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# Stir-Fry Pollo
Chemistry PPT 2010
by
## Barrett McCall
on 29 April 2010
Report abuse
#### Transcript of Stir-Fry Pollo
What Did I Eat? by Barrett McCall What Shalt Thy Cook? Hmm..I Know! Stir-Fry Chicken Makes 5 servings
2 boneless,skinless chicken breasts
11/2 cups white rice
2 cups carrots, chopped
1 spear brocolli, chopped
1/2 cup Yellow Bell Pepper - chopped
1 cup onion - chopped
1 ear corn
4 tablespoons extra virgin olive oil
3 tablespoons tamari sauce
Heat the olive oil in a skillet or wok over medium-high heat
Saute all vegetables about 15 min until crisp
Remove vegetables from skillet, reserving oil
Add the vegetables back into the skillet
Add tamari sauce and lower heat
Let simmer for 3 min
Cook rice according to directions in separate pot
Serve Immediately
How many calories (C) from carbohydrates are in this dish? 979.2 C How many calories (C) from protein are in this dish? 286.8 C How many calories (C) from fat are in this dish? 568.4 C How many total calories are contained in this dish? 1834.4C / 5 servings = 366.9C per serving How many Joules of energy are stored in your dish? 1834C x 4184J/C = 7.67x10^6J What percent of a 2,000 calorie (C) diet would your dish represent? 1834 C / 2000 C = 91.7%
367 C / 2000 C = 18.35% per person If 1000g of water absorbed the joules of energy stored in your dish, determine the change in temperature of the water.
7.67x10^6 J = 1.0x10^3g x 4.18J/gC x 1834C - entire dish or 1.53x10^6 J = 1.0x10^3g x 4.18J/gC x 367C per person
chg in T = 1834 C (dish) or chg in T = 367 C (per person) If you (a 140lb individual walking at a pace of 2 mph) burn 80 calories per mile of walking, how many miles would you have to walk to burn off the calories in your dish?
80 C / 1 mile = 367C/x mile
367 C / 80 C = 4.59 miles Discussion
This dish comes from an Asian method of quick cooking food in a small amount of fat, over a high heat, for a short period of time.
It became a popular method of cooking because the food cooked quickly and conserved fuel.
All the ingredients are readily available in any grocery store.
It was a relatively easy meal to prepare (once the vegetables were cut up), it was delicious and nutritious with only 367 calories per serving. Analysis Analysis Receipe THE END Ingredient Amount Carbs (g) Protein (g) Fat (g)
Chicken -2 skinless breasts 8 oz 0 47.0 5.0
Rice - Uncle Ben’s white 1 1/2 cups 222.0 24.0 0
Carrots - chopped 2 cups 4.0 0 0
Broccoli - chopped 1 spear 4.0 0 1.0
Bell Pepper-chopped 1/2 cup 3.8 0.7 0.15
Onion - chopped 1 cup 2.0 0 0
Corn - yellow 1 ear 3.0 0 1.0
Extra Virgin Olive Oil 4 tablespns 0 0 56.0
Tamari Sauce 3tablespns 6.0 0 0
Totals 244.8 71.7 63.15
How many calories (C) from carbohydrates are in this dish? 244.8 x 4 = 979.2 C
How many calories (C) from protein are in this dish? 71.7 x 4 = 286.8 C
How many calories (C) from fat are in this dish? 63.15 x 9 = 568.4 C
How many total calories are contained in this dish? 979C + 287C + 568C = 1834C or 1834C / 5 servings = 367C per serving Analysis How many Joules of energy are stored in your dish? 1834C x 4184J/C = 7.67x106J (dish)
367C x 4184J/C = 1.53x106J (per serving) What percent of a 2,000 calorie (C) diet would your dish represent? 1834 C / 2000 C = 91.7% (dish)
367 C / 2000 C = 18.35% (per serving) If 1000g of water absorbed the joules of energy stored in your dish, determine the change in temperature of the water.
q=mc∆T
7.67x106 J = 1.0x103g x 4.18J/g°C x 1834C (dish)
1.53x106 J = 1.0x103g x 4.18J/g°C x 367C (per serving)
∆T = 1834 C (dish) or ∆ T = 367 C (per serving) If you (a 140lb individual walking at a pace of 2 mph) burn 80 calories per mile of walking, how many miles would you have to walk to burn off the calories in your dish?
