``): ` ` ``vowel ``=` `Node(x); ` ` ``start ``=` `vowel; ` ` ``else``: ` ` ``vowel.``next` `=` `Node(x); ` ` ``vowel ``=` `vowel.``next``; ` ` ` ` ``else``: ` ` ``if` `(consonant ``=``=` `None``): ` ` ``consonant ``=` `Node(x); ` ` ``end ``=` `consonant; ` ` ``else``: ` ` ``consonant.``next` `=` `Node(x); ` ` ``consonant ``=` `consonant.``next``; ` ` ` ` ` ` ``head ``=` `head.``next``; ` ` ` ` ``# In case when there is no vowel in the incoming LL ` ` ``# then we have to return the head of the consonant LL ` ` ``if` `(start ``=``=` `None``): ` ` ``return` `end; ` ` ` ` ``# Connecting the vowel and consonant LL ` ` ``vowel.``next` `=` `end; ` ` ``return` `start; ` ` ` `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` ``head ``=` `None``; ` ` ``head ``=` `append(head, ``'a'``); ` ` ``head ``=` `append(head, ``'b'``); ` ` ``head ``=` `append(head, ``'c'``); ` ` ``head ``=` `append(head, ``'e'``); ` ` ``head ``=` `append(head, ``'d'``); ` ` ``head ``=` `append(head, ``'o'``); ` ` ``head ``=` `append(head, ``'x'``); ` ` ``head ``=` `append(head, ``'i'``); ` ` ` ` ``print``(``"Linked list before :"``); ` ` ``printlist(head); ` ` ` ` ``head ``=` `arrange(head); ` ` ` ` ``print``(``"Linked list after :"``); ` ` ``printlist(head); ` ` ` `# This code is contributed by Rajput-Ji `
C#
`/* C# program to arrange consonants and ` `vowels nodes in a linked list */` `using` `System; ` ` ` `public` `class` `GFG { ` ` ` ` ``/* A linked list node */` ` ``public` `class` `Node { ` ` ``public` `char` `data; ` ` ``public` `Node next; ` ` ` ` ``public` `Node(``char` `x) { ` ` ``this``.data = x; ` ` ``this``.next = ``null``; ` ` ``} ` ` ``}; ` ` ` ` ``/* Function to add new node to the List */` ` ``static` `Node append(Node headRef, ``char` `data) { ` ` ``Node new_node = ``new` `Node(data); ` ` ``Node last = headRef; ` ` ``if` `(headRef == ``null``) { ` ` ``headRef = new_node; ` ` ``return` `headRef; ` ` ``} ` ` ``while` `(last.next != ``null``) ` ` ``last = last.next; ` ` ``last.next = new_node; ` ` ``return` `headRef; ` ` ``} ` ` ` ` ``// utility function to print linked list ` ` ``static` `void` `printlist(Node head) { ` ` ``if` `(head == ``null``) { ` ` ``Console.Write(``"Empty List\n"``); ` ` ``return``; ` ` ``} ` ` ``while` `(head != ``null``) { ` ` ``Console.Write(head.data + ``" "``); ` ` ``if` `(head.next != ``null``) ` ` ``Console.Write(``"-> "``); ` ` ``head = head.next; ` ` ``} ` ` ``Console.WriteLine(); ` ` ``} ` ` ` ` ``/* ` ` ``* function to arrange consonants and vowels nodes ` ` ``*/` ` ` ` ``static` `Node arrange(Node head) { ` ` ``Node vowel = ``null``, consonant = ``null``, start = ``null``, end = ``null``; ` ` ``while` `(head != ``null``) { ` ` ``char` `x = head.data; ` ` ` ` ``// Checking the current node data is vowel or ` ` ``// not ` ` ``if` `(x == ``'a'` `|| x == ``'e'` `|| x == ``'i'` `|| x == ``'o'` `|| x == ``'u'``) { ` ` ``if` `(vowel == ``null``) { ` ` ``vowel = ``new` `Node(x); ` ` ``start = vowel; ` ` ``} ``else` `{ ` ` ``vowel.next = ``new` `Node(x); ` ` ``vowel = vowel.next; ` ` ``} ` ` ``} ``else` `{ ` ` ``if` `(consonant == ``null``) { ` ` ``consonant = ``new` `Node(x); ` ` ``end = consonant; ` ` ``} ``else` `{ ` ` ``consonant.next = ``new` `Node(x); ` ` ``consonant = consonant.next; ` ` ``} ` ` ``} ` ` ``head = head.next; ` ` ``} ` ` ` ` ``// In case when there is no vowel in the incoming LL ` ` ``// then we have to return the head of the consonant LL ` ` ``if` `(start == ``null``) ` ` ``return` `end; ` ` ` ` ``// Connecting the vowel and consonant LL ` ` ``vowel.next = end; ` ` ``return` `start; ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main(String[] args) { ` ` ``Node head = ``null``; ` ` ``head = append(head, ``'a'``); ` ` ``head = append(head, ``'b'``); ` ` ``head = append(head, ``'c'``); ` ` ``head = append(head, ``'e'``); ` ` ``head = append(head, ``'d'``); ` ` ``head = append(head, ``'o'``); ` ` ``head = append(head, ``'x'``); ` ` ``head = append(head, ``'i'``); ` ` ` ` ``Console.Write(``"Linked list before :\n"``); ` ` ``printlist(head); ` ` ` ` ``head = arrange(head); ` ` ` ` ``Console.Write(``"Linked list after :\n"``); ` ` ``printlist(head); ` ` ` ` ``} ` `} ` ` ` `// This code is contributed by Rajput-Ji `
Javascript
` `
TIME COMPLEXITY:- O(n)
SPACE COMPLEXITY:- O(n)
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https://dtkaplan.github.io/RforCalculus/representing-mathematical-functions.html | 1,579,813,987,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250613416.54/warc/CC-MAIN-20200123191130-20200123220130-00481.warc.gz | 407,639,445 | 11,538 | # Chapter 1 Representing mathematical functions
As the title suggests, in this book we are going to be using the R computer language to implement the operations of calculus along with related operations such as graphing and solving.
The topic of calculus is fundamentally about mathematical functions and the operations that are performed on them. The concept of “mathematical function” is an idea. If we are going to use a computer language to work with mathematical functions, we will need to translate them into some entity in the computer language. That is, we need a language construct to represent functions and the quantities functions take as input and produce as output.
If you happen to have a background in programming, you may well be thinking that the choice of representation is obvious. For quantities, use numbers. For functions, use R-language functions. That is indeed what we are going to do, but as you’ll see, the situation is just a little more complicated than that. A little, but that little complexity needs to be dealt with from the beginning.
## 1.1 Numbers, quantities, and names
The complexity mentioned in the previous section stems the uses and real-world situations to which we want to be able to apply the mathematical idea of functions. The inputs taken by functions and the outputs produced by them are not necessarily numbers. Often, they are quantities.
Consider these examples of quantities: money, speed, blood pressure, height, volume. Isn’t each of these quantities a number? Not quite. Consider money. We can count, add, and subtract money – just like we count, add, and subtract with numbers. But money has an additional property: the kind of currency. There are Dollars1, Renminbi, Euros, Rand, Krona and so on. Money as an idea is an abstraction of the kind sometimes called a dimension. Other dimensions are length, time, volume, temperature, area, angle, luminosity, bank interest rate, and so on. When we are dealing with a quantity that has dimension, it’s not enough to give a number to say “how much” of that dimension we have. We need also to give units. Many of these are familiar in everyday life. For instance, the dimension of length is quantified with units of meters, or inches, or miles, or parsecs, and so on. The dimension of money is measured with euros, dollars, renminbi, and so on. Area as a dimension is quantified with square-meters, or square-feet, or hectares, or acres, etc.
Unfortunately, most mathematics texts treat dimension and units by ignoring them. This leads to confusion and error when an attempt is made to apply mathematical ideas to quantities in the real world.
In this book we will be using functions and calculus to work with real-world quantities. We can’t afford to ignore dimension and units. Unfortunately, mainstream computer languages like R and Python and JavaScript do not provide a systematic way to deal with dimension and units automatically. In R, for example, we can easily write
x <- 7
which stores a quantity under the name x. But the language bawks at something like
y <- 12 meters
## Error: <text>:1:9: unexpected symbol
## 1: y <- 12 meters
## ^
Lacking a proper computer notation for representing dimensioned quantities with their units, we need some other way to keep track of things.
Here is what we will do. When we represent a quantity on the computer, we will use the name of the quantity to remind us what the dimension and units are. Suppose we want, for instance, to represent a family’s yearly income. A sensible choice is to use names like income or income_per_year or family_income. We might even give the units in the name, e.g. family_income_euros_per_year. But usually what we do is to document the units separately for a human reader.
This style of using the name of a quantity as a reminder of the dimension and units of the quantity is outside the tradition of mathematical notation. In high-school math, you encountered $$x$$ and $$y$$ and $$t$$ and $$\theta$$. You might even have seen subscripts used to be more specific, for instance $$x_0$$. Someone trained in physics will know the traditional and typical uses of letters. $$x$$ is likely a position, $$t$$ is likely a time, and $$x_0$$ is the position at time zero.
In many areas of application, there are many different quantities to be represented. To illustrate, consider a textbook about the turbulent flow of fluids: Turbulent Flows by Stephen B. Pope (ISBN 978-0521598866). In the book’s prefatory matter, there is a 14-page long section entitled “Nomenclature.” Figure 1.1 shows a small part of that section, just the symbols that start with the capital letters R and S.
Each page in the nomenclature section is dense with symbols. The core of most of them is one or two letters, e.g. $$R$$ or $$Re$$. But there are only so many letters available, even if Greek and other alphabets are brought into things. And so more specificity is added by using subscripts, superscripts, hats (e.g. $$\hat{S}_{ij}$$), bars (e.g. $$\bar{S}_{ij}$$), stars, and so on. My favorite on this page is $$Re_{\delta^\star}$$ in which the subscript $$\delta^\star$$ itself has a superscript $$^\star$$.
For experts working extensively with algebraic manipulations, this elaborate system of nomenclature may be unavoidable and even optimal. But when the experts are implementing their ideas as computer programs, they have to use a much more mundane approach to naming things: character strings like Reynolds.
There can be an advantage to the mundane approach. In this book, we’ll be working with calculus, which is used in a host of fields. We’re going to be drawing examples from many fields. And so we want our quantities to have names that are easy to recognize and remember. Things like income and blood_pressure and farm_area. If we tried to stick to x and y and z, our computer notation would become incomprehensible to the human reader and thus easily subject to error.
## 1.2 R-language functions
R, like most computer languages, has a programming construct to represent operations that take one or more inputs and produce an output. In R, these are called “functions.” In R, everything you do involves a function, either explicitly or implicitly.
Let’s look at the R version of a mathematical function, exponentiation. The function is named exp and we can look at the programming it contains
exp
## function (x) .Primitive("exp")
Not much there, except computer notation. In R, functions can be created with the key word function. For instance, to create a function that translates yearly income to daily income, we could write:
as_daily_income <- function(yearly_income) {
yearly_income / 365
}
The name selected for the function, as_daily_income, is arbitrary. We could have named the function anything at all. (It’s a good practice to give functions names that are easy to read and write and remind you what they are about.)
After the keyword function, there is a pair of parentheses. Inside the parentheses are the names being given to the inputs to the function. There’s just one input to as_daily_income which we called yearly_income just to help remind us what the function is intended to do. But we could have called the input anything at all.
The part function(yearly_income) specifies that the thing being created will be a function and that we are calling the input yearly_income. After this part comes the body of the function. The body contains R expressions that specify the calculation to be done. In this case, the expression is very simple: divide yearly_income by 365 – the (approximate) number of days in a year.
It’s helpful to distinguish between the value of an input and the role that the input will play in the function. A value of yearly income might be 61362 (in, say, dollars). To speak of the role that the input plays, we use the word argument. For instance, we might say something like, “as_yearly_income is a function that takes one argument.”
Following the keyword function and the parentheses where the arguments are defined comes a pair of curly braces { and } containing some R statements. These statements are the body of the function and contain the instructions for the calculation that will turn the inputs into the output.
Here’s a surprising feature of computer languages like R … The name given to the argument doesn’t matter at all, so long as it is used consistently in the body of the function. So the programmer might have written the R function this way:
as_daily_income <- function(x) {
x / 365
}
or even
as_daily_income <- function(ghskelw) {
ghskelw / 365
}
All of these different versions of as_daily_income() will do exactly the same thing and be used in exactly the same way, regardless of the name given for the argument.2 Like this:
as_daily_income(61362)
## [1] 168.1151
Often, functions have more than one argument. The names of the arguments are listed between the parentheses following the keyword function, like this:
as_daily_income <- function(yearly_income, duration) {
yearly_income / duration
}
In such a case, to use the function we have to provide all the arguments. So, with the most recent two-argument definition of as_daily_income(), the following use generates an error message:
as_daily_income(61362)
## Error in as_daily_income(61362): argument "duration" is missing, with no default
as_daily_income(61362, 365)
## [1] 168.1151
One more aspect of function arguments in R … Any argument can be given a default value. It’s easy to see how this works with an example:
as_daily_income <- function(yearly_income, duration = 365) {
yearly_income / duration
}
With the default value for duration the function can be used with either one argument or two:
as_daily_income(61362)
## [1] 168.1151
as_daily_income(61362, duration = 366)
## [1] 167.6557
The second line is the appropriate calculation for a leap year.
To close, let’s return to the exp function, which is built-in to R. The single argument to exp was named x and the body of the function is somewhat cryptic: .Primitive("exp").
It will often be the case that the functions we create will have bodies that don’t involve traditional mathematical expressions like $$x / d$$. As you’ll see in later chapters, in the modern world many mathematical functions are too complicated to be represented by algebraic notation.
Don’t be put off by such non-algebra function bodies. Ultimately, what you need to know about a function in order to use it are just three things:
1. What are the arguments to the function and what do they stand for.
2. What kind of thing is being produced by the function.
3. That the function works as advertised, e.g. calculating what we would write algebraically as $$e^x$$.
## 1.3 Literate use of arguments
Recall that the names selected by the programmer of a function are arbitrary. You would use the function in exactly the same way even if the names were different. Similarly, when using the function you can pick yourself what expression will be the value of the argument.
For example, suppose you want to calculate $$100 e^{-2.5}$$. Easy:
100 * exp(-2.5)
## [1] 8.2085
But it’s likely that the $$-2.5$$ is meant to stand for something more general. For instance, perhaps you are calculating how much of a drug is still in the body ten days after a dose of 100 mg was administered. There will be three quantities involved in even a simple calculation of this: the dosage, the amount of time since the dose was taken, and what’s called the “time constant” for elimination of the drug via the liver or other mechanisms. (To follow this example, you don’t have to know what a time constant is. But if you’re interested, here’s an example. Suppose a drug has a time constant of 4 days. This means that a 63% of the drug will be eliminated during a 4-day period.)
In writing the calculation, it’s a good idea to be clear and explicit about the meaning of each quantity used in the calculation. So, instead of 100 * exp(-2.5), you might want to write:
dose <- 100 # mg
duration <- 10 # days
time_constant <- 4 # days
dose * exp(- duration / time_constant)
## [1] 8.2085
Even better, you could define a function that does the calculation for you:
drug_remaining <- function(dose, duration, time_constant) {
dose * exp(- duration / time_constant)
}
Then, doing the calculation for the particular situation described above is a matter of using the function:
drug_remaining(dose = 100, duration = 10, time_constant = 4)
## [1] 8.2085
By using good, descriptive names and explicitly labelling which argument is which, you produce a clear and literate documentation of what you are intending to do and how someone else, including “future you,” should change things for representing a new situation.
## 1.4 With respect to …
We’ve been using R functions to represent the calculation of quantities from inputs like the dose and time constant of a drug.
But R functions play a much bigger role than that. Functions are used for just about everything, from reading a file of data to drawing a graph to finding out what kind of computer is being used. Of particular interest to us here is the use of functions to represent and implement the operations of calculus. These operations have names that you might or might not be familiar with yet: differentiation, integration, etc.
When a calculus or similar mathematical operation is being undertaken, you usually have to specify which variable or variables the operation is being done “with respect to.” To illustrate, consider the conceptually simple operation of drawing a graph of a function. More specifically, let’s draw a graph of how much drug remains in the body as a function of time since the dose was given. The basic pharmacokinetics of the process is encapsulated in the drug_remaining() function. So what we want to do is draw a graph of drug_remaining().
Recall that drug_remaining() has three arguments: dose, duration, and time_constant. The particular graph we are going to draw shows the drug remaining as a function of duration. That is, the operation of graphing will be with respect to duration. We’ll consider, say, a dose of 100 mg of a drug with a time constant of 4 days, looking perhaps at the duration interval from 0 days to 20 days.
In this book, we’ll be using operations provided by the mosaic and mosaicCalc packages for R. The operations from these packages have a very specific notation to express with respect to. That notation uses the tilde character, ~. Here’s how to draw the graph we want, using the package’s graphFun() operation:
graphFun(
drug_remaining(dose = 100, time_constant = 4, duration = t) ~ t,
tlim = range(0, 20))
A proper graph would be properly labelled, for instance the horizontal axis with “Time (days)” and the vertical axis with “Remaining drug (mg)”. You’ll see how to do that in the next chapter, which explores the function-graphing operation in more detail.
1. We need to be more specific about dollars, because there are US dollars, Australian dollars, New Zealand dollars, Fijian dollars, Namimbian dollars and many others.
2. Highly experienced R programmers will recognize that this statement is only partially correct. A complete statement involves a computer-language concept called “scoping.” We’ll have another footnote later when we introduce makeFun(). You experts might want to keep an eye out for it. | 3,407 | 15,416 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2020-05 | latest | en | 0.953197 |
https://www.convertunits.com/molarmass/Thorium(IV)+Iodide | 1,591,497,392,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348523476.97/warc/CC-MAIN-20200607013327-20200607043327-00162.warc.gz | 670,791,886 | 7,325 | # ››Thorium(IV) Iodide molecular weight
Molar mass of ThI4 = 739.65598 g/mol
Molecular weight calculation:
232.0381 + 126.90447*4
# ››Percent composition by element
Element Symbol Atomic Mass # of Atoms Mass Percent Thorium Th 232.0381 1 31.371% Iodine I 126.90447 4 68.629%
# ››Calculate the molecular weight of a chemical compound
Enter a chemical formula:
Browse the list of common chemical compounds.
In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together.
Finding molar mass starts with units of grams per mole (g/mol). When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula.
If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100.
Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. These relative weights computed from the chemical equation are sometimes called equation weights.
Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance.
The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. We use the most common isotopes. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass.
A common request on this site is to convert grams to moles. To complete this calculation, you have to know what substance you are trying to convert. The reason is that the molar mass of the substance affects the conversion. This site explains how to find molar mass. | 518 | 2,446 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-24 | latest | en | 0.904639 |
https://byjus.com/maths/multiplication-and-division/?replytocom=176574 | 1,606,804,562,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141652107.52/warc/CC-MAIN-20201201043603-20201201073603-00568.warc.gz | 224,234,350 | 123,581 | # Multiplication And Division Of Decimals
Decimals are numbers with a decimal point (e.g.2.35). They represent the fraction of something and ten is the base of the decimal number system. A decimal notation or point differentiates an integer part from fractional part (e.g. 2.35 = 2 + 7/20). Like whole numbers, decimals can also be added, subtracted, multiplied and divided. As compared to addition and subtraction of decimal numbers multiplication and division of decimals are much easy task. Both multiplication and division frequently connected in numerous daily computations. Even though concepts are confusing, it’s easy to solve them. Let’s go through multiplication and division of decimal fractions one by one.
## Multiplication of Decimals
When we multiply 10 by 2 it is similar to the addition of 10 two times:Â 10 x 10. This is same in the case of decimals as well. For example: 0.33 x 2= 0.33 + 0.33 + 0.33. Multiplication of decimal numbers is similar to a multiplication of whole numbers. Steps for multiplication of decimals are given below with an example.
Consider multiplication of two numbers, 2.32 and 3 for example.
Step 1: Take the count of the total number of places (digits) to the right of the decimal point in both numbers.
Here, in 2.32 there are two digits to the right of the decimal point and 3 is a whole number with no decimal point. Therefore, the total number of digits to the right of the decimal is 2.
Step 2: Now forget about the decimal point and just multiply the numbers without the decimal point.
Step 3: After multiplication, put the decimal point in answer 2 places (step 1) from the right i.e. answer (2.32 x 3) will be 6.96.
Simply, just multiply the decimal numbers without decimal points and then give decimal point in the answer as many places same as the total number of places right to the decimal points in both numbers.
## Division of Decimals
If you divide decimal numbers as it is with a decimal point is quite confusing and difficult. We can use the same trick we used in the multiplication of decimals i.e. remove the decimal points and divide the numbers like whole numbers.
Let’s divide 40.5 by .20. Methods to divide these decimal numbers are as follows:
Method 1:Â Convert the decimal numbers into whole numbers by multiplying both numerator and denominator by the same number. The denominator must be always a whole number.
(Multiply both the numerator and denominator by 5)
Method 2: Â Alternatively, one can convert decimal numbers into whole numbers by multiplying with numbers having powers of 10 (10, 100, 1000, etc.).
• Consider the denominator, count the number of places (digits) right to the decimal point.
Here, the denominator is 0.20 and the number of digits right to the decimal point is 2.
40.5 ÷ 0.20 = 40.5 / 0.20
• Take the power of 10 the same as the number of digits right to the decimal point e. 102 = 100
• Multiply both numerator and denominator by 100.
### Multiplication and Division of Decimal Numbers by 10, 100 and 1000
As we know decimals are nothing but another form of expressing fractions having 10 as its base. We can express above examples as a fraction and can multiply and divide it as a fraction. Multiplication and division of decimals by numbers that have powers of 10 is easier than that by a whole number. Rules for multiplication and division of decimal numbers by 10, 100 and 1000:
Arithmetic Operation Rule Example Multiply by 10 (101) the number will move one place value to the left 5.63 x 10 =56.3 Multiply by 100 (102) the number will move two places value to the left 5.63 x 100 =563 Multiply by 1000 (103) the number will move three places value to the left 5.63 x 1000 =5630 Divide by 10 (101) the number will move one place value to the right 56.3 ÷ 10 =5.63 Divide by 100 (102) the number will move two places value to the right 56.3 ÷ 100 =0.563 Divide by 1000 (103) the number will move three places value to the right 56.3 ÷ 1000 =0.0563
To solve more problems on multiplication and division of decimals, download BYJU’S – The Learning App from Google Play Store and watch interactive videos.
#### 2 Comments
1. I have problem in multiply by the digit in which have decimal.
Please!
Tell me how to multiply the decimal digit?
1. do normal multiplication . after that count the number of decimals places. then you put decimal point (in Answer)from right to left. | 1,069 | 4,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2020-50 | latest | en | 0.870334 |
https://tw.tek.com/support/faqs/how-do-i-measure-crosstalk-differential-lines | 1,638,677,101,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363135.71/warc/CC-MAIN-20211205035505-20211205065505-00593.warc.gz | 654,688,685 | 12,533 | # How do I measure the crosstalk in differential lines?
### 問題：
How do I measure the crosstalk in differential lines?
### 答案：
The crosstalk is the induced voltage on one conductor due to a changing current in another. The line that provides a coupling signal is called the “aggressor,” and the line where you measure the crosstalk is the “victim.” The crosstalk is usually measured at the victim’s ends. The induced noise at the near end of the victim is called a Near End crosstalk (NEXT), and the noise induced at the far end is called a Far End crosstalk (FEXT).
Since we measure the differential crosstalk in the victim, we need to measure a differential response produced due to the vicinity of the aggressor. Therefore, to measure the NEXT we need to stimulate the aggressor differentially, measure the responses at both the near ends of the victim, and then subtract the obtained victim's waveforms. Similarly, to measure the FEXT we need to measure the responses at the far ends of the victim and subtract the acquired victim's waveforms. Please, keep in mind that when you measure the NEXT, the far ends of the victim must be differentially matched, otherwise, the FEXT will be reflected back to the near end.
To obtain the crosstalk as a percentage of the offender voltage, take the difference of the induced voltages at the near (NEXT) or far (FEXT) end of the victim, and divide it by the difference between the positive and negative voltages on the offender. Finally, to get the frequency dependent crosstalk in IConnect, use the difference of the induced voltages on the victim as the DUT waveform and the difference of the aggressor voltages as the Step waveform in IConnect S-parameter computation. | 377 | 1,720 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-49 | latest | en | 0.955533 |
https://web2.0calc.com/questions/please-help_81514 | 1,590,627,810,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347396300.22/warc/CC-MAIN-20200527235451-20200528025451-00067.warc.gz | 593,960,865 | 7,248 | +0
-1
74
15
+95
Find all real solutions for in $$2\left(2^x-\:1\right)\:x^2\:+\:\left(2^{x^2}-2\right)x\:=\:2^{x+1}\:-2$$
Apr 11, 2020
#1
+109486
0
do you mean
$$2^{x^2}=2^{(x^2)}\qquad or \qquad 2^{x^2}=(2^x)^2$$
.
Apr 11, 2020
#2
+95
+1
Sorry for the confusion, it is the first one.
littlemixfan Apr 11, 2020
#3
+109486
0
Yes that is what I thought but I still do not know how to solve it.
Melody Apr 11, 2020
#4
+109486
0
Have you tried entering it into Wolfram|Alpha?
Apr 11, 2020
#5
+95
+1
I have the answers as 1,-1,0 but using WolframAlpha, they don't give steps. And I want the steps to getting to this answer.
littlemixfan Apr 11, 2020
#6
+95
+1
It says that I have to pay to get the steps.
littlemixfan Apr 11, 2020
#7
+109486
0
Yes, it gets expensive if you want the steps....
Apr 11, 2020
#8
+95
0
do you know a way to get the steps for free? I have tried on other websites such as Symbolab but they all come out as error, and that they cannot solve this equation.
littlemixfan Apr 11, 2020
#9
+109486
0
No, sorry, I have no idea.
Apr 11, 2020
#10
+95
0
If you ever find a way to solve it, could you please let me know
littlemixfan Apr 11, 2020
#11
+109486
0
Yes, will do.
That goes both ways if anyone finds an answer in the near future.
Melody Apr 11, 2020
#12
+109486
+2
Here is a graphical solution.
Maybe it can be used to get ideas.
Apr 11, 2020
#13
+95
0
Thank you! This problem is really making me on edge. :) :(
littlemixfan Apr 11, 2020
#14
+29978
+2
See how I approached this problem here: https://web2.0calc.com/questions/help-please_7362#r4
Apr 11, 2020
#15
+95
0
Thank you very much!!
littlemixfan Apr 11, 2020 | 654 | 1,682 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2020-24 | longest | en | 0.920513 |
http://stackoverflow.com/questions/13385977/maximum-subarray-variation/13520599 | 1,454,943,396,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701153585.76/warc/CC-MAIN-20160205193913-00138-ip-10-236-182-209.ec2.internal.warc.gz | 207,054,006 | 18,913 | # Maximum Subarray variation
I have to solve a problem much like the maximum subarray problem. I have to find the largest subarray whose average is bigger than k. I thought the following trick. I can transform my array A[] of size n to a B[] where B[i] = A[i] - k. So now the average must be >0. But average greater than zero doesnt simply mean sum greater than zero? So I can directly apply Kadane's algorithm. Am I right? (always under the constraint that there is 1 positive value)
-
no, kadane's algorithm will still find you the subarray with the biggest sum...i have to solve the same problem. so far i kave find that if we create the array B as you mentioned above and then make the array C which contains the partial sums of the array B,then the maximum interval (i,j) that we are lookink for has the same number for i and j!!! for example:
array A is: 1 10 -1 -1 4 -1 7 2 8 1 .....and the given k is 5 then array B is: -4 5 -6 -6 -1 -6 2 -3 3 -4 array C is:-4 1 -5 -11 -12 -18 -16 -19 -16 -20 so the subarray that we are looking for is [7,2,8], has length 3, and has the same first and last element which is -16!!!!
edit: i forgot to tell that we are searching for a O(n) or an O(n*logn) algorithm.... @lets_solve_it you are right but your algorithm is O(n^2) whitch is way to big for the data we want to handle. i 'm close to solve it with the function map in c++,whitch is something like a hash table. i thing this is the right diredtion because here the elements of the array C have direct relation with their indexes! Also our professor told us that another possible solution ,is to make again the array C and then take a (special?) pivot to do quicksort....but i don't totally understand what we expect from quicksort to do.
-
@panos7:
after you have created array C (partial sums array), you seek two values of C, Ci and Cj,
such that, Cj>=Ci, and, (j-i) is as "big" as possible. (j-i) --> MAX.
then return j-i.
in your example, -16>=-18 so you returned j-i=9-6=3 | 546 | 1,989 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-07 | latest | en | 0.900461 |
https://primer-computational-mathematics.github.io/book/d_geosciences/Physical%20Processes/Dimensional_Analysis.html | 1,708,813,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474569.64/warc/CC-MAIN-20240224212113-20240225002113-00482.warc.gz | 464,313,158 | 16,799 | # Dimensional Analysis#
Physical Processes
We will find that there are many important physical problems that we would like to understand, but where there are practical, financial, computational or scientific impediments that prevent us from solving the problem exactly. In such cases, we will need often to be able to provide an order of magnitude solution, and in many cases this will either prove to be adequate or it will help us towards finding a more accurate solution.
Dimensional analysis provides one method of solving physical problems approximately.
## Units vs dimensions#
Dimensions refer to types of physical properties that can be measured whilst units are the way we have chosen to assign a value or quantity to the amount of that variable, e.g. ‘length’ is a dimension whilst ‘$$m$$’ is the SI unit we have chosen to use to quantify length (our American colleagues might have chosen units of ‘feet’).
Most physical quantities have associated units, for example a distance might be measured in metres, miles or light years. Although great care is needed when mixing units within a single equation, it is nonetheless meaningful to consider a length that is equal to a metre plus a foot, or to say that $$10$$ centimetres is equal to $$100$$ millimetres. In contrast, it is not physically meaningful to consider a quantity that is equal to a metre plus a kilogram, nor to say that $$10$$ centimetres is equal to $$100$$ grams.
Equations relating physical quantities must be dimensionally homogeneous. That is, the dimensions (but not necessarily the units) must be the same:
• on both sides of an equality
• on both sides of a plus or minus sign.
In an analogous way, the dimensions of the arguments must disappear for trigonometric, exponential and similar functions – it does not make sense to take the cosine of ten kilograms or raise four to the power of half a metre.
This simple idea often has far reaching consequences. In many cases, if we know the variables that are important in a physical system, then we can discover the form of the equations that relate them together using only information about their dimensions.
All physical quantities have dimensions and units derived from $$7$$ base or primary physical quantities. Table 1 lists those quantities, the SI units and symbols and the symbols used for the dimensions of those quantities.
$$\qquad$$Table 1: Base quantities, SI units and symbols for dimensions
## Specifying dimensions#
The dimensions of a physical quantity are conventionally written using square brackets. For example, if $$\rho$$ is density, $$m$$ is mass, and $$r$$ is the radius of a sphere, then:
$[\rho]=\frac{[m]}{[r]^{3}}$
This equation does not imply that $$\rho=m/r^{3}$$, only that their dimensions are the same.
Many common variables will have dimensions that are combinations of mass, length, and time. We give each of these the symbols $$M$$, $$L$$, and $$T$$. Thus we can write the dimensions of density = mass / volume as $$[ML^{-3}]$$, and of force = mass × acceleration as $$[MLT^{-2}]$$.
In thermal problems, variables can also have temperature as a dimension written as $$[\Theta]$$ For example, the dimensions of thermal conductivity are $$[M LT^{-3}\Theta^{-1}]$$.
In problems that involve electricity or magnetism, variables can also depend on electric charge $$[Q]$$ which, for historical reasons, is not base quantity, and instead is written in base quantities as current × time with dimensions $$[IT]$$
## Indicial equations#
An indicial equation is an equation in which the power(s) are unknown.
The pressure $$P$$ at a depth $$z$$ below the surface of a static body of water with a constant density $$\rho$$ is:
$P=\rho g z$
Dimensionally, this is:
$[P]=[\rho]\times[g]\times[z]$
$[ML^{-1}T^{-2}]=[ML^{-3}]\times[LT^{-2}]\times[L]=[ML^{-1}T^{-2}]$
Suppose that we did not know this, and simply assumed that pressure was some function of density, acceleration due to gravity and depth:
$P=f(\rho,g,z)$
Now, for this to be true, there must be some combination of $$\rho$$, $$g$$ and $$z$$ that has the same dimensions as $$P$$.
Since we can see that the dimensions of $$\rho$$, $$g$$ and $$z$$ are all independent (that is, we cannot combine any two of these to make a quantity that has the same dimensions as the third), then the only way to balance the dimensions is to assume:
$P=k\rho^{a}g^{b}z^{c}$
where $$a$$, $$b$$, $$c$$ and $$k$$ are constants yet to be found. Note that $$k$$ has no dimensions – it is just a number. Dimensionally we have:
$[ML^{-1}T^{-2}]=[ML^{-3}]^{a}\times[LT^{-2}]^{b}\times[L]^{c}$
For this to balance, we require the powers of $$M$$, $$L$$ and $$T$$ to be separately equal on both sides of the equation. Thus, although this looks like one equation, it is really three separate simultaneous equations:
For $$M$$, $$[M]=[M]^{a}$$, so: $$$a=1$$$
For $$L$$, $$[L^{-1}]=[L^{-3}]^{a}\times[L]^{b}\times[L]^{c}$$, so: $$$-1=-3a+b+c$$$
For $$T$$, $$[T^{-2}]=[T^{-2}]^{b}$$, so: $$$-2=-2b$$$
Solving these simultaneous equations, we get, $$a = 1$$, $$b = 1$$, $$c = 1$$, and deduce that:
$P=k \rho g z$
where $$k$$ is a constant that does not have any dimensions.
Dimensional analysis on its own cannot tell us anything about the value of the constant $$k$$: this depends upon our choice of units. The advantage of using the SI system of units is that this system has been designed so that for most equations $$k~1$$. That is, they may be equal for example to $$1/2$$ or $$\pi^{2}$$, but are unlikely to be equal to $$10^{7}$$. In many problems, including the one above, the value of the constant is $$1$$. There are, however, certain fundamental physical constants that are very different from $$1$$, even when using SI units. These constants are believed to be universal (literally) in nature and in time. Table 2 lists the constants you are most likely to encounter in your studies together with their values in SI and units.
$$\qquad$$Table 2: Universal physical constants, symbol and values in SI (with units).
You may find that you can solve simple problems without writing out the indicial equations explicitly and solving them, but in general try to solve the problem formally anyway because an intuitive approach will become increasingly difficult with more complicated problems.
## Dimensionless variables#
Ordinary numbers, such as $$2$$ or $$\pi$$, have no dimensions associated with them; they are dimensionless. Some physical variables are also dimensionless – for example strain which is a ratio of two lengths (or a ratio of two areas, volumes or angles).
Dimensionless numbers and dimensionless variables play an important role in aiding our understanding of physical problems and are important in designing scale models and in parameterising experiments.
The Buckingham pi theorem states that any physical problem that can be described by $$N$$ variables using $$n$$ dimensions (in other words $$n$$ types of physical quantities, not the number of spatial dimensions!) can also be described using only $$N – n$$ dimensionless variables.
## Dimensions of angles, vectors, differentials, integrals, functions#
• Angles have no dimensions but they do have units (in radians or degrees). Their dimensions are $$[1]$$ since an angle is just a means of describing a ratio between two lengths.
• The dimensions of a vector are the same as the dimensions of its magnitude.
• The dimensions of ordinary and partial differentials and integrals are obtained straightforwardly by ignoring the differentiation and integration, and using simply the dimensions of the variables in the equation. For example, the dimensions of $$dx/dt$$ are the same as the dimensions of $$x/t$$, and the dimensions of $$∫dx$$ are the same as the dimensions of $$x$$.
• The dimensions of the vector differential operators $$\nabla$$ (grad), $$\nabla\cdot$$ (div), and $$\nabla\times$$ (curl), are $$[L^{-1}]$$, and of $$\nabla^{2}$$ (the laplacian) are $$[L^{-2}]$$, since $$\nabla\equiv \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z} \right)$$.
• The dimensions of trigonometric, inverse trigonometric, exponential and logarithmic functions are all $$[1]$$.
• The arguments of trigonometric and exponential functions must be dimensionless.
• The arguments of logarithms can have dimensions. However, care is needed to ensure that the resulting equation is dimensionally sensible, e.g. the equation $$2 \log x − \log y = 1$$ can be rewritten as $$x^{2}/y=10$$, so that $$x^{2}$$ must have the same dimensions as $$y$$ if the original equation is to make sense.
Note that you can often use these properties to check that mathematical results, and intermediate steps, are sensible. If the objects in your equations have (or could be assigned) dimensions, then all the stages must make dimensional sense. This will not ensure you’re your results are right, but it will often be sufficient to indicate when they are wrong.
# import relevant modules
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
# create our own functions
def dont_show_scale(ax):
ax.set_yticklabels([])
ax.set_xticklabels([])
ax.set_xticks([])
ax.set_yticks([])
return ax
## Problem 2 – Centripetal force#
The tension in a string as a mass is twirled around in a horizontal circle, or the gravitational force that keeps a satellite in a circular orbit, is called the centripetal force. Use dimensional analysis to determine how this force $$F$$ is related to the mass of the object $$m$$, its speed in its orbit $$v$$, and the radius of that orbit $$r$$. That is, find $$a$$, $$b$$ and $$c$$ if:
$F=km^{a}v^{b}r^{c}$
You should recover a result that you may already know for the magnitude of centripetal force. Check that your final answer is indeed dimensionally homogeneous.
In this problem, the constant $$k$$ is equal to one if the orbit is circular.
Find $$a$$, $$b$$ and $$c$$ if $$F=km^{a}v^{b}r^{c}$$.
Dimensionally this is:
$[MLT^{-2}]=[M]^{a}\times[LT^{-1}]^{b}\times[L]^{c}$
For $$M$$, $$1 = a$$.
For $$L$$, $$1=b+c$$.
For $$T$$, $$-2=-b$$.
Solving gives: $$a = 1, b = 2, c = −1$$, so:
$F=k\frac{mv^{2}}{r}$
In this problem, $$k = 1$$, but we cannot determine that using dimensional analysis.
# plot the relations between F and its independent variables (m, v, r)
fig, (ax1, ax2, ax3) = plt.subplots(1, 3, figsize=(15, 5))
fig.tight_layout(w_pad=4)
x = np.linspace(0,4,100)
ax1.plot(x, x, 'b')
ax1.set_xlabel('$m$', fontsize=16)
ax1.set_ylabel('$F$', fontsize=16)
ax1.set_title('$F \propto m$', fontsize=16)
dont_show_scale(ax1)
ax2.plot(x, x**2, 'b')
ax2.set_xlabel('$v$', fontsize=16)
ax2.set_ylabel('$F$', fontsize=16)
ax2.set_title('$F \propto v^{2}$', fontsize=16)
dont_show_scale(ax2)
x = np.delete(x, 0) # remove 0 as 1/0 is infinity
ax3.plot(x, x**-1, 'b')
ax3.set_xlabel('$r$', fontsize=16)
ax3.set_ylabel('$F$', fontsize=16)
ax3.set_title('$F \propto 1/r$', fontsize=16)
dont_show_scale(ax3)
<AxesSubplot:title={'center':'$F \\propto 1/r$'}, xlabel='$r$', ylabel='$F$'>
## Problem 4 – Tsunami waves#
Tsunamis are produced by very long-period waves in the ocean that are initiated by a sudden change in the position of the sea surface, caused for example by a submarine earthquake or landslide. Gravity acts to restore the sea-surface to the horizontal, and the resulting movement of water propagates as a wave.
We might guess that the velocity $$v$$ of a tsunami wave depends upon the density of seawater $$\rho$$, the acceleration due to gravity $$g$$, and the depth of the ocean $$z$$. Use dimensional analysis to discover the relationship.
Do you discover anything special about the importance of density in this problem?
The Krakatoa explosion in Indonesia generated a tsunami that travelled to Cape Town in South Africa at an average velocity of $$200$$ $$m/s$$. Assuming that the unknown constant is equal to $$1$$ (a simple laboratory experiment shows that it is), use your result to calculate the average depth of the Indian ocean.
Use dimensional analysis to discover the relationship:
$v = \sqrt{gz}$
Density is not involved – this is the only variable that contains mass, so there is no way to include it.
Calculate the average depth of the Indian ocean: $$200 = \sqrt{9.8z}$$, so $$z \approx 4100$$ m.
# plot the relationship above
z = np.linspace(0,5000,200)
def tsunami_velocity(depth):
return (9.8*depth)**0.5
def ocean_depth(water_wave_velocity):
return (water_wave_velocity**2)/9.8
plt.figure(figsize=(8,6))
plt.plot(z, tsunami_velocity(z), 'b')
plt.plot(ocean_depth(200), 200, 'ro', label='Indian ocean')
plt.xlabel('z (m)')
plt.ylabel('v (m/s)')
plt.title('A plot of wave velocity vs depth in the ocean', fontsize=14)
plt.legend(loc='best', fontsize=10)
<matplotlib.legend.Legend at 0x1d9e7f4d520>
## References#
• Lecture note and practical for Lecture 1 of the Physical Processes module | 3,262 | 12,941 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-10 | latest | en | 0.932702 |
https://republicofsouthossetia.org/question/consider-the-differential-equation-2y-8y-18y-0-3-6-0-infinity-verify-that-the-given-functions-fo-14762482-67/ | 1,642,525,592,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300934.87/warc/CC-MAIN-20220118152809-20220118182809-00317.warc.gz | 543,684,654 | 13,386 | ## Consider the differential equation x2y” − 8xy’ + 18y = 0; x3, x6, (0, [infinity]). Verify that the given functions form a fundamental set o
Question
Consider the differential equation x2y” − 8xy’ + 18y = 0; x3, x6, (0, [infinity]). Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval. The functions satisfy the differential equation and are linearly independent since W(x3, x6) = Incorrect: Your answer is incorrect. ≠ 0 for 0 < x < [infinity].
in progress 0
2 weeks 2022-01-01T17:20:00+00:00 1 Answer 0 views 0
1. Answer: The equation is differentiated implcitly, with respect to x and y
Step-by-step explanation: x2y – 8xy + 18y = 0
By implicit differentiation: 2y + 2xdy/dx – (8y + 8dy/dx) – 18dy/dx = 0
2y – 8y – 8dy/dx – 18dy/dx + 2xdy/dx = 0
-6y + 2xdy/dx – 26dy/dx = 0
2xdy/dx – 26dy/dx = 6y
dy/dx(2x – 26) = 6y
∴ dy/dx = 6y/2(x – 13) = 3y/(x – 13) | 354 | 941 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2022-05 | latest | en | 0.810242 |
https://www.topperlearning.com/ncert-solutions/cbse-class-6-science/ncert-science-vi/water | 1,679,833,803,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945472.93/warc/CC-MAIN-20230326111045-20230326141045-00531.warc.gz | 1,124,953,264 | 56,512 | Request a call back
# Class 6 NCERT Solutions Science Chapter 14 - Water
## Water Exercise 145
### Solution 1
(a) Evaporation
(b) Condensation
(c) Drought
(d) Floods
Concept insight: Recall water cycle in nature.
### Solution 2
(a) Condensation
(b) Evaporation
(c) Condensation
(d) Evaporation
(e) Evaporation
Concept insight: Recall the terms used in water cycle.
### Solution 3
Only (c) and (e) are true.
Concept insight: Important from Exam Point of View.
## Water Exercise 146
### Solution 4
Yes, if we spread our school uniform near an anghithi or heater it will dry quickly as it will provide heat to water and it will convert into vapours quickly thus it will dry up quickly.
Concept insight: Recall the process of evaporation.
### Solution 5
A puddle of water is formed around a cool bottle because the water vapours near the cold bottle collide with it and lose their heat and condense.
Concept insight: Recall the process of condensation.
### Solution 6
During exhalation, carbon dioxide is released along with water vapours. If one breathes out, the released water vapours collide with the surface of the glass, thereby making it cooler. As a result, the water vapours present in the air surrounding the glass condense and get attached to the glass surface. Consequently, the glass becomes wet.
Concept insight: Recall the process of condensation.
### Solution 7
As air moves up in the atmosphere, it cools down at a sufficient height. The air becomes so cool that water vapour present in it condense to form small water droplets which remain suspended in the atmosphere. These suspended water droplets appear as clouds.
Concept insight: Recall the formation of clouds
### Solution 8
Drought occurs when an area does not receive rainfall for a period of year or more. Due to this, soil continues to lose water by evaporation and transpiration and becomes dry.