80 C / 1 mile = 367C / x mile
367 C / 80 C = 4.59 miles Analysis Analysis
Full transcript | 1,354 | 4,215 | {"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-2018-39 | latest | en | 0.878433 |
https://mathskills4kids.com/adding-and-subtracting-decimals-worksheets-pdf-for-grade-5 | 1,713,094,717,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816879.25/warc/CC-MAIN-20240414095752-20240414125752-00563.warc.gz | 365,789,232 | 18,108 | Hello everyone! Do you know that most 5th graders still find adding and subtracting decimals more challenging than operations with natural numbers? Oh yes! In this article, we have a solution for those students. Our free and printable worksheets are perfect for practicing and enhancing addition and subtraction of decimals skills in Grade 5 students. These Grade 5 Adding and subtracting decimals worksheets consist of fun activities that will strengthen the kid’s mastery of place value and estimation concepts.
• Also, this article will help us to understand the importance of practicing adding and subtracting decimals. Moreover, teachers and parents will discover the Grade 5 curriculum expectations for adding and subtracting decimals and the different types of worksheets to use.
In addition, they will find tips for effective practice with grade 5 worksheets and common mistakes to avoid when adding and subtracting decimals. This article also includes fun and interactive activities to supplement worksheet practice and additional online resources for grade 5 worksheets on adding and subtracting decimals.
Above all, you will be taught how to track students’ progress and evaluate their proficiency with adding and subtracting decimals.
## Fun and effective practice with Grade 5 worksheets for adding and subtracting decimals
Enjoy this fantastic article about decimals addition, and subtraction with thrilling ideas that will make math learning a fun and effective adventure. This fun and effective practice with Grade 5 worksheets for adding and subtracting decimals is designed to engage young learners while reinforcing important math skills.
These worksheets perfectly balance entertainment and education with various exciting exercises and activities. From solving real-world word problems to completing engaging puzzles, our worksheets ensure that learning is effective and enjoyable. Each worksheet is carefully crafted to target specific skills and concepts, allowing students to build a solid foundation in decimal operations.
So, whether you're a teacher looking for supplementary materials or a parent wanting to support your child's learning at home, Mathskills4kids.com is your ideal website to download and print fun Grade 5 Adding and subtracting decimals worksheets.
• ### Importance of practicing adding and subtracting decimals
Practicing adding and subtracting decimals is crucial for fifth-grade students. This foundational math skill is essential for their current grade level but also sets the stage for more complex mathematical concepts in the future. By mastering the addition and subtraction of decimals, students develop a strong number sense and gain the ability to solve real-life problems involving money, measurements, and data analysis.
Regular practice with grade 5 worksheets for adding and subtracting decimals worksheets helps students improve their accuracy and speed in computations. It also enhances their problem-solving skills, logical thinking, and ability to apply mathematical operations in various contexts.
Students become more confident in their math abilities with consistent practice, increasing their engagement and enjoyment.
In the fifth-grade curriculum, students are expected to develop a solid understanding of decimal place value and the addition and subtraction of decimals. They learn to compare and order decimals, understand the relationship between fractions and decimals, and apply decimal operations to solve problems.
When using grade 5 worksheets for adding and subtracting decimals, aligning the practice with the specific curriculum expectations is essential. This ensures students work on the appropriate skills and concepts at their grade level. Our worksheets take into account the specific learning goals of the fifth-grade curriculum, providing targeted exercises that allow students to master each aspect of adding and subtracting decimals.
• ### Different types of worksheets for adding and subtracting decimals
Mathskills4kids Grade 5 worksheets offer various exercises and activities to keep students engaged and motivated in adding and subtracting decimals. Here are some of the different types of worksheets for adding and subtracting decimals available:
• Traditional Computation Worksheets: These worksheets focus on the fundamental skills of adding and subtracting decimals using the standard algorithm. They provide ample practice for students to master the step-by-step process of aligning decimal places, carrying over, and borrowing.
• Word Problem Worksheets: Word problems offer a practical application of adding and subtracting decimals, allowing students to see the relevance of these operations in real-life scenarios. Our word problem worksheets present students with engaging situations that require them to use their decimal computation skills to find solutions.
• Interactive Worksheets: To make learning more interactive and hands-on, our interactive worksheets incorporate digital elements such as drag-and-drop activities, clickable options, and instant feedback. These worksheets provide an engaging and dynamic learning experience, making math practice more enjoyable for students.
• Puzzle Worksheets: Puzzle worksheets add an element of fun to the practice of adding and subtracting decimals. Students solve math problems to unlock clues or complete puzzles, fostering a sense of achievement and motivation to progress through the worksheet.
• ### Tips for Effective Practice with Grade 5 Worksheets
To make the most of grade 5 worksheets for adding and subtracting decimals, here are some tips for effective practice:
• Establish a Routine: Set aside dedicated practice time daily or a few times a week. Consistency is key to building proficiency in decimal operations.
• Start with the Basics: Begin with simpler worksheets focusing on one concept at a time. Gradually progress to more complex worksheets as students gain confidence and mastery.
• Encourage Independent Thinking: Encourage students to solve problems independently before seeking assistance. This helps them develop problem-solving skills and boosts their confidence in their abilities.
• Provide Timed Practice: Set a timer to challenge students to complete worksheets within a specific timeframe. This helps improve their speed and accuracy in adding and subtracting decimals.
• Utilize Peer Collaboration: Incorporate group activities or pair students to work on worksheets together. Collaboration promotes discussion, critical thinking, and peer learning.
• Offer Rewards and Incentives: Recognize students' efforts and achievements by rewarding them with incentives such as stickers, certificates, or small prizes. This fosters a positive learning environment and motivates students to excel.
• ### Common mistakes to avoid when adding and subtracting decimals
When practicing adding and subtracting decimals, addressing common mistakes students mostly make is important. Educators and parents can provide targeted guidance and support by being aware of these mistakes. Some common mistakes to avoid when adding and subtracting decimals include:
• Misalignment of Decimal Places: Students often forget to align the decimal points when adding or subtracting decimals. Emphasize the importance of maintaining proper place value alignment to avoid errors.
• Forgetting to Carry Over or Borrow: Students may overlook the need to carry over or borrow when adding or subtracting decimals with multiple digits. Please encourage them to double-check their computations to ensure accuracy.
• Misreading or Misinterpreting Word Problems: Word problems can be challenging, especially when they involve decimals. Remind students to read the problem carefully, identify the necessary operations, and accurately translate the information into numerical expressions.
• Errors in Decimal Point Placement: Decimal point placement errors can lead to significant miscalculations. Encourage students to pay attention to the number of decimal places in the given numbers and ensure that the decimal point is correctly positioned in the final answer.
• Skipping the Estimation Step: Estimation is an essential skill in decimal operations that helps students check the reasonableness of their answers. Remind students to estimate their results before performing the actual computation.
By addressing these common mistakes and providing targeted guidance, students can better understand adding and subtracting decimals and improve their accuracy.
• ### Fun and interactive activities to supplement Grade 5 adding and subtracting decimals worksheet practice
In addition to Mathskills4kids’ grade 5 adding and subtracting decimals worksheets, incorporating fun and interactive activities can enhance students' understanding and engagement in adding and subtracting decimals. Here are some ideas to supplement worksheet practice:
• Math Games: Introduce math games that involve adding and subtracting decimals. Games such as "Decimal War," "Decimal Bingo," or "Decimal Jeopardy" make learning enjoyable and provide an opportunity for friendly competition.