Concept insight: What happens if it does not rain for a long period? | 458 | 1,988 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | longest | en | 0.878013 |
http://www.matlabtips.com/author/admin/ | 1,580,000,398,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251681625.83/warc/CC-MAIN-20200125222506-20200126012506-00437.warc.gz | 250,806,666 | 12,411 | ## Algorithmic Trading in MATLAB : WFAToolbox
Today, I would like to welcome Alex from WFAToolbox who will present their latest trading tool in Matlab. Let us introduce you WFAToolbox – MATLAB App that allows you to develop algorithmic trading strategies in minutes, not months. WFAToolbox was … Continue reading
## Rounding errors
I would like today to talk about one very important concept that is often overlooked when you learn to use a computer for data analysis : Rounding errors. In Matlab as in other languages, numbers are represented as a series … Continue reading
Posted in Beginners | 2 Comments
## Happy new 2015
It is an established tradition that blogs will look back on new year and discuss past achievements. I also felt it was a good opportunity to share with you some results from this website statistics and suggest what the next … Continue reading
Posted in Various | 1 Comment
## Debugging in Matlab
You’ve just finished writing your latest and greatest Matlab program. You’re sure this program will solve all the world’s (or at the very least all of your) problems. Just as you run your program, you see the dreaded red error … Continue reading
Posted in Beginners | 4 Comments
## Logical operations and logical indexing in Matlab
In this post, we talk about logical operations but also how to use your Matlab skills to analyze your shopping expenses (Yes, we are all absolute nerds and love it). You will learn how to use logical operations to search … Continue reading
## How and when to convert between data types
In this post, we talk about data types and extend on our previous post on the matter. You will learn how to convert between them but more importantly when and why you should consider such things. As often, we delve … Continue reading
Posted in Beginners | 3 Comments
## ICA demystified
This post has been on the back of my mind for quite some time, ever since I wrote about Principal Component Analysis. Independent Component Analysis or ICA is an algorithm to extract underlying structure hidden in multi-dimensional datasets. As for … Continue reading
Posted in Intermediate | 1 Comment
## Stucked? How to get some help.
While my goal here is to help you out with Matlab and other data analysis endeavors, I can’t cover the vast number of problem you may encounter. Still, I can provide you with a general approach that should help you … Continue reading
Posted in Beginners | 1 Comment
## Mathematical operations in Matlab and the point
The point of this post is to talk about doing mathematics in Matlab. This is an extension of our previous post on Kickstarting Matlab. You will learn how to use all mathematical operators and get to understand your Matlab second best … Continue reading | 569 | 2,756 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-05 | latest | en | 0.949653 |
https://www.queryhome.com/puzzle/7219/how-much-money-did-he-start-with | 1,542,088,136,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741219.9/warc/CC-MAIN-20181113041552-20181113063552-00426.warc.gz | 979,164,442 | 32,356 | 169 views
Jayant went to the hardware store to buy some needed supplies. He spent half of what he had plus \$2.00 in the first store; half of what he had left plus \$1.00 in the second store; half of what he had left plus \$1.00 in the third store; and in the last store half of all he had. Three dollars were left over.
posted May 4, 2015
I think he has started with 64
On first store=64/2 - 2=32-2=30
On second store = 30/2 - 1 =15-1= 14
On third store= 14/2-1=7-1=6
On last store= 6/2=3
So finally he remains with 3 dollars
64 is the correct answer but algebraic reach is as under
he spends x/2 +2 in the first store and is left with x/2-2 dollars
he spends half of this +1 means x/4-1+1 in the second store and is left with x/4-2
he spends half of this +1 means x/8-1+1 in the third store and is left with x/8-2
he spends half of this means x/16-1 in the last store and is left with x/16-1 which is 3 dollars giving x=64
Let us check
out of 64 spent in first store 34 left with 30
out of 30 spent in second store 16 left with 14
out of 14 spent in third store 8 left with 6
spends 3 in the last store and is left with 3
Similar Puzzles
+1 vote
Mark is throwing his son Stewart a surprise birthday party but he has limited funds.
He spent half of his money plus \$2.00 on the cake.
Half of what was left plus \$2.00 was spent on balloons and decorations.
Then he spent half of what he had left plus \$1.00 on candy.
Now he is out of money, how much did Mark start with?
In a game called flip it or dip it, three people flip coins, and the odd person out has to double what others already have. After three games,each person has lost once and has thirty six pounds. | 496 | 1,679 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | longest | en | 0.986062 |
http://www.mathisfunforum.com/viewtopic.php?pid=224568 | 1,397,937,943,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00016-ip-10-147-4-33.ec2.internal.warc.gz | 545,773,514 | 7,024 | Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #51 2012-06-26 21:04:32
ganesh
Moderator
Offline
### Re: Objective Type
Hi anonimnystefy,
The answers #34, #35, and #36 are correct. Excellent!
37. Given r = 7 cm and l = 10 cm, the curved surface area of the cone is __________________
(a)
(b)
(c)
(d)
38. The Greatest Common Divisor of
is _________________
(a)
(b)
(c)
(d)
Character is who you are when no one is looking.
## #52 2012-06-26 21:13:07
anonimnystefy
Real Member
Offline
### Re: Objective Type
Hi ganesh
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #53 2012-06-26 23:45:34
bobbym
Online
### Re: Objective Type
Hi ganesh;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #54 2012-06-27 20:18:32
ganesh
Moderator
Offline
### Re: Objective Type
Hi anonimnystefy and bobbym,
The answers 37 and 38 are a and b respectively.
Good work, anonimnystefy!
Well done, bobbym!
39. The remainder when x3 + x2 - 3x + 5 is divided by (x - 1) is ___________________
(a) 8
(b) 4
(c) -8
(d) -4
40. A point on 3x + 2y ≤ 10 is ___________________
(a) (1,1)
(b) (2,3)
(c) (3,2)
(d) (4,0)
Character is who you are when no one is looking.
## #55 2012-06-27 23:24:48
bobbym
Online
### Re: Objective Type
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #56 2012-06-27 23:34:53
anonimnystefy
Real Member
Offline
### Re: Objective Type
Hi ganesh
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #57 2012-06-28 22:47:29
ganesh
Moderator
Offline
### Re: Objective Type
Hi bobbym and anonimnystefy,
The answers 39 and 40 are correct. Well done!
41. The point of intersection of the lines x + 2y = 3 and y = 0 is
(a) (3,0)
(b) (3,1)
(c) (0,3)
(d) (0,0)
42. Given
,
= _________________
(a) 3/4
(b) 3
(c) 4
(d) 4/3
Character is who you are when no one is looking.
## #58 2012-06-28 23:03:31
anonimnystefy
Real Member
Offline
### Re: Objective Type
Hi ganesh
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #59 2012-06-28 23:29:18
bobbym
Online
### Re: Objective Type
Hi ganesh;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #60 2012-06-29 15:46:01
ganesh
Moderator
Offline
### Re: Objective Type
Hi anonimnystefy and bobbym,
The answers #41 and #42 are correct. Good work!
43. Slope of the line 2y = 3x + 5 is
(a) 2/3
(b) 3/2
(c) -2/3
(d) -3/2
44. The probability of an impossible event is _________________
(a) 0
(b) 1
(c) 2
(d) not defined
Character is who you are when no one is looking.
## #61 2012-06-29 18:29:13
bobbym
Online
### Re: Objective Type
Hi ganesh;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #62 2012-06-29 23:23:35
anonimnystefy
Real Member
Offline
### Re: Objective Type
Hi ganesh
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #63 2012-06-30 22:10:21
ganesh
Moderator
Offline
### Re: Objective Type
Hi bobbym and anonimnystefy,
The answers 43 and 44 are correct. Well done!
45. The radius of a circle is 5 cm and length of chord is 8 cm. The perpendicular distance of the chord from the center is _________.
(a) 3 cm
(b)
cm
(c)
cm
(d) None of these
46. In a cyclic quadrilateral ABCD, if angle A = 80o, then angle C = _______________
(a) 80°
(b) 100°
(c) 40°
(d) 160°
Character is who you are when no one is looking.
## #64 2012-06-30 23:21:00
bobbym
Online
### Re: Objective Type
Hi ganesh;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #65 2012-07-01 21:47:56
ganesh
Moderator
Offline
### Re: Objective Type
Hi bobbym,
The answers 45 and 46 are correct. Well done!
47.
= ______________
(a)
(b)
(c)
(d)
48. The variance of 5 values is 9 when each number is doubled. The new variance is _________________
(a) 9
(b) 18
(c) 45
(d) 36
Character is who you are when no one is looking.
## #66 2012-07-01 23:01:06
bobbym
Online
### Re: Objective Type
Hi ganesh;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #67 2012-07-02 00:20:28
anonimnystefy
Real Member
Offline
### Re: Objective Type
Hi ganesh
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #68 2012-07-02 22:31:30
ganesh
Moderator
Offline
### Re: Objective Type
Hi bobbym and anonimnystefy,
and
.
Excellent, bobbym! Both correct!
Well done, anonimnystefy!
49. In a Geometric Progression, given a = 1 and r = 2, t5 = ________________
(a) 16
(b) 32
(c) 10
(d) 2
50. The total surface area of a sphere of radius 10 cm is ______________
(a)
cm2
(b)
cm2
(c) 400 cm2
(d) 200 cm2
Character is who you are when no one is looking.
## #69 2012-07-02 22:48:06
bobbym
Online
### Re: Objective Type
Hi ganesh;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #70 2012-07-02 23:06:53
anonimnystefy
Real Member
Offline
### Re: Objective Type
Hi ganesh
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #71 2012-07-03 17:18:18
ganesh
Moderator
Offline
### Re: Objective Type
Hi bobbym and anonimnystefy,
The answers 49 and 50 are correct. Good work!
51. The Least Common Multiple of 15a6, 75a5 is ______________________
(a) 15a6
(b) 75a5
(c) 75a6
(d) None of these
52. The maximum value of the function Z = 4x + 7y is at ___________________
(a) (0,1)
(b) (2,3)
(c) (3,0)
{d) (1,3)
Character is who you are when no one is looking.
## #72 2012-07-03 17:27:33
bobbym
Online
### Re: Objective Type
Hi ganesh;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
## #73 2012-07-04 04:17:52
anonimnystefy
Real Member
Offline
### Re: Objective Type
Hi ganesh
Last edited by anonimnystefy (2012-07-04 04:18:37)
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
## #74 2012-07-04 22:19:36
ganesh
Moderator
Offline
### Re: Objective Type
Hi bobbym and anonimnystefy,
The answers 51 and 52 are correct. Well done!
53. The point of intersection of the line 2x + 5y = 10 with the y-axis is ____________________
(a) (0,0)
(b) (5,0)
(c) (0,2)
(d) (1,2)
54. The equation of a line parallel to y-axis is _______________
(a) x = 0
(b) y = 0
(c) x = c
(d) y = c
Character is who you are when no one is looking.
## #75 2012-07-04 22:32:51
bobbym
Online
### Re: Objective Type
Hi;
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | 2,956 | 9,415 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2014-15 | longest | en | 0.847758 |
https://mathematicalworksheets.com/skip-counting-by-2-worksheet/ | 1,679,630,966,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945242.64/warc/CC-MAIN-20230324020038-20230324050038-00103.warc.gz | 427,238,089 | 12,101 | # Skip counting by 2 worksheet
A technique for practicing adding numbers by 2 is to skip count by 2. To acquire the following number in the series, we skip counting by 2 by adding 2 to the previous number. This worksheet will help you practice how to skip counting by two.
## What is a “skip counting by 2 worksheet”?
“Skip counting by 2 worksheet” is a math practice sheet that incorporates application, reflection, assessment, and problem-solving with a challenge to aid students in creating and resolving their word problems. This activity will assist students in understanding the lecture, applying new knowledge, and utilizing prior knowledge.
This worksheet will help you better understand the concept behind skip counting by two and how to identify the following number.
## Instructions on how to use the “skip counting by 2 worksheet.”
Use this math worksheet to learn how to skip counting by 2. We provide a 15-item activity to help you practice the topic.
.
A reflection section is included at the end of this worksheet to encourage learners to evaluate their thoughts (metacognition) and how they did during the lesson.
### Conclusion
Skip counting by 2 is simple as long as you know how to compute them and are familiar with the process. | 257 | 1,260 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2023-14 | longest | en | 0.964538 |
http://nrich.maths.org/public/leg.php?code=32&cl=2&cldcmpid=7544 | 1,503,476,304,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886117911.49/warc/CC-MAIN-20170823074634-20170823094634-00254.warc.gz | 312,352,815 | 10,072 | # Search by Topic
#### Resources tagged with Multiplication & division similar to Tiles in the Garden:
Filter by: Content type:
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### There are 165 results
Broad Topics > Calculations and Numerical Methods > Multiplication & division
### Function Machines
##### Stage: 2 Challenge Level:
If the numbers 5, 7 and 4 go into this function machine, what numbers will come out?
### Sending Cards
##### Stage: 2 Challenge Level:
This challenge asks you to investigate the total number of cards that would be sent if four children send one to all three others. How many would be sent if there were five children? Six?
### Abundant Numbers
##### Stage: 2 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Carrying Cards
##### Stage: 2 Challenge Level:
These sixteen children are standing in four lines of four, one behind the other. They are each holding a card with a number on it. Can you work out the missing numbers?
### Next Number
##### Stage: 2 Short Challenge Level:
Find the next number in this pattern: 3, 7, 19, 55 ...
### Exploring Number Patterns You Make
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers?
##### Stage: 2 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### It Was 2010!
##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Exploring Wild & Wonderful Number Patterns
##### Stage: 2 Challenge Level:
EWWNP means Exploring Wild and Wonderful Number Patterns Created by Yourself! Investigate what happens if we create number patterns using some simple rules.
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Code Breaker
##### Stage: 2 Challenge Level:
This problem is based on a code using two different prime numbers less than 10. You'll need to multiply them together and shift the alphabet forwards by the result. Can you decipher the code?
### Month Mania
##### Stage: 1 and 2 Challenge Level:
Can you design a new shape for the twenty-eight squares and arrange the numbers in a logical way? What patterns do you notice?
### A Mixed-up Clock
##### Stage: 2 Challenge Level:
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from these ten statements?
### Amy's Dominoes
##### Stage: 2 Challenge Level:
Amy has a box containing domino pieces but she does not think it is a complete set. She has 24 dominoes in her box and there are 125 spots on them altogether. Which of her domino pieces are missing?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
##### Stage: 2 Challenge Level:
Use the information to work out how many gifts there are in each pile.
### Being Thoughtful - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that require careful consideration.
### Clock Face
##### Stage: 2 Challenge Level:
Where can you draw a line on a clock face so that the numbers on both sides have the same total?
### Route Product
##### Stage: 2 Challenge Level:
Find the product of the numbers on the routes from A to B. Which route has the smallest product? Which the largest?
### The Deca Tree
##### Stage: 2 Challenge Level:
Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
### Oranges and Lemons
##### Stage: 2 Challenge Level:
On the table there is a pile of oranges and lemons that weighs exactly one kilogram. Using the information, can you work out how many lemons there are?
### Throw a 100
##### Stage: 2 Challenge Level:
Can you score 100 by throwing rings on this board? Is there more than way to do it?
### Rocco's Race
##### Stage: 2 Short Challenge Level:
Rocco ran in a 200 m race for his class. Use the information to find out how many runners there were in the race and what Rocco's finishing position was.
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### Back to School
##### Stage: 2 Challenge Level:
Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong?
### Today's Date - 01/06/2009
##### Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### Four Goodness Sake
##### Stage: 2 Challenge Level:
Use 4 four times with simple operations so that you get the answer 12. Can you make 15, 16 and 17 too?
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Mystery Matrix
##### Stage: 2 Challenge Level:
Can you fill in this table square? The numbers 2 -12 were used to generate it with just one number used twice.
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Being Determined - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level that may require determination.
### All the Digits
##### Stage: 2 Challenge Level:
This multiplication uses each of the digits 0 - 9 once and once only. Using the information given, can you replace the stars in the calculation with figures?
### The Money Maze
##### Stage: 2 Challenge Level:
Go through the maze, collecting and losing your money as you go. Which route gives you the highest return? And the lowest?
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Napier's Bones
##### Stage: 2 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### Penta Post
##### Stage: 2 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
### Number Tracks
##### Stage: 2 Challenge Level:
Ben’s class were cutting up number tracks. First they cut them into twos and added up the numbers on each piece. What patterns could they see?
### Asteroid Blast
##### Stage: 2 Challenge Level:
A game for 2 people. Use your skills of addition, subtraction, multiplication and division to blast the asteroids.
### Multiplication Squares
##### Stage: 2 Challenge Level:
Can you work out the arrangement of the digits in the square so that the given products are correct? The numbers 1 - 9 may be used once and once only.
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
### Machines
##### Stage: 2 Challenge Level:
What is happening at each box in these machines?
### Cows and Sheep
##### Stage: 2 Challenge Level:
Use your logical reasoning to work out how many cows and how many sheep there are in each field.
### Sept03 Sept03 Sept03
##### Stage: 2 Challenge Level:
This number has 903 digits. What is the sum of all 903 digits?
### Forgot the Numbers
##### Stage: 2 Challenge Level:
On my calculator I divided one whole number by another whole number and got the answer 3.125. If the numbers are both under 50, what are they?
### Make 100
##### Stage: 2 Challenge Level:
Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Book Codes
##### Stage: 2 Challenge Level:
Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?
### Mobile Numbers
##### Stage: 1 and 2 Challenge Level:
In this investigation, you are challenged to make mobile phone numbers which are easy to remember. What happens if you make a sequence adding 2 each time? | 2,214 | 9,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2017-34 | latest | en | 0.885504 |
https://web2.0calc.com/questions/base-3-question | 1,591,217,297,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347435987.85/warc/CC-MAIN-20200603175139-20200603205139-00539.warc.gz | 587,420,467 | 6,792 | +0
Base 3 question
0
100
5
+287
How many zeroes do we write when we write all the integers from 1 to 243 in base 3?
My incorrect work: 1-242 inclusive in base 3 would be 1-22222_3. The possible zeroes will be 216. Then, account for 243, then add another 5 zeros to get 221 as my final answer. Please give a concise solution on how to do this problem. Thank you :D
Mar 24, 2020
#1
+1968
+1
Pattern:1,2,10,11,12,20,21,22,100,…
Last number in the sequence: 3^5, or 100,000 in base 3.
In this sequence, every third number has a zero, and every 9th number has 2 zeros, and so on.
If we add all of these together, we will get: 81+27+9+3+1=121 zeros.
Hope this helped!
Mar 24, 2020
#2
+483
+1
:/
https://www.quora.com/How-many-zeroes-do-we-write-when-we-write-all-the-integers-from-1-to-243-in-base-3
@EpicWater
This is a reference website, has the same (albeit more detailed) solution that CGT outlined for us
jfan17 Mar 24, 2020
edited by jfan17 Mar 24, 2020
edited by jfan17 Mar 24, 2020
#4
0
int = (1 - 100000 in base 3]. There are, as you can see at the bottom, 289 zeros.
a = 0; b=0; c=0; d=0; e=0; f=0; g=0; h=0; i=0; k=0 for n in (int): for j in str(n): if j == "0": a +=1 if j == "1": b +=1 if j == "2": c +=1 # Code to count number of digits between ANY two ranges and their total value # if j == "3": d +=1 if j == "4": e +=1 if j == "5": f +=1 if j == "6": g +=1 if j == "7": h +=1 if j == "8": i += 1 if j == "9": k +=1 print([a, b, c, d, e, f, g, h, i, k],"Total number=",(a+b+c+d+e+f+g+h+i+k)) print("Sum Value =", b*1 + c*2 + d*3 +e*4 + f*5 + g*6 + h*7 + i*8 + k*9)
[289, 406, 405] - These are counts of all 3 digits [0, 1, 2]
Guest Mar 24, 2020
#3
+287
+1
thx for the help and support :D
Mar 24, 2020 | 686 | 1,737 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-24 | latest | en | 0.760849 |
https://netsuiteprofessionals.com/blog/question/formula-reference/ | 1,708,575,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00842.warc.gz | 457,266,113 | 25,529 | # Formula Reference
I have a sublist called Expense and another called Item. Both contain fields called Amount. I’m trying to implement another field and insert a formula referencing Amount from the Expense sublist because when I do {amount} / {quantity} it does not work at all so I believe due to conflict. How do I reference the one from Expense thanks
Beginner Asked on October 17, 2019 in
What transaction are you working with? It sounds like you are working with a vendor bill and the expense sublist of a vendor bill does not have a quantity column.
on October 17, 2019.
Purchase Order
on October 21, 2019.
On the purchase order form, there is the Amount in the Item subtab and Expense subtab. I created the Quantity field and PPU field and I want the PPU field to return the price per unit by dividing the {amount} / {quantity}
on October 21, 2019.
In general, the quantity column in not available at expense sub-list. You should use a formula (numeric) field with this formula – for purchase order transaction. case when {item} is NULL then {amount} else round((nvl({quantity},0)-nvl({quantitybilled},0))*nvl({rate},{amount}),2) end This formula is a good reference to start with.
Beginner Answered on October 18, 2019.
On the purchase order form, there is the Amount in the Item subtab and Expense subtab. I created the Quantity field and PPU field and I want the PPU field to return the price per unit by dividing the {amount} / {quantity}
on October 21, 2019.
OK, regard the items subtab it should be fine just make sure to use this formula {amount}/NVL({quantity},1) to avoid dividing with zero. Regards,the expense subtab you don’t have a quantity column to calculate your PPU. Would you like that the PPU will be calculate only for the items subtab?
on October 22, 2019.
Based on the Expense subtab. Because both of them have the amount field. Thanks
on October 22, 2019.
As I said before your formula will not work properly at the expense subtab because you don’t have a value there for the quantity. Hence the PPU field won’t work there.
on October 22, 2019.
Formula reference didn’t work for me at all so I created a script and it worked
Beginner Answered on October 28, 2019. | 539 | 2,214 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2024-10 | latest | en | 0.911435 |
https://electronics.stackexchange.com/questions/12686/how-to-discharge-smoothing-capacitors | 1,653,508,863,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00308.warc.gz | 291,687,468 | 69,797 | # How to discharge smoothing capacitors?
I have a simple 12V 10 A power supply with just a transformer and a rectifier. After doing some research and simulations, I've added 3 10 mF capacitors in parallel to smooth out the output.
My problem is that after turning the supply off, capacitors remain charged for quite some time. I can get small sparks after shorting the output even 5 minutes after turning the supply off. Right now I only have a single LED connected to the capacitors and it takes more than 10 minutes for it to turn off completely after powering the supply down and the capacitors still aren't fully discharged when it turns off.
The most obvious way to solve the problem would be to put a resistor and a switch on the output and connect the resistor to the capacitors after turning off the supply by hand, but I'm hoping to get something a bit smarter and a bit safer.
Another point is that I want to use the supply's original case which has very little free volume now that I've added the capacitors, so just putting a ceramic 11 W resistor could be a problem because there would be very little free space around it for safe cooling.
• 3 mF or 3 µF? Apr 6, 2011 at 19:14
• @endolith 30 mF or 30000 µF . Apr 6, 2011 at 19:22
Appropriate bleeder resistors are the usual solution. They aren't usually switched, although they can be.
The value depends on the time you require to discharge the capacitors. The formula is
$$V_{t} = V_{0} \, e^{ -t / RC }$$
where $V_{t}$ is the voltage at time t and $V_{0}$ is the initial voltage at time 0. It's an exponential function, so I'd just assume 1/10 of the initial voltage.
It isn't a power function, as someone edited it!
You should find that the power taken by the bleeder resistors is negligible compared to the 120W capability of the supply.
• How would I determine correct value? Apr 6, 2011 at 18:07
• I used 1$k \Omega$ resistor and it's not on fire yet, so I guess this works. Apr 6, 2011 at 20:40
• 30mF and 1k ohm at 12 V would mean it takes ~69 seconds to discharge to 1.2 volts. And, of course, the leakage current is ~12 mA. stevenvh's MOSFET idea hit me as well and seems reasonable to me, though I'm sure that (unlike me) he's actually seen it used IRL :-) Jun 19, 2012 at 11:51
What you want is a switch which is open when the circuit is powered, and closed when it is switched off. When closed it should discharge the capacitor over a resistor. You don't want to short the capacitor; they don't like that. Two approaches I can think of (from the top of my head):
1. Use a depletion MOSFET as the switch. Depletion MOSFETs conduct when there's no voltage applied to the gate. Apply a voltage to switch it off. This voltage can not be derived from the capacitor you want to discharge! Otherwise the MOSFET would never be switched off. (You think about this, if you don't get it tell me, and I'll try to explain.)
2. Use an ordinary NPN transitor which you drive from the capacitor's voltage. As long as there's a voltage present, it will discharge. Pull the transistor's base to ground if the circuit is switched on. Again, the voltage to do this is from a separate power supply.
• hi Steven, I have a 3KW strobe light LED array that has 81mF capacitance so that it can sustain 120Amp discharge for 5ms without dropping more than 4-5V on the rail. This all works nicely, but when I turn off the system power switch just like the OP of this question, the capacitor banks stay charged. I have actually already tested in my design the use of 4 infineon BSS159N depletion mode N channel MOSFETS, quickly shown here dropbox.com/s/1jv57olhhryco12/example.JPG . My circuit doesnt quite work though, as my 5V pull up (to turn them 'off' on power up) is powered by the 24v rail.. Jun 26, 2014 at 7:13
• So my question is, what could I do to convert my circuit so that I can somehow separate my power supply (which is 24V input, goes to cap banks and into a 5V DCDC module, which gives me my 5V rail..). As the entire system input is turned off, it takes quite a while (only a few LEDs) to discharge, as my MOSFETs manage to keep their voltage at their pins because I unfortunately forgot that my 5V supply is fed by the 24V and will not drop-out until the 24V rail dips to ~5.2V. One option i have is my controller gets Power over Ethernet which is independant of the 24V input! perhaps I could connect Jun 26, 2014 at 7:15
• something to the external power switch which indicates to my PoE powered controller to pull the MOSFET gates to 0V? Jun 26, 2014 at 7:16
Such huge caps seems to be an overkill... If it's regulated (linear/pulse) you would need to tune it till ripple would be acceptable with much less output capacitor. If you have alot of high-freq noise - you would need to add several ceramic caps. Also, make sure that your inductor at the output is calculated correctly.
• There's no regulation and there's nothing to tune inside. The circuit is basically what I showed here plus a circuit beaker and LED on the output and fuse and a power switch on the input. I've had around 8 V of ripple on the output when the supply is unloaded, now I have around 0.1 V. +1 for mentioning high frequency noise. I totally forgot to measure the actual frequencies. Also, where would I add the inductor? Apr 7, 2011 at 9:40
• You just add DCDC converter, which will guarantee 0.01V ripple with 100uF output cap and will be much much better that this. They does not cost much. Apr 7, 2011 at 12:27
• That's my plan. I'll eventually have one at the output and solve the problems. I was concerned that horrible rectifier and transformer side will cause problems to the regulator, so I added some extra capacitors. Apr 7, 2011 at 12:31 | 1,471 | 5,715 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2022-21 | longest | en | 0.949124 |
https://wikivisually.com/wiki/1%2C024_%28number%29 | 1,519,340,902,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814292.75/warc/CC-MAIN-20180222220118-20180223000118-00186.warc.gz | 772,396,087 | 39,512 | # 1024 (number)
(Redirected from 1,024 (number))
← 1023 1024 1025 →
Cardinal one thousand twenty-four
Ordinal 1024th
(one thousand twenty-fourth)
Factorization 210
Divisors 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
Greek numeral ,ΑΚΔ´
Roman numeral MXXIV
Binary 100000000002
Ternary 11012213
Quaternary 1000004
Quinary 130445
Senary 44246
Octal 20008
Duodecimal 71412
Vigesimal 2B420
Base 36 SG36
1024 is the natural number following 1023 and preceding 1025.
1024 is a power of two: ${\displaystyle 2^{10}}$ (2 to the 10th power).[1] It is the lowest power of two requiring four decimal digits, and the lowest power of two containing the digit 0 in its decimal representation (excluding any leading zeroes).
It is also the square of 32: ${\displaystyle 32^{2}}$.
1024 is the smallest number with exactly 11 divisors (but note that there are smaller numbers with more than 11 divisors; e.g., 60 has 12 divisors) (sequence A005179 in the OEIS).
## Approximation to 1000
The neat coincidence that 210 is nearly equal to 103 provides the basis of a technique of estimating larger powers of 2 in decimal notation. Using 210a+b ≈ 2b103a is fairly accurate for exponents up to about 100. For exponents up to 300, 3a continues to be a good estimate of the number of digits.
For example, 253 ≈ 8×1015. The actual value is closer to 9×1015.
In the case of larger exponents the relationship becomes increasingly more inaccurate with errors exceeding an order of magnitude for ${\displaystyle a\geq 97}$, for example:
{\displaystyle {\begin{aligned}{\frac {2^{1000}}{10^{300}}}&=\exp \left(\ln \left({\frac {2^{1000}}{10^{300}}}\right)\right)\\&=\exp \left(\ln \left(2^{1000}\right)-\ln \left(10^{300}\right)\right)\\&\approx \exp \left(693.147-690.776\right)\\&\approx \exp \left(2.372\right)\\&\approx 10.72\end{aligned}}}
In measuring bytes, 1024 is often used in place of 1000 as the quotients of the units byte, kilobyte, megabyte, etc. In 1999, the IEC coined the term kibibyte for multiples of 1024, with kilobyte being used for multiples of 1000.
## Special use in computers
In binary notation, 1024 is represented as 10000000000, making it a simple round number occurring frequently in computer applications.
1024 is the maximum number of computer memory addresses that can be referenced with ten binary switches. This is the origin of the organization of computer memory into 1024-byte chunks or kibibytes.
In the Rich Text Format (RTF), language code 1024 indicates the text is not in any language and should be skipped over when proofing. Most used languages codes in RTF are integers slightly over 1024.
1024×768 pixels and 1280×1024 pixels are common standards of display resolution. | 795 | 2,704 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-09 | latest | en | 0.767991 |
https://number.academy/52451 | 1,660,806,386,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573172.64/warc/CC-MAIN-20220818063910-20220818093910-00731.warc.gz | 396,889,974 | 12,044 | # Number 52451
Number 52,451 spell 🔊, write in words: fifty-two thousand, four hundred and fifty-one . Ordinal number 52451th is said 🔊 and write: fifty-two thousand, four hundred and fifty-first. The meaning of number 52451 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 52451. What is 52451 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 52451.
## What is 52,451 in other units
The decimal (Arabic) number 52451 converted to a Roman number is (L)MMCDLI. Roman and decimal number conversions.
#### Weight conversion
52451 kilograms (kg) = 115633.5 pounds (lbs)
52451 pounds (lbs) = 23791.6 kilograms (kg)
#### Length conversion
52451 kilometers (km) equals to 32592 miles (mi).
52451 miles (mi) equals to 84412 kilometers (km).
52451 meters (m) equals to 172082 feet (ft).
52451 feet (ft) equals 15988 meters (m).
52451 centimeters (cm) equals to 20650 inches (in).
52451 inches (in) equals to 133225.5 centimeters (cm).
#### Temperature conversion
52451° Fahrenheit (°F) equals to 29121.7° Celsius (°C)
52451° Celsius (°C) equals to 94443.8° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
52451 seconds equals to 14 hours, 34 minutes, 11 seconds
52451 minutes equals to 1 month, 1 week, 1 day, 10 hours, 11 minutes
### Codes and images of the number 52451
Number 52451 morse code: ..... ..--- ....- ..... .----
Sign language for number 52451:
Number 52451 in braille:
Images of the number
Image (1) of the numberImage (2) of the number
More images, other sizes, codes and colors ...
## Mathematics of no. 52451
### Multiplications
#### Multiplication table of 52451
52451 multiplied by two equals 104902 (52451 x 2 = 104902).
52451 multiplied by three equals 157353 (52451 x 3 = 157353).
52451 multiplied by four equals 209804 (52451 x 4 = 209804).
52451 multiplied by five equals 262255 (52451 x 5 = 262255).
52451 multiplied by six equals 314706 (52451 x 6 = 314706).
52451 multiplied by seven equals 367157 (52451 x 7 = 367157).
52451 multiplied by eight equals 419608 (52451 x 8 = 419608).
52451 multiplied by nine equals 472059 (52451 x 9 = 472059).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 52451
Half of 52451 is 26225,5 (52451 / 2 = 26225,5 = 26225 1/2).
One third of 52451 is 17483,6667 (52451 / 3 = 17483,6667 = 17483 2/3).
One quarter of 52451 is 13112,75 (52451 / 4 = 13112,75 = 13112 3/4).
One fifth of 52451 is 10490,2 (52451 / 5 = 10490,2 = 10490 1/5).
One sixth of 52451 is 8741,8333 (52451 / 6 = 8741,8333 = 8741 5/6).
One seventh of 52451 is 7493 (52451 / 7 = 7493).
One eighth of 52451 is 6556,375 (52451 / 8 = 6556,375 = 6556 3/8).
One ninth of 52451 is 5827,8889 (52451 / 9 = 5827,8889 = 5827 8/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
52451
#### Is Prime?
The number 52451 is not a prime number. The closest prime numbers are 52433, 52453.
#### Factorization and factors (dividers)
The prime factors of 52451 are 7 * 59 * 127
The factors of 52451 are 1 , 7 , 59 , 127 , 413 , 889 , 7493 , 52451
Total factors 8.
Sum of factors 61440 (8989).
#### Powers
The second power of 524512 is 2.751.107.401.
The third power of 524513 is 144.298.334.289.851.
#### Roots
The square root √52451 is 229,021833.
The cube root of 352451 is 37,432709.
#### Logarithms
The natural logarithm of No. ln 52451 = loge 52451 = 10,867635.
The logarithm to base 10 of No. log10 52451 = 4,719754.
The Napierian logarithm of No. log1/e 52451 = -10,867635.
### Trigonometric functions
The cosine of 52451 is 0,514009.
The sine of 52451 is -0,857785.
The tangent of 52451 is -1,668813.
### Properties of the number 52451
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 52451 in Computer Science
Code typeCode value
52451 Number of bytes51.2KB
Unix timeUnix time 52451 is equal to Thursday Jan. 1, 1970, 2:34:11 p.m. GMT
IPv4, IPv6Number 52451 internet address in dotted format v4 0.0.204.227, v6 ::cce3
52451 Decimal = 1100110011100011 Binary
52451 Decimal = 2122221122 Ternary
52451 Decimal = 146343 Octal
52451 Decimal = CCE3 Hexadecimal (0xcce3 hex)
52451 BASE64NTI0NTE=
52451 SHA1f8e21e1473b0c3336469d173e8556055cf332b2b
52451 SHA2240566aa4087e6996e96a9a66117f64f2736e7ec447daca8a7c3953259
52451 SHA25628aaf2fb39d46513dbca4a22b42e8efe845df5a89589cb6a99860297b646f80c
52451 SHA3841aa2df269de273fc6452a1560bf06550c5285e42b4131ded28d0a18f5aef36e6f19c0720dea7de26d2699b37e921d516
More SHA codes related to the number 52451 ...
If you know something interesting about the 52451 number that you did not find on this page, do not hesitate to write us here.
## Numerology 52451
### Character frequency in number 52451
Character (importance) frequency for numerology.
Character: Frequency: 5 2 2 1 4 1 1 1
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 52451, the numbers 5+2+4+5+1 = 1+7 = 8 are added and the meaning of the number 8 is sought.
## Interesting facts about the number 52451
### Asteroids
• (52451) 1994 VU is asteroid number 52451. It was discovered by T. Kobayashi from Ōizumi Observatory on 11/3/1994.
## № 52,451 in other languages
How to say or write the number fifty-two thousand, four hundred and fifty-one in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 52.451) cincuenta y dos mil cuatrocientos cincuenta y uno German: 🔊 (Anzahl 52.451) zweiundfünfzigtausendvierhunderteinundfünfzig French: 🔊 (nombre 52 451) cinquante-deux mille quatre cent cinquante et un Portuguese: 🔊 (número 52 451) cinquenta e dois mil, quatrocentos e cinquenta e um Chinese: 🔊 (数 52 451) 五万二千四百五十一 Arabian: 🔊 (عدد 52,451) اثنان و خمسون ألفاً و أربعمائة و واحد و خمسون Czech: 🔊 (číslo 52 451) padesát dva tisíce čtyřista padesát jedna Korean: 🔊 (번호 52,451) 오만 이천사백오십일 Danish: 🔊 (nummer 52 451) tooghalvtredstusinde og firehundrede og enoghalvtreds Dutch: 🔊 (nummer 52 451) tweeënvijftigduizendvierhonderdeenenvijftig Japanese: 🔊 (数 52,451) 五万二千四百五十一 Indonesian: 🔊 (jumlah 52.451) lima puluh dua ribu empat ratus lima puluh satu Italian: 🔊 (numero 52 451) cinquantaduemilaquattrocentocinquantuno Norwegian: 🔊 (nummer 52 451) femti-to tusen, fire hundre og femti-en Polish: 🔊 (liczba 52 451) pięćdziesiąt dwa tysiące czterysta pięćdziesiąt jeden Russian: 🔊 (номер 52 451) пятьдесят две тысячи четыреста пятьдесят один Turkish: 🔊 (numara 52,451) elliikibindörtyüzellibir Thai: 🔊 (จำนวน 52 451) ห้าหมื่นสองพันสี่ร้อยห้าสิบเอ็ด Ukrainian: 🔊 (номер 52 451) п'ятдесят двi тисячi чотириста п'ятдесят одна Vietnamese: 🔊 (con số 52.451) năm mươi hai nghìn bốn trăm năm mươi mốt Other languages ...
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## Comment
If you know something interesting about the number 52451 or any natural number (positive integer) please write us here or on facebook. | 2,470 | 7,125 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2022-33 | latest | en | 0.626683 |
https://mathoverflow.net/questions/199471/is-the-group-hamm-omega-cap-iso-0m-g-compact | 1,656,306,711,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103328647.18/warc/CC-MAIN-20220627043200-20220627073200-00461.warc.gz | 428,724,737 | 25,307 | Is the group $Ham(M,\omega)\cap Iso_{0}(M,g)$ compact?
let $(M,J,g,\omega)$ be a compact K\"ahler manifold of complex dimension at least $2$. As usual $J$ is the complex structure, $\omega$ is the symplectic form, $g$ is the Riemannian metric and $$\omega(\cdot,\cdot)=g(J\cdot, \cdot)\,.$$
I denote with $Ham(M,\omega)$ the group of Hamiltonian symplectomorphisms w.r.t. the symplectic form $\omega$ and with $Iso_{0}(M,g)$ the identity component of the group of isometries w.r.t. the metric $g$.
Is the group $Ham(M,\omega)\cap Iso_{0}(M,g)$ compact? Maybe it is an easy question but i haven't figured out the answer yet. Any suggestion or hint is welcome and if the question is not well suited for this site i'll move it to MSE.
• The group of isometries that are also symplectomorphisms is closed, ok. But, a priori, this group is bigger than the one of hamiltonian isometries. How do you see that $Ham(M,\omega)$ is closed? Moreover, which topology you put on it to say that? The only topology i can think about is the metric one induced by the Hofer norm and still i cannot prove that the group of hamiltonian isometries is closed. Mar 9, 2015 at 16:46
• You should give a look at Chapter 10 of the McDuff-Salamon book: it is almost an exercise to prove that $Ham(M,\omega)$ is normal in $Symp(M,\omega)$. More interestingly, it corresponds to the kernel of a very interesting (continuous) morphism. At the end of the chapter there is a proof that $Ham(M,\omega)$ is a Lie group, whose Lie algebra is the algebra of Hamiltonian vector fields. Mar 9, 2015 at 17:13
• @student The Hofer topology on $Ham(M, \omega)$ is not particularly well-behaved. A better topology is the smooth compact-open topology (or sometimes called $C^{\infty}$-topology). Moreover, for Kähler manifolds a diffeomorphim which preserves two of the structures $g, \omega, J$ also preserves the third one -- for example an isometry of $g$ which preserves $\omega$ also preserves $J$. Mar 11, 2015 at 16:07 | 554 | 1,986 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2022-27 | latest | en | 0.91394 |
https://philoid.com/question/23546-a-verandah-of-width-225-m-is-constructed-all-along-he-outside-a-room-which-is-55-m-long-and-4-m-wide-find-i-the-area-of-the-vera | 1,726,404,986,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651630.14/warc/CC-MAIN-20240915120545-20240915150545-00750.warc.gz | 420,785,381 | 8,310 | ##### A verandah of width 2.25 m is constructed all along he outside a room which is 5.5 m long and 4 m wide.Find:(i) The area of the verandah(ii) The cost of cementing the floor of the verandah at the rate of Rs. 200 per m2.
(i) It is given in the question that,
Length of room = 5.5 m
Breadth of room = 4 m
We know that,
Area of rectangle = Length × Breadth
Area of room = 5.5 × 4
= 22 m2
From the figure it can be observed that the new length and breadth of the given room (When verandah is included) is 10 m and 8.5 m respectively
Area of room including verandah = 10 × 8.5
= 85 m2
Area of verandah = Area of the room including verandah – Area of room
= 85 – 22
= 63 m2
(ii) It is also given in the question that,
Cost of cementing 1 m2 area of the floor of the verandah = Rs 200
Cost of cementing 63 m2 area of the floor of the verandah = 200 × 63
= Rs 12600
8 | 287 | 883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-38 | latest | en | 0.948504 |
http://mathhelpforum.com/advanced-math-topics/36876-every-positive-number-has-unique-positive-square-root.html | 1,529,958,452,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267868876.81/warc/CC-MAIN-20180625185510-20180625205510-00233.warc.gz | 198,153,923 | 11,232 | # Thread: Every Positive Number Has A Unique Positive Square Root
1. ## Every Positive Number Has A Unique Positive Square Root
Hello,
I am currently studying Introduction to Analysis by Maxwell Rosenlicht. On page 28 -29 he gives this proof for Every Positive Number Has A Unique Positive Square Root. Can someone please help me understand a couple of parts of it? I have indicated in bold what I don't understand.
If $\displaystyle 0 < x_{1} < x_{2}$ then $\displaystyle x_{1}^2 < x_{2}^2$ . That is, bigger positive numbers have bigger square roots. Thus any given real number can have at most one positive square root.
It remains to show that if $\displaystyle a \in \Re$, a > 0, then $\displaystyle a$ has at least one positive square root. For this purpose consider the set
$\displaystyle S = \{ x \in \Re : x \geq 0 , x^2 \leq a \}$
This set is nonempty, since $\displaystyle 0 \in S$, and bounded from above, since if $\displaystyle x > max \{ a,1 \}$ we have $\displaystyle x^2 = x \cdot x > x \cdot 1 = x > a$.Hence, y = l.u.b. S exists.
I don't understand this part with the max function. Is the point of this to show that for all the elements of S there is a minimum $\displaystyle x^2$ that may or not be in S that is the least upper bound? Also, why is it the max {a,1} and not just x > a?
We proceed to show that $\displaystyle y^2 =a$. First, $\displaystyle y > 0$, for $\displaystyle min \{ 1 , a \} \in S$ , since $\displaystyle ( min \{1 ,a \} )^2 \leq min \{ 1 , a \} \cdot 1 = min \{ 1 , a \} \leq a$. Next, for any $\displaystyle \epsilon$ such that $\displaystyle 0 < \epsilon < y$ we have $\displaystyle 0 < y - \epsilon < y < y + \epsilon$, so
Also, why is there a min{a,1} here? Why not just drop the min {1,a}?
$\displaystyle ( y - \epsilon )^2 < y^2 < ( y + \epsilon ) ^2$
since bigger positive numbers have bigger squares. By the definition of $\displaystyle y$ there are numbers greater than $\displaystyle y - \epsilon$ in $\displaystyle S$ , but $\displaystyle y + \epsilon \notin S$. Again, using the fact that bigger positive numbers have bigger squares, we get
$\displaystyle ( y - \epsilon ) ^2 < a < ( y + \epsilon )^2$
Hence
$\displaystyle ( y - \epsilon )^ 2 - ( y + \epsilon ) ^2 < y^2 - a < ( y + \epsilon ) ^2 - ( y - \epsilon )^2$
so
$\displaystyle \mid y^2 - a \mid < ( y+ \epsilon )^ 2 - ( y - \epsilon )^2 = 4 y \epsilon$
The inequality $\displaystyle \mid y^2 - a \mid < 4 y \epsilon$ holds for any $\displaystyle \epsilon$ such that $\displaystyle 0 < \epsilon < y$, and by choosing $\displaystyle \epsilon$ small enough we can make $\displaystyle 4 y \epsilon$ less than any preassigned positive number. Thus $\displaystyle \mid y^2 - a \mid$ is less than any positive number. Since $\displaystyle \mid y^2 - a \mid \geq 0$ we must have $\displaystyle \mid y^2 - a \mid = 0$, proving $\displaystyle y^2 = a$.
Thank you.
2. Originally Posted by Famboozle
It remains to show that if $\displaystyle a \in \Re$, a > 0, then $\displaystyle a$ has at most [that should be "at least"] one positive square root. For this purpose consider the set
$\displaystyle S = \{ x \in \Re : x \geq 0 , x^2 \leq a \}$
This set is nonempty, since $\displaystyle 0 \in S$, and bounded from above, since if $\displaystyle x > max \{ a,1 \}$ we have $\displaystyle x^2 = x \cdot x > x \cdot 1 = x > a$.Hence, y = l.u.b. S exists.
I don't understand this part with the max function. Is the point of this to show that for all the elemnts of S there is a minimum $\displaystyle x^2$ that may or not be in S that is the least upper bound? Also, why is it the max {a,1} and not just x > a?
At this stage, we're looking for an upper bound for S. It doesn't have to be the least upper bound. So we're looking for a number that's bigger than anything in S. The difficulty is that if x>1 then $\displaystyle x^2>x$, whereas if x<1 then $\displaystyle x^2<x$. If $\displaystyle x>a$ and a≥1, then you can conclude that $\displaystyle x^2>a$. That doesn't work when a<1. For example (taking a=1/4), suppose that $\displaystyle x^2>1/4$. You can't conclude that x>1/4. (In fact, the best you can say is that x>1/2.) But if $\displaystyle x^2>1$ then certainly x>1, and 1 in turn is greater than a. The device of using max{a,1} is simply a mechanism for covering the two cases a≥1 and a<1 simultaneously.
Originally Posted by Famboozle
We proceed to show that $\displaystyle y^2 =a$. First, $\displaystyle y > 0$, for $\displaystyle min \{ 1 , a \} \in S$ , since $\displaystyle ( min \{1 ,a \} )^2 \leq min \{ 1 , a \} \cdot 1 = min \{ 1 , a \} \leq a$. Next, for any $\displaystyle \epsilon$ such that $\displaystyle 0 < \epsilon < y$ we have $\displaystyle 0 < y - \epsilon < y < y + \epsilon$, so
Also, why is there a min{a,1} here? Why not just drop the min {1,a}?
Similar sort of reason. It is necessary to show that y>0 (because the subsequent section of the proof requires that y–ε>0, for some ε>0). So we need to find a positive number whose square is less than a. If a≥1 then 1 is such a number; if a<1 then a itself is such a number. In either case, min{a,1} will do the job. | 1,565 | 5,125 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-26 | latest | en | 0.788539 |
https://forum.vectric.com/viewtopic.php?f=2&t=2001 | 1,585,972,338,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370519111.47/warc/CC-MAIN-20200404011558-20200404041558-00267.warc.gz | 458,237,685 | 11,850 | ## Z axis calibration
This forum is for general discussion regarding VCarve Pro
ianto36
Posts: 10
Joined: Tue Jul 17, 2007 7:28 pm
Location: Wales, UK
### Z axis calibration
Hi all
This might sound daft but how do you set the z axis for depth? I'm afraid of the tool cutting in too low.
If you position the router at the corner of the workpiece to be milled and set this as the z zero are all points calculated in relation to this?
How do you manage if you have to change tools and they are different lengths?
So many questions
Ian
den73160
Posts: 32
Joined: Tue Aug 23, 2005 8:37 pm
If you position the router at the corner of the workpiece to be milled and set this as the z zero are all points calculated in relation to this?
Yes, that would set the x,y,z to that point on the material to 0,0,0 and unless something happens would be the point where it returns HOME.
How do you manage if you have to change tools and they are different lengths?
This is always the thing that messes up most people. Unless you have a zero plate or an automatic tool changer everytime you change bits you will have to rezero the bit to the material.
lockeyone
Vectric Craftsman
Posts: 147
Joined: Thu Dec 07, 2006 4:45 am
Model of CNC Machine: other software 50-50 router, Mach3, Epilog 60W
Location: Schofield, Wisconsin, USA
Contact:
I place a slip of paper between bit and material and just bring the bit down till it tickles the piece of paper. Sometimes if the bits are the same length and depending on material I will leave the bit loose and run Z to the zero letting the bit set itself on the material. Make sure the bits the same length as the last one. If your afraid to damage the surface you could use something as a spacer and do the same thing and then offset Z the thickness of the spacer. The slip of paper is my most used method.
If you want to kill time, why not work it to death!
Rusty
Vectric Craftsman
Posts: 127
Joined: Mon Jul 03, 2006 4:50 am
Model of CNC Machine: Homebuilt
Ian:
I use Mach3 as my cnc controller software. You set your starting cutter position (center or wherever you pick) & set the "X" & "Y" axis to ZERO. Select & install the cutting tool, lower the "Z" down to the surface you want to reference as your starting point, then set the "Z" to ZERO. You are now ready to cut the file. I always lift up the "Z" a bit, just in case. Your "Z" readout has now changed reflecting how much you moved it. No worries, the "Z" still knows where the zero is (don't reset any of the axis). When that file finishes cutting your tool will move back to "X" zero, "Y" zero, & "Z" will be what ever you had set for safe retracted height (not sure if that is the right terminology) to be, it should not be zero. If you next file uses the same tool, then you do not change any settings unless that specific file is referencing a different starting height.
If you need to change tools, for the next file, then just lower the "Z" down (after tool is installed) to the starting height & reset ONLY the "Z" axis to ZERO. Ready to go. You can do as many tool changes as required, just remember to reset the "Z" axis to zero each time.
I was scared at first, but it didn't take long to understand what needed to be done. Just remember that the cnc controller software does & goes wherever it is instructed, so most of the booboos are from the operator. Just take to time to go thru the setup carefully before you press the start/go/run button.
Hope that helped a bit.
Good luck.
Rusty
TReischl
Vectric Wizard
Posts: 3360
Joined: Thu Jan 18, 2007 6:04 pm
Model of CNC Machine: 8020 Build 48X36X10 RP 2010 Screenset
Location: Leland NC
I stopped at a metal supply house and bought a short length of .5 inch aluminum bar stock. Cut it to about 3 inch lengths and use those for spacers. As all the guys mentioned above, I use it for a spacer. Once the tool "touches" the bar, I set my Z to .5 and presto, I am ready to go. In Mach, it is just a matter of typing .5 into the Z axis display and pressing enter.
I use the .5 bars because they allow me some room for error when I first rapid down. I use the aluminum because it is easy on the cutters.
One thing to remember, especially on home built machines is to always approach the setting height from the same direction. This takes care of any backlash that might be in the z axis screw.
I also do some editing of the cnc file, I take out the first and last move to X0.000Y0.000 and also remove the Z approach height on the first positioning move. My vcarve always sets about .23 something or other and I would rather it NOT be coming down as it approaches. I think I can change that somwhere in the files, maybe I will go look for that! I take out those moves to X0Y0 because I get tired of waiting for the machine while it drives around.
"If you see a good fight, get in it." Dr. Vernon Jones
CRFultz
Vectric Wizard
Posts: 1160
Joined: Tue Mar 28, 2006 4:21 pm
Location: Longview, Texas
Some tooling also has depth rings set at a predetermined height...therefore one set of the Z leaves any more changes at the same height irregardless where your axis may be located for the tool change...
This also helps when your previous Z zero surface has been removed from earlier cuts...
Chuck
GroBru
Posts: 17
Joined: Fri Aug 18, 2006 11:44 pm
### Z axis calibration
Use a small angle to set your cutting bit one end of the angle buts under the router and the other end supports the bit and you tighten your bit in place
and then you set your 0 z surface if you need to change bit use the angle again and you bits are always the same height if you need a pict i will post
one
Bruno T of Edmundston
TReischl
Vectric Wizard
Posts: 3360
Joined: Thu Jan 18, 2007 6:04 pm
Model of CNC Machine: 8020 Build 48X36X10 RP 2010 Screenset
Location: Leland NC
Not quite sure of your method, a pic would be a big help, thanks for offering!
"If you see a good fight, get in it." Dr. Vernon Jones
GroBru
Posts: 17
Joined: Fri Aug 18, 2006 11:44 pm
### Z axis calibration
Here are the picts hope it come out ok and its always comes out the same height the angle in 2 x 1 x 1" wide or whatever you need
Attachments
Bruno T of Edmundston
Peteg
Posts: 26
Joined: Sat Jan 11, 2020 11:14 pm
### Re: Z axis calibration
I tried this method when I was changing tools, but found it to not be precise enough (my sloppy work though) So I tied into my emergency stop circuit. I attach a probe to the cutter and a another to a 0.0015" feeler gauge. I lower the Z axis and as soon as they touch, the circuit is made and the machine instantly stops. Then all that is required is to zero the Z axis in Mach 3.
Cheers
Peter
TReischl
Vectric Wizard
Posts: 3360
Joined: Thu Jan 18, 2007 6:04 pm
Model of CNC Machine: 8020 Build 48X36X10 RP 2010 Screenset
Location: Leland NC
### Re: Z axis calibration
This is a very old thread, about 13 years worth of old. I have long since put a touch plate on my machine. In fact, I built an entirely new machine during those years.
"If you see a good fight, get in it." Dr. Vernon Jones
Xxray
Vectric Wizard
Posts: 2113
Joined: Thu Feb 17, 2011 8:47 am
Model of CNC Machine: CAMaster Stinger 1
Location: MI USA
### Re: Z axis calibration
Wow, I think this is the oldest thread resurrection I have ever seen but no matter, topic still relevant.
I used a touch plate for years until I started to use drag bits, which do not work with touch plates. So I went the paper method and never looked back, use it for everything now and sold my touch plate as it was just collecting dust. For me, zeroing with paper is just as accurate if nor more, just as easy and considerably quicker.
Doug | 1,999 | 7,654 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-16 | latest | en | 0.920608 |
https://betterlesson.com/lesson/resource/1966960/proportional-reasoning-test-smartboard-notebook | 1,511,158,291,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934805914.1/warc/CC-MAIN-20171120052322-20171120072322-00437.warc.gz | 578,484,437 | 19,885 | ## Proportional Reasoning Test Smartboard.notebook - Section 2: Explore
Proportional Reasoning Test Smartboard.notebook
# Proportional Reasoning Test - Time to Put it All Together!
Unit 2: Proportional Reasoning
Lesson 12 of 14
## Big Idea: Time to test! Let the students show you what they have learned about proportional relationships!
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Standards:
Subject(s):
Math, Number Sense and Operations, proportional relationships, constant of proportionality, proportional reasoning, 7th grade
60 minutes
### Heather Stephan
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Nothing to upload details close | 345 | 1,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2017-47 | latest | en | 0.815672 |
https://allbusinesshoursnow.com/do-hourly-employees-get-paid-overtime/ | 1,679,481,224,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296943809.22/warc/CC-MAIN-20230322082826-20230322112826-00138.warc.gz | 126,198,240 | 14,602 | # Do hourly employees get paid overtime
If you are searching for the Do hourly employees get paid overtime then must check out reference guide below.
## What is time and a half for \$15 an hour?
Once an eligible employee works 40 hours in a week, additional hours must be paid at a minimum overtime rate of time and a half, which is 1.5 times an employee’s regular hourly wage. For example, if Jess is typically paid \$15 per hour, that means she makes \$22.50 per hour with time and a half (\$15 × 1.5).
Calculating Overtime for Hourly Employees
Overtime pay is calculated: Hourly pay rate x 1.5 x overtime hours worked. Here is an example of total pay for an employee who worked 42 hours in a workweek: Regular pay rate x 40 hours = Regular pay, plus. Regular pay rate x 1.5 x 2 hours = Overtime pay, equals.
## How is overtime pay different from regular pay?
For each overtime hour worked you are entitled to an additional one-half the regular rate for hours requiring time and one-half, and to the full rate for hours requiring double time.
This law requires employers in the Commonwealth to pay an overtime wage rate of one and a half times the regular rate for all hours worked in excess of 40 hours in a workweek. Violations of this Act will be enforced by the Virginia Department of Labor and Industry.
## What is overtime pay for \$18 an hour?
Time and a half of 18. According to the Fair Labor Standards Act, most employers are required to pay you at least time and a half if you make \$18 an hour and work more than 40 hours a week.
## What is double time for \$17 an hour?
Interactive Overtime Chart
Overtime Conversion Chart
Regular Wage Time and a half
\$17.00 \$25.50
\$17.50 \$26.25
\$18.00 \$27.00
4s
## What’s time and a half of \$18 an hour?
What is time and a half for \$18 an hour? If you are paid \$18 per hour, you will make \$27 per hour when being paid time and a half (\$18 × 1.5) and \$36 when being paid double time.
## What is time and a half for \$20 an hour?
Assume an employee earns \$20 hourly during a 40-hour work week. Their time and a half pay would be \$20 x 1.5 for a total of \$30 an hour.
The rate for overtime work is 1.5x a worker’s hourly rate. In other words, the regular rate of pay plus another 50%. In the typical case of an hourly or non-exempt employee, if you are paying Sandy \$14 per hour, you would have to pay her 1.5 x \$14 = \$21 per hour for every overtime hour.
## How many hours can I legally work in a day?
Young people can’t work more than eight hours a day or more than 40 hours a week. Unlike adults, there is no opt out for this. If you work for two different employers on the same day, you still can’t work more than a total of eight hours. In England you must be in part-time education or training until your 18th …
## Is overtime taxed more?
Overtime pay and the tax bracket myth explained:
If you are in a higher federal income tax bracket, you may think that your overtime pay is taxed at a higher rate; this is not the case. Overtime pay is taxed at the same marginal rate as your base salary – regardless of your tax bracket.
Shift Differentials — Some employers pay a shift differential if an employee works at night, on Sundays, or on holidays. If the differential is less than 50% of the employee’s hourly rate of pay, it must be included in the calculation of the regular rate of pay used to compute the employee’s overtime rate.
## Who is exempt from overtime pay in Virginia?
Executives, administrators, and other professionals earning at least \$455 per week do not have to be paid overtime under Section 13(a)(1) of the Fair Labor Standards Act. External salespeople (who often set their own hours) are also exempted from VA overtime requirements, as are some types of computer-related workers.
## How many hours can I legally work in a day in Virginia?
Work is only permitted between the hours of 7 a.m. and 7 p.m. When school is in session, they may work a maximum of 3 hours a day, and 18 hours a week. Work is only permitted between the hours of 7 a.m. and 9 p.m.
If an employee works 8 or more consecutive hours, the employer must provide a 30-minute break and an additional 15 minute break for every additional 4 consecutive hours worked.
## What is time and a half for \$20 an hour?
Assume an employee earns \$20 hourly during a 40-hour work week. Their time and a half pay would be \$20 x 1.5 for a total of \$30 an hour.
## How do you calculate 1.5 overtime?
How to calculate overtime pay for hourly employees
1. \$10 x 40 hours = \$400 base pay.
2. \$10 x 1.5 = \$15 overtime rate of pay.
3. \$15 x 6 overtime hours = \$90 overtime pay.
4. \$400 + \$90 = \$490 total pay. | 1,152 | 4,694 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2023-14 | latest | en | 0.933206 |
http://en.wikipedia.org/wiki/Path-ordered_exponential | 1,404,978,603,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1404776405824.29/warc/CC-MAIN-20140707234005-00052-ip-10-180-212-248.ec2.internal.warc.gz | 45,703,733 | 11,059 | # Ordered exponential
(Redirected from Path-ordered exponential)
The ordered exponential (also called the path-ordered exponential) is a mathematical operation defined in non-commutative algebras, equivalent to the exponential of the integral in the commutative algebras. In practice the ordered exponential is used in matrix and operator algebras.
## Definition
Let A be an algebra over a real or complex field K, and a(t) be a parameterized element of A,
$a \mathrel{:} K \to A. \,$
The parameter t in a(t) is often referred to as the time parameter in this context.
The ordered exponential of a is denoted
$\operatorname{OE}[a](t) \equiv \mathcal{T} \left\{e^{\int_0^t a(t') \, dt'}\right\} \equiv \sum_{n = 0}^\infty \frac{1}{n!} \int_0^t \cdots \int_0^t \mathcal{T} \left\{a(t'_1) \cdots a(t'_n)\right\} \, dt'_1 \cdots dt'_n$
where $\mathcal{T}$ is a higher-order operation that ensures the exponential is time-ordered: any product of a(t) that occurs in the expansion of the exponential must be ordered such that the value of t is increasing from right to left of the product; a schematic example:
$\mathcal{T} \left\{a(1.2) a(9.5) a(4.1)\right\} = a(9.5) a(4.1) a(1.2).$
This restriction is necessary as products in the algebra are not necessarily commutative.
The operation maps a parameterized element onto another parameterized element, or symbolically,
$\operatorname{OE} \mathrel{:} \left(K \to A\right) \to \left(K \to A\right).$
There are various ways to define this integral more rigorously.
### Product of exponentials
The ordered exponential can be defined as the left product integral of the infinitesimal exponentials, or equivalently, as an ordered product of exponentials in the limit as the number of terms grows to infinity:
$\operatorname{OE}[a](t) = \prod_0^t e^{a(t') \, dt'} \equiv \lim_{N \rightarrow \infty} \left( e^{a(t_N) \Delta t} e^{a(t_{N-1}) \Delta t} \cdots e^{a(t_1) \Delta t} e^{a(t_0) \Delta t} \right)$
where the time moments {t0, …, tN} are defined as tii Δt for i = 0, …, N, and Δtt / N.
### Solution to a differential equation
The ordered exponential is unique solution of the initial value problem:
\begin{align} \frac{d}{d t} \operatorname{OE}[a](t) &= a(t) \operatorname{OE}[a](t) \text{,} \\ \operatorname{OE}[a](0) &= 1 \text{.} \end{align}
### Solution to an integral equation
The ordered exponential is the solution to the integral equation:
$\operatorname{OE}[a](t) = 1 + \int_0^t a(t') \operatorname{OE}[a](t') \, dt'.$
This equation is equivalent to the previous initial value problem.
### Infinite series expansion
The ordered exponential can be defined as an infinite sum,
$\operatorname{OE}[a](t) = 1 + \int_0^t a(t_1) \, dt_1+ \int_0^t \int_0^{t_1} a(t_1) a(t_2) \, dt_2 \, dt_1 + \cdots.$
This can be derived by recursively substituting the integral equation into itself. | 868 | 2,860 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2014-23 | latest | en | 0.758592 |
https://plugincafe.maxon.net/topic/12350/discussions-about-spline-ik/8 | 1,597,163,451,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738816.7/warc/CC-MAIN-20200811150134-20200811180134-00192.warc.gz | 445,901,653 | 20,566 | • Hello,
I rewritten spline ik due to some requirements, but I was stuck on the problem of rotation. I have tried to use up interpolation or spline trail, but there will still be their own problems. I also searched a lot paper, but did not find a clear solution, do you have any good ideas,or are these papers recommended?
welcome any ideas
• Hi @chuanzhen, thanks for reaching out us.
Although the topic is not strictly related to Cinema 4D plugin development can you please elaborate more on what's the "problem of rotation" you're referring to and what you mean by "up interpolation" or "spline trail"? This will avoid the whole community to do useless guess work and maybe provide based on your context better suggestions.
Cheers, R
• Are you talking about constructing a series of frames along a spline? At least that is what I get from your question. I assume you are using parallel transport (what you seem to call up-interpolation) to construct the frames and ran into the banking problem that comes with that task in general and the PTF algorithm in particular (what you are referring to as rotation).
Unfortunately there is no easy answer or fix to this problem, as there is no mathematical foundation for that what humans would consider a "logical" set of frames for a spline.
As you also talked about papers, you probably already found Rotation Minimizing Frames, which is the common solution for the shortcomings of PTF, but also a bit more involved. Could you elaborate on where you are stuck?
• @r_gigante @zipit
The detailed explanation is shown below
“up interpolattion”
Try use "spline trail"
I read some papers, but always been very slow, just only expanded know, such as Frenet–Serret formulas etc. But I did not directly find a better way to replace my current method, and I have been looking for it.
Computation of Rotation Minimizing Frames This seems to be a good paper, I will read it carefully.
• Hi,
1. As suspected, the method you are describing is usually called Parallel Transpport (the geometric problem in mathematics in general) or Parallel Transport Frame (PTF, the specific algorithm in Computer Graphics to use Parallel Transport to construct a smooth series of frames).
2. Just to be clear, as your image, you apparently put a lot of effort into, is not clear on that point for me: In PTF you construct the first component of the first frame in your series with an up-vector and the tangent of that element. For all other frames you use instead the component of the previous frame that replaced that up-vector in the first frame (i.e. the parallel transport part of the algorithm). Or as python style pseudo code for clarity:
``````# With a hypothetical type 'Frame' which has the attributes i, j, k
# for the components of a frame.
tangents = get_tangents()
frames = []
for n, t in enumerate(tangents):
# Select our 'up vector' based on n.
up_vec = SOME_UP_VECTOR if n is 0 else frames[n - 1].j
# This is the vector we construct diffrently based on wether
# we are constructing the zerost or nth (n != 0) element.
temp = up_vec x t
i = normalize(temp x t)
j = normalize(i x t)
k = normalize(i x j)
frames.append(Frame(i, j, k))
``````
1. I am still unclear on what you would consider wrong about your results. There are ways to adjust PTF sets for the banking problem (by smoothing the banking of the frames after they have been constructed for example) and RMF is very similar to PTF, with the major difference being that it also looks ahead and not only back. However both aPTF and RMF are normally only needed when you have closed splines (the first and last frame won't fit then with standard PTF) or you have drastic changes of tangency in the set of line segments, the spline, you are interpolating over. The results of aPTF and RMF should be very similar or identical to what is shown in your image (unless I am overlooking a problem here).
2. As you seem to aim for replicating Cinema's results: I might be wrong, as I cannot find it at the moment, but I think I have seen that Cinema does expose its PTF algorithm somewhere in the C++ SDK. Again, I might be wrong and mixing up SDKs here.
edit: Ah, I see now the marked problem in the second image. It is rather hard to just judge that from looking at an image, but at first glance I cannot see a reason, why a standard PTF should fail on that spline (as there are no drastic changes in tagency in your spline and its not closed either). You should double check your algorithm. If things remain unclear, check the PTF link I have posted above. However, be aware, that most cases "drastic tangency change" cannot properly be solved with PTF or a naive implemenation of aPTF that just smoothes the banking of your frames.
edit2: What I was probably also unclear about, is the term "banking problem". It refers to the orientation of your frames along the axis of tangency, i.e. what you call "rotation".
Cheers,
zipit
• @zipit Thank you for your patient reply, it really helped me a lot. I need to learn about PTF,This is a new point of knowledge for me, and I need some time to understand how it works. I will update the progress of the solution here.
Yes,rotation is banking ,in my method it points to the next object in the chain of objects，not use tangent
• in my method it points to the next object in the chain of objects，not use tangent
I am not quite sure how you mean that, but to be clear: If you have a chain of objects, i.e. an array of line segments, then the vector from the position of the current 'object' to the position of the next 'object' is the tangent of the line segment array at that point. Or in other words: The tangents in my example above are the vectors from the point n to the point n+1 in the list of points of a segementized spline.
Cheers
zipit
• @zipit Hi, I came back a bit late.
I read the paper and watched the video. The video is understandable and the paper is still obscure. The effect of Parallel Transport is similar to some functions of c4d python utils.SplineHelp ().
The Parallel Transport method is useful, but using the control object to control the rotation of the joint (rotate around z) is still a bit unclear (in the picture below). The paper seems to mention some solutions such as (Animating rotation with quaternion curves) and (Fiber bundle twist reduction.)
• I read the paper and watched the video. The video is understandable and the paper is still obscure.
Which paper are you referring to? The RMF paper? It has a chunk of pseudo-code you can straight up copy.
The Parallel Transport method is useful, but using the control object to control the rotation of the joint (rotate around z) is still a bit unclear (in the picture below).
I am still unclear on what is intended on what is not in your pictures. Is the twisting motion of your chain of objects intended or not? It would probably be best, if you just would generate an image of what your algorithm currently gives you and then create a second image, where you have fiddled your objects by hand into the orientation, you want them to have.
The paper seems to mention some solutions such as (Animating rotation with quaternion curves) and (Fiber bundle twist reduction.)
I am still not sure which paper you are talking about. There are other algorithms, especially a lot of hacks for PTF, but at least I would say that RMF will give you the best results, while being relatively simple, and PTF (or a naive version of aPTF like shown in the video) will give you good results, while being very simple. You have also to keep in mind that these all are hacks, i.e. algorithms for imitating what humans would consider an aesthetical set of frames placed on a curve. In reality the normals and bi-normals of a curve will flip based on the tangency of that curve.
Cheers,
zipit
• @zipit Thanks,this paper Parallel Transport Approach to Curve Framing
I will follow your suggestion to explore RMF. It may be that the language barrier prevents us from communicating well, but I still get a lot of help from your answer. I use Google translate | 1,822 | 8,074 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-34 | latest | en | 0.968033 |
https://solvedlib.com/n/find-the-area-of-the-region-bounded-by-the-graphs-of-the,18167809 | 1,657,016,981,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00477.warc.gz | 573,328,011 | 21,148 | # Find the area of the region bounded by the graphs of the given equations_ y = 11x,y=x2The area is
###### Question:
Find the area of the region bounded by the graphs of the given equations_ y = 11x,y=x2 The area is
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Homework 2: Problem 12 Previous Problem Problem List Next Problem point) Solve the initial value problem y' + 2y e8r y(0) = -8 y(z) Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining Email WeBWorK TA...
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https://www.calculatoratoz.com/en/central-angle-of-annulus-sector-given-inner-arc-length-and-breadth-calculator/Calc-38685 | 1,716,701,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058868.12/warc/CC-MAIN-20240526043700-20240526073700-00728.warc.gz | 598,578,205 | 62,144 | ## Central Angle of Annulus Sector given Inner Arc Length and Breadth Solution
STEP 0: Pre-Calculation Summary
Formula Used
Central Angle of Annulus Sector = Inner Arc Length of Annulus Sector/(Outer Circle Radius of Annulus-Breadth of Annulus)
Central(Sector) = lInner Arc(Sector)/(rOuter-b)
This formula uses 4 Variables
Variables Used
Central Angle of Annulus Sector - (Measured in Radian) - Central Angle of Annulus Sector is the angle whose apex (vertex) is the center of the concentric circles of Annulus and whose legs (sides) are radii intersecting the circles in four distinct points.
Inner Arc Length of Annulus Sector - (Measured in Meter) - Inner Arc Length of Annulus Sector is the distance between the two points along the inner curve of Annulus.
Outer Circle Radius of Annulus - (Measured in Meter) - Outer Circle Radius of Annulus is the radius of a larger circle of the two concentric circles that form its boundary.
Breadth of Annulus - (Measured in Meter) - Breadth of Annulus is defined as the shortest distance or measurement between outer circle and inner circle of Annulus.
STEP 1: Convert Input(s) to Base Unit
Inner Arc Length of Annulus Sector: 3 Meter --> 3 Meter No Conversion Required
Outer Circle Radius of Annulus: 10 Meter --> 10 Meter No Conversion Required
Breadth of Annulus: 4 Meter --> 4 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
Central(Sector) = lInner Arc(Sector)/(rOuter-b) --> 3/(10-4)
Evaluating ... ...
Central(Sector) = 0.5
STEP 3: Convert Result to Output's Unit
0.5 Radian -->28.6478897565466 Degree (Check conversion here)
28.6478897565466 28.64789 Degree <-- Central Angle of Annulus Sector
(Calculation completed in 00.020 seconds)
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## < 13 Central Angle of Annulus Sector Calculators
Central Angle of Annulus Sector given Diagonal
Central Angle of Annulus Sector = acos((Outer Circle Radius of Annulus^2+Inner Circle Radius of Annulus^2-Diagonal of Annulus Sector^2)/(2*Outer Circle Radius of Annulus*Inner Circle Radius of Annulus))
Central Angle of Annulus Sector given Diagonal and Inner Circle Radius
Central Angle of Annulus Sector given Diagonal and Outer Circle Radius
Central Angle of Annulus Sector given Perimeter
Central Angle of Annulus Sector = (Perimeter of Annulus Sector-(2*(Outer Circle Radius of Annulus-Inner Circle Radius of Annulus)))/(Outer Circle Radius of Annulus+Inner Circle Radius of Annulus)
Central Angle of Annulus Sector given Perimeter and Inner Circle Radius
Central Angle of Annulus Sector = (Perimeter of Annulus Sector-(2*Breadth of Annulus))/((2*Inner Circle Radius of Annulus)+Breadth of Annulus)
Central Angle of Annulus Sector given Perimeter and Outer Circle Radius
Central Angle of Annulus Sector = (Perimeter of Annulus Sector-(2*Breadth of Annulus))/((2*Outer Circle Radius of Annulus)-Breadth of Annulus)
Central Angle of Annulus Sector given Area and Outer Circle Radius
Central Angle of Annulus Sector = (2*Area of Annulus Sector)/(Breadth of Annulus*((2*Outer Circle Radius of Annulus)-Breadth of Annulus))
Central Angle of Annulus Sector given Area and Inner Circle Radius
Central Angle of Annulus Sector = (2*Area of Annulus Sector)/(Breadth of Annulus*((2*Inner Circle Radius of Annulus)+Breadth of Annulus))
Central Angle of Annulus Sector given Area
Central Angle of Annulus Sector = (2*Area of Annulus Sector)/(Outer Circle Radius of Annulus^2-Inner Circle Radius of Annulus^2)
Central Angle of Annulus Sector given Outer Arc Length and Breadth
Central Angle of Annulus Sector = Outer Arc Length of Annulus Sector/(Inner Circle Radius of Annulus+Breadth of Annulus)
Central Angle of Annulus Sector given Inner Arc Length and Breadth
Central Angle of Annulus Sector = Inner Arc Length of Annulus Sector/(Outer Circle Radius of Annulus-Breadth of Annulus)
Central Angle of Annulus Sector given Outer Arc Length
Central Angle of Annulus Sector = Outer Arc Length of Annulus Sector/Outer Circle Radius of Annulus
Central Angle of Annulus Sector given Inner Arc Length
Central Angle of Annulus Sector = Inner Arc Length of Annulus Sector/Inner Circle Radius of Annulus
## Central Angle of Annulus Sector given Inner Arc Length and Breadth Formula
Central Angle of Annulus Sector = Inner Arc Length of Annulus Sector/(Outer Circle Radius of Annulus-Breadth of Annulus)
Central(Sector) = lInner Arc(Sector)/(rOuter-b)
## What is an Annulus Sector?
An Annulus Sector, also known as the circular ring sector, is a cut piece from an Annulus that is joined by two straight lines from its center.
## What is Annulus?
In mathematics, an Annulus (plural Annuli or Annuluses) is the region between two concentric circles. Informally, it is shaped like a ring or a hardware washer. The word "annulus" is borrowed from the Latin word anulus or annulus meaning 'little ring'. The adjectival form is annular (as in annular eclipse).The area of an Annulus is the difference in the areas of the larger circle of radius R and the smaller one of radius r.
## How to Calculate Central Angle of Annulus Sector given Inner Arc Length and Breadth?
Central Angle of Annulus Sector given Inner Arc Length and Breadth calculator uses Central Angle of Annulus Sector = Inner Arc Length of Annulus Sector/(Outer Circle Radius of Annulus-Breadth of Annulus) to calculate the Central Angle of Annulus Sector, The Central Angle of Annulus Sector given Inner Arc Length and Breadth formula is defined as the angle whose apex (vertex) is the center of the concentric circles of Annulus and whose legs (sides) are radii intersecting the circles in four distinct points, calculated using inner arc length and breadth of Annulus Sector. Central Angle of Annulus Sector is denoted by Central(Sector) symbol.
How to calculate Central Angle of Annulus Sector given Inner Arc Length and Breadth using this online calculator? To use this online calculator for Central Angle of Annulus Sector given Inner Arc Length and Breadth, enter Inner Arc Length of Annulus Sector (lInner Arc(Sector)), Outer Circle Radius of Annulus (rOuter) & Breadth of Annulus (b) and hit the calculate button. Here is how the Central Angle of Annulus Sector given Inner Arc Length and Breadth calculation can be explained with given input values -> 1641.403 = 3/(10-4).
### FAQ
What is Central Angle of Annulus Sector given Inner Arc Length and Breadth?
The Central Angle of Annulus Sector given Inner Arc Length and Breadth formula is defined as the angle whose apex (vertex) is the center of the concentric circles of Annulus and whose legs (sides) are radii intersecting the circles in four distinct points, calculated using inner arc length and breadth of Annulus Sector and is represented as Central(Sector) = lInner Arc(Sector)/(rOuter-b) or Central Angle of Annulus Sector = Inner Arc Length of Annulus Sector/(Outer Circle Radius of Annulus-Breadth of Annulus). Inner Arc Length of Annulus Sector is the distance between the two points along the inner curve of Annulus, Outer Circle Radius of Annulus is the radius of a larger circle of the two concentric circles that form its boundary & Breadth of Annulus is defined as the shortest distance or measurement between outer circle and inner circle of Annulus.
How to calculate Central Angle of Annulus Sector given Inner Arc Length and Breadth?
The Central Angle of Annulus Sector given Inner Arc Length and Breadth formula is defined as the angle whose apex (vertex) is the center of the concentric circles of Annulus and whose legs (sides) are radii intersecting the circles in four distinct points, calculated using inner arc length and breadth of Annulus Sector is calculated using Central Angle of Annulus Sector = Inner Arc Length of Annulus Sector/(Outer Circle Radius of Annulus-Breadth of Annulus). To calculate Central Angle of Annulus Sector given Inner Arc Length and Breadth, you need Inner Arc Length of Annulus Sector (lInner Arc(Sector)), Outer Circle Radius of Annulus (rOuter) & Breadth of Annulus (b). With our tool, you need to enter the respective value for Inner Arc Length of Annulus Sector, Outer Circle Radius of Annulus & Breadth of Annulus and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Central Angle of Annulus Sector?
In this formula, Central Angle of Annulus Sector uses Inner Arc Length of Annulus Sector, Outer Circle Radius of Annulus & Breadth of Annulus. We can use 12 other way(s) to calculate the same, which is/are as follows -
• Central Angle of Annulus Sector = (2*Area of Annulus Sector)/(Outer Circle Radius of Annulus^2-Inner Circle Radius of Annulus^2)
• Central Angle of Annulus Sector = Outer Arc Length of Annulus Sector/(Inner Circle Radius of Annulus+Breadth of Annulus)
• Central Angle of Annulus Sector = (Perimeter of Annulus Sector-(2*(Outer Circle Radius of Annulus-Inner Circle Radius of Annulus)))/(Outer Circle Radius of Annulus+Inner Circle Radius of Annulus)
• Central Angle of Annulus Sector = (2*Area of Annulus Sector)/(Breadth of Annulus*((2*Outer Circle Radius of Annulus)-Breadth of Annulus))
• Central Angle of Annulus Sector = (2*Area of Annulus Sector)/(Breadth of Annulus*((2*Inner Circle Radius of Annulus)+Breadth of Annulus))
• Central Angle of Annulus Sector = (Perimeter of Annulus Sector-(2*Breadth of Annulus))/((2*Inner Circle Radius of Annulus)+Breadth of Annulus)
• Central Angle of Annulus Sector = (Perimeter of Annulus Sector-(2*Breadth of Annulus))/((2*Outer Circle Radius of Annulus)-Breadth of Annulus)
• Central Angle of Annulus Sector = acos((Outer Circle Radius of Annulus^2+Inner Circle Radius of Annulus^2-Diagonal of Annulus Sector^2)/(2*Outer Circle Radius of Annulus*Inner Circle Radius of Annulus))
• Central Angle of Annulus Sector = acos(1-((Diagonal of Annulus Sector^2-Breadth of Annulus^2)/(2*Inner Circle Radius of Annulus*(Inner Circle Radius of Annulus+Breadth of Annulus))))
• Central Angle of Annulus Sector = acos(1-((Diagonal of Annulus Sector^2-Breadth of Annulus^2)/(2*Outer Circle Radius of Annulus*(Outer Circle Radius of Annulus-Breadth of Annulus))))
• Central Angle of Annulus Sector = Outer Arc Length of Annulus Sector/Outer Circle Radius of Annulus
• Central Angle of Annulus Sector = Inner Arc Length of Annulus Sector/Inner Circle Radius of Annulus
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### “Part specification is longer than depth” warning in Position function [duplicate]
I'm working on a project which analyzes financial data. I download the price and dividend information for the SPY ETF which tracks the S&P 500: ... | 2,409 | 10,536 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-51 | latest | en | 0.897819 |
https://gerardnico.com/wiki/data_mining/discriminant_analysis | 1,505,898,847,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818686983.8/warc/CC-MAIN-20170920085844-20170920105844-00631.warc.gz | 672,731,347 | 17,690 | # Statistics Learning - Discriminant analysis
Discriminant analysis is a classification method.
In discriminant analysis, the idea is to:
1. model the distribution of X in each of the classes separately.
2. use what's known as Bayes theorem to flip things around to get the probability of Y given X. $Pr(Y|X)$
The Bayes theorem is a basis for discriminant analysis.
## 3 - Bayes theorem for classification
$$\begin{array}{rrl} Pr(Y = k|X = x) & = & \frac{\displaystyle Pr(X = x|Y = k) Pr(Y = k)}{\displaystyle Pr(X = x)} \\ & = & \frac{\displaystyle \pi_k f_k(x)}{\displaystyle \sum_{l=1}^K \pi_l f_l (x)} \end{array}$$
where:
• $\pi_k$ is the prior probability for class k.
• The marginal is the summing over all the classes. $\displaystyle \sum_{l=1}^K \pi_l f_l (x)$
• This formula is quite general where we can plug in any probability densities: $f_k(x) =Pr(Y = k|X = x)$ is the density for X in class k.
This approach is quite general, and other (distributions|density) can be used including non-parametric approaches. By altering the forms for $f_k(x)$, we get different classifiers (ie classification rule).
### 3.1 - Gaussian
When $f_k(x)$ are Gaussian densities, with the same covariance matrix in each class, this leads to linear discriminant analysis.
The two popular forms of linear discriminant analysis are:
• linear : same variance for all X
• and quadratic : different variance for all X
### 3.2 - Naive Bayes
When you have a large number of features (like 4,000), you really wouldn't want to estimate the large covariance matrices.
You will then assume that in each class the density factors into a product of densities. $$f_k(x) = \prod^p_{j=1} f_{jk}(x_j)$$ where:
• k is the class
• p is the number of parameters
ie that the variables are conditionally independent in each of the classes.
If we plug it into the above Bayes formula, we get something known as the naive Bayes classifier.
For linear discriminant analysis, this means that the covariances matrix $\sigma_k$ are diagonal. Instead of estimating the covariance matrix, if you've got p variables, we got P squared parameters that must be estimated.
Although the assumption seems very crude or wrong, the naive Bayes classifier is actually very useful in high-dimensional problems. We end up with maybe biased estimates for the probabilities
In classification, we're mainly concerned about which class has the highest probability and not whether we got the probabilities exactly right.
In terms of classification, as we just need to to classify to the largest probability, we can tolerate quite a lot of bias and still get good classification performance. What we get in return is much reduced variance from having to estimate far fewer parameters.
## 4 - Classify to the highest density
We classify a new point according to which density is highest.
On the right, when the priors are different we favor the pink class, the decision boundary has shifted to the left.
• When the classes are well-separated, the parameter estimates for the logistic regression model are surprisingly unstable. Linear discriminant analysis does not suffer from this problem.
• If the sample size N is small and the distribution of the predictors X is approximately normal in each of the classes, the linear discriminant model is again more stable than the logistic regression model.
• Linear discriminant analysis is popular when we have more than two response classes, because it also provides low-dimensional views of the data.
• If you have the right population model, Bayes rule is the best you can possibly do.
## 6 - Example
Suppose that in Ad Clicks (a problem where you try to model if a user will click on a particular ad) it is well known that the majority of the time an ad is shown it will not be clicked. What is another way of saying that?
• Ad Clicks have a low Prior Probability (Status: correct)
• Ad Clicks have a high Prior Probability.
• Ad Clicks have a low Density.
• Ad Clicks have a high Density
Whether or not an ad gets clicked is a Qualitative Variable. Thus, it does not have a density. The Prior Probability of Ad Clicks is low because most ads are not clicked. | 970 | 4,184 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2017-39 | longest | en | 0.878938 |
http://www.edupil.com/question/let-abc-be-a-triangle-in-which-ab-ac-let-l-be-the-locus-that-bx-cx/ | 1,477,273,196,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719463.40/warc/CC-MAIN-20161020183839-00550-ip-10-171-6-4.ec2.internal.warc.gz | 417,729,825 | 18,186 | # Let ABC be a triangle in which AB = AC. Let L be the locus that BX = CX
Let ABC be a triangle in which AB = AC. Let L be the locus that BX = CX. Which of the
following statements are correct?
1,L is a straight line passing through A and in-cetre of triangle ABC is on L.
2. L is a straight line passing through A and orthocentre of triangle ABC is a point on L.
3. L is a straight line passing through A and centroid of triangle ABC is a point on L.
Select the correct answer using the code given below.
(a) 1 and 2 only
(b) 2 and 3 only
(c) 1 and 3 only
(d) 1, 2 and 3
Anurag Mishra Professor Asked on 20th February 2016 in | 181 | 628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2016-44 | longest | en | 0.943593 |
http://www.physicsforums.com/showthread.php?p=3391465 | 1,386,851,042,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164583115/warc/CC-MAIN-20131204134303-00079-ip-10-33-133-15.ec2.internal.warc.gz | 491,870,521 | 19,665 | # How Transistor works - verifying
by Nikarus
Tags: transistor, verifying
P: 963
Quote by wbeaty Just two questions. First, might you yet recall which text you yourself had in your undergrad semiconductor physics course, the one that first covered transistors? Second question: I haven't clearly stated my reasoning yet, so do you want to hear it? That's my second question.
Kittel - solid state physics as undergrad.
Muller & Kamens - Device Fabrication For Integrated Electronics at grad school, MSEE.
Sze - Physics of Semiconductor Devices, grad school Ph.D.-EE.
If you have reasoning you'd like to explain, sure, by all means do so.
Claude
P: 963
Quote by Studiot Many thanks for this answer. I have now had time to review this website with the following results. Comparing this website with the one I linked to in post#21 of this thread I think that 'Ratch' and 'Ratchit' are one and the same person. I note he signs himself Ratch in cabraham's E-Tech thread, although his handle is 'ratchit'. In June 2008 he started the thread I linked to (post#21) by referring to your site (amasci) as proof that 'transistors are voltage controlled not current controlled'. There was significant discussion, including the nature of the term 'control', though not of the gargantuan proportions of the one in 2010 in cabraham's link. I apologise to claude if we had primed the pump for that argument in 2008. I felt that his summary in post#39 here was particularly good. However I would take issue with equations posted in post#50. My version of equation (3) has another term which is significant in certain types of transistor and reminds us that there are other agents that affect, and therefore can 'control', the collector current. 3) Ic = alpha*Ie + Ico go well
Thanks for your feedback. I am well aware of the additional "Ico" term in eqn 3), & what you've presented is correct. But when we describe quantities that "control" a device, we are not usually referring to leakage & other parasitic flaws.
In a bjt, the objective is to control collector current w/ some sort of input signal, as found in uctlrs, transducers, photodiodes, etc. The inherent leakage current Ico, which exists due to the non-ideal nature of the reverse biased c-b jcn, is present & varies greatly w/ temp.
Of course Ico plays a role in determining Ic, but it is not something we use to control Ic. It is an inherent property of the bjt, one which fortunately has a very small influence on bjt Ic behavior, for silicon material.
In the 1950's when germanium was the dominant bjt material, Ico was a real problem at medium to high temps. Designers had to account for the large Ico c-b leakage when employing Ge devices at temps above 50 or 75 C. The limit was around 100 C.
Then silicon replaced Ge around 1959, & Ico for Si is generally small enough to neglect. Again, it's there, but Si devices can operate to the mil temp range of 125 C & beyond w/o Ico being too large an eror. It is an error term for sure, & your eqn is more precise than the simplified version I presented.
But Ico, & I believe I'll get universal backing/concensus on this, is NOT a "control" quantity. It influences Ic for sure, but we don't control Ic by setting a value for Ico. I think this whole question revolves around the meaning of "control".
Ib, Vbe, Ie, Vbc, Ies, Vt, Ico, etc., all have influence over Ic depending how the device is driven. If the b-e jcn is driven by a true current source or voltage source w/ a large series resistor, then Vt, & Ies determine the Vbe value at a given temp. Also, Ico adds to whatever Ic value is obtained from eqn 1) or 3).
Which quantity are we adjusting to get a specific value of Ic? That is what we mean by control. Again, your version of eqn 3) is more precise than my simplified version. Ico exists indeed & influences Ic. But if Ico is 2.7 uA, & we bias the bjt at Ic value of 1.0 mA, the error is just 0.27%. For larger Ic value, the error is less.
Again, w/ Ge devices, the presence of large Ico values forced the designer to take it into consideration. Circuit topology was built around the need to mitigate large Ico values. EEs from the 1950's can give you insight into this practice. BR.
Claude
P: 5,462
I think this whole question revolves around the meaning of "control".
I have no doubt you have great knowledge of semiconductors and have already commented how worthwhile your technical analyses are.
But I also feel there is a lack of acceptance that others may also have something valid to say and a corresponding willingness to listen to them as well as to expound to them.
Here is a simple experiment to prove that it is possible to 'control' Ic with the base not even connected.
You can show this by allowing photons to enter the transistor, biased suitably between collector and emitter only. It is possible to switch on the transistor by this means.
As a matter of interest how do multiple emitter transistors fit into your control scheme? and how do you describe control by emitter injection?
P: 963
Quote by Studiot I have no doubt you have great knowledge of semiconductors and have already commented how worthwhile your technical analyses are. But I also feel there is a lack of acceptance that others may also have something valid to say and a corresponding willingness to listen to them as well as to expound to them. Here is a simple experiment to prove that it is possible to 'control' Ic with the base not even connected. You can show this by allowing photons to enter the transistor, biased suitably between collector and emitter only. It is possible to switch on the transistor by this means. As a matter of interest how do multiple emitter transistors fit into your control scheme? and how do you describe control by emitter injection?
Thanks again Studiot for your feedback. With photon stimulation, instead of forward biasing the base-emitter junction, photons provide the energy to transition valence electrons into conduction. Though the base lead is not brought out to the outside world, the base-emitter junction carries a conduction current due to photon stimulation.
There is no base lead terminal, but there is a base region internally. Emitter region emits electrons which are yanked into the collector via the E field of the reverse biased c-b jcn. Have I answered your question.
Your last question is a good one. "Control by emitter injection" is how I describe the bjt in general. Emitter electrons injected towards the base quickly become collector current. Base holes injected into the emitter recombine there w/ electrons in the emitter. But base injection does not directly produce collector current.
The reason the base is doped w/ acceptor ions (for a p type base, i.e. npn bjt device), is to improve c-b reverse breakdown voltage & leakage current. A heavy doping of acceptor atoms into the p base results in more base current for a given collector curret, an undesirable thing. But doing so reduces Ico, a bad thing, & improves c-b junction blocking voltage ability.
What function does the injection component of base current serve is as follows. THe lower the doping density in the base, the lower the base injection current & the higher the beta value. Superbeta bjt's at op amp inputs use these devices. But the c-b blocking voltage ability is a few volts, & the leakage current from c-b is horrendous.
For bjt devices that are to operate w/ a Vce of 50, 100, or more volts, w/ low leakage current c-b, i.e. Ico, the base must be doped heavier than is optimum for high beta. Thus current gain is sacrificed for low Ico & high c-b blocking voltage.
It's all about tradeoffs. Ic is produced by Ie. Ib is needed, but we wish to minimize it. If we minimize Ib too much, Ico goes through the roof, & Vce,blocking plummets. So base doping is optimized according to how high Vce must block, & how low Ico needs to be. Higher voltage devices have lower beta. Can't get around that. Did I help?
Claude
P: 5,462
Thanks again Studiot for your feedback.
When I look over this thread I see many of the same points and formulae that were raised in the 2008 thread I linked to.
In particular, since others may have missed them
1)What happens if you try to force current control via the base.
2)What happens if you force voltage control via the base.
3)What happens at different frequencies - a very important point since the bjt is not only a DC device.
4)Whether you are interested in the internal workings of the bjt or worings of a circuit using a bjt.
5)The role of the emitter.
The 2008 thread I linked to contains some further information, including references to orifinal articles and other practical demonstrations on how to vary Ic including by varying the input frequency only (= control of Ic by frequency) and a great deal more on the photon aspect.
I would agree the demonstrations were contrived to show that any one method only achieves partial control, since it can be subverted by another.
I also agree that the weight of practical experience of countless engineers and scientists ove the years have found current control configurations to be the most useful.
Thank you for some excellent insights.
P: 116
Quote by cabraham Kittel - solid state physics as undergrad. Muller & Kamens - Device Fabrication For Integrated Electronics at grad school, MSEE. Sze - Physics of Semiconductor Devices, grad school Ph.D.-EE.
Thanks! I wanted to make sure we were "on the same page," approximately. My own text titles I don't quite recall, but I think they're in a cellar box from 1979. I'm almost certain that the main one was Sze above.