• Real-life Applications: Connect adding and subtracting decimals to real-life situations. For example, create a pretend store where students can practice calculating the total cost of items with decimal prices.
• Math Manipulatives: Use manipulatives such as base-10 blocks, place value discs, or fraction strips to demonstrate concretely the addition and subtraction of decimals. This visual and tactile approach helps students better understand decimal operations.
• Online Interactive Tools: Explore online platforms that offer interactive tools and games specifically designed for adding and subtracting decimals. These resources provide a technology-driven learning experience that appeals to students' interests.
By incorporating these activities alongside grade 5 worksheets, students can experience a well-rounded and engaging approach to practicing adding and subtracting decimals.
• ### Tracking progress and evaluating proficiency with adding and subtracting decimals
To ensure students progress in their understanding and proficiency of adding and subtracting decimals, tracking their performance and evaluating their skills is essential. Here are some methods for tracking progress and evaluating proficiency with adding and subtracting decimals:
• Formative Assessments: Regularly administer formative assessments, such as quizzes or short tests, to gauge students' understanding of adding and subtracting decimals. Analyze the results to identify areas of improvement and adjust instruction accordingly.
• Observations: Observe students' problem-solving strategies during classwork and homework assignments. This provides insights into their thinking processes and allows for targeted feedback and support.
• Peer and Self-Assessment: Encourage students to assess their work and provide constructive feedback to their peers. Peer and self-assessment promote metacognitive skills and allow students to reflect on their progress.
• Cumulative Assessments: Administer cumulative assessments at regular intervals to assess students' overall proficiency in adding and subtracting decimals. These assessments can be in the form of unit tests or end-of-term evaluations.
By consistently tracking progress and evaluating proficiency, educators and parents can identify areas for improvement and provide additional support as needed.
### Bonus: Online Resources for Grade 5 adding and subtracting decimals worksheets
In today's digital age, numerous online resources are available to supplement grade 5 adding and subtracting decimals worksheets. These resources provide additional practice opportunities, interactive tools, and instructional videos to support students' learning.
Here are some reputable online platforms to explore:
• Decimals Addition & Subtraction Worksheets for Grade 5 | K5 Learning: This website has a variety of worksheets that cover adding and subtracting decimals up to 3 digits, with and without missing numbers, in columns, and with money notation.
• Adding and Subtracting Decimals activity for 5th grade - Liveworksheets.com: This website has an interactive activity that lets your students practice adding and subtracting decimals online. They will type the correct answers in the boxes and get instant feedback and scores. This activity is engaging and fun, and it can be used on any device with an internet connection.
• Digital Math Activity - Addition and Subtraction with Decimals - iKNOWit.com: This interactive, online math game from iKnowIt.com will help your fifth-grade students solve vertical and horizontal addition and subtraction problems with decimals. They will also learn how to line up the decimals, Add or subtract fractional parts, Add or subtract whole numbers, regroup when necessary, and solve word problems involving addition and subtraction with decimals. https://www.iknowit.com/lessons/e-addition-subtraction-decimals.html
•
Thank you for sharing the links of MathSkills4Kids.com with your loved ones. Your choice is greatly appreciated.
### Conclusion
Regular practice with grade 5 adding and subtracting decimals worksheets: Free and printable is perfect for student's mathematical development. Students will strengthen their decimal computation skills and develop essential problem-solving abilities and logical thinking by engaging in fun and effective practice.
Mathskills4kids Grade 5 worksheets provide a variety of exercises and activities that cater to different learning styles and preferences. Whether it's traditional computation worksheets, word problem worksheets, interactive worksheets, or puzzle worksheets, there is something for every student.
Supplementing worksheet practice with fun and interactive activities further enhances students' understanding and engagement. Math games, real-life applications, math manipulatives, and online resources offer additional opportunities for practice and exploration.
By tracking progress and evaluating proficiency, educators and parents can ensure that students are making steady progress in their mastery of adding and subtracting decimals.
With the right resources, guidance, and consistent practice, students can develop a strong foundation in decimal operations, setting them up for success in future math endeavors. So, why wait? Download now and start practicing.
• | 2,491 | 14,456 | {"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-2024-18 | latest | en | 0.913388 |
http://mathhelpforum.com/statistics/106880-help-stuck-impossible-probability-problem-print.html | 1,524,820,172,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524127095762.40/warc/CC-MAIN-20180427075937-20180427095937-00010.warc.gz | 205,221,854 | 3,419 | # Help! stuck on impossible probability problem...
• Oct 8th 2009, 12:49 PM
alexM0tts
Help! stuck on impossible probability problem...
I think I have to find out the net profit? I am not sure and I don't even know what formula to use. I definitely know it is not the complementary rules that apply. Here it goes: A 28-year-old man pays $165 for a one-year life insurance policy with coverage of$120,000. If the probability that he will live through the year is 0.9995, what is the expected value for the insurance policy?
Can someone enlighten me and show me how to get this. I'm pretty sure there will be plenty of these on my exam next week. I need some good tutoring, i'm not good in math. Thanks!
• Oct 8th 2009, 05:53 PM
mr fantastic
Quote:
Originally Posted by alexm0tts
i think i have to find out the net profit? I am not sure and i don't even know what formula to use. I definitely know it is not the complementary rules that apply. Here it goes: a 28-year-old man pays $165 for a one-year life insurance policy with coverage of$120,000. If the probability that he will live through the year is 0.9995, what is the expected value for the insurance policy?
can someone enlighten me and show me how to get this. I'm pretty sure there will be plenty of these on my exam next week. I need some good tutoring, i'm not good in math. Thanks!
(0.9995)(165) - (0.0005)(120,000)
• Oct 9th 2009, 02:20 AM
aidan
Quote:
Originally Posted by mr fantastic
(0.9995)(165) - (0.0005)(120,000)
Mr.F
If 10,000 men aged 28 buy a 1 year policy for $165, then the insurance company collects$1,650,000.
The 5 policies pay out $120,000 each for a loss of$600,000.
The difference is $1,050,000. Your formula yields$1,049,175.
Why is there a difference of $825 (= 5 * 165). It almost seems logical that the$165 collected is responsible for a $120,000 loss, but I'm not grasping the whole concept. Would you explain why the$165 would not be part of the total value?
• Oct 9th 2009, 04:17 AM
CaptainBlack
Quote:
Originally Posted by aidan
Mr.F
If 10,000 men aged 28 buy a 1 year policy for $165, then the insurance company collects$1,650,000.
The 5 policies pay out $120,000 each for a loss of$600,000.
The difference is $1,050,000. [SIZE=4]Your formula yields$1,049,175.
Why is there a difference of $825 (= 5 * 165). It almost seems logical that the$165 collected is responsible for a $120,000 loss, but I'm not grasping the whole concept. Would you explain why the$165 would not be part of the total value?
$\displaystyle E=-0.9995\times 165+0.0005(120000-165)$ | 729 | 2,557 | {"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} | 3.703125 | 4 | CC-MAIN-2018-17 | latest | en | 0.939946 |
https://javascriptinfo.com/view/129112/javascript-calculate-the-nth-root-of-a-number | 1,604,013,437,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00466.warc.gz | 375,591,558 | 6,336 | # JavaScript: Calculate the nth root of a number
I'm trying to get the nth root of a number using JavaScript, but I don't see a way to do it using the built in `Math` object. Am I overlooking something?
If not...
Is there a math library I can use that has this functionality?
If not...