Quote by cabraham If you have reasoning you'd like to explain, sure, by all means do so.
OK. But first...
Do you now understand that my article is not aimed at engineers? Right? Did you take note of who the original intended audience was?
P: 5,462
Did you take note of who the original intended audience was?
That too is a very good point.
P: 116
Quote by Studiot Comparing this website with the one I linked to in post#21 of this thread I think that 'Ratch' and 'Ratchit' are one and the same person. I note he signs himself Ratch in cabraham's E-Tech thread, although his handle is 'ratchit'.
Yep probably the same, but probably avoiding ban by mods. Following your links, I note a constant parade of odd questionable events there. Mods freely using personal attacks? And misusing moderator power while participating in battles with users? Multiplying sockpuppets and constant deceptive practice. The usual anti-trolls rule against namecalling is missing.
Quote by Studiot In June 2008 he started the thread I linked to (post#21) by referring to your site (amasci) as proof that 'transistors are voltage controlled not current controlled'.
<red sweaty face>BUT THEY ARE!!!</red sweaty face>
But seriously, my article originally was written in response to this question: what's a good way to explain the BJT to my grandmother? Or equivalently, how can we simplify the BJT down to the point where its operation becomes obvious to anyone?
Here's an appropriate comment: Simplifers and complexifiers
Of course that question is entirely different from this one: what's the single best transistor model, the One True Path which all Proper chip engineers should use? Never ask that one, since it's a recipe for endless "Swiftian Battles."
A Swiftian battle is where two populations are driven to nearly-murderous rage over disagreement regarding the One True Way to crack an egg. In Gulliver's Travels, the two Lilliputan countries were slaughtering each other because every Proper citizen knows that morning breakfast eggs must be cracked on the pointed end ...and those despicable Outsider Others who disagree, they're disgusting heretics who need prompt incineration. :) Johnathan Swift clearly was well acquainted with the academics of his time. Tiny little men? Best way to "crack an egg?" Obvious rage and a desire to silence the blasphemous opponents?
My transistor article inadvertently triggers one of these battles when I wrap the original goal inside another one: "How do they REALLY work." The flames break out. Even Win Hill of Art of Electronics arrived and put in his two cents.
P: 116
Quote by Studiot I also agree that the weight of practical experience of countless engineers and scientists ove the years have found current control configurations to be the most useful.
Ooo, here's an idea and a trick question inspired by the transistor war...
What if Ib was actually zero in BJTs?
What if hfe didn't exist? We'd be screwed, right? We could no longer use the linear relation between Ib and Ic. The diode exponential function would become part of everything. To prevent extreme distortion, only very small Vbe signals would be allowed.
Right? :)
P: 5,462 I think I have already described, here or in my references, the situation when Ib=0. It is not my purpose to discuss the inner working of other sites, except to observe that progress is generally facilitated by goodwill. With goodwill moderation is often self moderation. I have described, as has Claude, methods for calculating the variation of Ic with frequency by considering current or charge. I have further described, in later posts in my link, an experiment which shows how to set up a transistor so that this change in Ic with frequency is independent of base voltage. I asked Ratch for a derivation of equations showing how to calculate with the voltage only model in this case, with no response. You have the same opportunity here. I am not wedded to any particular transistor model. surely the 'best model' is the one which yields the required information with the least calculative effort. As such there is no best model, since models and calculations, correct at 2Hz will not suffice for 2GHz. A qualitative model such as the man cranking the variable resistor is very useful for achieving uderstanding but little use in real circuit calculations. I usually offer a man turning a tap (faucet) to control the flow. go well
P: 963
Quote by wbeaty Ooo, here's an idea and a trick question inspired by the transistor war... What if Ib was actually zero in BJTs? What if hfe didn't exist? We'd be screwed, right? We could no longer use the linear relation between Ib and Ic. The diode exponential function would become part of everything. To prevent extreme distortion, only very small Vbe signals would be allowed. Right? :)
If Ib was zero, it would NOT be a bjt! How can an n-p-n structure have Ib = 0 when b-e jcn is forward biased? An E field exists in b-e region oriented so that holes move from base to emitter & electrons move from e to b. The n-type emitter has an abundance of free electrons, they being the majority carrier w/ the jcn under fwd bias. Likewise the base has an abundance of free holes, same reasoning.
It is impossible for an E field to selectively impart motion upon the emitter electrons & NOT on the base holes. How can Ib ever be zero? We can minimize Ib by doping base region very lightly w/ acceptors, much lower than emitter doping of donors. This reduces no. of holes injected into the emitter under forward bias, so that Ie >> Ib.
If we reduced base acceptor doping density to zero, we still have base current. Intrinsic silicon still conducts current in the presence of an E field. The device is now no longer n-p-n, but rather n-i-n ( "i" = "intrinsic").
To reduce Ib to zero literally means we must make the base an insulator. Then Ib = 0. There are 2 problems. The device is no longer "bipolar". The word "bipolar" literally means "2 polarities". An insulator for the base region makes the device unable to conduct electrons emitted from the emitter. When b-e is fwd biased, how can emitted electrons get through the base & onward to collector if said base is an insulator?
A device w/ Ib = 0 is simply NOT BIPOLAR. A fwd biased p-n b-e jcn in close proximity to a rev biased n-p c-b jcn is why the device is called bipolar. An insulated input electrode that forces Ib to zero is not bipolar.
We have a transistor constructed with an insulated input electrode such that the input current dc value is zero. It is a MOSFET. The JFET is similar only that the input jcn is reverse biased resulting in near zero gate current at low freq. Neither of these devices is classified as bipolar. Both are voltage controlled. They need gate current to operate. But Vgs is the quantity directly controlled, with Ig being indirect & incidental, yet important.
If a device has Ib = 0, it is not a bjt. It's that simple. How can a b-e p-n jcn be fwd biased w/o base current? Bill you have to examine what you're saying & do a sanity check. An insulated input device is not a bjt! Period.
How can I make things any clearer. If Ib were zero how do I explain bjt action? Well, if we make Ib very low, as in a superbeta device, Ie still controls Ic. My current control model, straight from OEMs, uses Ie to control Ic, not Ib. Ib is a factor, but Ie is what Drs. Ebers & Molls used in their 1954 paper as the control quantity for Ic. I have the paper if you need it. I'll email it to anyone.
Another question is "what if Vbe was zero?" An ideal p-n jcn under forward bias should have zero voltage at any finite current value. One could say that Vbe can vanish as well. If every electron emitted from emitter entered the p type base region w/o a potential barrier being formed, the case for a perfect rectifying p-n jcn, Vbe equal zero, but Ic is still alpha*Ie. This requires a material w/ a band gap energy of zero.
One can theorize zero Vbe and/or zero Ib. Real world semiconductors have non-zero Ib & Vbe both. If both vanished, we still have Ic = alpha*Ie. That is still the transistor action law. An emitter resistor is connected to the bjt, & input signal is at the base. Ideally zero current enters the base, & Vbe is zero. The entire input voltage appears across emitter resistor Re. Ic = -Ib + Ie = 0 + Ie = Ie. The output voltage of this emitter follower is exactly Vin.
Hence we have an ideal buffer. A voltage gain of exactly one (zero Vbe drop), w/ infinite input impedance (Ib = 0). Ideally Ib & Vbe can both vanish. But we always have Ic = alpha*Ie = 1*Ie.
In reality, like it or not, Vbe & Ib are facts we must deal with. I'd love it if my emitter followers could swing to the rail (Vbe = 0). I'd love to drive a 25 amp motor w/ zero base current in the bjt buffer.
I can wish for zero Vbe & Ib all I want. Wishing isn't getting.
Any other questions? Best regards.
Claude
P: 116
Quote by cabraham If Ib was zero, it would NOT be a bjt!
OK. But first...
Do you now understand that my transistor article is not aimed at engineers? Right?
Did you take note of who the original intended audience was?
P: 116
Quote by cabraham Any other questions?
Do you now understand that my transistor article is not aimed at engineers?
When I stated who the original intended audience was, did you see that message?
P: 963
Quote by wbeaty Yes. Still looking for answers: Do you now understand that my transistor article is not aimed at engineers? When I stated who the original intended audience was, did you see that message?
Yes, I am well aware of the target audience. Offering explanations that differ from established canon is not the problem. Bringing in info that counters the canon is where I have trouble.
I realize that QM, charge control, doping density, band gap, etc., is more than a lay person can digest. But you made statements that were downright contrary to canon. That is why I made my remarks.
Everybody has their unique way of explaining things, & I make no claims that my explanations are superior. Personally, I agree with your method of examining the device as 2 back to back diodes opposing. Also, it is all important that we distinguish between 2 diodes & an actual functioning bjt.
The best method I used to teach this concept to non-electrical majors (civil, machanical, chemical) when I taught at uni, was as follows. Make the base region so wide, say 1.0 mm, that every e- (electron) emitted recombines in the base region with an h+ (hole). Here we have 2 back to back diodes & nothing more. There is no transistor action whatsoever.
But there is a potential barrier, a depletion zone, & an exponential I-V curve.
Ie = Ies*exp((Vbe/Vt) - 1).
Ic = alpha*Ie = 0, since alpha is zero for such an arrangement.
We have no transistor action at all, nothing but a fwd biased b-e jcn diode, & a rev biased c-b jcn diode. The logarithmic/exponential relation holds steadfast because it is a diode relation. We still have no bjt functionality.
Reducing the base region width results in an increase in collector current Ic since alpha is now increasing. When the base region is very thin, on the order of 1.0 micron, Ic is almost the value of Ie.
In the 1st case we had near zero Ic, the 2nd case shows Ic almost equal to Ie. What changed? NOT the potential barrier relation, not the I-V "Ohm's Law" relation, not the Shockley diode eqn. What changed is alpha. When alpha is near zero, we have a pair of back to back diodes per the exponential eqn above.
Ie is exponentially related to Vbe, Ic = 0. In the 2nd case, alpha approaches 1 & Ic = alpha*Ie = alpha*Ies*exp( ).
Bill I would recommend approaching the explanation of bjt action when addressing lay people in terms of the base being too thin to allow recombination. I personally like your 2 diode approach. If you start w/ a p-n jcn diode, explaning recombination, you can then advance to bjt. The difference is that the e- from the emitter pass through the super thin base so quickly, there is not enough time for recombination. The current from base to emitter literally overshoots the base & gets yanked into the collector.
Anyway, I believe that something along those lines might work for lay people. Again, I don't claim to be the best teacher at all. I get your point about the target audience. Thanks for your feedback & interest. BR.
Claude
P: 116
Quote by cabraham Yes, I am well aware of the target audience.
OK, thanks! And yes, comparing narrow base to extremely wide base is an excellent teaching strategy of which I previously hadn't heard.
Regarding canon, earlier you stated that the Ebers-Moll model is a CC model, and their original paper employed current-controlled current sources. Correct?
Are you certain that you're not misinterpreting something? N.b. that your assertion regarding Ebers-Moll being a CC model goes entirely contrary to numerous undergrad Uni books which give the following CV (large signal transconductance) equation as the central feature of Ebers-Moll model:
Ic=Is*(exp(Vbe/Vt)-1) (for large hfe of course, alpha ~=1)
Is this not canon? If the above CV equation isn't "Ebers-Moll model," then you've discovered a vast flaw in an enormous number of textbooks.
P: 5,462 Cabraham's approach in stressing the width of the interdepletion region is excellent and fundamental - or at least it was when I learned about transistors back in the 1960s. (post#68) Pierce devotes several pages to calculations of the effect of base width on carrier density, potential diagrams etc and has an excellent diagram on page 101, which I have appended. For expansion of these formulae (posts #68 & 69) I also recommend the monograph by E H Cooke-Yarborough of the Atomic energy Research Establishment at Harwell An introduction to transistor circuits (1957) go well Attached Thumbnails
P: 963
Quote by wbeaty OK, thanks! And yes, comparing narrow base to extremely wide base is an excellent teaching strategy of which I previously hadn't heard. Regarding canon, earlier you stated that the Ebers-Moll model is a CC model, and their original paper employed current-controlled current sources. Correct? Are you certain that you're not misinterpreting something? N.b. that your assertion regarding Ebers-Moll being a CC model goes entirely contrary to numerous undergrad Uni books which give the following CV (large signal transconductance) equation as the central feature of Ebers-Moll model: Ic=Is*(exp(Vbe/Vt)-1) (for large hfe of course, alpha ~=1) Is this not canon? If the above CV equation isn't "Ebers-Moll model," then you've discovered a vast flaw in an enormous number of textbooks.
Send me a message, & I'll email you the Ebers-Moll 1954 paper. It shows the bjt collector current being controlled by emitter current, i.e. a CCCS. Alpha is the parameter which determines the Ic value from Ie. They show a schematic w/ a CCCS controlling Ic.
The exp( ) relates the Vbe to Ic, but alpha is all important. The exp( ) covers the relation between Vbe & Ie. To get Ic we need alpha. The diode equation for the b-e jcn is Ie = Ies*exp((Vbe/Vt) - 1)> But Ic = alpha*Ie. So Ic = alpha*Ies*exp( ). The collector current is controlled by alpha*Ie. But Ie has a direct relation w/ Vbe as well as Ib.
All 3 eqns are relevant. A bjt offers both current gain as well as voltage gain. To compute current gain we use eqn 1) for the common emitter & emitter follower topologies. For the common base current gain, we use eqn 3). To compute voltage gain we use eqn 2). Again, all 3 eqns come into play when thoroughly analyzing bjt behavior.
We cannot make more than one quantity the controlling quantity. We generally control Ie, & Ic = alpha*Ie. Sometimes we control Ib, w/ Ic = beta*Ib. Usually this is not good due to beta dependency. We want networks that rely on well defined parameters like alpha & resistor values. Resistors have tight tolerances.
Send me a message, & I'll email you the E_M paper, the horses' mouth on the E-M eqns. BR.
Claude
P: 116
Quote by cabraham
Quote by wbeaty I'm trying to understand the details of Ebers-Moll. Based on your semiconductor expertise, would you say that the Ebers-Moll model depicts the BJT as a current-controlled device?
Yes, that is what figure 5 on page 1765 depicts.
I'll take a look.
Beware, this could be a case similar to Maxwell's Equations. Go to Maxwell's original work and you won't find his four equations. Maxwell never actually wrote those four, and probably wasn't aware they existed. Maxwell attacked the problem in an obscure way, employing magnetic vector potential, quaternions, and writing twenty equations. Later scientists came in and revised everything, producing the four equations known today. If you rely on Maxwell as the "horse's mouth," then you'd be in big trouble. (Officially the four equations are today called the Hertz/Heaviside equations. But Maxwell discovered the original mathematical form which describes EM fields.) See 2008 Microwave Journal, 23 years: Acceptance of Maxwell Theory http://bit.ly/qRQNCH
Related Discussions Electrical Engineering 9 Precalculus Mathematics Homework 13 Engineering Systems & Design 13 Calculus & Beyond Homework 3 General Math 4 | 6,254 | 26,814 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2013-48 | longest | en | 0.970246 |
https://www.teacherspayteachers.com/Product/Subtracting-Decimals-Activity-Pack-3121741 | 1,531,824,317,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589634.23/warc/CC-MAIN-20180717090227-20180717110227-00272.warc.gz | 1,007,239,120 | 17,555 | # Subtracting Decimals Activity Pack
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This subtracting decimals activity pack is full of supplemental resources perfect for reinforcing skills taught in 6th grade math. Let students choose their work based on their different learning styles or use multiple activities in differentiated stations.
★WHAT'S INCLUDED:★
• a 9 problem solve and sketch activity
• a 8 problem cut and paste activity
• a 10 problem mistory lib activity on the life of Annie Oakley
• a 10 problem coloring page activity
• a 10 piece adding decimals flap book (perfect for a later review)
• answer keys for each activity
★WHAT IS A SOLVE AND SKETCH ACTIVITY:★
In a solve and sketch activity, students will solve a problem and circle the solution. They will then find the problem block on a blank answer document page. There they will sketch in the picture assigned to the solution. Once all 9 blocks have been sketched in, there will be a large, put together image revealed.
★WHAT IS A MISTORY LIB:★
A mistory lib is a fill in the blank story that can be completed by correctly solving each math problem. When the story is complete you will have learned a little history about a particular person or topic. Basically a math activity with a history twist = mistory lib.
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Sign Up | 569 | 2,426 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-30 | latest | en | 0.870968 |
https://brownmath.com/swt/books.htm | 1,695,768,374,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00072.warc.gz | 153,796,381 | 4,526 | BrownMath.com → Stats w/o Tears → Recommended Books
# Stats without TearsRecommended Statistics Books
Updated 15 Nov 2021
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These pages change automatically for your screen or printer. Underlined text, printed URLs, and the table of contents become live links on screen; and you can use your browser’s commands to change the size of the text or search for key words. If you print, I suggest black-and-white, two-sided printing.
## Statistics for Citizens
For more practical applications, written in a non-technical way, I recommend:
• Lewis, H.W. (1997). Why Flip a Coin? John Wiley & Sons.
Lots of illustrations of how thinking in a statistical way can help you make decisions. Applications include voting, gambling, war, and the stock market. A very enjoyable read.
• Mlodinow, Leonard. (2008). The Drunkard’s Walk: How Randomness Rules Our Lives. Pantheon.
If you’re like most people, your intuitions about anything having to do with probability are usually wrong. Without using formulas, Mlodinow gets you to think more clearly about probability, with zillions of real-world examples from all areas of everyday life.
• Malkiel, Burton G. (2003). A Random Walk down Wall Street. W.W. Norton & Company.
Malkiel has done exhaustive statistical analysis of the stock market, to help you make wise decisions. He doesn’t just tell you what to do, he shows you the statistical evidence and explains what the statistics mean.
You’ll probably find a later edition; Malkiel updates the book every couple of years.
• Vickers, Andrew (2010). What Is a p-Value Anyway? 34 Stories to Help You Actually Understand Statistics. Addison-Wesley.
Statistics really is stories, and these stories are short and fun, but each with a point behind it. Stats can seem like a bunch of formulas, but really it’s about learning to think logically, and Vickers does a great job of leading you to that way of thinking.
• Dewdney, A.K. (1993). 200% of Nothing. John Wiley & Sons.
An excellent and highly readable tour through probability. Dewdney presents lots of situations, many from advertising, and helps you see how to use statistical thinking (educated common sense, really) to avoid being taken in.
The book runs out of steam toward the end, but the first nine or ten chapters are excellent — I particularly recommend “The Great Pepsi Challenge” (page 24), the lottery discussion (pages 56–59), and the decision whether to buy a store’s extended warranty (page 91).
• Gigerenzer, Gerd. (2002). Calculated Risks. Simon & Schuster.
For ordinary people making legal or medical decisions. Some applications include AIDS counseling and DNA evidence. A very enjoyable read.
• Latzko, William J., and David M. Saunders. (1995). Four Days with Dr. Deming. Addison-Wesley.
Dr. W. Edwards Deming is famous for teaching first Japanese and then American businesses to apply statistical methods to management, especially to quality. This book is in the form of a CEO’s “thought journal” at one of Dr. Deming’s seminars, as he finds that pretty much everything he knew about managing his company is wrong. It’s kept lively with lots of pictures and conversations, plus of course quotes from Dr. Deming’s lectures. You’ll want to read this book slowly and let the ideas expand in your mind.
“It’s so simple,” says Dr. Deming, and he’s right. You don’t actually need a statistical background to understand this book, but you’ll recognize that the key ideas come from our week 9, sample variability.
If you can’t find these in your library for regular checkout, ask a librarian to get them from another library for you, or get them from a bookseller.
## Textbooks
The best textbook I’ve seen is DeVeaux, Velleman, and Bock’s Intro Stats (Pearson Addison Wesley, 2009). It’s written in a breezy, conversational style, perfect for self study. I really like the many sections “What can go wrong?” because you should always have possible pitfalls in mind.
Another excellent textbook is Freedman, Pisani, and Purves, Statistics (Norton, 2007). There’s not as much eye candy — no color at all, for instance — but it says what needs to be said and spends adequate time on the philosophy behind the methods.
If you’d like something a little less formal than a textbook, I recommend Larry Gonick & Woollcott Smith, The Cartoon Guide to Statistics. (HarperPerennial, 1993, ISBN 0-06-273102-5). Despite its lighthearted appearance, this is actually a pretty good statistics book. Its advantages include high readability, brief explanations, and low cost (under \$12 new at Amazon in March 2004). On the down side, it presents things in a different order from our course, it doesn’t cover data types or χ², and you need to look elsewhere for practice problems. Still I think it’s great value for the money.
## What’s New?
• 15 Nov 2021: Update Dewdney’s URL.
• 27 Nov 2015: Add The Drunkard’s Walk.
• (intervening changes suppressed)
• 10 May 2004: New article.
Because this textbook helps you, | 1,160 | 4,986 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2023-40 | latest | en | 0.896655 |
https://geometrycoach.com/triangle-congruence-by-sss-and-sas/ | 1,721,605,909,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517796.21/warc/CC-MAIN-20240721213034-20240722003034-00843.warc.gz | 249,493,966 | 23,660 | # Triangle Congruence by SSS and SAS
Before you can ever start with proofs your students need to have a clear understanding of what makes sides and angles of triangles congruent. This lesson on Triangle Congruence by SSS and SAS is one of the more memorization based lessons to teach. With that said the only way to memorize something and master is it to see it multiple times in multiple ways. Here are a couple of things to help you out!
## Congruent Figures
Two figures are said to be congruent if they have same shape and the same size. The corresponding angles of the two congruent figures are equal. The corresponding sides are congruent.
The two figures shown below are congruent.
The sides of two rectangles have a ratio of 1: 1. The rectangles are congruent.
## Congruent Triangles
### Triangle Congruence by Side Side Side
The congruence of triangles can be proved using its three sides and three angles.
#### If all the three sides of one triangle are congruent to all the three corresponding sides of another triangle, the two triangles are said to be congruent.
The two triangles shown below are congruent by SSS Postulate.
### Triangle Congruence by Side Angle Side
#### If the two sides and their included angle of one triangle is congruent to the two sides and their included angle of another triangle, then the two triangles are said to be congruent.
The included angle is the angle made at the point where two sides of a triangle meet. The two triangles shown below are congruent by SAS Postulate.
## Triangle Congruence by SSS and SAS Worksheet – Docs & PowerPoint
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## Triangle Congruence by SSS and SAS – PDFs
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4-2 Slide Show (FREEBIE) | 491 | 2,120 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-30 | latest | en | 0.920568 |
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Fearliss
Registered: 31.03.2003
From:
Posted: Saturday 30th of Dec 08:26 Hi dudes , It’s been more than a week now and I still can’t figure out how to solve a few math problems on lowest denominator calculator . I have to submit this work by the end of next week. Can someone show me the way to get started? I need some help with percentages and side-angle-side similarity. Any sort of guidance will be appreciated.
IlbendF
Registered: 11.03.2004
From: Netherlands
Posted: Sunday 31st of Dec 15:58 Hi! I guess I can help you out on how to solve your problem. But for that I need more details. Can you give details about what exactly is the lowest denominator calculator problem that you have to work out. I am quite good at solving these kind of things. Plus I have this great software Algebrator that I downloaded from the internet which is soooo good at solving algebra assignment. Give me the details and perhaps we can work something out...
thicxolmed01
Registered: 16.05.2004
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Posted: Monday 01st of Jan 13:16 I know a couple of professors who use Algebrator themselves to teach students.
humimar
Registered: 17.02.2005
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Posted: Wednesday 03rd of Jan 09:19 I appreciate your help. Now I’m so curious to check out this new software. Can someone post a website address where I can order a copy for myself ?
Homuck
Registered: 05.07.2001
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https://www.wyzant.com/resources/answers/21411/an_experienced_painter_can_paint_a_house_in_6hrs_while_it_takes_an_apprentice_painter_9hrs_to_paint_the_same_size_house_how_long_would_it_take_for_2_of_them | 1,527,465,841,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794870497.66/warc/CC-MAIN-20180527225404-20180528005404-00202.warc.gz | 874,792,935 | 16,337 | 0
# an experienced painter can paint a house in 6hrs while it takes an apprentice painter 9hrs to paint the same size house, how long would it take for 2 of them
together to paint a similar house?
### 1 Answer by Expert Tutors
Betsy S. | Mathematics TutorMathematics Tutor
5.0 5.0 (1 lesson ratings) (1)
0
Hi Tanisha,
This one took me some time to figure out the best way to complete.
First think about the rate it takes each of them to work in what fraction of a house they can complete in an hour.
profession: 1/6 houses/hr
apprentice: 1/9 houses/hr
Next, since they will be working together combine their painting speed:
(1/6)+(1/9) = 5/18
That means working together they can complete 5 houses in 18 hours.
So 1 house can be done in 18/5 or 3.6hrs.
Note: Since time doesn't uses decimals 0.6hrs is the equivalent of 36 minutes, so the answer can be written as 3 hours and 36 minutes.
I hope this is understandable, let me know if you need more help. | 264 | 964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2018-22 | latest | en | 0.951412 |
https://antennamediatorino.eu/en/what-are-some-mathematical-fields-that-can-be-useful-to-philosophers/ | 1,656,784,095,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104189587.61/warc/CC-MAIN-20220702162147-20220702192147-00696.warc.gz | 141,503,703 | 20,503 | # What are some mathematical fields that can be useful to philosophers?
## How is math used in philosophy?
Mathematical knowledge and the ability to use it is the most important means of tackling quantifiable problems, while philosophical training enhances the ability to analyse issues, question received assumptions and clearly articulate understanding.
## Do philosophers study math?
Mathematics is the philosophical language nature prefers, and science is the only truly effective means we have for connecting our philosophy to reality. Thus maths and science are crucial for good philosophy – for getting things right. Truth is not always intuitive or comfortable.
## What is the most useful field of math?
Arithmetic also involves more complex concepts of mathematics such as limits, exponents, etc. This is the simplest and the most essential branch of mathematics since it’s used in our everyday life and also at the same time, used for computation, etc.
## Who are the 10 philosophers of mathematics?
And, famously, he didn’t eat beans.
• Hypatia (cAD360-415) Hypatia (375-415AD), a Greek woman mathematician and philosopher. …
• Girolamo Cardano (1501 -1576) …
• Leonhard Euler (1707- 1783) …
• Carl Friedrich Gauss (1777-1855) …
• Georg Cantor (1845-1918) …
• Paul Erdös (1913-1996) …
• John Horton Conway (b1937) …
• Grigori Perelman (b1966)
## How are philosophy and math connected?
Mathematics is quantitative in nature, whereas Philosophy is qualitative. Mathematics is about numbers; Philosophy is about ideas. The key link then between the two subjects is logical problem solving. The mathematical proof and philosophical argument bear a strong resemblance.
## Why is mathematics part of philosophy?
Then mathematics could be defined as one of the branches of philosophy in which theories are built on definitions and axioms and the results are proven and physics can be thought of as some kind of philosophical theory of laws of nature (you know the full Latin name of Newton´s book Principia) that are seeked both …
## Who invented 0?
Brahmagupta
“Zero and its operation are first defined by [Hindu astronomer and mathematician] Brahmagupta in 628,” said Gobets. He developed a symbol for zero: a dot underneath numbers.
## Who is the No 1 mathematician of the world?
Isaac Newton is a hard act to follow, but if anyone can pull it off, it’s Carl Gauss. If Newton is considered the greatest scientist of all time, Gauss could easily be called the greatest mathematician ever.
## Who is the best mathematician alive?
Ten Most Influential Mathematicians Today
• Ian Stewart.
• John Stillwell.
• Bruce C. Berndt.
• Timothy Gowers.
• Peter Sarnak.
• Martin Hairer.
• Ingrid Daubechies.
• Andrew Wiles.
## Who invented infinity?
infinity, the concept of something that is unlimited, endless, without bound. The common symbol for infinity, ∞, was invented by the English mathematician John Wallis in 1655. Three main types of infinity may be distinguished: the mathematical, the physical, and the metaphysical.
## Who invented 1?
Hindu-Arabic numerals, set of 10 symbols—1, 2, 3, 4, 5, 6, 7, 8, 9, 0—that represent numbers in the decimal number system. They originated in India in the 6th or 7th century and were introduced to Europe through the writings of Middle Eastern mathematicians, especially al-Khwarizmi and al-Kindi, about the 12th century.
## Who invented school?
Horace Mann invented school and what is today the United States’ modern school system. Horace was born in 1796 in Massachusetts and became the Secretary of Education in Massachusettes where he championed an organized and set curriculum of core knowledge for each student.
## Who invented homework?
Roberto Nevelis
Roberto Nevelis of Venice, Italy, is often credited with having invented homework in 1095—or 1905, depending on your sources.
## Who taught the first teacher?
Of course, if we were to believe Greek mythology, it was the god Chiron who taught the first teacher, seeing as that the centaur was known for his abilities to impart knowledge. | 926 | 4,070 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2022-27 | longest | en | 0.901708 |
http://forums.wolfram.com/mathgroup/archive/2010/Jul/msg00110.html | 1,575,676,950,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540491491.18/warc/CC-MAIN-20191206222837-20191207010837-00010.warc.gz | 59,351,515 | 7,974 | Re: FindRoot with parameterized interpolated function from NDSolve
• To: mathgroup at smc.vnet.net
• Subject: [mg110759] Re: FindRoot with parameterized interpolated function from NDSolve
• From: Peter Pein <petsie at dordos.net>
• Date: Mon, 5 Jul 2010 06:01:52 -0400 (EDT)
• References: <i0pmlc\$slv\$1@smc.vnet.net>
```Am Sun, 4 Jul 2010 10:09:48 +0000 (UTC)
schrieb Ulvi Yurtsever <a at b.c>:
> I have an interpolated function obtained
> as the solution to a system of ODEs via NDSOlve.
> The system and the solution depend on a number
> of parameters. I then want to plug this function into
> FindRoot to find the numerical solution of a system
> of equations which depend on both the dependent variable
> of the ODEs and the parameters. Mathematica barfs at
> the use of parameters as in the following example:
>
> First, what works as expected:
>
>
> In[135]:= solnn =.
>
> In[142]:=
> solnn[a_] :=
> NDSolve[{x'[t] == a *y[t], y'[t] == -x[t], x[0] == 1, y[0] == 0}, {x,
> y}, {t, 0, Pi}]
>
> In[144]:= FindRoot[(x /. solnn[1][[1]])[t] - (y /. solnn[1][[1]])[
> t] == 0, {t, 2}]
>
> Out[144]= {t -> 2.35619}
>
>
> however:
>
>
> FindRoot[{(x /. solnn[a][[1]])[t] - (y /. solnn[a][[1]])[t] == 0,
> a - 1 == 0}, {{a, 0}, {t, 2}}]
>
>
> produces, instead of the expected
>
> {a->1., t->2.355619},
>
> lots of error messages to the effect that NDSolve has encountered non-
> numerical initial values etc
>
> Is there any other way to use FindRoot for the purpose I am trying
> to use it?
>
Define an auxilliary function which evaluates iff a _and_ t have got
numeric values (I changed solnn a bit for better handling):
In[1]:= Clear[solnn,x,y];
solnn[a_?NumericQ] :=
Block[{t},
First[
NDSolve[{x'[t] == a*y[t], y'[t] == -x[t], x[0]==1, y[0] == 0},
{x, y}, {t, 0, Pi}]
]]
In[3]:= FindRoot[x[t] - y[t] == 0 /. solnn[1], {t ,2}]
Out[3]= {t->2.35619}
In[4]:= target[a_?NumericQ, t_?NumericQ] := x[t] - y[t] /. solnn[a]
In[5]:= FindRoot[{target[a, t] == 0, a - t/3 == 0},
{{a, 0}, {t, 2, 0, Pi}}]
Out[5]= {a->0.860285,t->2.58085}
Peter
```
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• Next by thread: Re: FindRoot with parameterized interpolated function from NDSolve | 850 | 2,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2019-51 | longest | en | 0.642239 |
https://help.scilab.org/docs/6.0.1/en_US/penlaur.html | 1,721,270,809,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514816.43/warc/CC-MAIN-20240718003641-20240718033641-00541.warc.gz | 254,961,970 | 3,861 | Change language to:
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Please note that the recommended version of Scilab is 2024.1.0. This page might be outdated.
See the recommended documentation of this function
Scilab Help >> Linear Algebra > Matrix Pencil > penlaur
# penlaur
Laurent coefficients of matrix pencil
### Syntax
```[Si,Pi,Di,order]=penlaur(Fs)
[Si,Pi,Di,order]=penlaur(E,A)```
### Arguments
Fs
a regular pencil `s*E-A`
E, A
two real square matrices
Si,Pi,Di
three real square matrices
order
integer
### Description
`penlaur` computes the first Laurent coefficients of `(s*E-A)^-1` at infinity.
`(s*E-A)^-1 = ... + Si/s - Pi - s*Di + ...` at `s` = infinity.
`order` = order of the singularity (order=index-1).
The matrix pencil `Fs=s*E-A` should be invertible.
For a index-zero pencil, `Pi, Di,...` are zero and `Si=inv(E)`.
For a index-one pencil (order=0),`Di` =0.
For higher-index pencils, the terms `-s^2 Di(2), -s^3 Di(3),...` are given by:
`Di(2)=Di*A*Di`, `Di(3)=Di*A*Di*A*Di` (up to `Di(order)`).
### Remark
Experimental version: troubles when bad conditioning of `so*E-A`
### Examples
```F=randpencil([],[1,2],[1,2,3],[]);
F=rand(6,6)*F*rand(6,6);[E,A]=pen2ea(F);
[Si,Pi,Di]=penlaur(F);
[Bfs,Bis,chis]=glever(F);
norm(coeff(Bis,1)-Di,1)```
### See also
• glever — inverse of matrix pencil
• pencan — canonical form of matrix pencil
• rowshuff — shuffle algorithm
Report an issue << pencan Matrix Pencil quaskro >>
Copyright (c) 2022-2024 (Dassault Systèmes)Copyright (c) 2017-2022 (ESI Group)Copyright (c) 2011-2017 (Scilab Enterprises)Copyright (c) 1989-2012 (INRIA)Copyright (c) 1989-2007 (ENPC)with contributors Last updated:Mon Feb 12 19:26:47 CET 2018 | 575 | 1,701 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-30 | latest | en | 0.581407 |
https://rebab.net/1-liter-is-how-many-cups/ | 1,659,996,800,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570879.1/warc/CC-MAIN-20220808213349-20220809003349-00199.warc.gz | 453,398,922 | 6,294 | How many cups in a liter? We will answer this question and help you master precise measuring so that you never fail at any recipe again!
Cooking might seem simple, especially to those who do not practice it on their own.
You are watching: 1 liter is how many cups
It seems like you only need to read the recipe, follow a few straightforward instructions, and that’s it. How come then so many people fail at it?
I’ll tell you a secret. It is not as simple as it seems. You can have the best recipe in your hands and still end up with a messy kitchen and your meal in a trash can.
Why is that?
Numerous things can go wrong, but today I’ll deal with measuring issues.
Let’s paint the picture: you have found a great recipe online, read the reviews, and made sure that it turns out great.
Unfortunately, there is a problem: the recipe lists measurements you are not familiar with – for example, cups instead of milliliters.
In such situations, metric, imperial, and the US conversion tables come in handy as they guide you and help you prepare the recipe the right way.
I’ve also created Cooking Measurements and Unit Conversion Table to help you out.
Or you could opt out of other conversion calculator app.
Quite often, recipes call for cups of water or milk, while some other recipes require you to measure your liquid ingredients in milliliters and liters.
The question arises: How many cups in a liter?
I will resolve this dilemma for you…:)
Also, check out my special kitchen cheat sheet.
## How Many Cups in a Liter
Let us cut right to the chase!
We all know that recipes require precision and that we have to measure the ingredients accurately if we want to be successful.
Besides a reliable kitchen scale, it would be a good idea to have a set of measuring cups as well.
As we said, recipes often call for cups of milk or water, but there are different kinds of cups, and it is sometimes safer to measure liquid ingredients in millimeters, deciliters, and liters.
In general, and quite roughly, one liter is usually considered equivalent to about four average cups.
However, the measurement differs somewhat depending on which type of cups are used.
There are following types of cups:
For us, it is most important to know the metric, US, and the UK or Imperial system as that is what we encounter in our day to day life. So, here we go:
### 1. The Metric System
In the metric system, 1 liter equals 1000 mL. One metric cup equals 250 mL. You can easily calculate how many cups there are in one liter: 1000 / 250 = 4. Therefore, there are four cups in one liter in the metric system.
### 2. For Imperial System (UK)
We often find recipes written for the UK audience – it is logical as we share the same language. Unfortunately, we do not share the same measuring system as well. British people rely on the Imperial system.One imperial cup has a capacity of 284.131 mL or 10 imperial fluid ounces. One liter equals 1000 mL or 35.1951 imperial fluid ounces. That means that math is as follows: 35.1951 / 10 = 3.51951. As you can see, there are 3.51951 imperial cups in one liter in the UK system.
### 3. The US system
In the United States, we rely on the so-called US cup when it comes to measuring liquids. 1 US cup has a capacity of 236.58 mL or 8 US fluid ounces. One liter equals 1000 mL or 33.814 US fluid ounces.The math is the same as in the previous example: 33.814 / 8 = 4.22675. Hence, there are 4.22675 cups in one liter in the US system.
### Word of Caution
Nowadays, people in the UK use the metric system rather than the imperial one.
In translation, the modern recipes will deal with the metric cups rather than the traditional imperial cups.
How can you know whether the person who wrote the recipe meant the imperial cup or the metric one?
You can never be 100% sure, but if you find the recipe in the old recipe book, you can assume that the conversion you need is from the imperial system (3.52 cups in 1 liter).
If you find the recipe online, which is more likely, the answer is probably 4 cups per liter as given in the metric system.
Related article: How Many Ounces There Are In A Shot; How Many Tablespoons Are In One Ounce
## What to Do When You Need to Convert More than One Liter into Cups?
It is quite simple, you will use the same logic and the same math formula and multiply the number of liters with the number of cups – of course, this number will be different depending on the system you are referring to.
Let us label the number of liters with an “x,” the math would then be as follows:
The US system: X multiplied by 4.22675The Imperial system: X multiplied by 3.51951The metric system: X multiplied by 4
Here are some examples that will clarify everything even further:
If you are interested in finding out how many US cups there are in a liter and a half, you will simply multiply 1.5 by 4.22675 and find out that you need 6.34 US cups.
Likewise, you can use this volume units conversion tool to convert between imperial, US, and metric cups and liters.
## Volume Units in Short + 2 Practical & Easy to Follow Reference Lists
### 1. Liter
It is a metric system volume unit, and the abbreviation used to label it is “L.”
1L = 1000 mL = 33.814 US fluid ounces = 35.1951 imperial fluid ounces.
### 2. Cup
Cup is another volume unit used to measure both liquids and dry ingredients (with some differences discussed later). The abbreviation used to mark one cup is “c.”
1 US c = 8 US fluid ounces1 imperial c = 10 imperial fluid ounces1 metric cup = 250 mL.
### Word of Caution: Dry cups vs. Liquid cups
The US liquid cups are different from US dry cup measurements, and you should acknowledge that before you start converting the measurements provided in your recipe.
If you need to measure dry ingredients such as flour or sugar, you should use a dry measuring cup. In this way, the measurement will be far more precise as it is often too hard to level off the dry ingredients in the liquid measuring cup.
After all, dry cups convert to grams and ounces and cannot directly translate to milliliters and liters, and the same goes the other way around.
See more: Why Does Atomic Radius Increase Down A Group ? Periodic Trends
Further Reading: How Many Ounces There Are In A Quart
### Bonus
Practical reference list for the metric measurement system:
1/4 cup: 60 mL1/3 cup: 70 mL1/2 cup: 125 mL2/3 cup: 150 mL3/4 cup: 175 mL1 cup: 250 mL1 1/2 cups: 375 mL2 cups: 500 mL4 cups: 1 liter
Practical reference list for the Metric to Imperial conversion:
25 ml: 1 fl oz50 ml: 2 fl oz75 ml: 2 1/2 fl oz100 ml: 3 1/2 fl oz125 ml: 4 fl oz150 ml: 5 fl oz175 ml: 6 fl oz200 ml: 7 fl oz225 ml: 8 fl oz250 ml: 9 fl oz300 ml: 10 fl oz350 ml: 12 fl oz400 ml: 14 fl oz425 ml: 15 fl oz450 ml: 16 fl oz500 ml: 18 fl oz600 ml: 1 pint700 ml: 1 1/4 pints850 ml: 1 1/2 pints1 liter: 1 3/4 pints
## Conclusion
Learning how many cups in a liter will make you one step closer to mastering precise measuring and perfecting your cooking skills.
If you have anything to add, feel free to do so. If not, share the knowledge! | 1,749 | 7,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2022-33 | latest | en | 0.966537 |
https://math.answers.com/questions/What_is_700_divided_by_9 | 1,642,978,616,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304309.59/warc/CC-MAIN-20220123202547-20220123232547-00022.warc.gz | 449,417,154 | 73,305 | 0
What is 700 divided by 9?
Wiki User
2016-03-18 01:15:28
77.7778
Melvin Bayer
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2021-02-26 23:42:31
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2016-03-18 01:15:28
77.7778
700 divided by 9 is 77 with remainder 7 or 77.7777777778
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View results | 195 | 515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-05 | latest | en | 0.859878 |
http://www.sdcpublications.com/multimedia/978-1-58503-767-4/files/krb/krb_pr_page3.htm | 1,527,106,636,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865809.59/warc/CC-MAIN-20180523200115-20180523220115-00314.warc.gz | 467,085,026 | 1,807 | # Pure Rotation
### Gears and belts
• Belts: The velocity (v) and tangential acceleration (at) of the belt must be the same at any point along the belt. The normal acceleration will change depending on the pulley radius.
• Gear: The velocity (v) and tangential acceleration (at) of the contacting point on both gears are equal. The normal acceleration will change depending on the gear radii. The gear radii ratio of two meshing gears is dependent on the ratio of the number of gear teeth (N).
### Gear and belt equations
Based on the gear and belt properties listed above, we can derive the following equations.
• Angular velocity: v = r1ω1 = r2ω2
•
• Gear teeth relationship: N1ω1 = N2ω2
•
• Angular acceleration: at = r1α1 = r2α2 | 185 | 739 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-22 | latest | en | 0.837248 |
https://resources.orcad.com/schematic-capture-circuit-simulation/2020-tips-for-finding-current-in-a-circuit | 1,627,320,610,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152144.81/warc/CC-MAIN-20210726152107-20210726182107-00640.warc.gz | 491,236,283 | 43,644 | # Tips for Finding Current in a Circuit
March 5, 2020
One of my previous trips abroad almost ended in a disaster. As a meticulous person, I’ve always kept my passport by my side. Before you wonder, no, I didn’t lose the passport but rather the arrival slip that I ought to have returned to immigration before boarding the flight.
It took me 30 minutes of frantic searching before finding the slip in a heap of receipts. I’ve no idea how it ended among the receipts, nor how I’ve ended with spending almost all my money that results in that thick pile of receipts.
Thankfully, I didn’t have to go through the hassle of hours of interrogation by the immigration department. Similarly, you can get into various trouble if you forgot to calculate the current in your circuit, particularly when it’s crucial to the functionality of the design.
## Finding Current In A Circuit With Ohm’s Law
You can’t discuss calculating current without bringing up the most basic laws of electronics: Ohm’s Law. Ohm’s Law denotes the relation between voltage, current, and resistance. You ought to be familiar with the equation V = IR.
From the equation, you can easily get the value of the current by dividing the voltage with the resistance or:
I = V/R.
That’s basic electronics that you ought to know before even drafting a circuit. The calculation looks simple when you have a simple circuit with a single voltage and resistor element. In a more complex circuit where you have a nest of series and parallel configurations, you’ll need to analyze the circuit between nodes.
A voltage divider calculation will be handy in determining the voltage that falls between two nodes. It also helps to understand that electric current splits into different branches according to the resistance of the individual branch.
### Current Calculation For Capacitor And Inductor
As useful as it is, Ohm’s law only applies for resistive loads in the circuit. If you have a capacitor and inductor, you’ll need to approach the circuit analysis differently. In a DC circuit, where the voltage source is a DC signal, the following rules apply:
• All inductors are short circuit.
• All capacitors are open circuit.
However, the above assumptions are no longer true when AC signals are applied to capacitors and inductors.
The current flowing through a capacitor from a sinusoidal voltage is stored as a charge and it’s given by the following formula:
I = C (dv/dt)
From the equation, it can be seen that the capacitor goes through a continuous series of charge and discharge when an AC voltage is applied.
Capacitors behave differently in DC and AC analysis.
As for the inductor, the calculation is slightly more complicated due to the characteristic of the component. Inductors produce an opposing force to the current that flows through it and the current is calculated by:
I = 1/L ∫vdt
When you plot the values of current against voltage for capacitors and inductors, you’ll notice that both parameters are 90 degrees out of phase with each other.
### Determining Current For Non-Linear Circuits
It’s a mistake to overlook the current calculation for non-linear components. Resistors, capacitors, and inductors are linear components, which means the resulting current is proportionate to the voltage applied.
Non-linear components have an I-V curve that is not proportionate, in which you can’t determine the value of a current with a single formula. For example, a diode is a non-linear component as very little current passes through when the voltage across it is less than its forward voltage (Vf). As the voltage passes Vf, the forward current increases drastically and it’s only limited by the maximum limit of the component.
Diodes are non-linear components where current is not proportionate to the voltage applied.
Transistors, varistors, and Zener diodes are some examples of non-linear components. When these components are placed in a circuit, you’ll need to consider how the non-linear I-V curve will affect the flow of the current to other components.
It’s important to know the current that is passing through in a circuit if you’re working on a PCB. The amount of current will determine the thickness and width of the PCB trace. This is crucial when you’re dealing with high-power designs.
Working with the suite of layout tools and analysis options from Cadence can make any of your current design challenges a breeze. However, manually calculating the current can be tedious, and even experienced designers do make mistakes. Therefore, it helps if you’re using a PCB software like OrCAD PCB Designer that’s equipped with a SPICE simulator. Verifying the calculation will minimize the errors. | 958 | 4,698 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-31 | latest | en | 0.935963 |
https://metanumbers.com/38119 | 1,627,455,650,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153531.10/warc/CC-MAIN-20210728060744-20210728090744-00589.warc.gz | 402,960,688 | 10,831 | ## 38119
38,119 (thirty-eight thousand one hundred nineteen) is an odd five-digits prime number following 38118 and preceding 38120. In scientific notation, it is written as 3.8119 × 104. The sum of its digits is 22. It has a total of 1 prime factor and 2 positive divisors. There are 38,118 positive integers (up to 38119) that are relatively prime to 38119.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 22
• Digital Root 4
## Name
Short name 38 thousand 119 thirty-eight thousand one hundred nineteen
## Notation
Scientific notation 3.8119 × 104 38.119 × 103
## Prime Factorization of 38119
Prime Factorization 38119
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 38119 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.5485 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 38,119 is 38119. Since it has a total of 1 prime factor, 38,119 is a prime number.
## Divisors of 38119
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 38120 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 19060 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 195.241 Returns the nth root of the product of n divisors H(n) 1.99995 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 38,119 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 38,119) is 38,120, the average is 19,060.
## Other Arithmetic Functions (n = 38119)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 38118 Total number of positive integers not greater than n that are coprime to n λ(n) 38118 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 4023 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 38,118 positive integers (less than 38,119) that are coprime with 38,119. And there are approximately 4,023 prime numbers less than or equal to 38,119.
## Divisibility of 38119
m n mod m 2 3 4 5 6 7 8 9 1 1 3 4 1 4 7 4
38,119 is not divisible by any number less than or equal to 9.
## Classification of 38119
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Prime Power
• Square Free
## Base conversion (38119)
Base System Value
2 Binary 1001010011100111
3 Ternary 1221021211
4 Quaternary 21103213
5 Quinary 2204434
6 Senary 452251
8 Octal 112347
10 Decimal 38119
12 Duodecimal 1a087
20 Vigesimal 4f5j
36 Base36 tev
## Basic calculations (n = 38119)
### Multiplication
n×i
n×2 76238 114357 152476 190595
### Division
ni
n⁄2 19059.5 12706.3 9529.75 7623.8
### Exponentiation
ni
n2 1453058161 55389124039159 2111378019248701921 80483618715741268526599
### Nth Root
i√n
2√n 195.241 33.6548 13.9729 8.24571
## 38119 as geometric shapes
### Circle
Diameter 76238 239509 4.56492e+09
### Sphere
Volume 2.32013e+14 1.82597e+10 239509
### Square
Length = n
Perimeter 152476 1.45306e+09 53908.4
### Cube
Length = n
Surface area 8.71835e+09 5.53891e+13 66024
### Equilateral Triangle
Length = n
Perimeter 114357 6.29193e+08 33012
### Triangular Pyramid
Length = n
Surface area 2.51677e+09 6.52767e+12 31124
## Cryptographic Hash Functions
md5 8dd831e4a369590fd04f354d960b2967 39fc045cb7b0e600559908491ff05091db2e75a6 ec2163be05eab497abf7af5e82c9b48a7182800bf838dda822803cb43a0de394 9bb2a75d12a50e61ced19635459356e7f21e234e4fef27871647e8213340c579048b81a53db9f6172bde433205fa8d67df9924e8a034ce3c4770fd7583a78b28 80a97766c2ab84adbaf2d662984b7bc593639878 | 1,429 | 4,142 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2021-31 | latest | en | 0.814342 |
http://www.fixya.com/support/t22437621-bar122hgla_weather_station_change_main | 1,532,334,681,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676595531.70/warc/CC-MAIN-20180723071245-20180723091245-00360.warc.gz | 460,386,648 | 32,444 | Question about Oregon Scientific Lwb0997110113004 Thermometer, Tht312/Blrw Clock Radio
# I have a BAR122HGLA weather station,how do I change the main unit to Celsius from fahrenheit
How do I change the main unit to Celsius?
Posted by Anonymous on
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## 2 Suggested Answers
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Posted on Jan 02, 2017
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• 1 Answer
Based on a suggestion from another thread, I removed the batteries from the base unit, waited until the display went blank,then replaced the batteries. After a few seconds the outside temp started displaying again.
Posted on Dec 16, 2009
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1 Answer
### How to change from Celsius to fehrenheit
On the Celsius scale, the freezing and boiling points of water are exactly 100 degrees apart, thus the unit of the Fahrenheit scale, a degree Fahrenheit, is 5/9 of a degree Celsius. The Fahrenheit scale coincides with the Celsius scale at -40 °F, which is the same temperature as -40 °C.
Sep 11, 2017 | Cars & Trucks
Tip
### C++ Source code for Celsius to Fahrenheit
#include <cstdio>
#include <cstdlib>
#include <iostream>
using namespace std;
int main(float nNumberofArgs, char* pszArgs[])
{
double celsius;
cout << "Enter the temperature in Celsius:";
cin >> celsius;
double factor;
factor = 212 - 32;
double fahrenheit;
fahrenheit = factor * celsius/100 + 32;
cout << "Fahrenheit value is:";
cout << fahrenheit << endl;
system("PAUSE");
return 0;
}
on Jun 16, 2010 | Computers & Internet
1 Answer
### How do I change temperature reading from cecius to fahreenheit?
Is there a button marked ?C/?F: ?
?C/?F:
Press and release the "?C/?F" button to switch between Celsius and Fahrenheit mode.
Jun 21, 2016 | Acu-Rite Acurite Wireless Indooroutdoor...
1 Answer
1. ### Can I still switch my temperature from Fahrenheit to Celsius
feedback.weather.com/.../24423-weather-com-can-... The Weather Channelweather.com - Local Forecast / Maps / Video. Yes, you can switch the temperaturescale for any location on weather.com from Fahrenheit to Celsius, and back ...
Jul 22, 2015 | Measuring Tools & Sensors
1 Answer
### Have a Nexxtech weather station. How to I change temp reading from fare height to celsius?
Has anyone fixed this? I have the same problem. The outside unit is in C but the indoors one in F and I can't work out how to change it
Oct 24, 2013 | ATOMIC Measuring Tools & Sensors
1 Answer
### How to I change from Celsius to Fahrenheit on my eddie bauer flashlight alarm clock?
When the clock is in time mode the temperature is in the lower right hand corner. In this mode pressing the down arrow button will cycle from celsius to fahrenheit to seconds and then back to celsius.
Dec 05, 2010 | Eddie Bauer Measuring Tools & Sensors
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http://forums.wolfram.com/mathgroup/archive/2007/Oct/msg00546.html | 1,719,279,926,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198865545.19/warc/CC-MAIN-20240625005529-20240625035529-00590.warc.gz | 12,621,441 | 8,279 | Re: Re: How to do count for sub list?????
• To: mathgroup at smc.vnet.net
• Subject: [mg82203] Re: [mg82160] Re: [mg82095] How to do count for sub list?????
• From: DrMajorBob <drmajorbob at bigfoot.com>
• Date: Sun, 14 Oct 2007 06:20:40 -0400 (EDT)
• References: <200710120650.CAA04282@smc.vnet.net> <16695328.1192265948343.JavaMail.root@m35>
• Reply-to: drmajorbob at bigfoot.com
If all the numbers are 0 or 1, you can Map Total:
list = {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0,
0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0,
0}};
Total /@ list
{0, 0, 2, 1, 0, 0, 1, 1}
The more general situation can be reduced to that using Unitize:
list = RandomInteger[{-1, 1}, {10, 10}]
{{-1, 0, 1, 1, 0, 0, 0, 0, -1, 1}, {0, 1, 0, 1, 1, -1, -1,
0, -1, -1}, {-1, 0, 1, 0, 1, -1, -1, -1, 0, 0}, {1, 0, 0, 1, 0,
0, -1, 1, 1, -1}, {0, 1, 0, 1, -1, -1, -1, 1, 0, 0}, {1, 1, 1, 0, 1,
1, -1, 0, -1, 0}, {0, -1, 1, 0, 1, 1, -1, -1, 0, -1}, {1, 0, 1, 0,
0, -1, 0, 1, -1, 1}, {-1, -1, 0, -1, -1, 0, 0, 1, 1, -1}, {0, -1, 1,
0, 0, 1, -1, -1, 1, -1}}
Total /@ Unitize@list
{5, 7, 6, 6, 6, 7, 7, 6, 7, 7}
I suspect this is faster than using Count.
This counts non-zeroes by columns:
Total@Unitize@list
{6, 6, 6, 5, 6, 7, 7, 7, 7, 7}
Bobby
On Sat, 13 Oct 2007 02:59:38 -0500, Ricardo Samad <resamad at gmail.com>
wrote:
> Hi,
>
> name your list:
>
> l = {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0,
> 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0,
> 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0,
> 0}, {0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0,
> 0}};
>
> and use
>
> Table[Length[l[[n]]] - Length[Position[l[[n]], 0]], {n, 1, Length[l]}]
>
> or
>
> Length[#] - Length[Position[#, 0]] & /@ l
>
>
> By the way, the result is {0, 0, 2, 1, 0, 0, 1, 1} and not
> {0,0,1,1,0,0,1,1=
> }
> as written in your e-mail.
>
> Ricardo
>
>
>
>
> On 10/12/07, wangzhen0829 at gmail.com <wangzhen0829 at gmail.com> wrote:
>>
>> Hallo, everyone
>>
>> I am new to Mathematica, and now I have a problem with it, hope to get
>> some help from here.
>>
>> I have a
>>
>> {{0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {
>> 0, 0, 0, 1, 0, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {
>> 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}, {
>> 0, 0, 0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 1, 0, 0, 0, 0, 0}}
>>
>>
>>
>> I need to get a list which represent the number of NON-zero number in
>> each sublist.
>>
>> I should get a list like:
>>
>> {0,0,1,1,0,0,1,1}
>>
>> Thank you!
>>
>>
>>
>
>
> --
> ____________________________________
> Ricardo Elgul Samad
>
> tel: (+55 11) 3133-9372
> fax: (+55 11) 3133-9374
>
> Centro de Lasers e Aplica=E7=F5es
> IPEN/CNEN-SP
> AV. Prof. Lineu Prestes 2242
> 05508-000
> S=E3o Paulo - SP
> Brazil
> ____________________________________
>
>
>
--
DrMajorBob at bigfoot.com
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• Next by thread: Re: How to do count for sub list????? | 1,836 | 3,304 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-26 | latest | en | 0.559925 |
https://silvercityindiaonline.com/qa/what-is-an-estimate-in-econometrics.html | 1,619,050,895,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039554437.90/warc/CC-MAIN-20210421222632-20210422012632-00420.warc.gz | 546,926,347 | 8,712 | # What Is An Estimate In Econometrics?
## What is an example of an estimate?
To find a value that is close enough to the right answer, usually with some thought or calculation involved.
Example: Alex estimated there were 10,000 sunflowers in the field by counting one row then multiplying by the number of rows..
## What is the point estimate of the difference between the means?
A point estimate for the difference in two population means is simply the difference in the corresponding sample means.
## What is purpose of estimate?
The purpose of an estimate has a different meaning to different people involved in the process. To the owner, it provides a reasonable, accurate idea of the costs. This will help him or her decide whether the work can be undertaken as proposed, needs to be modified, or should be abandoned.
## What are the types of econometrics?
There are two branches of econometrics: theoretical econometrics and applied econometrics. The former is concerned with methods, both their properties and developing new ones. It is closely related to mathematical statistics, and it states assumptions of a particular method, its properties.
## What are the types of estimate?
There are five types of estimates based on accuracy:Order of Magnitude. Also called Rough Order of Magnitude (ROM) or Rough Cost Estimate, or Conceptual Estimate, this type of estimate is used for project screening, or deciding which among several projects to proceed with. … Feasibility. … Preliminary. … Substantive. … Definitive.
## How do you calculate an estimate?
The general rule for estimating is to look at the digit to the right of the digit you want to estimate. Estimating or rounding to the nearest whole number means looking at the digit to the right of the decimal. If you see a digit greater than 5, round up, and if it’s less than 5, round down.
## What does estimate mean?
Verb. estimate, appraise, evaluate, value, rate, assess mean to judge something with respect to its worth or significance. estimate implies a judgment, considered or casual, that precedes or takes the place of actual measuring or counting or testing out.
## What is a calculated estimate called?
To approximate is to calculate the value of something based on informed knowledge. As a verb, approximate means “to estimate.” Unlike the word guess, approximate implies the use of a logical or mathematical method. …
## What is the difference between a point estimate and a confidence interval?
Point estimation gives us a particular value as an estimate of the population parameter. … Interval estimation gives us a range of values which is likely to contain the population parameter. This interval is called a confidence interval.
## What is estimation econometrics?
• An estimator is a function of the data (sample), a random. variable, a statistic. • An estimate is a particular realization of an estimator. • Analog principle: replacing population distribution in the.
## What is a reasonable estimate?
A reasonable estimate does not exceed the original numbers in a problem. For instance, in a division problem like 424 ÷ 4 − 10, we can check is our answer is reasonable or not, by estimating an approximate answer or by plugin gin using the division formula.
## What are two ways to estimate?
There are different methods for estimation that are useful for different types of problems. The three most useful methods are the rounding, front-end and clustering methods.
## Why is it good to estimate?
In real life, estimation is part of our everyday experience. … For students, estimating is an important skill. First and foremost, we want students to be able to determine the reasonableness of their answer. Without estimation skills, students aren’t able to determine if their answer is within a reasonable range.
## Does estimate mean Guess?
An estimate is the resulting calculation or judgment. A related term is approximation, meaning close or near. In between a guess and an estimate is an educated guess, a more casual estimate. An idiomatic term for this type of middle-ground conclusion is ballpark figure.
## What is the difference between an estimator and an estimate?
Originally Answered: What is the difference between an estimator and an estimate? An estimator is a function that maps samples into your parameter space. An estimate is the value of that function taken on a particular sample.
## What is an estimate in statistics?
In statistics, estimation refers to the process by which one makes inferences about a population, based on information obtained from a sample.
## What is an instance when you might want to find an estimate?
Instances that require estimated answers instead of exact answers are those in which a high level of accuracy is not required. This may be due to the value being calculated being very large. In order to estimate the value, we simply round the value after a certain number of digits after the decimal. | 979 | 4,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-17 | latest | en | 0.93548 |
https://www.jiskha.com/questions/82992/hydrogen-chloride-is-added-to-a-buffer-solution-of-ammonia-nh3-and-ammonium-chloride | 1,610,818,249,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703506832.21/warc/CC-MAIN-20210116165621-20210116195621-00586.warc.gz | 852,780,240 | 4,735 | # Science
Hydrogen chloride is added to a buffer solution of ammonia, NH3, and ammonium chloride, NH4Cl. What is the effect on the concentration of ammonia? On the concentration of ammonium chloride?
1. 👍
2. 👎
3. 👁
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4. ### CHEMISTRY
If one added 130 mL of 2 M NaOH to 1 L of a buffer composed of 2.3 M NH3 and 3.2 M NH4Cl, what would be the pH of the resulting solution? The Kb of NH3 is 1.8 Γ 10β5. | 923 | 3,015 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2021-04 | latest | en | 0.87962 |
https://community.infiniteflight.com/t/when-is-the-best-time-to-start-descending/350464 | 1,611,663,713,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00383.warc.gz | 270,156,814 | 6,408 | # When is the best time to start descending?
What time should I descend?
My favourite flight attitude is FL350
1 Like
Its diffrent for each flight level.
3 Likes
3 Likes
Probably before you land 😉
8 Likes
The top of descent (Tod) is never a constant number. It changes from flight to flight. However, using the variables from each flight, you can figure out when to descend using a simple mathematical equation:
TOD (in nm from the airport) = Flight Level Of Cruise - Desired Flight Level / 3
3 Likes
If you don’t wanna do all the maths, just start descending with around 30 minutes ETA.
1 Like
This is how i do it:
cruise flight level -( airport elevation + 3000) : V/S= minutes before i have to descent
2 Likes
Too early. Just take the ALT of the airfield and subtract that from your cruise. And then half your cruise then start decent at that time.
Example
Cruise: FL350
Airfield ATL: 2000
35,000-2000 = 33,000
33/2 = 16.5
So in that case you’d start descent around 16 minutes out.
The 2 is VS speed of 2000:)
I use it every time and works pretty well, just remember the lower you get you should slow down:)
4 Likes
This is a very unrealistic way of doing it.
What do you mean is not realistic its just diffrent way… What would be the realistic way?
Enlighten me on a “Realistic” Way:) Always open to new ideas/advice👍 and i know it’s not realistic. But it works and is probably one of the easiest and safest ways on IF
1 Like
Here’s another way https://fpltoif.com/ make a FLP using this website, i believe it’s gives you a descent profile and a climbing profile as well
It’s long to explain but in short :
1. Throttle idle
2. VS down
3. Monitor speed and keep it inside a window (m. 78 - m.70 later 300ias - 250ias)
4. Adjust VS and keep throttle to idle or low power.
5. If over 250 at FL100, stop at it and continue (you should be around 30nm of your destination)
you should be much lower the FL100 when 30 NM from your destination.
That’s with a lot of calculations. If you descend from FL300 and have the nose level constantly level then you’ll be at 3500ft close to the airport.
Stil doesnt explain when to descent
1 Like
It’s pretty basic, takes less time then most of the other ways to find TOD. And like i’ve stated already
I don’t go deep in calculations
Let’s say I’m at FL350
35×3 = TOD (105nm)
35×3.5 on heavys
What makes you say that?
And thats the realistic way? | 628 | 2,416 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-04 | latest | en | 0.905708 |
https://www.theflatearthsociety.org/forum/index.php?topic=63157.msg1671857 | 1,627,588,563,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153892.74/warc/CC-MAIN-20210729172022-20210729202022-00329.warc.gz | 1,095,174,273 | 10,028 | On the subject of a curve
• 21 Replies
• 2090 Views
Scroto Gaggins
• 671
• Hobbiton represent
On the subject of a curve
« on: March 19, 2015, 01:37:43 AM »
Lots of FE'rs, trolls or not, cite the common 'proof' of the earths flatness as the fact that one cannot see curvature on the ground. People have then tried to explain that the earth is too big to notice any difference. People like JRowe have been very adamant that a curved line cannot look flat.
They're wrong. Reasons why:
Consider a Cartesian plane. On it, an exponential graph of the formula y=2x
I cant post one, but a quick google search will show what i mean.
It is quite clearly a curve, any asshole can see that.
But if one looks at the 'tail' end of the line (i.e. x<-2 or so) then it begins to look rather flat.
When you look at the coordinates, it is most assuredly a curve.
Similar principal with the curvature on the horizon.
tl;dr- the horizon can be curved without it being noticeable. Maths
They are taking the hobbits to Isengard.
?
herewegoround
• 286
Re: On the subject of a curve
« Reply #1 on: March 19, 2015, 03:58:06 AM »
Lots of FE'rs, trolls or not, cite the common 'proof' of the earths flatness as the fact that one cannot see curvature on the ground. People have then tried to explain that the earth is too big to notice any difference. People like JRowe have been very adamant that a curved line cannot look flat.
They're wrong. Reasons why:
Consider a Cartesian plane. On it, an exponential graph of the formula y=2x
I cant post one, but a quick google search will show what i mean.
It is quite clearly a curve, any asshole can see that.
But if one looks at the 'tail' end of the line (i.e. x<-2 or so) then it begins to look rather flat.
When you look at the coordinates, it is most assuredly a curve.
Similar principal with the curvature on the horizon.
tl;dr- the horizon can be curved without it being noticeable. Maths
Curves lines are locally straight, curved surfaces are locally flat; both approximately. It's a well understood concept in mathematics. It can be tested easily. Draw a large circle a few metres in diameter on the ground. Mark off two point on the circle about 1 cm apart. Rub out the rest of the circle and the bit left will look straight.
FalseProphet
• 3696
• Life is just a tale
Re: On the subject of a curve
« Reply #2 on: March 19, 2015, 04:28:39 AM »
If curved lines can look straight, why can straight lines never look curved?
Scroto Gaggins
• 671
• Hobbiton represent
Re: On the subject of a curve
« Reply #3 on: March 19, 2015, 04:46:59 AM »
Lots of FE'rs, trolls or not, cite the common 'proof' of the earths flatness as the fact that one cannot see curvature on the ground. People have then tried to explain that the earth is too big to notice any difference. People like JRowe have been very adamant that a curved line cannot look flat.
They're wrong. Reasons why:
Consider a Cartesian plane. On it, an exponential graph of the formula y=2x
I cant post one, but a quick google search will show what i mean.
It is quite clearly a curve, any asshole can see that.
But if one looks at the 'tail' end of the line (i.e. x<-2 or so) then it begins to look rather flat.
When you look at the coordinates, it is most assuredly a curve.
Similar principal with the curvature on the horizon.
tl;dr- the horizon can be curved without it being noticeable. Maths
Curves lines are locally straight, curved surfaces are locally flat; both approximately. It's a well understood concept in mathematics. It can be tested easily. Draw a large circle a few metres in diameter on the ground. Mark off two point on the circle about 1 cm apart. Rub out the rest of the circle and the bit left will look straight.
I know, thats why i posted
They are taking the hobbits to Isengard.
Scroto Gaggins
• 671
• Hobbiton represent
Re: On the subject of a curve
« Reply #4 on: March 19, 2015, 04:49:59 AM »
If curved lines can look straight, why can straight lines never look curved?
Because a straight line by definition, cant be curved.
Get it right, Noriega!
They are taking the hobbits to Isengard.
FalseProphet
• 3696
• Life is just a tale
Re: On the subject of a curve
« Reply #5 on: March 19, 2015, 04:53:40 AM »
If curved lines can look straight, why can straight lines never look curved?
Because a straight line by definition, cant be curved.
Get it right, Noriega!
Not even locally?
Scroto Gaggins
• 671
• Hobbiton represent
Re: On the subject of a curve
« Reply #6 on: March 19, 2015, 04:56:39 AM »
If curved lines can look straight, why can straight lines never look curved?
Because a straight line by definition, cant be curved.
Get it right, Noriega!
Not even locally?
A mathematical straight line can never be anything but.
A straight line in nature will eventually be somewhat curved, due to the nature of molecular structure, I guess.
They are taking the hobbits to Isengard.
FalseProphet
• 3696
• Life is just a tale
Re: On the subject of a curve
« Reply #7 on: March 19, 2015, 05:09:34 AM »
If curved lines can look straight, why can straight lines never look curved?
Because a straight line by definition, cant be curved.
Get it right, Noriega!
Not even locally?
A mathematical straight line can never be anything but.
Oh... But can you prove that?
?
Zekaria
• 9
• Woohoo!
Re: On the subject of a curve
« Reply #8 on: March 19, 2015, 05:40:24 AM »
If curved lines can look straight, why can straight lines never look curved?
Because a straight line by definition, cant be curved.
Get it right, Noriega!
Not even locally?
A mathematical straight line can never be anything but.
Oh... But can you prove that?
Now, now, FalseProphet, don't be difficult.
y=x+1
Plug any numbers into x and you will see that you will get a straight line.
How do you know that you know that you know?
Mikey T.
• 2783
Re: On the subject of a curve
« Reply #9 on: March 19, 2015, 05:42:37 AM »
Yes.
Lets use a horizontal line to make it simpler. So lets use x as the horizontal plane and y as the vertical plane. Lets plot that line at y=1. For this line X can equal any point positive or negative to infinity. The y coordinate never changes. So mathematically this horizontal (from our perspective) is never curved. That's a 2 dimensional representation. To add a 3rd dimension you just add a perpendicular z axis and plot the line at, say z=5. If z and y never change and x = negative infinity to infinity, this is a mathematically straight line that never curves.
For a curved line, we give a y=1 and z= 5, but say at x= 1 million and at x = negative 1 million, y = 0 and z = 4. This is a mathematically curved line. But looking at the plot line for x=1 and x= -1 y and z are essentially 1 and 5. Depending on your perspective the line would look straight at a more local level than being able to see most of the line and then possibly being able to see a curve.
Mikey T.
• 2783
Re: On the subject of a curve
« Reply #10 on: March 19, 2015, 05:44:00 AM »
See I wasted that whole paragraph. Zekaria gives one formula makes mine look long winded lol.
?
Zekaria
• 9
• Woohoo!
Re: On the subject of a curve
« Reply #11 on: March 19, 2015, 06:08:32 AM »
Well Mikey, you probably have a better understanding of math than I do haha
How do you know that you know that you know?
FalseProphet
• 3696
• Life is just a tale
Re: On the subject of a curve
« Reply #12 on: March 19, 2015, 06:09:06 AM »
Now, now, FalseProphet, don't be difficult.
y=x+1
Plug any numbers into x and you will see that you will get a straight line.
That's not a proof, it's just a formula. It's math, man, you do not get it cheap.
FalseProphet
• 3696
• Life is just a tale
Re: On the subject of a curve
« Reply #13 on: March 19, 2015, 06:10:36 AM »
Yes.
Lets use a horizontal line to make it simpler. So lets use x as the horizontal plane and y as the vertical plane. Lets plot that line at y=1. For this line X can equal any point positive or negative to infinity. The y coordinate never changes. So mathematically this horizontal (from our perspective) is never curved. That's a 2 dimensional representation. To add a 3rd dimension you just add a perpendicular z axis and plot the line at, say z=5. If z and y never change and x = negative infinity to infinity, this is a mathematically straight line that never curves.
For a curved line, we give a y=1 and z= 5, but say at x= 1 million and at x = negative 1 million, y = 0 and z = 4. This is a mathematically curved line. But looking at the plot line for x=1 and x= -1 y and z are essentially 1 and 5. Depending on your perspective the line would look straight at a more local level than being able to see most of the line and then possibly being able to see a curve.
That's not even a formula, it's just Drawing for Kids
?
Zekaria
• 9
• Woohoo!
Re: On the subject of a curve
« Reply #14 on: March 19, 2015, 06:19:00 AM »
Now, now, FalseProphet, don't be difficult.
y=x+1
Plug any numbers into x and you will see that you will get a straight line.
That's not a proof, it's just a formula. It's math, man, you do not get it cheap.
So tell me then, what would be proof that a mathematical straight line is actually straight? A graph?
How do you know that you know that you know?
FalseProphet
• 3696
• Life is just a tale
Re: On the subject of a curve
« Reply #15 on: March 19, 2015, 06:47:05 AM »
Now, now, FalseProphet, don't be difficult.
y=x+1
Plug any numbers into x and you will see that you will get a straight line.
That's not a proof, it's just a formula. It's math, man, you do not get it cheap.
So tell me then, what would be proof that a mathematical straight line is actually straight? A graph?
A mathematical proof is a deductive demonstration of a mathematical statement.
"a line defined by the function f(x)=x+1 is always straight regardless of the chosen value for x". That's your statement
Even if I would try 1 million values for x, and even if I would draw a graph 1 lightyear long it would prove nothing mathematically.
You have to prove me your statement in a deductive way, so that I see at once, it must work for any x. Otherwise it is no mathematical proof.
?
Zekaria
• 9
• Woohoo!
Re: On the subject of a curve
« Reply #16 on: March 19, 2015, 08:26:21 AM »
The equation I gave you is in fact a linear equation. It has a constant slope. I would say that most mathematicians would agree that a linear equation would give you a straight line.
Let’s say we have 3 points: (0,1), (74,75), & (99,100)
If slope can be represented as M in an equation, and we assume the line is straight, we will see that slope for a linear equation is:
M = Y2 – Y1
X2 – X1
If we plug in the first two points [(0,1) & (74,75)] into the equation above, we will get
M1 = (75 - 1) = 1 = 1
(74 - 0) 1
Now, if the slope between points 1 & 3 or 2 & 3 are any different from what we calculated, than it will be proof that the line isn’t linear.
Now, let’s finish this thing and figure out if the line is in fact straight.
M2 = (100 - 1) = 1 = 1
(99 - 0) 1
M3 = (100 - 75) = 25 = 1
(99 - 74) 25
Now, do the slopes equal each other?
They do! Who would have thought…
How do you know that you know that you know?
?
Zekaria
• 9
• Woohoo!
Re: On the subject of a curve
« Reply #17 on: March 19, 2015, 08:27:45 AM »
Aaaaaaand I just wasted a bunch of time explaining something that should be pretty obvious.
How do you know that you know that you know?
FalseProphet
• 3696
• Life is just a tale
Re: On the subject of a curve
« Reply #18 on: March 19, 2015, 08:33:53 AM »
The equation I gave you is in fact a linear equation. It has a constant slope. I would say that most mathematicians would agree that a linear equation would give you a straight line.
Let’s say we have 3 points: (0,1), (74,75), & (99,100)
If slope can be represented as M in an equation, and we assume the line is straight, we will see that slope for a linear equation is:
M = Y2 – Y1
X2 – X1
If we plug in the first two points [(0,1) & (74,75)] into the equation above, we will get
M1 = (75 - 1) = 1 = 1
(74 - 0) 1
Now, if the slope between points 1 & 3 or 2 & 3 are any different from what we calculated, than it will be proof that the line isn’t linear.
Now, let’s finish this thing and figure out if the line is in fact straight.
M2 = (100 - 1) = 1 = 1
(99 - 0) 1
M3 = (100 - 75) = 25 = 1
(99 - 74) 25
Now, do the slopes equal each other?
They do! Who would have thought…
Yes, it can be proved by Differential Calculus. You win.
FalseProphet
• 3696
• Life is just a tale
Re: On the subject of a curve
« Reply #19 on: March 19, 2015, 08:35:11 AM »
Aaaaaaand I just wasted a bunch of time explaining something that should be pretty obvious.
Welcome to the FE Forum!
?
Zekaria
• 9
• Woohoo!
Re: On the subject of a curve
« Reply #20 on: March 19, 2015, 08:48:17 AM »
Thanks, I feel right at home
How do you know that you know that you know?
Mikey T.
• 2783
Re: On the subject of a curve
« Reply #21 on: March 19, 2015, 09:22:53 AM »
Yep. welcome. Nice job | 3,706 | 13,103 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-31 | latest | en | 0.941039 |
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Calculus Calculus Math Forum
November 11th, 2012, 05:04 PM #1 Newbie Joined: Nov 2012 Posts: 29 Thanks: 0 Newton's Law (conceptual problem)... 17. Show that Newton's method applied to f(x) in Eq. 6 leads to xsub1 = r + h if xnought = r - h and to xsub1 = r-h if xnought = r+h. Interpret the result geometrically. Eq 6 is f(x) = (2x+1)^1/2 - (x + 4)^1/2 I know how to do Newton's method. I just don't understand where this talk about the height and radius comes into play. I don't know how to go about solving this problem. Thank you.
November 11th, 2012, 08:16 PM #2 Newbie Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... please.
November 11th, 2012, 08:27 PM #3 Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Newton's Law (conceptual problem)... I just took a look at it, spent some time with it, and do not get the intended result. Are you sure the function is copied correctly?
November 11th, 2012, 09:03 PM #4 Newbie Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... i forgot to add. for equation six, a=2 and b=4. it is copied correctly otherwise.
November 11th, 2012, 09:04 PM #5 Newbie Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... that represents that there is a root between 2 and 4.
November 12th, 2012, 07:35 AM #6 Newbie Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... nothing?...
November 12th, 2012, 08:47 AM #7
Math Team
Joined: Sep 2007
Posts: 2,409
Thanks: 6
Re: Newton's Law (conceptual problem)...
Quote:
Originally Posted by nicoleb 17. Show that Newton's method applied to f(x) in Eq. 6 leads to xsub1 = r + h if xnought = r - h and to xsub1 = r-h if xnought = r+h. Interpret the result geometrically. Eq 6 is f(x) = (2x+1)^1/2 - (x + 4)^1/2 I know how to do Newton's method. I just don't understand where this talk about the height and radius comes into play. I don't know how to go about solving this problem. Thank you.
There is nothing said about "height and radius". If you are refering to r and h, they are just arbitrary numbers.
Quote:
i forgot to add. for equation six, a=2 and b=4.
? There are no "a" and "b" in equation six.
Newtons method for solve f(x)= 0 is to construct a sequence of numbers recursively defined by $x_{n+1}= x_n- \frac{f(x_n)}{f#39;(x_n)}$
Here $f(x)= (2x+ 1)^{1/2}- (x+ 4)^{1/2}$ and $f'(x)= (2x+ 1)^{-1/2}- \frac{1}{2}(x+ 4)^{-1/2}$.
You have already been told, by MarkFL, that what you are trying to prove is not true.. To see that, let's use specific numbers: r= 2, h= 1. Then [latex]x_0= r- h= 1[latex]. $f(1)= \sqrt{3}- \sqrt{5}=-0.5040$ and $f'(1)= \frac{1}{\sqrt{3}}- \frac{1}{2\sqrt{5}}= 0.3537$. $\frac{f(1)}{f'(1)}= \frac{-.5040}{.3537}= -0.3537$. That makes $x_1= 1- 0.3537= 0.6463$ which is NOT equal to r+ h= 3.
November 12th, 2012, 11:01 AM #8 Newbie Joined: Nov 2012 Posts: 29 Thanks: 0 Re: Newton's Law (conceptual problem)... Well, that's the problem. I don't know what else to tell you.
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# If $y=f(x)=\large\frac{2x-1}{x-2}$ then $f(y)=$
$\begin{array}{1 1}(A)\;1 \\(B)\;y \\(C)\;2y-1 \\ (D)\;y-2 \end{array}$
If $y=f(x)=\large\frac{2x-1}{x-2}$ then $f(y)=1$
Hence A is the correct answer. | 115 | 219 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.140625 | 3 | CC-MAIN-2018-22 | latest | en | 0.228466 |
https://juicystudio.com/services/readability.php?url=research12b.blogspot.com | 1,656,251,037,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103269583.13/warc/CC-MAIN-20220626131545-20220626161545-00253.warc.gz | 386,072,762 | 4,980 | ## Contents
Gunning Fog, Flesch Reading Ease, and Flesch-Kincaid are reading level algorithms that can be helpful in determining how readable your content is. Reading level algorithms only provide a rough guide, as they tend to reward short sentences made up of short words. Whilst they're rough guides, they can give a useful indication as to whether you've pitched your content at the right level for your intended audience.
## Interpreting the Results
This service analyses the readability of all rendered content. Unfortunately, this will include navigation items, and other short items of content that do not make up the part of the page that is intended to be the subject of the readability test. These items are likely to skew the results. The difference will be minimal in situations where the copy content is much larger than the navigation items, but documents with little content but lots of navigation items will return results that aren't correct.
Philip Chalmers of Benefit from IT provided the following typical Fog Index scores, to help ascertain the readability of documents.
Typical Fog Index Scores
Fog Index Resources
6 TV guides, The Bible, Mark Twain
8 - 10 Most popular novels
10 Time, Newsweek
11 Wall Street Journal
14 The Times, The Guardian
Over 20 Only government sites can get away with this, because you can't ignore them.
Over 30 The government is covering something up
The following table contains the readability results for http://research12b.blogspot.com.
Summary Value
Total sentences 506
Total words 1615
Average words per Sentence 3.19
Words with 1 Syllable 854
Words with 2 Syllables 353
Words with 3 Syllables 145
Words with 4 or more Syllables 263
Percentage of word with three or more syllables 25.26%
Average Syllables per Word 1.89
Gunning Fog Index 11.38
## Gunning-Fog Index
The following is the algorithm to determine the Gunning-Fog index.
• Calculate the average number of words you use per sentence.
• Calculate the percentage of difficult words in the sample (words with three or more syllables).
• Add the totals together, and multiply the sum by 0.4.
• Algorithm: (average_words_sentence + number_words_three_syllables_plus) * 0.4
The result is your Gunning-Fog index, which is a rough measure of how many years of schooling it would take someone to understand the content. The lower the number, the more understandable the content will be to your visitors. Results over seventeen are reported as seventeen, where seventeen is considered post-graduate level.
The following is the algorithm to determine the Flesch Reading Ease.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of syllables per word multiplied by 84.6 and subtract it from the average number of words multiplied by 1.015.
• Subtract the result from 206.835.
• Algorithm: 206.835 - (1.015 * average_words_sentence) - (84.6 * average_syllables_word)
The result is an index number that rates the text on a 100-point scale. The higher the score, the easier it is to understand the document. Authors are encouraged to aim for a score of approximately 60 to 70.
The following is the algorithm to determine the Flesch-Kincaid grade level.
• Calculate the average number of words you use per sentence.
• Calculate the average number of syllables per word.
• Multiply the average number of words by 0.39 and add it to the average number of syllables per word multiplied by 11.8.
• Subtract 15.50 from the result.
• Algorithm: (0.39 * average_words_sentence) + (11.8 * average_syllables_word) - 15.9
The result is the Flesch-Kincaid grade level. Like the Gunning-Fog index, it is a rough measure of how many years of schooling it would take someone to understand the content. Negative results are reported as zero, and numbers over twelve are reported as twelve. | 882 | 3,885 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2022-27 | latest | en | 0.887572 |
https://classroom.thenational.academy/lessons/explore-adding-on-hours-and-half-hours-6njpad?activity=video&step=2&view=1 | 1,621,077,806,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991801.49/warc/CC-MAIN-20210515100825-20210515130825-00584.warc.gz | 196,231,486 | 39,407 | # Explore adding on hours and half hours
In this lesson, you will be adding either one hour or half an hour onto a given time.
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 4
Q1.What part of the day do you eat breakfast?
1/4
Q2.What part of the day do you go to school?
2/4
Q3.What time does the boy go to bed?
3/4
Q4.What time does she eat her dinner at?
4/4
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
## Question 4
Q1.What part of the day do you eat breakfast?
1/4
Q2.What part of the day do you go to school?
2/4
Q3.What time does the boy go to bed?
3/4
Q4.What time does she eat her dinner at?
4/4
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
Quiz:
# Explore adding on hours and half hours
Time
## Question 3
Q1.What time does Little Red Riding Hood speak to the wood cutter?
1/3
Q2.What time does Little Red Riding Hood get to her Grandma's house?
2/3
Q3.If Little Red Riding Hood gets to the market at 10 o'clock and leaves the market at 11 o'clock. How long did she take at the market?
3/3
Quiz:
# Explore adding on hours and half hours
Time
## Question 3
Q1.What time does Little Red Riding Hood speak to the wood cutter?
1/3
Q2.What time does Little Red Riding Hood get to her Grandma's house?
2/3
Q3.If Little Red Riding Hood gets to the market at 10 o'clock and leaves the market at 11 o'clock. How long did she take at the market?
3/3
# Lesson summary: Explore adding on hours and half hours
### Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Climb stairs
On the spot:
Chair yoga
### Take part in The Big Ask.
The Children's Commissioner for England wants to know what matters to young people. | 630 | 2,469 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | longest | en | 0.882505 |
https://www.physicsforums.com/threads/anything-missing-or-redundant-about-this-one-sided-limit-proof.1010691/ | 1,708,786,958,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474541.96/warc/CC-MAIN-20240224144416-20240224174416-00197.warc.gz | 928,984,972 | 20,660 | • mcastillo356
In summary: Actually, it doesn't matter if the function is increasing or not. The error lies in assuming that for any ##\epsilon>0##, there exists a single value of ##\delta## that works for all ##x## in the interval ##(0, \delta)##. This is not true for all functions, as the counterexamples you mentioned (absolute value and square root functions) show.
#### mcastillo356
Gold Member
Homework Statement
Prove ##\lim{(x^{2/3})}## when ##x\rightarrow{0^{+}}## is 0
Relevant Equations
##\forall{\epsilon>0}##, find ##\delta>0## such that ##0<x<\delta\Rightarrow{|f(x)|<\epsilon}##
Hi, PF
In a Spanish math forum I got this proof of a right hand limit:
"For a generic ##\epsilon>0##, in case the inequality is met, we have the following: ##|x^{2/3}|<\epsilon\Rightarrow{|x|^{2/3}}\Rightarrow{|x|<\epsilon^{3/2}}##. Therein lies the condition. If ##x>0##, then ##|x|=x##; therefore, if the following holds: ##0<x<\epsilon^{3/2}\Rightarrow{|f(x)|<\epsilon}##, eventually, we can state: ##\forall{\epsilon>0}\;\exists{\delta>0}## s.t. ##0<x<\delta\Rightarrow{|f(x)|<\epsilon}##. In conclusion, the ##\delta## sought is epsilon elevated to three means, ##\delta=\epsilon^{3/2}##."
What is your opinion? It's right, yes, but... Something tells me it shall be improved.
Love
I'm going to click, no preview.
mcastillo356 said:
In conclusion, the δ sought is epsilon elevated to three means, δ=ϵ3/2."
Epsilon raised to the three-halves power. Since the purported limit is 0, it would be nice, but not essential to start with ##|x^{2/3} - 0| < \epsilon##, but that's a very minor nit. Otherwise, I don't see anything wrong with the proof.
mcastillo356
Proof of left handed limit for ##x^{3/2}##, when ##x\rightarrow{0^{-}}##
For a generic ##\epsilon>0##, in case the inequality mets (or inequation? Is it an algebraic inequality?), we have the following: ##|x^{2/3}|<\epsilon\Rightarrow{|x|^{2/3}< \epsilon}\Rightarrow{|x|<\epsilon^{3/2}}##. Therein lies the condition. If ##x<0##, then ##|x|=-x##; therefore, if the following holds: ##-\epsilon^{3/2}<x<0\Rightarrow{|f(x)|<\epsilon}##, eventually, we can state, ##\forall{\epsilon>0}\;\exists{\delta>0}## s.t. ##-\delta<x<0\Rightarrow{|f(x)|<\epsilon}##. In conclusion, the ##\delta## sought is epsilon raised to the three-halves power, ##\delta=\epsilon^{3/2}##.
Right?
Last edited by a moderator:
mcastillo356 said:
Proof of left handed limit for ##x^{3/2}##, when ##x\rightarrow{0^{-}}##
For a generic ##\epsilon>0##, in case the inequality mets (or inequation? Is it an algebraic inequality?), we have the following: ##|x^{2/3}|<\epsilon\Rightarrow{|x|^{2/3}< \epsilon}\Rightarrow{|x|<\epsilon^{3/2}}##. Therein lies the condition. If ##x<0##, then ##|x|=-x##; therefore, if the following holds: ##-\epsilon^{3/2}<x<0\Rightarrow{|f(x)|<\epsilon}##, eventually, we can state, ##\forall{\epsilon>0}\;\exists{\delta>0}## s.t. ##-\delta<x<0\Rightarrow{|f(x)|<\epsilon}##. In conclusion, the ##\delta## sought is epsilon raised to the three-halves power, ##\delta=\epsilon^{3/2}##.
Right?
No.
##x^{3/2}## isn't defined for x < 0 if you're limited to real output values.
Think about it -- if x = -0.1, say, then you have to evaluate either ##\sqrt{-.1^3}##, or ##(\sqrt{-.1})^3##, neither of which is real.
The reason for the right-hand limit of the first post is precisely because the function isn't defined for negative x, so the left-hand limit doesn't exist, either.
mcastillo356
I disagree. We deal with absolute values
mcastillo356 said:
I disagree. We deal with absolute values
Only a sith deals in absolutes.
But seriously, what does this mean? Is your function actually ##|x|^{3/2}##?
mcastillo356
mcastillo356 said:
I disagree. We deal with absolute values
That's not the limit you wrote in post #3:
mcastillo356 said:
Proof of left handed limit for ##x^{3/2}##, when ##x\rightarrow{0^{-}}##
Again, ##f(x) = x^{3/2}## is defined (as a real valued function) only for ##x \ge 0##.
mcastillo356
I meant ##f(x)=x^{2/3}##
Sorry, I will try to mend it.
I think the proof is fine then.
mcastillo356
Hi, PF
Mark44 said:
No.
##x^{3/2}## isn't defined for x < 0 if you're limited to real output values.
Think about it -- if x = -0.1, say, then you have to evaluate either ##\sqrt{-.1^3}##, or ##(\sqrt{-.1})^3##, neither of which is real.
The reason for the right-hand limit of the first post is precisely because the function isn't defined for negative x, so the left-hand limit doesn't exist, either.
##\lim{(x^{3/2})}## when ##x\rightarrow{0^{+}}## is ##0##
Proof
##\forall{\epsilon>0}\;\exists{\delta>0}## s.t. ##0<x<\delta\Rightarrow{|f(x)|<\epsilon}##
Pick a ##\delta=\epsilon^{2/3}##. This way, ##f(x)## can be as closer to ##0## as desired, getting ##x## close enough to ##0##, ##0<x<\delta=\epsilon^{2/3}##
This right hand limit equals ##f(0)=0##, but it will never be continous, as @Mark44 suggests, and the graph shows:
Right? I've wrote not checking.
Last edited by a moderator:
mcastillo356 said:
Hi, PF
##\lim{(x^{3/2})}## when ##x\rightarrow{0^{+}}## is ##0##
Proof
##\forall{\epsilon>0}\;\exists{\delta>0}## s.t. ##0<x<\delta\Rightarrow{|f(x)|<\epsilon}##
Pick a ##\delta=\epsilon^{2/3}##. This way, ##f(x)## can be as closer to ##0## as desired, getting ##x## close enough to ##0##, ##0<x<\delta=\epsilon^{2/3}##
You should show explicitly how ##\delta=\epsilon^{2/3}## implies that ##|x^{3/2} - 0 | < \epsilon##.
mcastillo356 said:
This right hand limit equals ##f(0)=0##, but it will never be continous, as @Mark44 suggests, and the graph shows:
No, that's not what I suggested. The function ##f(x) = x^{3/2}## is continuous on its domain, ##[0, \infty)##, and is right-continuous at 0.
mcastillo356 said:
View attachment 294897
Right? I've wrote not checking.
mcastillo356
Re reading this thread, I think there is actually a piece that is not spelled out specifically enough. As written, you can basically replace ##f(x)=x^{2/3}## with any invertible function, and it's not super obvious which step stops working. Do you know?
mcastillo356
Office_Shredder said:
Re reading this thread, I think there is actually a piece that is not spelled out specifically enough. As written, you can basically replace ##f(x)=x^{2/3}## with any invertible function, and it's not super obvious which step stops working. Do you know?
Sorry, can you explain further?
mcastillo356 said:
Sorry, can you explain further?
Let ##f## be an invertible function with ##f(0)=0##. Then for any ##\epsilon>0##, let ##\delta=f^{-1}(\epsilon)##. Then ##0<x< f^{-1}(\epsilon)## implies ##f(x)<\epsilon##. Hence ##\lim_{x\to 0^+} f(x)=0##.