What's the best algorithm to do this myself?
Can you use something like this?
``````Math.pow(n, 1/root);
``````
eg.
``````Math.pow(25, 1/2) == 5
``````
The `n`th root of `x` is the same as `x` to the power of `1/n`. You can simply use `Math.pow`:
``````var original = 1000;
var fourthRoot = Math.pow(original, 1/4);
original == Math.pow(fourthRoot, 4); // (ignoring floating-point error)
``````
Use Math.pow()
Note that it does not handle negative nicely - here is a discussion and some code that does
http://cwestblog.com/2011/05/06/cube-root-an-beyond/
``````function nthroot(x, n) {
try {
var negate = n % 2 == 1 && x < 0;
if(negate)
x = -x;
var possible = Math.pow(x, 1 / n);
n = Math.pow(possible, n);
if(Math.abs(x - n) < 1 && (x > 0 == n > 0))
return negate ? -possible : possible;
} catch(e){}
}
``````
You could use
``````Math.nthroot = function(x,n) {
//if x is negative function returns NaN
return this.exp((1/n)*this.log(x));
}
//call using Math.nthroot();
``````
The `n`-th root of `x` is a number `r` such that `r` to the power of `1/n` is `x`.
In real numbers, there are some subcases:
• There are two solutions (same value with opposite sign) when `x` is positive and `r` is even.
• There is one positive solution when `x` is positive and `r` is odd.
• There is one negative solution when `x` is negative and `r` is odd.
• There is no solution when `x` is negative and `r` is even.
Since `Math.pow` doesn't like a negative base with a non-integer exponent, you can use
``````function nthRoot(x, n) {
if(x < 0 && n%2 != 1) return NaN; // Not well defined
return (x < 0 ? -1 : 1) * Math.pow(Math.abs(x), 1/n);
}
``````
Examples:
``````nthRoot(+4, 2); // 2 (the positive is chosen, but -2 is a solution too)
nthRoot(+8, 3); // 2 (this is the only solution)
nthRoot(-8, 3); // -2 (this is the only solution)
nthRoot(-4, 2); // NaN (there is no solution)
``````
For the special cases of square and cubic root, it's best to use the native functions `Math.sqrt` and `Math.cbrt` respectively.
As of ES7, the exponentiation operator `**` can be used to calculate the nth root as the 1/nth power of a non-negative base:
``````let root1 = Math.PI ** (1 / 3); // cube root of ?
let root2 = 81 ** 0.25; // 4th root of 81
``````
This doesn't work with negative bases, though.
``````let root3 = (-32) ** 5; // NaN
``````
Here's a function that tries to return the imaginary number. It also checks for a few common things first, ex: if getting square root of 0 or 1, or getting 0th root of number x
``````function root(x, n){
if(x == 1){
return 1;
}else if(x == 0 && n > 0){
return 0;
}else if(x == 0 && n < 0){
return Infinity;
}else if(n == 1){
return x;
}else if(n == 0 && x > 1){
return Infinity;
}else if(n == 0 && x == 1){
return 1;
}else if(n == 0 && x < 1 && x > -1){
return 0;
}else if(n == 0){
return NaN;
}
var result = false;
var num = x;
var neg = false;
if(num < 0){
//not using Math.abs because I need the function to remember if the number was positive or negative
num = num*-1;
neg = true;
}
if(n == 2){
//better to use square root if we can
result = Math.sqrt(num);
}else if(n == 3){
//better to use cube root if we can
result = Math.cbrt(num);
}else if(n > 3){
//the method Digital Plane suggested
result = Math.pow(num, 1/n);
}else if(n < 0){
//the method Digital Plane suggested
result = Math.pow(num, 1/n);
}
if(neg && n == 2){
//if square root, you can just add the imaginary number "i=?-1" to a string answer
//you should check if the functions return value contains i, before continuing any calculations
result += 'i';
}else if(neg && n % 2 !== 0 && n > 0){
//if the nth root is an odd number, you don't get an imaginary number
//neg*neg=pos, but neg*neg*neg=neg
//so you can simply make an odd nth root of a negative number, a negative number
result = result*-1;
}else if(neg){
//if the nth root is an even number that is not 2, things get more complex
//if someone wants to calculate this further, they can