Do you see the error in this proof for an arbitrary function? I just copied what you did for ##x^{2/3}##
mcastillo356
Let me see... It only works because it's fractional, even, and positive power function. No. Counterexample: absolute value function for real numbers; square root function. Haven't checked, give me some time
mcastillo356 said:
Let me see... It only works because it's fractional, even, and positive power function. No. Counterexample: absolute value function for real numbers; square root function. Haven't checked, give me some time
It might help to think of an invertible function ##f:[0,\infty) \to \mathbb{R}## which is *not* continuous at 0 (this is not super hard, but not a totally trivial thing to do)
mcastillo356
Office_Shredder said:
Let ##f## be an invertible function with ##f(0)=0##. Then for any ##\epsilon>0##, let ##\delta=f^{-1}(\epsilon)##. Then ##0<x< f^{-1}(\epsilon)## implies ##f(x)<\epsilon##. Hence ##\lim_{x\to 0^+} f(x)=0##.
Do you see the error in this proof for an arbitrary function? I just copied what you did for ##x^{2/3}##
But it is an increasing function. ##f(x)<\epsilon\;\forall{\epsilon>0}##
mcastillo356 said:
But it is an increasing function.
But you did not note the dependence on that in your proof, let alone prove it.
mcastillo356
haruspex said:
But you did not note the dependence on that in your proof, let alone prove it.
Incisive remark. Definitely, there was something missing. I'm working on it. | 2,471 | 8,093 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-10 | latest | en | 0.724921 |
https://brainmass.com/math/basic-algebra/259074 | 1,397,939,168,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00642-ip-10-147-4-33.ec2.internal.warc.gz | 795,242,513 | 7,737 | # Fundamental operations on polynomials and factoring
A.Find p(x) + 3q(x)
p(x) = 2x^4 + 15x^3 - 10x^2 + 14
q(x) = 35x^4 - 16x^2 + x - 11
b.Find P(-1/2) if P(x) = 4x^4 -2x^3 + 17
c.Simplify: (-3x + 4x^2 + 6x^3) - (4 - x + 2x^3) - (-6x^2 + y)
d.Add: (10x^2 + 3y^2 - 6z^2 + xy + 4yz + 2xz) + (2x^2 - y^2 + 5z^2 - 2xy + 7yz - 3xz)
e.Multiply: (2x - 3y)^2
f.Multiply: (6x - 7) (6x + 7)
g.Divide: (3x^3 - x^2 + 10x - 4) รท (x + 3)
B. Factor completely:
a.x^2 - 5x - 7x + 35
b.2x^2 + 11x + 5
c.x^4y - 16y
C. Solve the following problems involving applications of polynomials.
a. A photo is 4 inches longer than it is wide. A 3-inch border is placed around the photo making the total area of the photo and border 165 square inches. What are the dimensions of the photo?
b. A rectangular patio is 7 ft longer than it is wide. Determine the dimensions of the patio if it measures 13 ft diagonally.
c. Three consecutive even integers are such that the square of the third is 100 more than the square of the second. Find the three integers.
I have completed most of these but was a little confused. I had the hardest problems with the word problems.
#### Solution Summary
Step by step solutions to all the problems are provided. | 457 | 1,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2014-15 | latest | en | 0.859951 |
https://oeis.org/A076301 | 1,725,767,181,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00389.warc.gz | 429,150,945 | 3,979 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A076301 Related to number of labeled partially ordered sets. 1
1, 1, 3, 15, 96, 720, 6120, 57960, 604800, 6894720, 85276800, 1137628800, 16286054400, 249080832000, 4053790540800, 69960578688000, 1276290183168000, 24542432538624000, 496183962193920000, 10522301185363968000 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS G. C. Greubel, Table of n, a(n) for n = 0..445 M. Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013 FORMULA E.g.f.: -(2-4*x + 3*x^2)/(-6*x^2 - 2 + 6*x + 2*x^3). a(n) = (4 - n + n^2)*n!/4. - G. C. Greubel, May 04 2018 MAPLE seq(1/4*(4-n+n^2)*n!, n=0..30); MATHEMATICA Table[(4 -n +n^2)*n!/4, {n, 0, 30}] (* G. C. Greubel, May 04 2018 *) PROG (PARI) for(n=0, 30, print1((4 -n +n^2)*n!/4, ", ")) \\ G. C. Greubel, May 04 2018 (Magma) [(4 -n +n^2)*Factorial(n)/4: n in [0..30]]; // G. C. Greubel, May 04 2018 CROSSREFS Sequence in context: A306027 A354412 A304072 * A112913 A109283 A370210 Adjacent sequences: A076298 A076299 A076300 * A076302 A076303 A076304 KEYWORD nonn AUTHOR Detlef Pauly (dettodet(AT)yahoo.de), Mar 14 2003 STATUS approved
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Last modified September 7 23:15 EDT 2024. Contains 375749 sequences. (Running on oeis4.) | 616 | 1,601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.58414 |
https://issuu.com/abigaylez/docs/final_final_final_6e6ea5abd54396 | 1,506,333,931,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818690591.29/warc/CC-MAIN-20170925092813-20170925112813-00231.warc.gz | 693,401,272 | 22,999 | ABBY ZAPLAN 638164 Semester 2/2013 Virtual Environments Group 11
Module 4 _Reflection Student Journal
module 1 _IDEATION
Introduction to material system
In this module, I was initially assigned with an inflatable armband, which fell under the inflate material system.
I was required to familiarize myself with the object – its material system in particular – by exploring the inflatable in terms of its form, structure and dimensions.
module 1 _IDEATION
The reading ‘300 years of industrial design’ (Heath et. al., 2000) gave an interesting insight and emphasis on the importance of measured drawings. Heath (2000) expresses how measuring an object in detail is important in terms of getting an idea of how the object might have been constructed and the reasons behind its design.
Relevance
Image from Barnes and Noble 2013
‘Inside Rhinoceros’ outlined the three main types of 3D computer models and some key functions of Rhino.
Image from Barnes and Noble 2013
These two readings along with Miralles’ ‘How to lay out a croissant’ (1988/1991) – where it was demonstrated the possible logic behind the measuring process of an object – assisted me in understanding why we were required to do a measured drawing set of our object, how I would have to go about physically measuring and modelling the object digitally, and most importantly – how and why an inflatable armband worked as a material system.
Image from Word Press 2012
module 1 _IDEATION
Application
In order to apply the knowledge I gained from the readings and tutorial tasks, I constructed a small sketch model imitating the inflate material system, using plastic table cloth and tape.
Deflated inflatable sketch model
Inflated inflatable sketch model
Seam of inflatable sketch model
From this activity I was forced to think about what material would be most suitable to prevent air escaping and I was able to develop my skills in creating seams. Furthermore I was able to see how air affected the movement and shape of the plastic model.
module 1 _IDEATION
Sommer’s (1969) ‘Personal space: the behavioural basis of design’, provided a great introduction to the idea of personal space – “an area with invisible boundaries surrounding a person’s body into which intruders may not come” (p 26). The reading also introduced the idea that “invisible boundaries” differed for each individual and how going into these boundaries results in varying reactions depending on the individual.
Personal space Taking inspiration from this reading and the inflatable armband, I produced some sketch designs for the second skin.
Image from Open Library 2008
From this process of generating ideas, it became a challenge to decide whether I wanted to protect the personal space of every aspect of the body or if I wanted to focus on a particular body part.
module 1 _IDEATION
module 2 _DESIGN
Collaboration of ideas
This module was the beginning of group work. I was assigned with another two students with whom I was also now to do the combined material system of inflate AND panel and fold.
Panel and fold image, Harriet Craig, 2013
My group began by first working in collaboration to come up with a starting design from which we can use to work on and develop from. Instead of looking at the whole human body, a group agreement was made to focus on the arm area in terms of creating a 2nd skin design.
Nerve endings on forearm, hand and fingertips. Image by Abbott Laboratories. 2010
Research was done on the hand and arm areas, and it was discovered that the hand was one of the most sensitive parts of the body (Oracle Education Foundation, 2013) – the forearm being the least sensitive and the fingertips highest in sensitivity (Abbott Laboratories, 2010). We developed our design based on these facts.
module 2 _DESIGN
Idea development
We proposed to have the base to be the inflatable component with the triangular spikes being the panel and fold component. Although at this point we were not sure how we would attach the spikes to the base without the use of tape.
Precedent examples. Pinterest.com. Unknown Date.
Sketch model of initial design. Photo by Harriet Craig, 2013.
module 2 _DESIGN
Here we also focused greatly on precedent examples relevant to our material system and further developed and refined our unique idea of personal space.
Challenging conventions Taking advice from Lecture number 4, Thomas Heatherwick’s video and taking inspiration from the precedents, we experimented with swapping the material systems of the base and spikes to try something a little different.
Thomas Heatherwick. Image from Wookmark, 2013.
Rhino model from different views
First prototype. Three photos by Yingli Liu, 2013.
However this did not work out too well. Our first prototype was not as successful as we wanted it to be for we learnt that the paper we were proposing to use was too stiff and looked quite awkward when worn. It also impeded movement on the arm.
module 1 _IDEATION
Further development
We tried, as a group, various times to experiement with the two material systems and how we would use them together. We even attempted to separate the material systems into sections. While this was a form of development it still did not feel like it was the right design. Not only did this design stray away from the our original idea of the spatial variation of sensitivity and how it relates to a need for personal space in the forearm or hand areas, it also did not seem realistic to construct consdering the panel and fold was paper and the inflateable, plastic.
module 2 _DESIGN
Uniting material systems
In an attempt to combine our material systems into one, (instead of looking at them as two sepearate components), we developed a new design which made the base and the spikes both made of plastic and both capable of inflating.
Rhino model from different views
Prototype of first model integrating both systems as one
module 2 _DESIGN
module 3 _FABRICATION
Materiality During this module we experimented with a different types of plastic.As we further developed our second skin , the first type of different plastic that we used to make our prototype with was book cover plastic. Previously we were using black table cover plastic to construct our prototypes which is opaque and quite flimsy.
Tape being used to make seams in the prototype
We found that using clear plastic for a change better enhances our idea of personal space as it is supposedly “invisible� but very much still there.
Book covering plastic used for prototype
The clear plastic allowed the lower arm and hand to also be fully covered which fully fulfills the protection of the hand, as we had been proposing from the very start. Our initial design only sat above the hand and not around it therefore not granting full safety. The use of the clear plastic allows the hand to be seen symbolizing its vulnerability - but showing that if one comes too close or touches it unapprovingly, that there is a defense mechanism.
module 3 _FABRICATION
Tools In replacement of the clear tape that we used before to make the seams of the inflatable prototypes, we ultimately resorted to using a soldering iron instead.
Soldering iron producing a clean -cut edge
We found that even though the tape was clear - and therefore blends in with the book covering plastic - using the Soldering iron saves a lot of time and produces more professional and neat looking seams.
Soldetring iron being used to seal an edge
module 3 _FABRICATION
The Making Process Iwamoto’s ‘Digital fabrications: architectural and material techniques’, extesively highlights the different uses of digital technology in turning a design into a fabrication.
Although Iwamoto mentions different processes which bring convenience to a designer when it comes to its physical construction due to technology, comparing it to how my group has to fabricate our design, it has made me realize that technology can only be relied on to on an extent.
In our case where the shape of our design is prone to change due to potential movement caused by incoming and outgoing air and the material we’re using, we can not really tell if the design that we have constructed digitally will look like what we physically have to create ourselves. Digital fabrications: architectural and material techniques. Image from Urban A&O, 2009.
module 3 _FABRICATION
Prototype development Each week we tried a different plastic to see if it would better lead an aeathetically as well as functionally successful inflatable system.
Rhino models of developed design
After trying book covering plastic, we found that it was too rigid and therefore the prototype already looks “inflated� to begin with without yet any air flow.
Prototype worn in context
We wanted to demonstrate movement in our second skin design. Therefore we searched for a less rigid clear plastic.
module 3 _FABRICATION
Prototype using book covering plastic
Materiality The next type of plastic we used for prototyping was clear cellophane.
Cellophane for prototyping
With this material, we found that it was too fragile and easily ripped. When air was blown inside the prototype we had to be careful not to put too much air in or else it might have burst. The cellophane was also prone to holes especially at the seams where the soldering iron had sealed the edges together. Prototype made using prototype
We once again went and looked for a more suitable plastic which was not as thick as book covering plastic but not as thin as cellophane.
module 3 _FABRICATION
Materiality/final fabrication
Ait source pointed into two plastic tubes
The last and final plastic we tried and used for our final fabrication was paint masking plastic.
Hand inserted inside second skin design
First we used paint masking plastic to also construct the tubes leading to the base where air would be blown through. However this presented issues in holes forming due to force of the air. To ensure that the tubes connected to the base would be strong, we used book covering plastic for these components of the second skin.
Prototype fabricated using paint masking plastic
Second skin fully worn
module 3 _FABRICATION
This was the most successful material for it was not too rigid nor was it or too flimsy.
module 4 _REFLECTION
Reflection
Virtual Environments has been a challenging subject which has both taught me new things about designing and also new skills regarding the fabrication process of a design. Regarding the reading ‘Building the Future: Recasting Labor in Architecture’ (Bernstein et. al., 2008) my group and I can definitely serve as good examples of risk takers when it came to our second skin design project. Each week we would take the risk of changing our designs even though we may not know if it will work in real life. Furthermore, and especially in the testing of which material would best suit our physical model. Without knowing what would be the outcome, each week would still work hard in constructing our model and each week we would learn why one thing worked and another thing did not. Although my group went through many hardships in the development process of our design, it was a worhtwhile experience to learn together. Because with each failure, I knew that it was a step closer to success. It is indeed very evident in our final fabricated design. If one would compare the first prototype we had created to the final one we ended up with there are so many changes apparent. I used to be afraid of change when it came to design because there was a level of uncertainty and risk associated with it, and not to mention the time I thought I was wasting if i happened to develop the design into one that was worse after all. From Virtual Environements, I have learned that it’s okay to stray off for a while or even many times - because as long you have a will to improve your design, you would always be lead to a better path.
module 4 _REFLECTION
It is so easy when designing to just go with the conventional ideas because again, that is the safe way and you know you can not go wrong. I can recount during the semester when I watched a video by Thomas Heatherwick and was fascinated and was in awe at how different and innovative his ideas were. Ever since I watched that and through out the lectures where it has been reminded how I should not conform to a preconceived ideas of an object when it comes to designing, I have began to have a fresher view of what it means to design. Designing may to a certain degree imitate or simply extend what is previously made, but it’s mostly about doing something different and challenging. If it were not for this, there woulld be no innovation. Virtual environments has taught me how to think like a designer and how a designer should not be afraid to struggle. Working in groups for the project has really allowed me to develop my communication skills when it comes to getting tasks done and I hope to develop this in my future environments subjects and also, take with me the many valuable experiences, skills and lessons I’ve learnt while doing this subject.
module 4 _REFLECTION
References Abbott Laboratories. 2010. Why is Alternative Site Testing Less Painful? In: Alternative Site Testing. Available: http://www.abbottdiabetescare.ca/adc_ca/ url/content/en_CA/20.10.30:30/general_content/General_Content_0000321.htm. Accessed: August 21 2013. Building the Future: Recasting Labor in Architecture/ Philip Bernstein, Peggy Deamer. Princeton Architectural Press. c2008. pp 38-42 Barnes and Noble, 2013. Available: http://www.barnesandnoble.com/w/inside-rhinoceros-4-ron-k-c-cheng/1100210085. Accessed: November 4 2013. Barnes and Noble, 2013. Available: http://www.barnesandnoble.com/w/300-years-of-industrial-design-adrian-heath/1003829040. Accessed: November 4 2013. Cheng, R. (2008). Inside Rhinoceros 4 / Ron K.C. Cheng. Clifton Park, NY : Thomson/Delmar Learning, c2008. Digital fabrications: architectural and material techniques / Lisa Iwamoto. New York : Princeton Architectural Press, c2009. Enric Miralles,Carme Pinos, “How to lay out a croissant� El Croquis 49/50 Enric Miralles, Carme Pinos 1988/1991, En Construccion pp. 240-241 Heath, A., Heath, D., & Jensen, A. (2000). 300 years of industrial design : function, form, technique, 1700-2000 / Adrian Heath, Ditte Heath, Aage Lund Jensen. New York : Watson-Guptill, 2000. Open Library, 2008. https://openlibrary.org/books/OL5680150M/Personal_space_the_behavioral_basis_of_design. Accessed: November 4 2013. Oracle Education Foundation. 2013. Sense-Sational Facts. In: Your sense of touch. Available: http://library.thinkingquest.org/3750/touch/touch.html. Accessed: August 21 2013. Precedent examples: Available: http://pinterest.com/pin/268456827759104763/ http://pinterest.com/pin/268456827759104732/. Authors are unknown. Accessed: August 27 2013. Urban A&O, 2009. Avaialble: http://urbanao.com/news/?paged=5. Accessed: November 4 2012. Visual theme of Module 2: Facebook. Year unknown. Available: https://www.facebook.com/pages/Anatomy-In-Motion/147107135344108?directed_target_id=0. Accessed: August 21 2013. Sommer, R. (1969). Personal space : the behavioral basis of design / Robert Sommer. Englewood Cliffs, N.J. : Prentice-Hall, c1969. Wookmark, 2013. Available: http://www.wookmark.com/image/147669/thomas-heatherwick-portrait-smart-urban-stage. Acessed: November 4 2013.
Final M4 Journal Submission
Abigayle Zaplan 638164 | 3,515 | 15,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2017-39 | latest | en | 0.938217 |
https://convertcalculators.com/12-in-binary/ | 1,675,422,435,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500044.66/warc/CC-MAIN-20230203091020-20230203121020-00772.warc.gz | 204,333,667 | 13,821 | # 12 In Binary
By Desk Incharge - 3 February 2023
12 in binary for quick calculations, along with a large collection of calculators on math, finance, fitness, and 12 in binary, each with in-depth information. 12 in binary is a beautiful, free online scientific calculator with advanced features for evaluating percentages, fractions, exponential functions, logarithms, trigonometry. However, 12 in binary The Digital Unit Converter 12 in binary, currency, density, energy, force, length, mass, power, pressure, speed, temperature.
### 1. 12 in Binary - How to Convert 12 from Decimal to Binary?
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12 in binary is 1100. To find decimal to binary equivalent, divide 12 successively by 2 until the quotient becomes 0. The binary equivalent can be obtained by ...
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This adding of additional hexadecimal digits to convert both decimal and binary numbers into an Hexadecimal Number is very easy if there are 4, 8, 12 or 16 ...
## 5. Decimal to Binary (Definition, Conversion, Table and Examples)
https://byjus.com/maths/decimal-to-binary/
For example, if 1210 is a decimal number then its equivalent binary number is 11002. Thus, it is easy to convert the given decimal to binary using simple tricks ...
#### 7. A 12 minute Orbital Period Detached White Dwarf Eclipsing Binary
https://arxiv.org/abs/1107.2389
Jul 12, 2011 ... Upon contact the systems may merge yielding a rapidly spinning massive WD, form a stable interacting binary, or possibly explode as an ... | 380 | 1,565 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-06 | latest | en | 0.850432 |
https://123deta.com/document/zpww45ry-renormalized-phase-dynamics-in-oscillatory-media.html | 1,627,687,939,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154032.75/warc/CC-MAIN-20210730220317-20210731010317-00166.warc.gz | 84,173,605 | 39,534 | # Renormalized phase dynamics in oscillatory media
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## Renormalized Phase Dynamics in Oscillatory Media
Naofumi Tsukamoto,1,*Hirokazu Fujisaka,1,†and Katsuya Ouchi2
1Department of Applied Analysis and Complex Dynamical Systems, Graduate School of Informatics, Kyoto University, Kyoto 606-8501, Japan
2Kobe Design University, Kobe 651-2196, Japan
(Received 27 December 2006; published 27 September 2007)
Based on the complex Ginzburg-Landau equation (CGLE), a new mapping model of oscillatory media is proposed. The present dynamics is fully determined by an effective phase field renormalized by amplitude. The model exhibits phase turbulence, amplitude turbulence, and a frozen state reported in the CGLE. In addition, we find a state in which the phase and amplitude have spiral structures with opposite rotational directions. This state is found to be observed also in the CGLE. Thus, one concludes that the behaviors observed in the CGLE can be described by only the phase dynamics appropriately constructed. DOI:10.1103/PhysRevLett.99.134102 PACS numbers: 05.45.a, 82.40.Bj
Various spatiotemporal patterns are observed in non-equilibrium systems, such as fluid systems, chemical reac-tion, and nonlinear optics [1]. Such dynamics in spatially extended systems are often described by nonlinear partial differential equations. The complex Ginzburg-Landau equation (CGLE) is one of the most-studied nonlinear par-tial differenpar-tial equations. The CGLE is a universal equa-tion which describes slow spatiotemporal variaequa-tion near a supercritical Hopf bifurcation [2] and exhibits a rich vari-ety of dynamical behaviors [3–5] in spite of its simple form
_
Ar; t A 1 ic2jAj2A D1 ic1r2A; (1) where c1, c2, and D>0 are real parameters and A is the complex order parameter. Almost 20 years ago, for effi-ciency of numerical simulations, some partial differential equations were approximated by mapping models [6,7]. Mapping models of the CGLE have been also obtained and investigated as an approximation [3,8–13].
In this Letter, we consider a mapping model of the CGLE in order to extract the essential natures of the CGLE. We use the mapping model constructed by the method proposed in Refs. [12,13], which is similar to those in Refs. [3,8–11]. There are two steps to obtain mapping models: (i) splitting the time evolution of the CGLE into two parts and (ii) recombining them. One of the divided two parts is a local part which consists of the first two terms in the right-hand side of Eq. (1), and the other is a nonlocal part which is the spatial coupling term in Eq. (1). The time evolution of each part can be solved analytically as shown below.
The time evolution of the local part is described by the Stuart-Landau equation _A A 1 ic2jAj2A, which is obtained by omitting the spatially coupling term in Eq. (1). Integrating the Stuart-Landau equation over time width and setting t eic2tAt, we obtain
t F t; (2)
where F f1 e2j j2 e2g1ic2=2. The
time evolution of the nonlocal part is described by the complex diffusion equation _ D1 ic1r2 , which is obtained by omitting the local parts in the right-hand side of Eq. (1). Integrating the equation over time width , we have
r; t LD r; t; (3)
where LD is the linear operator defined by LDfr
R
KDr r0fr0dr0. Here KDr f4D1 ic1gd=2ejrj
2=4D1ic
1, and d is the spatial
dimensional-ity. Recombining the divided time evolutions, we obtain the mapping model
n1r LDF nr: (4)
This equation is a model based on the CGLE with a control parameter and coincides with Eq. (1) in the limit ! 0. Therefore, Eq. (4) for small is expected to show approxi-mately the dynamics in the CGLE. Actually, the spatio-temporal dynamics in the CGLE have been investigated by using mapping models similar to Eq. (4) as an approxima-tion [3,8–11].
In this Letter, we take the opposite limit ! 1 by keeping D 1 fixed. In this limit, the local map F reduces to
F F1
j j1ic2 0;
0 0; (5)
which maps an arbitrary state onto either a state on the limit cycle (j j 1) or the unstable fixed point ( 0) of the Stuart-Landau equation. Thus, the limit ! 1 re-moves the relaxation process to the limit cycle. By setting L LD1 e1ic1r
2
, the above procedure leads to the complex Ginzburg-Landau map (CGLM)
n1r LF nr: (6)
We will show that the CGLM (6) exhibits the
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poral behaviors reported in the CGLE and discuss the essence of the CGLE dynamics.
It should be noted that the limit ! 1 allows us to renormalize the amplitude component into the phase able as follows. Introducing the renormalized phase vari-able n arg n c2logj nj for n 0, we define the
phase field with the phase singular point as znr F nr, that is, znr einrfor nr 0 and znr
0 for nr 0. Here zn 0 represents the phase singular
point, and n describes an isochron of the Stuart-Landau equation. Equation (6) can be written as n1r Lznr,
which indicates that the time evolution of nis determined by zn. In addition, we obtain the mapping system of the phase field with the phase singular point as
zn1r FLznr; (7)
which is the phase description of the CGLM. Because there is no approximation through the derivation of Eq. (7) from Eq. (6), this phase description is valid even when the CGLM exhibits a state in which the amplitude plays an important role in dynamics. In the case that there is no phase singular point, Eq. (7) can be reduced to the phase map
ein1r Leinr=jLeinrj1ic2: (8)
First, we carry out the linear stability analysis of plane wave solutions of the CGLM
^
nr ejqj
2
expifq r c1 c2jqj2ng ; (9) with a constant vector q. Because the solutions have no phase singular point ( ^n 0), we use the phase
map (8), in which the plane wave solutions can be written in the form ^nr q r c1 c2jqj2n const. By substituting n ^n n into Eq. (8), the linearized
equation for n is obtained as n1r <1 ic2e1ic1r
22iqr
nr . Therefore, the Fourier
trans-form ~nk of n obeys ~ n1k eqk~nk; (10) where qk jkj2 2ic1q k lnf1 ic2=2 eic1jkj22qk 1 ic 2=2 eic1jkj 22qk g. If <fqkg > 0,
the plane wave with the wave number q is linearly unstable against the perturbation with a wave number k. Expansion of qk to fourth order in jkj leads to
qk iVgjkj D2jkj2 i gjkj3 D4jkj4; (11) with Vg 2c1 c2qk, D2 1 c1c2 21 c22q2k, g2 31 c 2 23c1 4c2q2kqk, and D4161 c 2 2 f3c2 1 24c1c2q2k 81 3c22q4kg. Here qk q k=jkj.
In the case that D2< 0, that is, the wave number q satisfies
jqj2> q2
E 1 c1c2=21 c22, the plane wave solu-tion is linearly unstable against long-wavelength perturba-tions. This instability is identical to the Eckhaus instability observed in the CGLE [4]. In particular, the spatially
uniform state loses its stability at 1 c1c2 0, which is the same as the Benjamin-Feir-Newell criterion in the CGLE, and the Benjamin-Feir instability occurs for 1
c1c2< 0. Then D2< 0 is satisfied for arbitrary wave
num-bers, and thus all plane wave solutions are linearly un-stable. It is expected that the absolute instability of the plane wave solutions [14] can be also investigated.
Next, we show numerical results of the CGLM in a 2D system with the linear size L 128. We have numeri-cally confirmed that Eq. (7) exhibits the same behavior as that in Eq. (6). Numerical cost to solve Eq. (6) is almost the same as that for Eq. (7), and zncan be obtained from nbut not vice versa. Therefore, we use Eq. (6) instead of Eq. (7) for the numerical simulations.
Slightly below the critical point 1 c1c2 0 of the Benjamin-Feir instability, the phase turbulence arises from initial conditions n 1. As shown in Fig. 1(a), the amplitude j nj has the cellular structure as well as in the CGLE [15]. The phases arg n and nhave the spatial structure similar to that of the amplitude, and its spatial mean slowly increases.
Now we prove that the phase turbulence observed in the CGLM coincides with that in the CGLE. Because this turbulence has no phase singular point, we can use the phase map (8) in this proof. As shown in the linear stability analysis given above, only the long-wavelength modes are destabilized slightly below the critical point 1 c1c2 0. Therefore, the spatial scale of the variation of the phase variable nis expected to be large near the critical point. By letting the spatial scale be of order 1=2with a small-ness parameter , the term r2ein is estimated to be of order , and the linear operator L is expanded as L
e1ic1r2 1 1 ic
1r2121 ic12r2r2 O3. Substituting this expansion into Eq. (8) and setting n1
n! @n=@n, we obtain the Kuramoto-Sivashinsky (KS)
equation: @n @n 1 c1c2r 2 n c2 c1rn2 1 2c 2 1 2c1c2 1r2r2n: (12)
It should be noted that the coefficients of r2
nand rn2
in Eq. (12) coincide with those in the KS equation derived
FIG. 1. Snapshots of j nj of (a) phase turbulence for c1; c2
1; 1:2 and (b) amplitude turbulence for c1; c2 1; 0:6.
In (b), there are 544 defects. 134102-2
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from the CGLE by the phase reduction method, while the coefficients of r2r2
n derived from the CGLM and the
CGLE are different [2,16]. This deviation may be carried by the difference of the definitions of the ‘‘phases,’’ nfor the CGLM and arg for the CGLE. Note that the deviation vanishes at the critical point 1 c1c2 0 and that, in the limit 1 c1c2 ! 0, the amplitude fluctuation vanishes (j nj ! 1), and thus n approaches arg n. These facts
imply that the phase turbulences in the CGLM and the CGLE are quantitatively the same near the critical point. One may find that the coefficient of r2r2
n is different
from D4in Eq. (11) for q 0. This is because the growth rate of Eq. (10), ~n1 ~n eq 1 ~n, is not qk
but [eqk 1], whose long-wavelength expansion for q 0 gives the coefficients of Eq. (12).
In addition to the phase turbulence, we observe the amplitude turbulence for c1; c2 1; 0:6 [Fig.1(b)], which is characterized by the exponential decay of the spatial correlation [Fig.2(a)]. In this state, a lot of phase singular points (defects) exist over the whole space. The temporal evolution of the number N of defects [Fig.2(b)] shows that pairs of defects are created and annihilate in time. We find that the correlation length c defined in
Fig.2(a)is approximately equal to the mean distance dm
between a defect and its nearest neighbor: c dm 9:0.
Because the defect turbulence observed in the CGLE also has these characteristics [17], the amplitude turbulence in the CGLM is expected to be identical to the defect turbu-lence in the CGLE.
For c1; c2 1:0; 0:4, after the transient amplitude turbulent state, spiral waves tend to appear and eventually cover the whole space. The amplitude shows no temporal evolution, as shown in Fig.3(a), while the phase exhibits the spiral waves, which rotate at a constant speed. This state is called either the frozen state (FS) or the vortex glass state [9]. For c1; c2 3:0; 0:4, although spiral waves also occur, the amplitude is not frozen, and both the phase and the amplitude have spiral structures [Fig.3(b)]. Here-after we call this state the amplitude spiral state (ASS).
Figure4depicts details of the spatial structures near the spiral core in the FS and the ASS. There are at least three
qualitative differences between the FS and the ASS. First, the FS has an ordered spiral structure in the phase [Fig. 4(a)] and a rotationally symmetric hole structure in the amplitude [Fig.4(b)]. However, the ASS has a distorted spiral in the phase [Fig. 4(c)] and an ordered spiral in the amplitude [Fig.4(d)]. In the ASS case, the spiral structures in the amplitude and the phase have opposite rotational directions. Second, the position of the spiral core is mo-tionless in the FS, while the core position rotates in the ASS. For example, the core in Fig. 4(d) rotates counter-clockwise. Third, far from the spiral core, the FS exhibits the plane wave described by Eq. (9), while the ASS ex-hibits the plane wave with amplitude modulations. As shown below, the latter seems identical to the modulated amplitude wave (MAW) observed in the 1D CGLE, which is described as x; t ax vteix!t[18,19]. In the
1D CGLM, we found a solution [Fig. 5(a)] satisfying the relation nx ax vneix!n, which was
demon-strated by the fact that nxeix!n exhibits the trans-lational motion with keeping its profile [Fig. 5(b)]. The
(a) (b) r Correlation λc 0 0.2 0.4 0.6 0.8 1 0 10 20 30 5000 500 550 600 650 n N
FIG. 2. Statistics of amplitude turbulence for c1; c2
1; 0:6. (a) Spatial correlation <h nr n0i=hj n0j2i, where hi denotes time averaging. The correlation length c is defined in the same way as that in Ref. [17]. (b) Temporal evolution of the number N of defects in the whole space. The dashed line indicates the average number hNi 572:9.
FIG. 3. Snapshots of j nj of (a) the FS for c1; c2 1; 0:4
and (b) the ASS for c1; c2 3; 0:4.
FIG. 4 (color online). Snapshots of (a),(b) the FS for c1; c2
2; 0:4 in a subsystem of the linear size l 12:5 and (c),(d) the ASS for c1; c2 3; 0:5 in a subsystem of the
linear size l 31:25. (a),(c) Phase field arg n; (b),(d) amplitude field j nj with isophase curves (red: < n 0; green: = n 0).
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amplitude pattern of this solution [Fig.5(c)] is quite similar to those of the plane wave of the ASS [Fig.5(d)] and the MAW in the 1D CGLE [Fig.5(e)]. It should be noted that, depending on initial conditions, both the FS and the ASS can be observed (and, hence, both are stable) for parameter values in a certain range in the c1; c2 space.
We found the ASS in the present Letter. In addition, we numerically obtained the ASS in the 2D CGLE (not the CGLM) from specific initial conditions [20]. Hence, the ASS is a stable state of the CGLE, and this fact implies that it is possible to observe the ASS in other oscillatory media. It was found that the long-wavelength modulated spiral can be observed in the CGLE with heterogeneity [21] and the Belousov-Zhabotinsky reaction [22] and that the CGLE with a constant term exhibits a state in which both the amplitude and the phase have spiral structures [23]. To investigate the relevance of these states with the ASS is a future problem.
The main difference between the CGLE and the CGLM is the presence or absence of the relaxation process to the limit-cycle attractor (j j 1). In spite of the difference, the results presented in this Letter reveal that the CGLM can well reproduce the dynamics observed in the CGLE. Therefore, we believe that the relaxation process does not play an important role in the CGLE and that the CGLE can be well described by only the phase dynamics appropri-ately constructed. These facts may give new insight into the understanding of oscillatory media.
We thank A. S. Mikhailov and N. Fujiwara for valuable comments. This study was partially supported by the 21st Century COE Program ‘‘Center of Excellence for Research and Education on Complex Functional Mechanical
Systems’’ at Kyoto University.
*tsuka@acs.i.kyoto-u.ac.jp †Deceased.
[1] M. C. Cross and P. C. Hohenberg, Rev. Mod. Phys. 65, 851 (1993).
[2] Y. Kuramoto, Chemical Oscillations, Waves, and
Turbu-lence (Springer-Verlag, Berlin, 1984); Chemical Oscilla-tions, Waves, and Turbulence (Dover, New York, 2003).
[3] T. Bohr, M. H. Jensen, G. Paladin, and A. Vulpiani,
Dynamical Systems Approach to Turbulence (Cambridge
University Press, Cambridge, England, 1998).
[4] I. S. Aranson and L. Kramer, Rev. Mod. Phys. 74, 99 (2002).
[5] H. Chate´ and P. Manneville, Physica (Amsterdam) 224A, 348 (1996).
[6] Y. Oono and S. Puri, Phys. Rev. Lett. 58, 836 (1987). [7] M.-N. Chee, R. Kapral, and S. G. Whittington, J. Chem.
Phys. 92, 7315 (1990).
[8] T. Bohr, A. W. Pedersen, M. H. Jensen, and D. Rand, in
New Trends in Nonlinear Dynamics and Pattern-Forming Phenomena, edited by P. Coullet and P. Huerre (Plenum,
New York, 1990); T. Bohr, A. W. Pedersen, and M. H. Jensen, Phys. Rev. A 42, 3626 (1990).
[9] G. Huber, P. Alstrøm, and T. Bohr, Phys. Rev. Lett. 69, 2380 (1992).
[10] X.-G. Wu and R. Kapral, J. Chem. Phys. 94, 1411 (1991). [11] A. Torcini, H. Frauenkron, and P. Grassberger, Phys. Rev.
E 55, 5073 (1997).
[12] S. Uchiyama and H. Fujisaka, Phys. Rev. E 56, 99 (1997). [13] T. Watanabe, Y. Tsubo, and H. Fujisaka, Phys. Rev. E 65,
026213 (2002).
[14] I. S. Aranson, L. Aranson, L. Kramer, and A. Weber, Phys. Rev. A 46, R2992 (1992).
[15] P. Manneville and H. Chate´, Physica (Amsterdam) 96D, 30 (1996).
[16] N. Tsukamoto, H. Fujisaka, and K. Ouchi, Prog. Theor. Phys. 116, 669 (2006).
[17] P. Coullet, L. Gil, and J. Lega, Phys. Rev. Lett. 62, 1619 (1989).
[18] A. Torcini, Phys. Rev. Lett. 77, 1047 (1996).
[19] L. Brusch, M. G. Zimmermann, M. van Hecke, M. Ba¨r, and A. Torcini, Phys. Rev. Lett. 85, 86 (2000).
[20] We observed the ASS in the 2D CGLE with the initial condition for the field of the ASS obtained in the CGLM. The parameter values used in the numerical simulation of Eq. (1) are c1 3:0, c2 0:5, and D
1:0. For these parameters, ‘‘usual’’ initial conditions lead to the FS.
[21] L. Brusch, A. Torcini, and M. Ba¨r, Phys. Rev. Lett. 91, 108302 (2003).
[22] L. Q. Zhou and Q. Ouyang, Phys. Rev. Lett. 85, 1650 (2000).
[23] A. S. Mikhailov (private communication); N. Fujiwara, Ph.D. thesis, Kyoto University, 2007.
FIG. 5. Temporal evolutions of (a) < nx and (b) < nxeix!n ( 0:125, ! 0:2504) in the 1D CGLM. (c) –(e) Snapshots of the amplitude patterns. The pa-rameter values of both the CGLM and the CGLE are c1; c2
3; 0:5, and D 1:0 for the CGLE.
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# The cart started at t=0,its acceleration varies with time as shown in figure. Find the distance travelled in 30 seconds and draw the position time graph.
2 years ago
Susmita
425 Points
In t=0 to 10 s,acceleration f=10 m/s2.Initial velocity u=0 at t=0.Final velocity v at t=10s can be calculated from
v=u+ft
Or,v=10×10=100 m/s
Distance s travelled in this time can be calculated from
v2=u2+2fs
or,s1=v2/2f=1000/2=500m
In the time interval t=10 to 20s,f=0.So particle moves with constant speed v=100m/s.So distance travelled is s2= vt=100×10=1000m
In the time interval t=20 to 30 second,f=-10m/s2.Initial velocity is u=100m/s and final velocity
v=u+ft
Or,v=100-100=0m/s
Distance travelled
v2=u2+2fs3
Or,s3=500 m.
So total distance travelled is 2000.
To draw the graph for the 1st segment,
f=10
Or,d2x/dt2=10
Integrating
dx/dt=10t+c
Velocity is zero at t=0.so c=0.
Again integrating
x=10t2/2=5t2
It is a parabola
For the second segment,
f=0
d2x/dt2=0
Or,dx/dt=constant=say c
Integrating
x=ct+c1
This is a straight line.
For the 3rd segment
x=-5t2+c2t+c3
It is also a parabola.
Please try to draw these in sequentially in a graph as I have no option of attaching file from my device.
Hope this helps.
2 years ago
Susmita
425 Points
In both cases draw half the parabola.Beacuse t is only positive here.
In the 1st case the parabola is of the form
y=x2.Draw only the right side of it starting from origin.
In the 3rd segment it is of the form y=-x2.In this case also draw only the right half of the parabola.
2 years ago
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## #1 Re: This is Cool » My New Twin Prime Numbers » 2015-07-02 19:04:10
By the way, here is what I used. Just brute force.
``perl -Mntheory=:all -MMath::GMPz -E 'for my \$pt (1..11) { for my \$n (1 .. 50000) { my(\$sign,\$s)=(1,0); forprimes { \$s += \$sign* Math::GMPz->new(\$_)**\$pt; \$sign=-\$sign } nth_prime(\$n); if (\$n < \$s && is_provable_prime(\$s-\$n) && is_provable_prime(\$s+\$n)) { say "\$pt: \$s +/- \$n"; last; } } }'``
Takes about 1.5 minutes. Generating the proofs vs. plain BPSW makes no noticeable timing difference.
``````1: 8 +/- 5
2: 20 +/- 3
3: 106 +/- 3
4: 560 +/- 3
5: 59779076418421204640 +/- 1281
6: 4255809704937643329920 +/- 609
7: 63835705531104194020747282 +/- 711
8: 5199099913899300539585745185120 +/- 957
9: 86486076326025141392980473635379192344452 +/- 4035
10: 1287183127776263297776662121192361760476608760 +/- 3723
11: 101678827728094895209692286827415375382 +/- 435``````
I imagine this could be translated into Pari/GP with little effort, and its isprime is APR-CL with modern versions. Mathematica would probably not translate directly but should be easy enough. ProvablePrimeQ according to documentation uses various methods including ECPP, but I have no idea how fast it is.
## #2 Re: This is Cool » My New Twin Prime Numbers » 2015-06-30 01:48:14
I didn't before (just using something similar to Mathematica's PrimeQ), but these numbers are small so I changed to is_provable_prime. They're provable primes (if they weren't we'd have found something that was a BPSW + extra M-R tests counterexample, which would be astonishing).
## #3 Re: This is Cool » My New Twin Prime Numbers » 2015-06-28 10:15:51
Oh, I see. You're restricting n to both the number of terms and the +/- unit.
I believe for Pt=5, n=1281; Pt=6, n=609; Pt=7, n=711; Pt=8, n=957; Pt=9, n=4035. Double check them.
## #4 Re: This is Cool » My New Twin Prime Numbers » 2015-06-27 16:20:43
Isn't that a solution for Pt=5? Similarly n=3 Pt=6 => +/-9, n=3 Pt=7 => +/- 63
## #5 Re: This is Cool » Max Number of Consecutive Primes » 2014-12-31 17:23:38
Stangerzv wrote:
Thanks danaj for the input. Can you list the 10 consecutive primes.
It's in post 15: p=247752179
``247752179: 5286145759 5286145811 5286145829 5286145831 5286145841 5286145873 5286145951 5286145981 5286145999 5286146029``
You can do this with Mathematica like:
``Table[Prime[247752179+i-1],{i,10}]``
or Perl/ntheory as:
``perl -Mntheory=:all -E 'say nth_prime(247752179+\$_) for 0..9;'``
or Pari/GP (slow) as:
``for(i=0,9,print(prime(247752179+i)))``
or the faster but still much slower than the others Pari/GP:
``s=prime(247752179); for(i=0,9,print(s);s=nextprime(s+1))``
## #6 Re: This is Cool » Max Number of Consecutive Primes » 2014-12-30 19:08:33
PrimeQ uses, from the best documentation I know of, BPSW + a base 3 strong probable prime test. BPSW is deterministic for 64-bit numbers like we're using, much less adding another M-R test. There should be no reason to use ProvablePrimeQ here. Or another way of looking at it: use PrimeQ, then for the few found examples, use ProvablePrimeQ -- if something comes up false then you have a great example of Mathematica doing something wrong.
Pari/GP has isprime() (like I put in the example) which is deterministic (BPSW then APR-CL if large), or ispseudoprime for just BPSW. They're identical for 64-bit inputs. Perl/ntheory has is_provable_prime() to do a proof, but similarly it is identical to is_prime or is_prob_prime for 64-bit inputs.
"So far, 8 is the largest." There is an example of 10, but I did not find anything higher than that.
## #7 Re: This is Cool » Max Number of Consecutive Primes » 2014-12-30 13:42:05
This takes the 'p' values and computes the 8 's' values corresponding to prime[p+0], prime[p+1], ..., prime[p+7]. Of course the 's' values are prime and sequential, but we can also check that for each one, 34+3*s is also prime. I will note that the 34+3s results are not sequential primes -- it's hard to tell if this is expected in the original post, but your posted results don't have them sequential.
These are just the values for 8. The runs of 9 and 10 would also be runs of 8.
I'm surprised Mathematica isn't faster. It uses a BPSW+1MR for large inputs, and may drop the extra MR for these small numbers where BPSW is deterministic. That would match Pari and Perl/ntheory. The forprime() call does help since we can easily loop through the primes.