//i'm just going to stop at *n?-1 (times the nth root of -1)
//you should also check if the functions return value contains * or ?, before continuing any calculations
result += '*'+n+?+'-1';
}
return result;
}
``````
Well, I know this is an old question. But, based on SwiftNinjaPro's answer, I simplified the function and fixed some NaN issues. Note: This function used ES6 feature, arrow function and template strings, and exponentation. So, it might not work in older browsers:
``````Math.numberRoot = (x, n) => {
return (((x > 1 || x < -1) && n == 0) ? Infinity : ((x > 0 || x < 0) && n == 0) ? 1 : (x < 0 && n % 2 == 0) ? `\${((x < 0 ? -x : x) ** (1 / n))}\${"i"}` : (n == 3 && x < 0) ? -Math.cbrt(-x) : (x < 0) ? -((x < 0 ? -x : x) ** (1 / n)) : (n == 3 && x > 0 ? Math.cbrt(x) : (x < 0 ? -x : x) ** (1 / n)));
};
``````
Example:
``````Math.numberRoot(-64, 3); // Returns -4
``````
Example (Imaginary number result):
``````Math.numberRoot(-729, 6); // Returns a string containing "3i".
``````
I have written an algorithm but it is slow when you need many numbers after the point:
https://github.com/am-trouzine/Arithmetic-algorithms-in-different-numeral-systems
``````NRoot(orginal, nthRoot, base, numbersAfterPoint);
``````
The function returns a string.
E.g.
``````var original = 1000;
var fourthRoot = NRoot(original, 4, 10, 32);
console.log(fourthRoot);
//5.62341325190349080394951039776481
`````` | 1,798 | 5,685 | {"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} | 3.5 | 4 | CC-MAIN-2020-45 | latest | en | 0.790013 |
https://e-classifieds.me/how-to-calculate-support-and-resistance-levels-for-stock/ | 1,695,911,564,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00686.warc.gz | 243,357,713 | 10,241 | # How To Calculate Support And Resistance Levels For Stock
Support and resistance form the foundation of all Technical Analysis.
Support – Any point where a falling stock price meets enough buyers to arrest the fall.
Resistance – Any point where a rising stock price meets enough sellers to stop the rising stock price.
Let me give some pictures to illustrate the difference between support and resistance
Technical Analysis: How to Calculate Support
As you can see in the picture above, the red line represents the support. As you can see with the circles, the stock price has bounced off of this support line 4 times in the last 4 months. The sellers have been unable to beat the buyers when the stock price approaches the support level.
Here’s another example:
In this example, you see it bounce 5 times in the last 4 months.
As the stock price falls, demand for the shares increases because of group psychology. This means that when the stock approaches the support line, enough buyers have the same opinion that together they are causing the price to reverse at the same point. The more times a stock bounces off of a support or resistance, the stronger this line is.
Both the support and resistance lines are not absolute points on a chart. When a stock approaches either line, maybe a bounce happens a few cents from the support or resistance. For example, let’s say support is drawn at \$4.50 and the stock is approaching this line. It could drop to \$4.45 and then bounce again when buyers overthrow the sellers. The point is, these lines are not hard drawn lines on a stock chart.
Technical Analysis: How to Calculate Resistance
So just like the support, the resistance acts as a line where the sellers take over the buyers and push the stock price down.
Here’s another example:
Okay. You get the point.
Now, if a stock breaks the resistance line, we could see it then act as a support.
Here is an example:
As you can see, the buyers overthrew the sellers the third time the stock approached the resistance. Then a month later, the stock price came back to that line but bounced off as a support. So these lines can be used in very long term plays, they will always be on the chart.