``for my \$s (qw/13155307 23373712 28619826 206480632 430788664 451840108 723650223 778689176 922591383 943068103 1005001898 1083424548 1406260752 1528839396 1653992654 1884407047 2112863733 2420339382 2444147088 2467436354 2559176982 3141752981 3239363657 3314161732 3346438597 3510021953/) { say "\$p: ", join(" ",map { nth_prime(\$p+\$_) } 0..7); }'``
``````13155307: 239878543 239878571 239878579 239878603 239878621 239878649 239878663 239878673
23373712: 440460583 440460589 440460593 440460619 440460623 440460659 440460673 440460703
28619826: 545465579 545465593 545465633 545465659 545465731 545465743 545465749 545465759
206480632: 4365959711 4365959713 4365959753 4365959779 4365959789 4365959809 4365959879 4365959909
430788664: 9442018519 9442018559 9442018583 9442018589 9442018601 9442018609 9442018613 9442018643
451840108: 9926055071 9926055079 9926055131 9926055133 9926055143 9926055163 9926055169 9926055173
723650223: 16255022843 16255022851 16255022863 16255022869 16255022939 16255022941 16255022969 16255023019
778689176: 17551241549 17551241551 17551241563 17551241579 17551241581 17551241593 17551241611 17551241639
922591383: 20958804241 20958804271 20958804283 20958804323 20958804341 20958804343 20958804371 20958804413
943068103: 21445686761 21445686781 21445686791 21445686793 21445686811 21445686839 21445686851 21445686901
1005001898: 22921075381 22921075409 22921075421 22921075423 22921075433 22921075451 22921075493 22921075501
1083424548: 24795040741 24795040759 24795040801 24795040903 24795040931 24795040943 24795040951 24795040969
1406260752: 32567701541 32567701561 32567701571 32567701609 32567701631 32567701643 32567701693 32567701709
1528839396: 35540355871 35540355881 35540355889 35540355901 35540355911 35540355923 35540355953 35540355959
1653992654: 38585987383 38585987399 38585987413 38585987441 38585987509 38585987519 38585987549 38585987591
1884407047: 44218726703 44218726723 44218726763 44218726793 44218726813 44218726829 44218726859 44218726913
2112863733: 49832772751 49832772763 49832772791 49832772793 49832772811 49832772851 49832772943 49832772961
2420339382: 57428667449 57428667473 57428667509 57428667529 57428667613 57428667671 57428667701 57428667799
2444147088: 58018595399 58018595419 58018595431 58018595443 58018595461 58018595483 58018595501 58018595519
2467436354: 58595995259 58595995273 58595995279 58595995319 58595995331 58595995349 58595995363 58595995439
2559176982: 60872272801 60872272819 60872272829 60872272859 60872272889 60872272913 60872272931 60872272933
3141752981: 75403452713 75403452721 75403452749 75403452781 75403452799 75403452811 75403452833 75403452881
3239363657: 77849739503 77849739521 77849739529 77849739589 77849739619 77849739659 77849739673 77849739719
3314161732: 79726451263 79726451293 79726451303 79726451353 79726451393 79726451443 79726451459 79726451503
3346438597: 80536791493 80536791509 80536791521 80536791553 80536791583 80536791611 80536791623 80536791659
3510021953: 84648776909 84648776911 84648776921 84648776939 84648776941 84648776963 84648776999 84648777019``````
## #8 Re: This is Cool » Max Number of Consecutive Primes » 2014-12-30 08:57:58
I can send you my perl script if you'd like. It takes 15s to find three 8-consecutives for s < 1e9. Searching to s < 1e10 takes about 2.5 minutes:
``````7 1718020 2492520 2671928 3036055 3562655 3594150 10654019 15191513 15626334 21063558 21752023 22152282 25059807 27161939 29545494 32921098 37188348 44503753 46206335 51087603 51960322 54481455 57923734 63665188 65256256 65443263 68573108 69790405 69888940 72402885 89135244 91447387 96231989 96907585 98870702 104486689 106724222 108468644 110963047 123035310 126428987 133783990 140775016 153601104 156826152 158202292 163519132 166930965 179742892 184518605 185994353 188583675 192392346 203631792 205873490 212190182 218841344 232714800 256292473 261209858 276867166 283985400 289567855 304492904 305351642 319820383 322860650 323803818 335518221 344398882 361505496 366930942 385324698 395261357 399795913 415012199 420673304 422144423 426839424 435086906 441592465 453986477
8 13155307 23373712 28619826 206480632 430788664 451840108
9 62888993
10 247752179``````
Searching to s < 1e11 (100,000,000,000) took 21 minutes:
``````7 1718020 2492520 2671928 3036055 3562655 3594150 10654019 15191513 15626334 21063558 21752023 22152282 25059807 27161939 29545494 32921098 37188348 44503753 46206335 51087603 51960322 54481455 57923734 63665188 65256256 65443263 68573108 69790405 69888940 72402885 89135244 91447387 96231989 96907585 98870702 104486689 106724222 108468644 110963047 123035310 126428987 133783990 140775016 153601104 156826152 158202292 163519132 166930965 179742892 184518605 185994353 188583675 192392346 203631792 205873490 212190182 218841344 232714800 256292473 261209858 276867166 283985400 289567855 304492904 305351642 319820383 322860650 323803818 335518221 344398882 361505496 366930942 385324698 395261357 399795913 415012199 420673304 422144423 426839424 435086906 441592465 453986477 466621403 466955611 470479122 475281381 478099131 486581146 488574100 492974241 495709419 503242169 514290308 524353559 526210953 547238431 555377530 560568673 563819394 564050336 565117578 567697430 574333477 581719259 604397264 605285517 609953548 611506702 619023097 627335360 644365306 664376162 665037413 686007851 718559335 725439993 735943665 743431923 746057515 748880752 750680263 794532893 797034046 807908141 817185739 830061428 838793990 844170390 847996148 863088773 868761819 872583472 875894406 880427154 885068607 885730110 886707614 897000052 922799091 930122687 949561295 953536846 965574106 970670184 971616015 984136807 999904292 1032348520 1032626596 1038415609 1042590507 1063551577 1071083510 1084818946 1092401906 1092937465 1095569057 1097103079 1121692213 1127021960 1141442850 1142159307 1167453552 1179332930 1198709921 1207536623 1225783001 1226020529 1226973918 1230881242 1255337127 1266567073 1269833145 1288913145 1301900559 1315347414 1330558720 1335151552 1336167535 1336470638 1357529972 1366303467 1367532201 1394632344 1415994612 1425615667 1468126216 1479052712 1484197484 1484664232 1499162590 1510892825 1525540560 1538768858 1541431708 1547113639 1561569771 1573285197 1575338467 1608245600 1620892699 1636594726 1639912926 1642785097 1649448543 1658387196 1659492894 1669878049 1673030859 1680288649 1689009936 1705139805 1707877965 1720548461 1725560049 1727975655 1740450178 1757696409 1767977017 1774252457 1783722479 1833272684 1835641302 1864180009 1875220655 1894374764 1914950297 1925191608 1930188541 1944046812 1947734807 1950943878 1961537003 1968415550 1978950200 1983147463 1992503523 2003383667 2011759321 2036328796 2074864705 2083105576 2097268421 2120562842 2131335524 2173046960 2174549806 2197537096 2206149990 2239254808 2243414356 2349463495 2368648878 2388262838 2397241591 2398126033 2403896576 2417795626 2421180508 2427214723 2433644912 2460192669 2464829434 2466386056 2470597216 2473662702 2477156833 2481924935 2486632950 2535173281 2562687057 2577546536 2594427982 2602427582 2615924025 2627636732 2648467667 2681906133 2708967552 2721761110 2741304277 2746549938 2750335395 2753208097 2763210973 2766258139 2777676867 2787956664 2817979773 2820157908 2837602399 2869185904 2878113684 2882066301 2884645265 2912256157 2935732944 2939577836 2946296457 2955639898 2961631157 2989076982 2996581210 3004860881 3019444513 3024664264 3026879459 3092518506 3094303338 3099197655 3127207507 3128262874 3140049442 3157014173 3175544783 3193904988 3194803790 3202819376 3217890148 3225417795 3229520764 3232636927 3244256924 3258840570 3277320913 3278576127 3285898935 3296994608 3312205351 3332431095 3342688030 3367721960 3380264281 3384164887 3396537088 3402359710 3403327218 3406882404 3419374349 3453540047 3473030524 3478573403 3487003920 3489342936 3541544528 3553821171 3563989241 3570263926 3579929644 3600057302 3606082748 3608868228 3612932294 3628577024 3631111020 3639429262 3648181186 3664377198 3676997253 3694484527 3733723540 3750029130 3770734294 3792743589 3807294468 3809322646 3818970943 3826632839 3852382997 3864241176 3937518769 3964472367 3968680328 3978909813 3980535308 3989589875 4032468634 4043649533 4045499779 4059220413 4091714016 4092344452 4097893987 4103216975 4104895899
8 13155307 23373712 28619826 206480632 430788664 451840108 723650223 778689176 922591383 943068103 1005001898 1083424548 1406260752 1528839396 1653992654 1884407047 2112863733 2420339382 2444147088 2467436354 2559176982 3141752981 3239363657 3314161732 3346438597 3510021953
9 62888993 2337819127 2805299994 3543590679 3649181083
10 247752179``````
Here is a Pari/GP version. It runs slower than Perl's ntheory for this, but it's not too bad (it needs Pari 2.8):
``s=0;sr=0;cr=0;forprime(p=1,1e9,s++;if(isprime(34+3*p),cr++,if(cr-sr>=7,print(cr-sr," ",sr));cr=s+1;sr=s+1))``
## #9 Re: This is Cool » Max Number of Consecutive Primes » 2014-12-27 13:17:52
length and p, where s = nth_prime(p) and 34+3s is prime; length the first run of consecutive p values
6 1373
7 1718020
8 13155307
9 62888993
10 247752179
For 82+5s:
6 131582
7 245075654
8 7358058514
Simple brute force walking primes looking at primality of 34+3s or 82+5s for s < 1e12. No easy answers past these for this search method.
## #10 Re: Euler Avenue » What had you learned today that you found interesting? » 2014-12-03 01:02:42
http://docserv.uni-duesseldorf.de/servlets/DocumentServlet?id=26247
and
http://arxiv.org/abs/1409.1780
Improvements to the Dusart (2010) bounds on prime count and nth prime. I also started playing around with nth prime limits via binary search on prime count bounds, which tentatively looks like a nice improvement (indicating that the prime count bound formulas are tighter than the nth prime bound formulas).
Also, someone finished computation of Pi(10^26) via combinatorial methods.
## #11 Re: Jokes » Experimental Mathematics Troll » 2014-10-23 19:16:27
I did it this way, which is longer but lets one see all the results. Perl:
``````use ntheory qw/:all/;
my @I = qw(A a B b C c); # lower case = child, upper case = guardian
forperm {
my \$s = "@I[@_]"; # the permuted string
say \$s unless \$s =~ /a.*A|b.*B|c.*C/; # print unless child precedes its guardian
} 6;``````
which prints out the 90 cases as: "A a B b C c" , "A a B C b c", "A a B C c b", etc. There are lots of ways to structure the restriction (e.g. a hash of positions, then do position comparisons).
## #12 Re: This is Cool » My New Twin Prime Numbers » 2014-10-20 17:10:23
Having 1024 nodes with 4 2880-core GPUs each doesn't solve anything if there isn't an application to run on it. Do you have something in mind?
Trial division would be trivial to parallelize, but is basically worthless. I see a paper by Asaduzzaman et al. from 2014 saying they have sped it up on GPUs, but (1) that's not hard, and (2) their comparison CPU code is 100x slower than the single-core CPU code I use, making their 90x speedup claim dubious. One thing that bothered me at GTC was the uncertainty around speedup numbers -- it's easy to get 100x speedup if you write cruddy CPU code in a weekend and then compare to GPU code you had four people working 6 months on.
AKS should be pretty easy to run in parallel (each of the s tests are independent), and the code that takes 99% of the time is pretty simple. You need to do lots and lots and lots of s += (a * b) % n type operations. Some problems:
• The growth rate difference means ECPP or APRCL on a single host will beat the hundreds of thousands of GPU cores running AKS. If the serial algorithm takes many trillions of years, how many cores will we need to make it run in an hour? Do you have that many GPUs?
• There is no certificate, so realistically your saying you ran this is no better than just saying you used well-known software to run BPSW + a Frobenius test + some number of random-base M-R tests. Less for me because I've seen too many broken AKS implementations. It doesn't matter how fast it is if it isn't correct.
There may be other tricks for AKS that can help, but you have to make major changes to it to get it to the same growth rate as APR-CL or ECPP. There are definitely some research opportunities here.
The BLS75 methods involve factoring, so won't scale to usefully large numbers.
I don't know how well APRCL would parallelize. I think there was a thesis about the subject, but it didn't give much detail.
ECPP certainly could benefit if you could write it. If you do, please make it open source and publicize it. That would be a major accomplishment of use to many people. There are quite a few places where things could run in parallel (the fanout is extremely large if you run speculatively).
## #13 Re: This is Cool » My New Twin Prime Numbers » 2014-10-20 03:28:08
Here is an example graph showing some times for primality proving. Single threaded on a 4.3GHz machine (more details available if you'd like). I have not seen a faster AKS implementation than the B+V version here, but they could be out there. Pari/GP's APR-CL (the isprime command) runs a bit slower than WraithX's APRCL 1.1. I have no idea how fast Mathematica is (timings from this thread make me think its BPSW is quite a bit slower than the graph here, but that is independent from its proof times). Primo is the most common program used for proving large general primes.
100 digit primes can be proved extremely fast (a few milliseconds). 1000 digits will take ~2-8 minutes. For ECPP and APR-CL the time increases at about the 4th power of the number of digits (for AKS it's the 6th power). By 20k digits, expect 3 months using all cores of a 4- or 6-core machine.
I'm not sure about the various DR-restricted numbers, but the first set of unrestricted (post 109) are fairly small. E.g. p=29 => n=932651 is 213 digits, p=31 => n=2547137 is 242 digits. They grow of course, as p=59 => n=4183583 is 469 digits.
## #14 Re: Help Me ! » Permutation Problems » 2014-09-06 21:05:21
Since I just added a permutation iterator to my module yesterday, this is a convenient time to try it. Using the hash to not count duplicates.
``````my %h;
my @m = split "", "mathematics";
forperm {
my \$s = join "", @m[@_];
undef \$h{\$s} if \$s =~ /a.*e.*a.*i/;
# alternate: undef \$h{\$s} if \$s =~ tr/aeiou//cdr eq "aeai"
} scalar(@m);
say scalar keys %h;``````
Result match's Bobbym's analytic answer. Thanks for the fun!
## #15 Re: This is Cool » Find all the prime numbers in a given range.... » 2014-08-28 20:07:16
What is the advantage of this over the well known Sieve of Eratosthenes?
I think you meant to say "up to a limit" (e.g. find all primes up to 29) instead of "in a given range" (e.g. find all primes in the range 5,208,079 and 6,206,197).
## #16 Re: This is Cool » Tests of Divisibility- Simple tricks » 2014-08-07 16:37:12
Many more, alternate methods, and examples: Wikipedia: Divisibility rule
## #17 Re: Coder's Corner » Just another perl hacker » 2014-07-18 03:58:03
(p.s. the talk at YAPC was by me, so we have a little Perl representation)
I didn't know I was supposed to ask a question... Will we have a proof of the Riemann Hypothesis by 2059?
## #18 Re: Coder's Corner » Just another perl hacker » 2014-07-17 00:57:23
Indeed there are.
There was even a "Perl and Scientific Programming" panel at YAPC this year, and someone gave a lightning talk about a Perl number theory package.
## #19 Re: Coder's Corner » AKS performance » 2014-06-07 08:51:56
Sorry for not including the link. It was here as well: http://www.mathisfunforum.com/viewtopic … 75#p273775.
Re implementation, Darn -- I was hoping to find something new! Thanks for looking.
## #20 Coder's Corner » AKS performance » 2014-06-07 07:06:15
danaj
Replies: 3
A year ago, bobbym wrote:
bobbym wrote:
They are right about AKS, it takes about an hour to prove the primality of a 1000 digit number on an old desktop.
Can you point me to the software you're using? This is much faster than any AKS implementation I am aware of. It's millions of times faster than stated performance numbers in papers about AKS implementations, e.g.
- Crandall and Papadopoulos (2003) "A 30-decimal-digit prime this took, [...], about a day to be resolved."
- Rotella (2005), "An Efficient Implementation of [AKS]" takes a minute for 4 digit numbers, and ends with "[...]the AKS algorithm, while it is elegant and relatively simple, should be viewed as an algorithmic curiosity rather than an algorithm to be used for actual primality proving."
- Salembier and Southerington (2006) show a time of 30 minutes for a 15 digit number.
- Brent (2010) estimates 840 years for a 308 digit prime using a version with some improvements.
My code/machine is running about 200 times faster than Brent's numbers, and I come up with en estimate of about 4200 years for a 1000 digit number. APR-CL and ECPP are about 2-5 minutes (single threaded) for 1k digits on the same computer.
## #21 Re: Help Me ! » Intersections of sets of primes » 2014-03-18 17:24:01
There are plenty of other not-uncommonly-discussed sets as well, e.g. Safe, Sophie Germain, Lucas, Fibonacci, Lucky, Palindromic, Pillai, Good, Cuban (y+1, y+2), Primorial (+1, -1), Euclid, Circular, Panaitopol. (Re digressing about Euclid and Primorial, one wants to distinguish between P#+/-1 and Pn#+/-1 -- that is primes up to P or the Pn-th prime). I'm sure one could come up with another hundred sets. Surveying OEIS would bring up lots of them, for example.
For intersections, I found a useful practical method to be listing the allowable subsets of n mod 210 for each type. Some don't have a pattern, but many do (e.g. cuban1, cuban2, twin, triplet, quadruplet, cousin, sexy, safe, sophie, panaitopol to name some). Not only does this help for single cases, but one can then do the intersection of the two (or 3 or more) types. Sometimes this yields a null set once past tiny numbers (e.g. quadruplet cuban primes, or safe quadruplets other than 5 and 11).
This doesn't really help answer the original questions however. For my reading of the first question, certainly we can find examples where sets intersect, e.g. twin sexy primes, or titanic cousin primes (e.g. 10^1000 + 744843).
I believe the answer to your second question is no in general. In some of the cases, the number of intersections from A to B between two sets will not necessarily follow the same trend as between A to B+delta. In other cases (what I believe most people are discussing) both sets will be infinite, though maybe it would be possible to prove that the number of elements of set P in an interval is always greater than the number of elements of set Q in the same interval, or compare asymptotic densities, or something along those lines (e.g. the density of titanic primes in an interval starting after 10^1000 is going to be at least as dense as any of the restricted sets in the same interval).
## #22 Re: This is Cool » Large Prime » 2014-03-14 17:56:17
I'm not sure if I should post, (1) it is somewhat necro-posting, (2) there seems to be another conversation going on about DH, and (3) I type way too much. But regarding generating large primes, this is a good paper about generating n-bit random primes: "Close to Uniform Prime Number Generation With Fewer Random Bits" by Fouque and Tibouchi available at eprint.iacr.org/2011/481. For crypto, you may want to also look at FIPS 186-4.
PRIMEINC: generate a random number in the range, run next_prime on it. Simple and fast, but bad distribution.
Trivial: Aka Monte Carlo method as in post 3. Perfectly uniform, but uses excessive randomness and is very slow for large sizes. A comment about the post 3 code: for sizes over 2 bits, you'd want to take n-2 random bits and then set bits n-1 and 0 -- there is no need to waste time and entropy testing even numbers. A well written test will return almost instantly, but you wasted time getting all those bits -- if you're on an isolated headless server using /dev/random you may wait hours to get the next set.
Modified: Methods like Fouque and Tibouchi A1 or A2; Joye-Paillier, etc. I use a somewhat similar method for odd ranges (e.g. for ranges not a power of 2). These sacrifice some uniformity for big increases in speed and less entropy consumption. (the entropy consumption may or may not matter to you, but sometimes it is important).
Provable primes, typically Shawe-Taylor or Maurer's FastPrime. Common for crypto use. You can find Shawe-Taylor in FIPS 186-4; Maurer's algorithm in his publicly available paper, Menezes's book, or various open source implementations. These work by generating a small random prime using the trivial method, then generating a larger one by iterating with additional random input until proven with either Pocklington or its improved version from BLS theorem 3. Then recurse to make successively larger primes. They only generate a subset of the primes in the range (10% for Maurer's FastPrime, substantially less for Shawe-Taylor) but that typically doesn't matter.
openssl or other software. Good and bad. OpenSSL likely doesn't make some mistakes you may make. But you have to make sure your platform has a recent version, your program actually calls the right executable, you send it the right arguments, you interpret the output and exceptions properly, you're ok with its decisions on randomness sources, you're OK with it not meeting FIPS 186-4 standards for primality testing, and know it is slower for sizes > 1024 than what can be done with GMP meeting the standards.
There are lots of other considerations. E.g. where are you getting your randomness? Are you using a good enough primality test for your purpose? Do you need provable primes? Strong primes? Writing your own code is fun, but may have bugs -- do lots of testing.
Whether these are "quick" or not depends on the size, which method you choose, your implementations, and what you think quick means. For 1024-bit primes, my older machine using Perl code generates them in 0.06s each (F&T algorithm 1, ISAAC seeded from /dev/random, BPSW + 1 M-R probable primes). Add 0.006s for 3 additional random M-R plus a Frobenius test. Add not much more for enough extra M-R tests to make FIPS 186-4 happy. For smaller sizes sometimes running a primality proof on the n-bit prime is faster than a constructive method, but your mileage may vary. At 1024 bits, my Maurer routine takes 0.65s, while Shawe-Taylor using FIPS 186-4 + SHA256 takes 0.24s. The generation code is in Perl so it could be faster, but the heavy work ends up being in C+GMP.
## #23 Re: Help Me ! » Is it any pseudo pattern gives prime number in proper range? » 2014-03-07 18:57:26
In the example for "is 281 prime" you did:
1. find squre root of number near about it is 17
2 17*=Pn*=510510
3 Ip17 =number not divisible by 2,3,5,7,9,...17
then: prime= Ip17-Pn* where IP17 started at 510513, and Pn* was 510510.
For the example "111,111,111,111,111" (decimal), I get a square root of 10540925. Step 2 as well as the prime generation says to use Pn*=10530925#. That came out to 4,576,061 digits for me. The number in "No of Ipn sub series" is going to be similarly titanic. What is different about this example from the 281 example?
You have to either store these or calculate them, and either way I don't see how you can say "don't worry about time [...] to find them." Why not just precalculate the primes directly if we get to discount the time and/or space?
## #24 Re: Help Me ! » Is it any pseudo pattern gives prime number in proper range? » 2014-03-06 19:06:41
510510 = 17# = primorial(17) = 2*3*5*7*11*13*17.
92160 = euler_phi(17#)
I think I'm with everyone else trying to figure out where Ipn is derived. The obvious way (I'm just making a guess) is that it is the primorial plus the previously generated primes with the single new prime sieved out. But that would make the primorial redundant. Then it reduces to "just use this sequence, which was made by sieving p (max p <= sqrt(n)) from the previous sequence. Which was made by sieving out prev_prime(p) from its previous sequence, etc."
I think Bobby's response to the earlier question does kind of get to the point. For the small 15 digit example you have to generate a 4.6 million digit primorial, then somehow come up with the sequence. Take a prime in the range of 2^512, which would be useful for an old RSA key. These take APR-CL 0.1 second to prove on a fast computer, or 200uS for BPSW (not proven, but it would be big news if it was composite since you'd be the first one to find a counterexample). For your method we would need to generate primorial(about 10^77) to get started. Ouch.
Taking a stab at your earlier question #2, which I will take as generating primes in a range, here are some ideas. If you just want something to run, look for primesieve. If you're looking for prime counts, there are fast algorithms that don't involve generating primes, so that's a different question.
1, either a very small range, or a very large base (e.g. well over 10^20). Pick one of:
a) Primality test with a good method that quickly throws out obvious composites.
b) Sieve to some convenient limit, then primality test the remainder.
2, the usual case. Use a segmented Sieve of Eratosthenes.
Notes:
- You may find reference to the Sieve of Atkin. If your SoA is faster than your SoE, I will bet you coded the SoE wrong. It's not as egregious as AKS, but it's another case where it sounds good in theory but it's not so hot in practice.
- there are a lot of ways to optimize the sieve internals. It's important to at least start with the correct 4-line SoE, but you can go crazy unrolling, presieving, cache-blocking, parallelizing, etc.
- especially in the large range, Oliveira e Silva's work on fast segmented sieves can be useful. I don't use it, but primesieve does and it is definitely better in some situations.
- For case 1b sieving, for example T.R. Nicely's prime gap verification program does this. Personally I just worked on making the primality test fast for weeding out composites, but partly because it made things simpler. As the range grows the sieving would be more and more advantageous. For cases like next/prev prime, the range is typically small and you're guessing at it, so it's more of a tossup.
- For bases over 3k digits there are some additional ideas.
## #25 Re: Help Me ! » Is it any pseudo pattern gives prime number in proper range? » 2014-03-06 04:53:52
This is a wheel sieve. See, for example, "Wheel factorization" on Wikipedia. There are some good papers by Pritchard, Quesada & van Pelt, and others. Crandall and Pomerance's book is available online if you don't have a copy, and Section 3.1.2 to 3.1.4 discusses some of this.
A lot of sieves use it to 30, as then one has 8 candidates per 30 numbers, which nicely packs into a byte, and you've got most of the gain. I believe primesieve uses it to 210. It starts getting unwieldy before long for very little additional gain. At some point, especially given caches, you're better off doing the extra small mod (or adding one more number to GCD if using bigints) than walking a giant table. Oh, another thing a lot of sieves do is a presieve of the range -- take the wheel-30 as an example, now make a constant memory chunk of size 7*11*13 (only 1001 bytes), and fill the sieve range with it appropriately rotated. Presto -- clearing the memory for the sieve just filled in all multiples of 7, 11, and 13 without any computation, so you can start sieving at 17. | 10,400 | 32,282 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2016-36 | longest | en | 0.688158 |
http://betterlesson.com/lesson/resource/2470600/exit-slip-simplifying-square-roots-docx | 1,487,711,559,000,000,000 | text/html | crawl-data/CC-MAIN-2017-09/segments/1487501170839.31/warc/CC-MAIN-20170219104610-00473-ip-10-171-10-108.ec2.internal.warc.gz | 27,775,416 | 21,109 | ## EXIT SLIP-simplifying square roots.docx - Section 4: CLOSURE
EXIT SLIP-simplifying square roots.docx
EXIT SLIP-simplifying square roots.docx
Unit 4: Powers and Exponents
Lesson 11 of 16
## Big Idea: In this lesson, students use previously acquired knowledge of square roots to simplify square roots completely.
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Subject(s):
Math, Algebra, Simplifying Equations and Expressions, Exponents and Exponential Functions, 8th grade math, square root, simplifying roots, perfect squares
50 minutes
### Mauricio Beltre
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http://oeis.org/A072403 | 1,596,625,543,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735939.26/warc/CC-MAIN-20200805094821-20200805124821-00341.warc.gz | 87,832,485 | 4,232 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A072403 Numerator of the Reingold-Tarjan sequence, denominator=A072404. 2
1, 2, 5, 4, 11, 10, 1, 8, 23, 22, 7, 20, 19, 2, 17, 16, 47, 46, 5, 44, 43, 14, 41, 40, 13, 38, 37, 4, 35, 34, 11, 32, 95, 94, 31, 92, 91, 10, 89, 88, 29, 86, 85, 28, 83, 82, 1, 80, 79, 26, 77, 76, 25, 74, 73, 8, 71, 70, 23, 68, 67, 22, 65, 64, 191, 190, 7 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS The Reingold-Tarjan sequence is based on the following function defined on even positive integers and range of the rational numbers: f(2*n) = if n is even then 2*f(n)/3 else (f(n+1)+f(n-1))/3 for n>1, f(2*1)=1. f(2*n) = a(n)/A072404(n) for n>1, a(1)=1 and A072404(1)=1. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 J.-P. Allouche and J. Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197; preprint. See Example 33. E. M. Reingold and R. E. Tarjan, On a greedy heuristic for complete matching, SIAM J. Computing 10 (1981), 676-681; Semantic Scholar. FORMULA a(n) / A072404(n) = 1 - Sum_{k=1..n} (1 / A000244(A029837(k)). - Reinhard Zumkeller, Jan 01 2013 PROG (Haskell) import Data.Ratio ((%), denominator) a072404 n = a072404_list !! (n-1) a072404_list = map denominator \$ scanl1 (-) \$ map ((1 %) . a000244) \$ a029837_list -- Reinhard Zumkeller, Jan 01 2013 CROSSREFS Cf. A000244, A029837, A072404 (denominators). Sequence in context: A100710 A069913 A326002 * A010078 A074639 A319525 Adjacent sequences: A072400 A072401 A072402 * A072404 A072405 A072406 KEYWORD nonn,frac AUTHOR Reinhard Zumkeller, Jun 16 2002 STATUS approved
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Last modified August 5 06:46 EDT 2020. Contains 336209 sequences. (Running on oeis4.) | 797 | 2,126 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-34 | latest | en | 0.545781 |
http://michaelnielsen.org/polymath1/index.php?title=Probabilistic_formulation_of_Hadwiger-Nelson_problem&oldid=10690&diff=prev | 1,579,390,938,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250593994.14/warc/CC-MAIN-20200118221909-20200119005909-00261.warc.gz | 107,604,524 | 14,774 | # Difference between revisions of "Probabilistic formulation of Hadwiger-Nelson problem"
This is a part of Polymath16 - for the main page, see Hadwiger-Nelson problem.
Suppose for sake of contradiction that we have a 4-coloring $c: {\bf C} \to \{1,2,3,4\}$ of the complex plane with no unit edges monochromatic, thus
$c(z) \neq c(w) \hbox{ whenever } |z-w| = 1. \quad (1)$
We can create further such colorings by composing $c$ on the left with a permutation $\sigma \in S_4$ on the left, and with the (inverse of) a Euclidean isometry $T \in E(2)$ on the right, thus creating a new coloring $\sigma \circ c \circ T^{-1}: {\bf C} \to \{1,2,3,4\}$ of the complex plane with the same property. This is an action of the solvable group $S_4 \times E(2)$.
It is a fact that all solvable groups (viewed as discrete groups) are amenable, so in particular $S_4 \times E(2)$ is amenable. This means that there is a finitely additive probability measure $\mu$ on $S_4 \times E(2)$ (with all subsets of this group measurable), which is left-invariant: $\mu(gE) = \mu(E)$ for all $g \in S_4 \times E(2)$ and $E \subset S_4 \times E(2)$. This gives $S_4 \times E(2)$ the structure of a finitely additive probability space. We can then define a random coloring ${\mathbf c}: {\bf C} \to \{1,2,3,4\}$ by defining ${\mathbf c} := {\mathbf \sigma} \circ c \circ {\mathbf T}^{-1}$, where $({\mathbf \sigma},{\mathbf T})$ is the element of the sample space $S_4 \times E(2)$. Thus for any complex number $z$, the random color ${\mathbf c}(z)$ is a random variable taking values in $\{1,2,3,4\}$. The left-invariance of the measure implies that for any $(\sigma,T) \in S_4 \times E(2)$, the coloring $\sigma \circ {\mathbf c} \circ T^{-1}$ has the same law as ${\mathbf c}$. This gives the color permutation invariance
${\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(z_1) = \sigma(c_1), \dots, {\mathbf c}(z_k) = \sigma(c_k) )\quad (2)$
for any $z_1,\dots,z_k \in {\bf C}$, $c_1,\dots,c_k \in \{1,2,3,4\}$, and $\sigma \in S_4$, and the Euclidean isometry invariance
${\bf P}( {\mathbf c}(z_1) = c_1, \dots, {\mathbf c}(z_k) = c_k ) = {\bf P}( {\mathbf c}(T(z_1)) = c_1, \dots, {\mathbf c}(T(z_k)) = c_k. \quad (3)$
(In probabilistic language, this means that the random coloring is a stationary process with respect to the action of $S_4 \times E(2)$. The extraction of a stationary process from a deterministic object is an example of the Furstenberg correspondence principle.)
## Bounds on $p_d$ for 4-colourings
A class of correlations that is of particular interest is that of vertex pairs at some distance $d$. Accordingly, define
$p_d := {\bf P}( \mathbf{c}(0) = \mathbf{c}(d) ).$
distance Lower bound Lower-bounding graph/method Upper bound Upper-bounding graph Notes
$\geq 1/2$ 1/2 Spindle
$\geq 1/(n \sqrt{3})$ $(3n-2)/3n$ (n+1)-gon with one edge length $1/\sqrt{3}$ and the rest d
1 0 Unit edge 0 Unit edge
$\frac{1}{\sqrt{3}}$ 1/28 Unit diamond plus centres of triangles, together with H, Corollary 16 1/3 Unit triangle plus its centre
2 1/6 H, Corollary 16 1/2 Spindle
${\sqrt{3}}$ 1/4 H, Corollary 16 1/2 Spindle
${\frac{\sqrt{5}+1}{2}}$ 1/5 regular pentagon with unit side length 2/5 regular pentagon with unit side length
${\frac{\sqrt{5}-1}{2}}$ 1/5 regular pentagon with ${\frac{\sqrt{5}-1}{2}}$ side length 2/5 regular pentagon with ${\frac{\sqrt{5}-1}{2}}$ side length
${\sqrt{11/3}}$ 1/118 $G_{40}$ 1/2 Spindle computer-verified
8/3 1 $V \oplus V \oplus H$ 1/2 Spindle computer-verified; leads to contradiction
## Bounds on ${\bf P}( \mathbf{c}(0) = \mathbf{c}(d_1) \mid \mathbf{c}(0) \neq \mathbf{c}(d_0) )$ for 4-colourings
$d_1$ $d_0$ Lower bound Lower-bounding graph Upper bound Upper-bounding graph Notes
$1+(-1)^{1/3}$ $2$ 1/2 H Equals $p_{\sqrt 3}/(1-p_2)$
$1+(-1)^{1/3}$ $1+(-1)^{-1/3}$ 1/2 H
## Proofs of bounds
One can compute some correlations of the coloring exactly:
### Lemma 1
Let $z,w \in {\bf C}$ with $|z-w|=1$. Then
${\bf P}( \mathbf{c}(z) = c ) = \frac{1}{4}\quad (4)$
for all $c=1,\dots,4$,
${\bf P}( \mathbf{c}(z) = \mathbf{c}(w) ) = 0\quad (5),$
and
${\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c' ) = \frac{1}{12} \quad (6)$
for any distinct $c,c' \in \{1,2,3,4\}$. If $u$ is at a unit distance from both z and w, then
${\bf P}( \mathbf{c}(z) = c; \mathbf{c}(w) = c'; \mathbf{c}(u) = c'' ) = \frac{1}{24} \quad (6')$
Proof By color invariance (2), the four probabilities in (4) are equal and sum to 1, giving (4). The claim (5) is immediate from (1). From (5) and color invariance, the 12 probabilities in (6) are equal and sum to 1, giving (6). The same argument gives (6').$\Box$
### Lemma 2
(Spindle argument) Let $|d| \geq 1/2$. Then
$p_d \leq \frac{1}{2} \quad (7).$
Proof We can find an angle $\theta$ with $|de^{i\theta}-d|=1$, then $\mathbf{c}(de^{i\theta}) \neq \mathbf{c}(d)$ almost surely. This means that at least one of the events $\mathbf{c}(0) = \mathbf{c}(d)$, $\mathbf{c}(0) = \mathbf{c}(d e^{i\theta})$ occurs with probability at most 1/2. The claim now follows from isometry invariance (3). $\Box$
### Lemma 3
(Using the K graph) We have
$52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) + {\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} ) \geq 1 \quad (8).$
Proof Consider the 61-vertex graph $K$ from de Grey's paper. It has 26 (isometric) copies of H, and thus 52 copies of the triangle $(1, e^{2\pi i/3}, e^{4\pi i/3})$. With probability at least $1 - 52 {\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) )$, none of these triangles are monochromatic. By the argument in that paper, this implies that the three linking diagonals $(-2, +2)$, $(-2 e^{2\pi i/3}, 2e^{2\pi i/3})$, $(-2 e^{4\pi i/3}, e^{-4\pi i/3})$ are monochromatic. This gives the claim. $\Box$
### Corollary 4
(Existence of monochromatic $\sqrt{3}$-triangles) We have ${\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) \geq \frac{1}{104}$.
Proof The probability ${\bf P}( \mathbf{c}(-z) = \mathbf{c}(z) \hbox{ for } z = 2, 2e^{2\pi i/3}, 2e^{4\pi i/3} )$ is at most ${\bf P}( \mathbf{c}(-2) = \mathbf{c}(2)) = p_4$, which by Lemma 2 is at most 1/2. The claim now follows from Lemma 3. $\Box$
### Computer-verified Claim 5
(Using the graph M) One has ${\bf P}( \mathbf{c}(1) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = 0$ (Note this contradicts Corollary 4).
Proof This simply reflects the fact that there is no 4-coloring of the 1345-vertex graph M from de Grey's paper with its central copy of H containing a monochromatic triangle. One can use other graphs for this purpose, such as the 278-vertex graph $M_1$ or $V \oplus V \oplus H$. $\Box$
### Computer-verified Claim 6
(Using the graph $V \oplus V \oplus H$) One has $p_{8/3} = 1$ (note this contradicts Lemma 2).
Proof This reflects the fact that every 4-coloring of $V \oplus V \oplus H$ must assign the same color to 0 and 8/3. There is also a 745-vertex subgraph of $V \oplus V \oplus H$ with the same property. $\Box$
### Lemma 7
(Using $G_{40}$) We have
$59 p_{\sqrt{11/3}} + p_{8/3} \geq 1$.
Proof This reflects the fact that every 4-coloring of the 40-vertex graph $G_{40}$ from Exoo-Ismaolescu in which none of the 59 pairs of vertices at distance $\sqrt{11/3}$ apart, will assign the same color to 0 and 8/3. (This is presumably human-verifiable.) $\Box$
### Corollary 8
One has
$p_{\sqrt{11/3}} \geq \frac{1}{118}$.
Proof Combine Lemma 7 and Lemma 2. $\Box$
### Lemma 9
(Using $G_{49}$) One has
$18 {\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq p_{\sqrt{11/3}} .$
Proof This reflects the fact that every 4-coloring of the 49-vertex graph $G_{49}$ from Exoo-Ismaolescu in which 0 and $\sqrt{11/3}$ have the same color, at least one of the 18 copies of $(1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3)$ is monochromatic. This is potentially human-verifiable. $\Box$
### Corollary 10
(Existence of monochromatic $1/\sqrt{3}$-triangles) One has ${\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) \geq \frac{1}{2124}.$
Proof Combine Corollary 8 and Lemma 9. $\Box$
### Computer-verified claim 11
One has ${\bf P}( \mathbf{c}(1/3) = \mathbf{c}(e^{2\pi i/3}/3) = \mathbf{c}(e^{4\pi i/3}/3) ) = 0$. (This contradicts Corollary 10).
Proof This reflects the fact that the 627-vertex graph $G_{627}$ from Exoo-Ismaolescu does not have any 4-colorings with $1/3, e^{2\pi i/3}/3, e^{4\pi i/3}/3$ monochromatic. $\Box$
For certain special distances d, one can improve the bound in Lemma 2:
### Lemma 12
If $k \geq 1$ is a natural number, $j\in\mathbb{Z}$, $\gcd(j,2k+1)=1$, and $r = \frac{1}{2} \csc\left(\frac{j\pi}{2k+1}\right)$ then
$p_r \leq \frac{k}{2k+1},$
thus for instance
$p_{\frac{1}{\sqrt{3}}} \leq \frac{1}{3}.$
Proof Observe that the regular 2k+1-polygon $r, re^{2\pi i/(2k+1)}, r e^{4\pi i/(2k+1)}, \dots, r^{4k\pi i/(k+1)}$ has unit side lengths. By the pigeonhole principle, we conclude that at most k of these vertices can have the same color as the origin, and the claim follows. $\Box$
On the other hand, for $k=2,j=1$ we also know from the regular pentagon of unit sidelength that
$\frac{1}{5} \leq p_{\frac{\sqrt{5}+1}{2}} \leq \frac{2}{5} \quad (9)$
since any 4-coloring of that pentagon has either one or two monochromatic diagonals.
Similarly, for $k=2,j=2$ we also know from the regular pentagon of $\frac{\sqrt{5}-1}{2}$ sidelength that
$\frac{1}{5} \leq p_{\frac{\sqrt{5}-1}{2}} \leq \frac{2}{5}$
since any 4-coloring of that pentagon has either one or two monochromatic edges.
### Lemma 13
We have
$7 p_{\frac{1}{\sqrt{3}}} \geq p_{\sqrt{3}}$.
Proof Consider the unit rhombus $0, 1, e^{i\pi/3}, e^{-i\pi/3}$ together with the centers $e^{i\pi/6}/\sqrt{3}, e^{-i\pi/6}/\sqrt{3}$. With probability $p_{\sqrt{3}}$, the two far vertices $e^{i\pi/3}, e^{-i\pi/3}$ are the same color, and then 0,1 will be two other colors. This forces either one of the centers $e^{i\pi/6}/\sqrt{3}$ of a triangle to have a common color with one of the vertices of that triangle, or the two centers must have the same color. Thus in any event one of the seven edges of distance $1/\sqrt{3}$ is monochromatic, giving the claim. $\Box$
### Corollary 14
We have $p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{728}$.
This slightly improves upon the lower bound of 1/2124 coming from Corollary 10.
Proof Combine Corollary 4 and Lemma 13. $\Box$
### Lemma 15
One has
$p_{\sqrt{3}} + p_2 \geq \frac{2}{3}$ and $2 p_{\sqrt{3}} + p_2 \geq 1$.
Proof As noted in de Grey's paper, there are essentially four 4-colorings of H. H has six edges of length $\sqrt{3}$ and three of length $2$. If we let a denote the number of monochromatic $\sqrt{3}$ edges and b the number of monochromatic $2$-edges, we see from inspection of all four colorings that $(a,b)$ is either $(6, 0), (4,0), (2, 1)$, or $(0,3)$. In particular, one always has $\frac{a}{6} + \frac{b}{3} \geq \frac{2}{3}$ and $2\frac{a}{6} + \frac{b}{3} \geq 1$. Taking expectations, we obtain the claim. $\Box$
### Corollary 16
We have $p_2 \geq \frac{1}{6}$, $p_{\sqrt{3}} \geq \frac{1}{4}$ and $p_{\frac{1}{\sqrt{3}}} \geq \frac{1}{28}$.
Proof Combine Lemma 2, Lemma 15, and Lemma 13. $\Box$
### Lemma 17
If $a,b,c \gt 0$ are the lengths of a triangle, then ${\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c$.
Proof Consider a triangle of side lengths $a,b,c$. Let $\left|z_2\right|=b,\left|a-z_2\right|=c$. If the c side is not monochromatic, then at least one of the other two sides must fail to be monochromatic also: $\mathbf{c}(a)\neq\mathbf{c}(z_2)\Rightarrow[\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)]$. $[A\Rightarrow B]\Rightarrow {\bf P}(A)\leq{\bf P}(B)$ thus
${\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) \geq {\bf P}(\mathbf{c}(a) \neq \mathbf{c}(z_2)) = 1-p_c$
${\bf P}(A\lor B) +{\bf P}(A\land B)={\bf P}(A)+{\bf P}(B)$, so
${\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)) + {\bf P}(\mathbf{c}(0)\neq\mathbf{c}(z_2)) - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))$
${\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\lor\mathbf{c}(0)\neq\mathbf{c}(z_2)) = 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))$
$1-p_c \leq 2 - p_a - p_b - {\bf P}(\mathbf{c}(a)\neq\mathbf{c}(0)\neq\mathbf{c}(z_2))$
Using the law of cosines: $z_2=b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)$
${\bf P}\left(\mathbf{c}(a)\neq \mathbf{c}(0)\neq \mathbf{c}\left(b\exp\left(i\arccos\left(\frac{a^b+b^2-c^2}{2ab}\right)\right)\right) \right) + p_a + p_b \leq 1 + p_c$ $\Box$
Note that Lemma 2 follows from the a=b, c=1 case of this lemma. Iterating this lemma starting with Lemma 2 we can also obtain slightly nontrivial upper bounds on $p_a$ for small values of a, e.g. $p_a \leq 1 - 2^{-k}$ when $a \geq 2^{-k}, k\in\mathbb{Z}^+$. Further, we can generalise the a=b case to one in which the triangle is replaced by a (k+1)-gon of which one edge is 1 and the others are all equal, leading to the stronger result $p_a \leq 1 - 1/k$ when $a \geq 1/k, k\in\mathbb{Z}^+$. Further strenthening is achieved by using $1/\sqrt{3}$ as the long edge, given Lemma 12.
### Lemma 18
Whenever $d\gt0$, one has the inequalities
$|p_{\phi d} - p_d| \leq \frac{2}{5}, p_{\phi d} + p_d \geq \frac{1}{5}, 2p_d - p_{\phi d} \leq 1, 2 p_{\phi d} - p_d \leq 1$
where $\phi := \frac{1+\sqrt{5}}{2}$ is the golden ratio. Also we have
$p_{d/\sqrt{3}} \leq \frac{1}{3} + p_d, \frac{1}{2} + \frac{1}{2} p_d.$
Note that this generalises (9), as well as a special case of Lemma 12.
Proof Consider the regular pentagon with sidelength $d$, so it also has 5 diagonals of length $\phi d$. Let $a \in \{0,1,2,3,4,5\}$ denote the number of monochromatic edges and let $b \in \{0,1,2,3,4,5\}$ denote the number of monochromatic diagonals. Observe:
• $a,b$ cannot both be zero (pigeonhole principle).
• $a$ cannot be 4. Similarly, $b$ cannot be 4.
• If $a=5$, then $b=5$, and conversely.
• If $a=0$, then $b=1,2$; similarly, if $b=0$, then $a=1,2$.
From this we observe the inequalities
$|\frac{a}{5}-\frac{b}{5}| \leq \frac{2}{5}; \frac{a}{5} + \frac{b}{5} \geq \frac{1}{5}; 2 \frac{a}{5} - \frac{b}{5} \leq 1; 2\frac{b}{5} - \frac{a}{5} \leq 1$
and on taking expectations we obtain the first claim. Similarly, if one considers the colorings of an equilateral triangle of sidelength $d$ together with its center, and counts the numbers $a,b \in \{0,1,2,3\}$ of monochromatic edges of length $d$ and $d/\sqrt{3}$ respectively, one observes that one always has $\frac{b}{3} \leq \frac{1}{3} + \frac{2}{3} \frac{a}{3}, \frac{1}{2} + \frac{1}{2} \frac{a}{3}$, and on taking expectations one obtains the claim.$\Box$
The hexagon $H$ has essentially four distinct colorings: the coloring $\hbox{2tri}$ with two triangles, the coloring $\hbox{1tri}$ with one triangle, the coloring $\hbox{axisym}$ that is symmetric around an axis, and the coloring $\hbox{centralsym}$ that is symmetric around the central point. This gives four probabilities $p_{H = 2tri}, p_{H = 1tri}, p_{H = axisym}, p_{H = centralsym}$ that sum to 1. By counting the number of monochromatic edges of length $\sqrt{3}, 2$ respectively, one also obtains the identities
$p_{\sqrt{3}} = p_{H = 2tri} + \frac{2}{3} p_{H = 1tri} + \frac{1}{3} p_{H = axisym}; \quad p_2 = \frac{1}{3} p_{H=axisym} + p_{H=centralsym}$
which reproves Lemma 15. Also
${\bf P}( \mathbf{c}(0) = \mathbf{c}(e^{2\pi i/3}) = \mathbf{c}(e^{4\pi i/3}) ) = p_{H = 2tri} + \frac{1}{2} p_{H=1tri}.$
Any 4-coloring of L contains at least one triangle within one of its 52 copies of H, thus
$p_{H = 2tri} + \frac{1}{2} p_{H=1tri} \geq \frac{1}{52}$
which reproves Corollary 4.
### Lemma 19
(Hubai) One has $p_{H = 1tri} + p_{H = axisym} \geq \frac{1}{10}$.
Proof Consider five copies of H centred at 0,1,2,3,4. With probability at least $1 - 5( p_{H = 1tri} + p_{H = axisym} )$, none of these copies of H are colored 1tri or axisym, and so must be colored 2tri or centralsym. One can check then that if one of the copies is colored 2tri, then so is any adjacent copy; thus all five copies are colored 2tri, or all five are colored centralsym. In either case we see that -1 and 5 are colored the same color. Comparing with Lemma 2 then gives the claim. $\Box$
### Theorem 20
One has $p_{H = 1tri} \gt 0$.
Proof Suppose for contradiction that $p_{H = 1tri} = 0$. Consider the triangular lattice ${\bf Z}[e^{\pi i/3}]$. We consider the dual graph, in which each element of the triangular lattice is the center of a hexagon in the dual graph, which is dual to the copy of H centered at that element. Color an edge $e^\perp$ in the dual lattice black if the original edge $e$ is monochromatic, and white otherwise. Thus, each hexagon has one of three coloring patterns up to rotation, depending on the coloring type of H:
• If H is colored 2tri, then all six edges of the hexagon are black.
• If H is colored axisym, then two opposing edges of the hexagon are black, and the other two are white.
• If H is colored centralsym, then none of the edges of the hexagon are black.
In particular, if one hexagon is colored 2tri, then either all adjacent hexagons are colored 2tri, or else all adjacent hexagons are colored axisym. If one follows an axisym hexagon along its axis of symmetry, the adjacent hexagons are either again axisym, or 2tri; thus, one either gets an infinite chain of axisym hexagons, a ray of axisym hexagons terminating at a 2tri hexagon, or an interval of axisym hexagons terminated at both sides by a 2tri hexagon. In the latter case one can continue the coloring to find that the 2tri hexagons are arranged in a lattice, with consecutive 2tri hexagons joined by chains of axisym hexagons. This leads to only a small number of possible hexagon colorings in the lattice:
1. Case 1: Every hexagon centred at an element ${\bf Z}[ e^{\pi i/3} ]$ is colored centralsym.
2. Case 2: Every hexagon centred at an element ${\bf Z}[ e^{\pi i/3} ]$ is colored 2tri.
3. Case 3.k: For some natural number $k \geq 2$, every hexagon centred in a coset of $k \cdot {\mathbf Z}[ e^{\pi i/3} ]$ is colored 2tri, the hexagons connecting any two adjacent hexagons in this coset are colored axisym, and all other hexagons are colored centralsym.
4. Case 4: Each horizontal row of hexagons either consists entirely of centralsym, or consists entirely of axisym (with the axis of symmetry horizontal).
5. Case 5: Each northwest row of hexagons consists either entirely of centralsym, or consists entirely of axisym (with axis of symmetry northwest).
6. Case 6: Each northeast row of hexagons consists either entirely of centralsym, or consists entirely of axisym (with axis of symmetry northeast).
7. Case 7: A single hexagon colored 2tri, with six rays of hexagons colored axisym emanating from this hexagon; all other triangles are centralsym.
Technically, Case 1 is contained in Cases 4,5,6 as written above, but this will not be an issue.
In the first case, the coloring is periodic with periods $2, 2 e^{\pi i/3}$. In the second case, it is periodic with periods $3, 3 e^{\pi i/3}$. In the third case, it is periodic with periods $3k, 3k e^{\pi i/3}$. Also note that for each k, one can check if Case 3.k holds by inspecting the coloring at a finite number of vertices. Thus the event that Case 3.k holds is "measurable" in the sense that a meaningful probability can be assigned. (But Cases 1,2,4,5,6 are not measurable events, they require an infinite number of points to be inspected, and the probability measure we are using is only finitely additive rather than infinitely additive.) In Case 4, the coloring is periodic with period 2; also, every coset of $2 \cdot {\bf Z}[e^{\pi i/3}]$ is 2-colored. Similarly for Case 5 and 6 (where the periods are $2 e^{2\pi i/3}$ and $2 e^{4\pi i/3}$ respectively.)
Let $\alpha_k$ be the probability that Case 3.k holds for the given value of $k$. Then $\sum_{k=2}^K \alpha_k \leq 1$ for any k, hence $\sum_{k=2}^\infty \alpha_k \leq 1$. In particular, we can find $K_1$ such that $\sum_{k={K_1}}^\infty \alpha_k \leq 0.1$ (say). Let $P$ be six times the least common multiple of $1,2,\dots,K_1$. Then the coloring is P- and $P e^{\pi i/3}$-periodic for Case 1, Case 2, and all Case 3.k with $k \leq K_1$. On the other hand, if $K_2$ is sufficiently large depending on $P$, and Case 3.k holds for some $k \geq K_2$, then almost all of the hexagons are colored centralsym, which makes the coloring "almost $P, P e^{\pi i/3}$-periodic" in the sense that
${\bf c}(z+P e^{\pi i j/3}) = {\bf c}(z) \hbox{ for } j=0,1,2,3,4,5$
will hold for at least $0.9$ of the lattice points $z \in {\bf Z}[e^{\pi i/3}]$ with $|z| \leq K_2$. Similarly for Case 7 (which is sort of a $k=\infty$ limiting case of Case 3.k.) Thus, with the probability $\geq 1 - \sum_{k=K_1}^{K_2} \alpha_k \geq 0.9$, the coloring of the seven vertices ${\bf c}(0), {\bf c}(P e^{\pi ij/3}, j=1,\dots,6$ is (up to rotation and recoloring) one of the three patterns of the central and linking vertices in Figure 3 of Aubrey's paper, namely
• ${\bf c}(0) = {\bf c}(P) = {\bf c}(P e^{\pi i/3}) = {\bf c}(P e^{2\pi i/3}) = {\bf c}(P e^{3\pi i/3}) = {\bf c}(P e^{4\pi i/3}) = {\bf c}(P e^{5\pi i/3})$;
• ${\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{3\pi i/3})$; ${\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3})$ for some $P' = P e^{\pi i j/3}$, $j=1,\dots,6$;
• ${\bf c}(0) = {\bf c}(P') = {\bf c}(P' e^{\pi i/3}) = {\bf c}(P' e^{2\pi i/3}) = {\bf c}(P' e^{3\pi i/3})$; ${\bf c}(P' e^{4\pi i/3}) = {\bf c}(P' e^{5\pi i/3})$ for some $P' = P e^{\pi i j/3}$, $j=1,\dots,6$.
Using the spindling argument from Aubrey's paper, we conclude that the third possibility must in fact hold with probability at least 0.8; on the other hand, from Lemma 2 this scenario can only occur with probability at most 1/2, giving the required contradiction. $\Box$
One should be able to refine this argument to show that $p_{H = 1tri} \gt c$ for an absolute constant $c\gt0$.
## Simplification rules for triplets of points in the complex plane
Deduced from the rule ${\bf P}(A\land B)+{\bf P}(A\land \lnot B)={\bf P}(A)$.
${\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) )$
${\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) + {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) = {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) )$
${\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) = {\mathbf c}(z_2) ) - {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) ) = {\bf P}( {\mathbf c}(z_0) = {\mathbf c}(z_1) ) - {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )$
${\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) \neq {\mathbf c}(z_0) ) + {\bf P}( {\mathbf c}(z_1) \neq {\mathbf c}(z_2) = {\mathbf c}(z_0) ) = {\bf P}( {\mathbf c}(z_0) \neq {\mathbf c}(z_1) \neq {\mathbf c}(z_2) )$
## Proofs of bounds for conditional probabilities
The trivial case, valid where $\left|d\right|\neq 1$:
$x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) = {\mathbf c}(d) )=0$
Trivial plus Baye's Theorem, valid where $d\neq 0$:
$x\in\mathbb{R}\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) )=\frac{{\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )}\leq 1$
Rearrange:
${\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{ix}) )+{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1$
Spindle method: for $\left|d\right|=d\geq 1/2$ and $\theta=2\text{arcsin}\left(\frac{1}{2d}\right)$:
${\bf P}( {\mathbf c}(0) = {\mathbf c}(d+e^{i\theta}) \mid {\mathbf c}(0) \neq {\mathbf c}(d) ) = \frac{1}{1-{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )} - 1\leq 1$
which is another way to see $\left|d\right|=d\geq 1/2\Rightarrow{\bf P}( {\mathbf c}(0) = {\mathbf c}(d) )\leq 1/2$. | 9,226 | 24,387 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2020-05 | latest | en | 0.792961 |
http://www.chemical-ecology.net/java/match1t.htm | 1,519,203,632,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891813602.12/warc/CC-MAIN-20180221083833-20180221103833-00379.warc.gz | 418,558,866 | 5,258 | Memory Matching Expectations
Given an even number of squares (N) that are filled with paired objects (O), where O is N/2, and then covered so we can't see the objects, then how many blind (random) tries (consisting of two choices of different squares) are expected in order to match all the objects. The objects are matched by choosing two different squares at random (a try) and if the objects in these two squares match then they remain uncovered, but if they do not then both are covered again and their placement is forgotten. Certainly a person using their memory can match the paired objects faster, but we want to investigate the expected number of tries to complete the matching of all objects without memory.
Now Enter the number of squares:
Enter a guess for the number of expected tries:
Now click the button to see the correct answer:
Number of expected Tries = to match all
The formula is Tries = 2 SUM (r=0 to r=n/2-1) n-2r-1
JavaScript © 1997-2003 by John A. Byers | 227 | 982 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-09 | latest | en | 0.945362 |
https://www.maplesoft.com/support/help/maple/view.aspx?path=DEtools%2Fclairautsol | 1,660,902,700,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573630.12/warc/CC-MAIN-20220819070211-20220819100211-00186.warc.gz | 738,742,638 | 29,881 | clairautsol - Maple Help
DEtools
clairautsol
find solutions of a Clairaut first order ODE
Calling Sequence clairautsol(lode, v)
Parameters
lode - first order differential equation v - dependent variable of the lode
Description
• The clairautsol routine determines if the first argument is a first order ODE of Clairaut type and, if so, returns a solution to the equation.
• The first argument is a differential equation in diff or D form and the second argument is the function in the differential equation.
• This function is part of the DEtools package, and so it can be used in the form clairautsol(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[clairautsol](..).
Examples
> $\mathrm{with}\left(\mathrm{DEtools}\right):$
> $\mathrm{ode}≔z\left(t\right)=t\mathrm{diff}\left(z\left(t\right),t\right)+{\mathrm{diff}\left(z\left(t\right),t\right)}^{3}$
${\mathrm{ode}}{≔}{z}{}\left({t}\right){=}{t}{}\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right){+}{\left(\frac{{ⅆ}}{{ⅆ}{t}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{z}{}\left({t}\right)\right)}^{{3}}$ (1)
> $\mathrm{clairautsol}\left(\mathrm{ode},z\left(t\right)\right)$
$\left\{{z}{}\left({t}\right){=}{-}\frac{{2}{}\sqrt{{-}{3}{}{t}}{}{t}}{{9}}{,}{z}{}\left({t}\right){=}\frac{{2}{}\sqrt{{-}{3}{}{t}}{}{t}}{{9}}{,}{z}{}\left({t}\right){=}{{\mathrm{_C1}}}^{{3}}{+}{t}{}{\mathrm{_C1}}\right\}$ (2)
> $\mathrm{ode}≔z\left(t\right)=t\mathrm{D}\left(z\right)\left(t\right)+{\mathrm{D}\left(z\right)\left(t\right)}^{2}-{\mathrm{D}\left(z\right)\left(t\right)}^{3}$
${\mathrm{ode}}{≔}{z}{}\left({t}\right){=}{t}{}{\mathrm{D}}{}\left({z}\right){}\left({t}\right){+}{{\mathrm{D}}{}\left({z}\right){}\left({t}\right)}^{{2}}{-}{{\mathrm{D}}{}\left({z}\right){}\left({t}\right)}^{{3}}$ (3)
> $\mathrm{clairautsol}\left(\mathrm{ode},z\left(t\right)\right)$
$\left\{{z}{}\left({t}\right){=}\frac{{t}}{{3}}{+}\frac{{2}}{{27}}{-}\frac{{2}{}\sqrt{{27}{}{{t}}^{{3}}{+}{27}{}{{t}}^{{2}}{+}{9}{}{t}{+}{1}}}{{27}}{,}{z}{}\left({t}\right){=}\frac{{t}}{{3}}{+}\frac{{2}}{{27}}{+}\frac{{2}{}\sqrt{{27}{}{{t}}^{{3}}{+}{27}{}{{t}}^{{2}}{+}{9}{}{t}{+}{1}}}{{27}}{,}{z}{}\left({t}\right){=}{-}{{\mathrm{_C1}}}^{{3}}{+}{{\mathrm{_C1}}}^{{2}}{+}{t}{}{\mathrm{_C1}}\right\}$ (4) | 925 | 2,357 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2022-33 | latest | en | 0.486335 |
https://www.physicsforums.com/threads/hermitian-vs-self-adjoint-operators.259127/ | 1,484,943,248,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280872.69/warc/CC-MAIN-20170116095120-00395-ip-10-171-10-70.ec2.internal.warc.gz | 970,332,567 | 17,035 | # Hermitian vs. self-adjoint operators
1. Sep 25, 2008
### Heirot
Hello,
what's the difference between Hermitian and self-adjoint operators? Our professor in Group Theory made a comment once that the two are very similar, but with a subtle distinction (which, of course, he failed to mention )
Thanks!
2. Sep 25, 2008
### Heirot
Oops... I found the answer. Sorry
3. Sep 25, 2008
### morphism
Well, what's this subtle distinction?
4. Sep 25, 2008
### Heirot
5. Sep 25, 2008
### morphism
Just make sure you're aware that it's a matter of convention. Most people use the two words to mean the same thing.
6. Sep 25, 2008
### George Jones
Staff Emeritus
While this is true for bounded operators, all the books I have checked on my shelf make a distinction for unbounded operators. According to the books that I have checked, a Hermitian operator is a bounded self-adjoint operator, i.e., all Hermitian operators are self-adjoint, but not all self-adjoint operators are Hermitian.
7. Sep 25, 2008
### atyy
Is this an example of the difference you are talking about, or is it yet another difference?
http://mathworld.wolfram.com/BerryConjecture.html:"...physicists define this operator to be Hermitian, mathematicians do not."
8. Sep 26, 2008
### George Jones
Staff Emeritus
Let $A$ be a densely defined linear operator on a Hilbert space. One of the defining properties, *, for all cases is
$$\left<Ax,y\right> = \left<x,Ay\right>, \ *$$
where $x$ and $y$ are elements of the Hilbert space.
The text that was used for the functional analysis course that I took as a student makes the following definitions:
$A$ is Hermitian if $A$ is bounded and * is true for every $x$ and $y$ in the Hilbert space;
$A$ is symmetric if * holds for for every $x$ and $y$ in the domain of $A$;
$A$ is self-adjoint if $A$ is symmetric and the domain of $A$ equals the domain of $A^\dagger$.
According to these defintions, every Hermitian operator is and self-adjoint, but not all self-adjoint operators are Hemitian. Some books leave off the first definition and call symmetric operators Hermitian. Then, every self-adjoint operator is Hermitian, but not all Hermitian operators are self-adjoint.
A symmetric (or Hemitian, depending on the terminology used) operator is self-adjoint iff it has trivial deficiency indices. Some physicists and physics references do not distinguish between the above defintions, and use the terms Hermitian and self-adjoint interchangeably to refer to all cases.
If $A$ is unbounded, then the Hilbert space is necessarily infinite-dimensional, and the domain of $A$ need not be all of the Hilbert space. If this is the case and if $A$ is symmetric, then the domain of $A$ is a subset of the domain of $A^\dagger$, which I have not defined.
In physics, the canonical commutation relation is important. If self-adjoint operators $A$ and $B$ satisfy such a relation, then it is easy to show that at least one of the operators must be unbounded. Suppose it is $A$. The Hellinger-Toeplitz theorem states any linear operator that satisfies * for all elements of the Hilbert space must bounded. Since $A$ is self-adjoint and unbounded, the domain of physical observable $A$ cannot be all of Hilbert space!
For one consequence of these concepts, see
Last edited: Dec 17, 2013
9. Sep 26, 2008
### morphism
That's a very interesting thread, George. Thanks for sharing!
By the way, in your references, are unbounded operators ever referred to as (unqualified) operators?
10. Sep 27, 2008
### atyy
11. Dec 16, 2008 | 943 | 3,551 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2017-04 | longest | en | 0.92492 |
https://stats.stackexchange.com/questions/533599/is-it-possible-that-x-j-and-x-k-are-independent-of-each-other-conditioni | 1,709,493,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476397.24/warc/CC-MAIN-20240303174631-20240303204631-00432.warc.gz | 545,790,559 | 39,365 | # is it possible that $X_{j}$ and $X_{k}$ are independent of each other conditioning on $Z = f(X_1,\cdots, X_N)$?
Suppose I have $$N$$ random variables $$\{X_j\}_{j=1}^N$$ and they are mutually independent. Also, I define $$Z = f(X_1,\cdots,X_N)$$ for some function $$f()$$. And I want to know that if it is possible that $$X_j$$ is independent of $$X_k$$ conditioning on $$Z$$ for any $$j\ne k$$.
I have asked a similar question where $$N=2$$ and $$f(X_1,X_2) = X_1+X_2$$, and the statement is false. I want to know if the statement is possibly true for general $$N$$ and some function $$f(\cdot)$$. Here I assume $$\{X_j\}$$ are not constant random variables.
Suppose $$X_i$$ are Rademacher variables ($$\pm 1$$ with equal probability) and $$Z=\sum_i X_i^2$$. Then $$Z$$ is constant, so conditioning on it has no effect.
Less trivially, suppose $$Z=\prod_i X_i$$ and $$N>2$$. For any specific $$j, k$$, $$P(Z=1|X_j,X_k)=1/2$$ so the distribution of $$(X_j,X_k)|Z$$ is the same as their unconditional distribution. In this situation, any $$N$$ of $$(X_1,X_2,\dots,X_N,Z)$$ are mutually independent but the full set is not.
As a slight extension, suppose $$X_i\sim N(0,1)$$, and $$Z=\prod_i \mathrm{sign} X_i$$. By the same argument as before, $$Z$$ is independent of any set of fewer than $$N$$ $$X_i$$s.
I can't think of an example where the $$X_i$$ are all continuous and $$Z$$ is a continuous, non-constant function of all of them. I'd be a bit surprised if there was one, but not very surprised. | 466 | 1,503 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 31, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-10 | latest | en | 0.928672 |
https://math.stackexchange.com/questions/4791353/n-points-in-the-plane-can-be-connected-with-n-1-clockwise-non-intersecting | 1,725,894,856,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651103.13/warc/CC-MAIN-20240909134831-20240909164831-00171.warc.gz | 353,057,203 | 44,573 | # $n$ points in the plane can be connected with $n-1$ clockwise, non-intersecting line segments from any starting point
This conjecture is based on a mobile game that I've published. The object of the game is:
• Given $$n ≥ 3$$ points in the Cartesian plane in general position (no $$3$$ of those points are a straight line):
• Connect all the points with $$n - 1$$ line segments, drawn continuously (without lifting pencil from paper). Drawing lines in this way forms a permutation of the set of $$n$$ points, with each line segment being defined by the $$n-1$$ subsequences of $$2$$ consecutive points in that permutation. (No line segment is drawn between the first and last points in the permutation.)
• Every subsequence of $$3$$ consecutively connected points must be clockwise oriented in the order that they were connected.
• No two of the line segments may intersect.
Equations that precisely define clockwise orientation and line intersection are provided under the "Insights" section of this post.
I've already proven that every instance of this game is solvable if the player gets to choose the point at which they start: Start at a point on the convex hull of the set of points, then continue making connections in a clockwise spiral to minimize each successive angle between line segments until every point has been reached.
Example solution:
However, I conjecture that the game is also solvable by using any arbitrary point as the starting point, even if it's not on the convex hull of all the points.
I have not yet manually created a counterexample to this conjecture, and I've also searched millions of possible levels using a computer program without finding a counterexample. However, I have yet to formally prove the conjecture.
Clarification:
Here's another wording of the problem:
Given a set of $$n ≥ 3$$ points $$\{P_1, P_2, ..., P_n\}$$ in general position, and given an arbitrary point $$Q$$ from that set, does there always exist a permutation of that set satisfying the following conditions:
• The first element of the permutation is $$Q$$
• None of the following line segments intersect with each other: $$\overline{P_i P_{i+1}}$$ for each $$i \in [1, n-1]$$
• Referring to $$P_i$$, $$P_{i+1}$$, and $$P_{i+2}$$ as $$(A_x, A_y)$$, $$(B_x, B_y)$$, and $$(C_x, C_y)$$ for each $$i \in [1, n-2]$$, the following condition is always satisfied: $$\begin{vmatrix}(B_x - A_x)&(B_y - A_y)\\(C_x - A_x)&(C_y - A_y)\end{vmatrix}<0$$
Alternative strategy:
If an alternative conjecture can be proven where $$Q$$ is the last element of the permutation instead of the first element, then the original conjecture also holds.
Reasoning: If the original conjecture holds for a set of points $$S$$, then the alternative conjecture holds for $$S'$$, where $$S'$$ has the points from $$S$$ reflected across a given arbitrary line (such as the y-axis).
Insights:
Whether two line segments intersect depends upon whether trios of their points are clockwise-oriented. This correspondence may be usable to generalize the problem, though I am not yet sure how.
Let $$c(A, B, C)$$ be true iff A, B, and C are clockwise-oriented and false otherwise. Let $$i(A, B, C, D)$$ be true iff $$\overline{AB}$$ and $$\overline{CD}$$ intersect with each other and false otherwise. Then:
$$c(A, B, C) = (\begin{vmatrix}(B_x - A_x)&(B_y - A_y)\\(C_x - A_x)&(C_y - A_y)\end{vmatrix}<0)$$
$$i(A, B, C, D) = (c(A, B, C) \oplus c(A, B, D)) \land (c(C, D, A) \oplus c(C, D, B))$$
The following identities hold on the function $$c$$:
$$c(A, B, C) = c(B, C, A) = c(C, A, B)$$
$$c(A, B, C) \oplus c(C, B, A)$$ always evaluates to true
The function $$f(N)$$ that gives the number of sets of N points that are distinct (defined below) is given by A000930, where $$f(N)$$ is the Nth element of that sequence. (Obviously, $$f(1)$$ and $$f(2)$$ are meaningless in the context of this problem.) For example, there is only one distinct set of $$3$$ points — they form a triangle. For $$4$$ points, there are $$2$$ distinct sets — one with a convex hull of 4 points and one with a single point within a convex hull of 3 other points.
Distinctness for the purpose of this result is defined by:
• the number of times the convex hull of the set of points can be removed before no points remain (i.e. the number of nested convex hulls in the initial set of points), and
• the numbers of points on each of the nested convex hulls.
I believe, but have not yet formally proven, that the solution for any set of points $$S$$ is generalizable to any other set of points $$T$$ if $$S$$ and $$T$$ are not distinct by the above definition.
• Carlyle proved that for a set of points $$S$$, a solution can always be found if the initial point is in $$T$$ or $$U$$, where $$T$$ is the convex hull of $$S$$, and $$U$$ is the convex hull of $$(S - T)$$.
• It follows that a solution can be found for any initial point in a set of $$6$$ or fewer points.
Thank you so much to everyone who's left comments! Here is a summary of the ideas and insights derived from my and others' comments.
• There have been misconceptions about the meaning of "clockwise". This term is precisely defined in the "clarification" and "insights" sections of my question.
• Observe the manner in which the following set of points is connected according to the rules in the problem statement: https://i.sstatic.net/gOMgp.png.
• The above set of points does not generalize to all sets of points with 3 convex hulls within each other (e.g. 3 convex hulls inscribed within concentric circles). Why does this not generalize? Consider the case of 4 convex hulls inscribed within concentric circles. Here is the general solution (with an arbitrarily large number of points on each). Here is a specific case, with a different solution than the general case.
Mobile game screenshots:
These screenshots illustrate the conjecture in the context of the mobile game ("Clockwise!" by me, Roy Sianez, available on the iOS App Store within the US). I'm attaching them as an image because the App Store link may not show the product page outside of the US.
• @Moti that gives you $n$ segments, not $n-1$. Moreover, from the phrasing of the question ("first" and "last" points) I think OP intends that the points are connected with the $k$-th segment joining points $k$ and $k+1$.
– M W
Commented Oct 22, 2023 at 5:26
• My apologies, the link to the game may only work within the United States. The game is called "Clockwise!" and is available on the iOS App Store within the US. I'll update my post with a screenshot of the App Store product page for illustration. Commented Oct 22, 2023 at 13:31
• @AlexRavsky Yes, your set of points can be connected like so: i.sstatic.net/gOMgp.png Commented Oct 26, 2023 at 19:11
• @AlexRavsky The circles of points generalize well to cases where each convex hull has a suitably large number of points along the hull's polygon. However, if I remember correctly, there are cases that can be solved one way with a sufficient number of points along each convex hull, but that must be solved another way when there are fewer points along the same convex hulls. This may mean that the construction of concentric circles is (unfortunately) not generic to all problems in the domain. I will try to post an example to illustrate this later tonight. Commented Oct 26, 2023 at 23:20
• [Following up on my last comment] Here is a general solution for a sufficiently large number of points around 4 concentric circles, starting from the second-to-innermost: i.sstatic.net/zcNCS.png. Here is an instance of 16 points that have 4 square convex hulls inscribed within the same concentric circles: i.sstatic.net/7sGIQ.png. The specific case has a different solution from the general case, which illustrates the limitations of generalizing with concentric circles. Commented Oct 27, 2023 at 17:47
Slight improvement: It is always possible to start on the second nested convex hull as well.
Firstly, by second nested convex hull, I mean the convex hull of the points obtained by removing all the points in the convex hull of the original set. (All the trivial cases such as a set with only one point not on the convex hull can be dealt with separately)
It is fairly straightforward to see why this is the case. Given a set of points $$S$$ and a point $$x$$ on the second nested convex hull of $$S$$, we can draw a line through $$x$$ that bisects $$S$$ into a region containing only points from the convex hull, and another region containing the rest of $$S$$. Below is an illustration of such a line drawn through such a point (please help me center this image):
The starting point is circled, and the line drawn is a line that bisects the plane into a region containing only points from the original convex hull, and the rest.
From here on, you can simply go to the most anti-clockwise point on the convex hull above the bisection line, call it $$y$$:
and complete the spiral as if you were completing it for the set of points $$S\setminus\{x\}$$ starting on the point $$y$$ which is on its convex hull, and this completion is guaranteed to not intersect the segment connecting $$x$$ to $$y$$, since none of the non-convex hull points of $$S$$ are above the bisection line, while the line segment connecting $$x$$ and $$y$$ is above the bisection line.
In general, if one starts on a non-convex hull point, then any path through the points satisfying the conditions of the problem will necessarily contain all but one line segment on the convex hull. This partial result came from trying to think how one might cleverly choose which line segment to omit, so that one of the spirals going in from one of the points adjacent to the omitted segment starts on the required point, since the path can then be completed in a way analogous to what is described above (I am leaving out some details mostly because I am still struggling to formalise this) but perhaps one can do a sort of induction on the number of nested convex hulls, to show that you can always cleverly choose an edge to omit, but my attempts at this have failed
**Edit, description of an algorithm I believe works, but have not been able to verify:
Take the starting point and find the nested hull that it lies on, draw a line that passes through the two line segments adjacent to the starting point (connecting it to the other points on its hull) now take the orientation of this line and start at one side of the set of points, and start sweeping it across, whenever the line sweeps past a point, neglect that point and calculate the new hulls with the remaining points. I believe that this procedure will always have a step where the starting point is isolated inside the most central hull, at this step, take a point just above the line, on the outermost hull and spiral inwards, anti-clockwise. This spiral is guaranteed to end on the starting point. Then take all those points below the line, and start on the rightmost one on the outermost hull just below the line, and spiral clockwise, then simply connect the two starting points of the spiral, and you have a solution. The struggle is proving that at some step the starting point is isolated in the innermost hull. Here are some diagrams showing the algorithm in action, hopefully someone can make it rigorous:
First, draw the nested convex hulls,and find a line going through the line segments adjacent to the starting point (the starting point is circled):
Now, shift this line all the way down (away from the point) and start sweeping it across:
Each time it passes a point, neglect that points, and draw the nested hulls for the points that remain, the "centers" of the nested hulls start collapsing towards the starting point, and eventually surpass it, but we are interested in finding the step at which the starting point appears isolated in the central hull:
At this point, take the rightmost point above the line, and spiral in anti-clockwise from that point (circled in red), also, complete the spiral below the line, starting with the rightmost point below the line:
Now connect the spirals, to obtain a full solution:
• This seems like sound reasoning to me. The strategy indicated by your last paragraph — dividing the plane by separating 2 points from the convex hull, then solving the remainder of the puzzle by starting from the convex hull of the remaining points — will always work when starting on the "second nested convex hull" as described in your answer, but likely won't generalize to a third, fourth, etc. nested convex hull. Commented Oct 31, 2023 at 0:19
• Nice idea for the 2nd nested hull. Can you please provide an illustration for why you believe the algorithm is true? I find it difficult to follow. Commented Nov 1, 2023 at 18:54
• @BenjaminWang I have added in some diagrams, I hope it is clear now? It is not supposed to be a proof, just a tried and tested idea Commented Nov 1, 2023 at 19:41
• I notice in your picture the starting point is already a member of the inner most hull. Does this still generally seem to work if you put the starting point in one of the middle hulls?
– M W
Commented Nov 1, 2023 at 19:47
• @LucaT.Castrillón you can also use the idea of one of the deleted answers to just get a non-self-intersecting path - pick any point not colinear with the ones in $S$, as a sort of center point, then rotate a ray around that center point, and connect the points in $S$ in the order that the ray hits them.
– M W
Commented Nov 1, 2023 at 23:01
(this is supposed to be a comment, but I couldn't figure out how to add pictures in the comment sections).
@Carlyle's idea is very nice. I just want to point out one thing: the choice of the sweeping line cannot be arbitrary.
Consider this example of $$6$$ points. Suppose we want to start from $$D$$. Then one cannot choose the sweeping line to be close to parallel to $$EF$$, as then $$B$$ will be removed first, and $$D$$ would lie on the convex hull. Instead, one must carefully choose the line so that $$C$$ and $$F$$ is removed.
• I think here one cannot even choose anything close to parallel to $EF$. Commented Nov 1, 2023 at 20:00
• Firstly, I cannot take credit for the idea, it was actually a friend of mine that came up with it, but he's not on stack exchange. Secondly, thank you for this example, it gives a lot of insight, because this means we will not be able to prove it using only the information given in my description, since the orientation of the line is too arbitrary Commented Nov 1, 2023 at 20:04 | 3,523 | 14,527 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 75, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2024-38 | latest | en | 0.952949 |
https://www.studypool.com/discuss/530727/what-are-the-intercepts-of-the-equation-18x-9y-3z-18?free | 1,481,276,384,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542693.41/warc/CC-MAIN-20161202170902-00170-ip-10-31-129-80.ec2.internal.warc.gz | 1,000,332,707 | 14,172 | ##### What are the intercepts of the equation 18x - 9y + 3z = 18 ?
Algebra Tutor: None Selected Time limit: 1 Day
What are the intercepts of the equation 18x - 9y + 3z = 18 ?
a. (1, 0, 0), (0, 2, 0), (0, 0, 6)
b. (6, 0, 0), (0, –3, 0), (0, 0, 1)
c. (6, 0, 0), (0, 3, 0), (0, 0, 1)
d. (1, 0, 0), (0, –2, 0), (0, 0, 6)
Without graphing, classify the following system as independent, dependent, or inconsistent.
4x - 2y = 14 , y = 2x -6
a. independent
b. inconsistent
c. dependent
Graph the system of inequalities y > -3 , y < - |x+2|. Which two quadrants does the solution lie in?
a. 1 and 4
b. 1 and 2
c. 2 and 3
d. 3 and 4
May 12th, 2015
1=d. (1, 0, 0), (0, –2, 0), (0, 0, 6)
2=
y = 2x + 7
y = 2x - 6
Inconsistent System:
The lines are parallel. There is no point of intersection. The system has no solution.
3=c. 2 and 3
May 12th, 2015
...
May 12th, 2015
...
May 12th, 2015
Dec 9th, 2016
check_circle | 424 | 925 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2016-50 | longest | en | 0.70063 |
https://physics.stackexchange.com/questions/776059/i-don-t-understand-noether-s-theorem-there-is-nothing-to-prove | 1,701,936,098,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100650.21/warc/CC-MAIN-20231207054219-20231207084219-00380.warc.gz | 521,294,973 | 46,223 | # I don’t understand Noether’s theorem… there is nothing to prove?
I don’t understand Noether’s theorem… there is nothing to prove?
If I understand Noether’s theorem correctly it says: if there is coordinate where the Lagrangian is invariant, then the conjugate momentum is conserved. However, this follows almost immediately from the Euler–Lagrange equation:
$$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}=\frac{\partial L}{\partial q}$$
If the Lagrangian is does not change in the direction of the coordinate $$q$$, we can describe this as:
$$\frac{\partial L}{\partial q}=0$$
According to the Euler–Lagrange equation, we then have;
$$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}=\frac{\partial L}{\partial q}=0$$
$$\frac{d}{dt}\frac{\partial L}{\partial\dot{q}}=0$$ $$\frac{d}{dt}p_q=0$$ $$p_q=\mathrm{constant}$$
In other words we have already shown that if the Lagrangian does not change in change the coordinate $$q$$, the related momentum to $$q$$ which is $$p_q$$ is conserved. But isn't this then already Noether's theorem?
A problem here is if you choose a coordinate system that does not contain a coordinate basis vector that points in the direction where the Lagrangian is invariant. We could have chosen a wrong basis. However, you can just change the basis and the proof is similar.
Now imagine we have a coordinate system $${q_1,q_2,\ \ldots,\ q_i}$$ where for every direction, the Lagrangian isn’t invariant. Then we have
$$\frac{\partial L}{\partial q_1}\neq0$$ $$\frac{\partial L}{\partial q_2}\neq0$$ $$\ldots$$
So there isn't any conserved conjugate momentum, or any conserved Noether’s charge. Now we are told that there is a certain direction $${\hat{q}}^\ast\$$ that the Lagrangian is invariant. However, our coordinate system $${q_1,q_2,\ \ldots,\ q_i}$$ does not have a unit vector that points in that exact direction.
Then you can just perform a change of basis in such a way that we have $${q^\ast,q_2^\prime,q_3^\prime,..}$$ and then perform the following again.
Then how to find the form of the Noether’s charge you can just project the total momentum vector in the first q system onto the direction of $$\hat{q}$$: $$Q=\vec{p}\cdot{\hat{q}}^\ast$$
$$Q=\begin{bmatrix} p_x \,\\ p_y \\ \ldots \end{bmatrix}^T \cdot {\hat{q}}^\ast$$
So my problem is basically that the conclusion of Noether's theorem really trivially falls out of setting the Euler-Lagrange equations equal to $$0$$. In the worst case you would have to do a coordinate transformation before you can do that. This suggests that I am missing something important, because this seems too easy for a proof.
I am probably misunderstanding something fundamental of the theorem. Where does my reasoning fail?
• If there was nothing to prove, why so long a text full of implications? Also, Noether's theorem allows for Noether symmetries involving combination of time $t$ and coordinates $q_k$. arxiv.org/abs/1812.03682 Aug 14 at 10:15
• Noether's theorem allows for broader class of symmetries than just $\partial L/\partial q_k = 0$ (Qmechanic's 2nd point). Aug 14 at 12:54
• So if I understand correctly I am just looking at a special simple case and the importance of Noether's theorem lies in the fact that it also is true for broader symmetries that cannot be expressed in cyclic coordinates? Aug 14 at 13:25
• Yes yes yes yes. Aug 14 at 13:35
• One key property of Noether's theorem that makes it so useful is that it is constructive. If you know the the symmetry transformation (which can be complicated), then Noether's theorem tells you exactly what the associated conserved charge is -- not just that it exists. The general procedure that goes from an arbitrary symmetry transformation to the corresponding charge is meat of the proof. Aug 16 at 0:46
1. Noether's theorem is just one method to determine conservation laws. If you have another, that's totally fine.
2. Not all quasi-symmetry transformations (which in principle can depend functionally on both generalized positions $$q^j$$, generalized velocities $$\dot{q}^k$$ and time $$t$$) can be reduced to the case of a cyclic/ignorable coordinate via a pertinent coordinate transformation, cf. OP's trick.
Counterexamples:
• Technically speaking, time $$t$$ is never a cyclic/ignorable coordinate, so OP's trick does not apply to the case of energy conservation. (Although it is a consequence of the Beltrami identity.)
• The conservation of the Laplace-Runge-Lenz vector in the Lagrangian (as opposed to the Hamiltonian) formulation of the Kepler problem. In this case Noether's theorem uses a velocity-dependent quasi-symmetry transformation.
3. Yes, Noether's theorem requires a proof, see e.g. the original 1918 paper.
• Thank you for your comment. Where is my reasoning false? Could you help me understand this? I hope this is possible in undergraduate level physics Aug 14 at 12:37
• There exist more general quasi-symmetry transformations. Aug 14 at 12:58
• Thanks for your reaction. I see that I did not incorporate that. But even leaving those quasi-symmetry transformations out and just evaluating the regular symmetries that can be made cyclic with a proper coordinate transformation... the proof seems so trivial and short. It's just equationg EL equation to $0$. I seem to be missing something even in this case. Aug 14 at 13:03
• @bananenheld One effect that also could play a role here is that often when an important result is discovered, later formalisms and teaching are restructured around that result, making it a central piece. So in an intellectual tradition that chose Noether's theorem as a key result, we wouldn't be surprised if Noether's theorem looks quite obvious in retrospect, even if it was hard to see for people educated in the tradition of early 1900s. (I know too little about the relevant history to claim this actually is the case, just noting that it could be based on analogies from elsewhere). Aug 14 at 17:53
• historical aside: the Noether theorems are often presented in simplified form, for teaching purposes. Unfortunately numerous authors published "generalisations" of these simplifications, without reading Noether's original papers. It wasn't until the 1970s that actual generalisations were published (Kosmann-Schwarzbach 2011). Hence the Noether theorems are deeper than is often assumed Aug 17 at 1:52
## Invariant Lagrangian and Conservation Laws:
You're correct that if the Lagrangian is invariant under a transformation, it leads to a conserved quantity according to Noether's theorem. However, the Lagrangian's invariance doesn't directly imply conservation of the conjugate momentum (often referred to as generalized momentum). Instead, it leads to a conserved current, which, when integrated over space or time, gives rise to a conserved charge. The conserved charge is related to the conserved quantity you're looking for.
## Change of Coordinates:
Your point about changing coordinates is somewhat correct, but there's a subtlety. Noether's theorem deals with symmetries in the action (the integral of the Lagrangian over time). If you change coordinates, it might alter the form of the Lagrangian, but if the action remains invariant under the coordinate transformation, Noether's theorem can still be applied. Your idea of changing coordinates to align with the invariant direction is useful, but remember that the transformed Lagrangian should still exhibit the same invariance.
## Conservation Laws in Absence of Invariance:
Your assertion that if the Lagrangian is not invariant in any direction, then there are no conserved quantities, might not always hold. It's true that Noether's theorem is directly applicable when there's an explicit symmetry in the Lagrangian, but there are situations where conservation laws can arise even without an explicit symmetry, through other mathematical considerations.
## Projection onto Invariant Direction:
While projecting the total momentum vector onto the invariant direction is a reasonable approach, remember that Noether's theorem is more general and doesn't always involve just momentum conservation. For example, in a gauge theory, the conserved current can be more complex than simple momentum, involving other variables as well.
## Physical Significance:
Finally, Noether's theorem not only provides a mathematical framework for deriving conservation laws from symmetries but also carries important physical implications. It links fundamental symmetries (like translations, rotations, gauge symmetries, etc.) to the preservation of physical quantities, contributing to our understanding of the underlying principles governing physical systems.
By these points you can understand it.
• Thank you very much for your clear explanation. So if I understand correctly that if the Lagrangian isn't invariant, but the action IS invariant under that transformation, there is still a conserved noether current? So I often see that the invariance of the lagrangian leads to conserved noether charge, but that is not strictly the correct condition of the theorem? Aug 16 at 5:40
• Is this answer written by AI? The way it is written and the way it explains things reminds me of AI. Aug 19 at 13:37
• None, I doesn't written by aI Aug 19 at 13:41 | 2,160 | 9,243 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2023-50 | longest | en | 0.86384 |
https://catfishlake.org/blog/how-much-is-100-kgs | 1,696,045,260,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510575.93/warc/CC-MAIN-20230930014147-20230930044147-00050.warc.gz | 183,055,308 | 23,117 | # How Much Is 100 kgs Weigh? | Weight Conversion Calculator
Do you ever find yourself wondering how much is 100 kgs? In today’s blog post, we will answer that question and more. Learn about the historical context of measurements of weight and density, discover what units are most commonly used to measure mass, and understand why different contexts might require specific measurement tools. We’ll also provide some useful conversion tools so you can accurately convert between kilograms or any other unit if necessary. Read on for an informative journey into measuring the amount of something on a large scale.
## Definition Of 100 Kgs And Its Relationship With The Metric System
Kilograms, or kg for short, are a unit of mass and weight in the Metric System. Originally defined as the mass of one liter (1L) of water at 4° Celsius, 1 kilogram is now officially equal to the international prototype kilogram which is kept at the International Bureau of Weights and Measures in France. One kilogram is equal to 1000 grams, and it is the base unit of mass in the International System of Units (SI).
## Benefits Of Understanding How Much 100 Kgs?
Knowing how much is 100 kgs is important when it comes to everyday applications such as the shipping of goods, calculating food portions for large groups of people, or even setting up a gym workout routine. Understanding the weight of items can help you determine a safe carrying capacity in any given situation and ensure accuracy in your measurements. Additionally, knowing how much is 100 kgs is essential when it comes to international trade and commerce. With this knowledge, you can ensure that your business transactions are conducted with proper accuracy and precision.
## How Much Is 100 Kgs?
Let’s answer the question of how much is 100 kgs.
• 100 kgs is equal to 220.46 lbs, 3,527.15 ounces, or 35,271.50 grams
• 100 kgs is equivalent to 33.06 US gallons of water
• 100 kgs is the approximate weight of a Bengal Tiger
• 100 kgs is approximately the same as a large suitcase filled with items
## Different Ways Of Measuring 100 Kgs
When measuring 100 kgs, there are several different ways to do so. The most common methods include using scales or weight plates. Scales and weight plates are accurate tools that can accurately measure the mass of an item in kilograms. Additionally, you can also measure the object’s volume to determine its mass by subtracting the volume of water displaced by the object. This method is most commonly used when measuring the mass of large objects, such as shipping containers or vehicles.
## Tips To Calculate And Convert 100 Kgs Correctly
1. Understand the basics of the Metric System and the different units used to measure mass.
2. Use a scale or weight plate to accurately measure the mass of an item in kilograms.
3. Determine the object’s volume if necessary by subtracting the volume of water displaced by the object.
4. Utilize our convenient conversion calculator above to convert between different units and accurately measure the mass of an item.
5. Make sure that you understand the context of your measurements before attempting any calculations or conversions.
## Common Mistakes When Calculating And Converting 100 Kgs
1. Not recognizing the importance of accuracy when measuring and converting between different units.
2. Making calculations without taking into account the context of the measurement. For example, a pound is not equal to the same amount of kilograms in all contexts.
3. Measuring an object’s mass inaccurately because you are using faulty tools or methods.
4. Assuming that one unit is equal to another without taking into account the necessary conversion factors.
## Examples Of How Many Kilograms Are In A Variety Of Everyday Item
• 1 apple weighs approximately 0.2 kgs
• A small bottle of water weighs approximately 0.6 kgs
• An average sized car battery weights approximately 17 kgs
• A large suitcase filled with items can weigh up to 100kgs
• A fully grown Bengal tiger can weigh up to 300kgs
## Real-World Applications Of Understanding The Weight Of 100 Kgs
• Shipping products with precision accuracy by taking into account the item’s mass.
• Calculating food portions for large groups of people.
• Setting up a gym workout routine based on individual weights and measurements.
• Determining the safe carrying capacity of any given situation or object.
• Accurately converting between different units of measurement for international trade and commerce.
• Understanding the importance of accuracy when it comes to measuring mass on a large scale.
## Conclusion: How Much Is 100 Kgs?
The answer to the question “how much is 100 kgs” depends on the context in which it is being asked. However, generally speaking 100 kgs is equal to 220.46 lbs, 3,527.15 ounces, or 35,271.50 grams. Additionally, 100 kgs is equivalent to 33.06 US gallons of water and can be used to measure the mass of a variety of everyday items, from apples to cars. Understanding how to accurately measure and convert between different units is a skill that can be applied in a variety of real-world scenarios, so make sure you understand the basics before attempting any calculations or conversions.
## FAQ: 100 kgs
### How much does 100 kgs weigh in pounds?
100 kgs is equal to 220.46 pounds.
### What is the equivalent weight of 100 kgs in stones?
100 kgs is equivalent to 15.79 stones.
### What is the mass of 100 kgs in grams?
100 kgs is equal to 100,000 grams.
### How many 100 kg bags can fit into a ton?
10 100 kg bags can fit into a ton.
### Is 100 kgs considered a healthy weight for an adult?
It depends on the individual’s body type. Generally, a healthy weight range for an adult is between 18.5 to 24.9 body mass index (BMI). To determine if 100 kgs falls within that range, it is important to take into account other factors such as height and age. A medical professional can best assess what constitutes a healthy weight for an individual.
### What is the approximate volume of 100 kgs of sand or gravel?
100 kgs of sand or gravel would be equal to approximately 125 liters. The exact volume will depend on the size and shape of the individual grains of sand or gravel.
### How many 100 kgs of textbooks can fit on a standard bookshelf?
This will depend on the size of the bookshelf, as well as the size and number of textbooks. On average, a standard bookshelf can hold up to approximately 400 kgs or 4,000 liters of textbooks. So depending on their size, this would be equivalent to 40 100 kg bags.
### What is the energy content of 100 kgs of various fuels?
The energy content of 100 kgs of fuel will depend on the type of fuel.
### What is the recommended weight capacity a 100 kgs-rated ladder?
A 100 kgs-rated ladder can typically support up to 120 kgs. It is important to check the manufacturer’s guidelines for the specific model in order to ensure that it can safely support the weight of any loads or equipment being used on the ladder. Always follow safety guidelines when using ladders and make sure to use appropriate protective clothing.
### Is 100kg hard to bench?
This will depend on the individual’s strength and fitness level. Generally, it is considered a good goal to be able to bench press one’s own body weight. However, many professional weightlifters can bench press more than 100 kgs. | 1,609 | 7,365 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2023-40 | latest | en | 0.92246 |
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