You have to remember that prices are driven by the actions of large numbers of people. So the likelihood of you being the only person that sees a support or resistance line is very slim because like it says in the first sentence in this post, “Support and Resistance forms the foundation of Technical Analysis.” Once again, these tools are not definite points on charts, so trade cautiously around them. | 532 | 2,614 | {"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} | 3.71875 | 4 | CC-MAIN-2023-40 | longest | en | 0.918484 |
http://www.typologycentral.com/forums/myers-briggs-and-jungian-cognitive-functions/10898-muddling-3.html | 1,513,505,867,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948595342.71/warc/CC-MAIN-20171217093816-20171217115816-00733.warc.gz | 475,887,594 | 16,753 | # Thread: The P and J Muddling
1. Originally Posted by dissonance
Nocap -- this is quite an issue, do you agree?
Maybe you could work some function definitions into your enlightening post...
I could do it if you don't want to...
As a matter of fact, that's all I've gotten to so far.
I haven't even built the functions into their J and P categories yet.
I wouldn't be able to write this thing without doing it.
2. Please no functions definitions by diss @nocap
Else, my beautiful little world is going to hell
3. Originally Posted by entropie
Please no functions definitions by diss @nocap
Else, my beautiful little world is going to hell
Yeah, nocap is a better/more concise writer than I, so I was hoping he'd do it
4. Originally Posted by dissonance
Hmm..... watch out for that, though. None of the tests are weighted correctly (in fact, I don't know if I've even seen a weighted test at all), and no one can accurately answer all the questions.
I think this is the reason so many people get confused -- they'll score really high on one function but they really prefer it in the opposite direction, or they might attribute T things to N things and vice versa...
The last test I took (the new cogprocesses one), I scored highest on Ti... but I know it's not in the same ballpark of usage as my Ni.
I absolutely agree with the latter portion that we may misunderstand the question(s) being asked or infer it to have a different connotation. However as for your former statement, I think it's safe to say that even if you are taking a test that merely results in dichotomy selections, you should still result in your dominant function (ie introversion/thinking=Ti). Furthermore if you consistently test the same way then it's safe to say that you are most likely that function. After that, it may be a free-for-all even with the auxiliary function. When I hear people say that they're introverts but score higher on their auxiliary, I am always perplexed. If you're scoring higher on any extraverted function then you are most likely an extravert. Myers-Briggs says however it's usually introverted men who confuse themselves as extraverts more than the other way around.
5. Originally Posted by Jack Flak
To question mark: Yeah, what dissonance said. Considering the fact that he and I almost never agree on type shit, you better listen.
Function tests don't tell you what you're doing with your brain more often, only what you think you're good at. To top it off, the test questions can be odd, and the functions themselves are poorly defined.
Whoa.... who ever said anything about function tests? I have always argued it's the worse means of determining type and the results will constantly change because they can only render which function you may be using at the present based on most likely a need to adapt to your environment or whatever has your attention at the time. I could take one of them today and actually come up with some extraverted function if I answered the questions honestly.
6. Apparently I did not make myself clear so let me recapitulate that my post coincides with my original statement that may have confused “whatever”. The whole point being made is that any discussion of dichotomies is rudimentary, however when we focus on J/P in particular. As Linda V. Berens says that if we spend too much time talking about J, before long we're really talking about SJ (extreme SJ!)... and if we spend too much time talking about P, we're really talking about NP (extreme NP!). So it's not good to single out and focus on J/P alone for drawing lots of Type conclusions.
MBTI Practitioner, Vicky Jo also reminds that whenever she see’s a conversation deteriorate to the point that J and P are the only letters being discussed, then she realizes that the discussion is not about type theory anymore, but a discussion of bias and stereotype. Vicky Jo says that clearly the participants don't know enough about type theory overall to keep the conversation going properly.
This is not a slam against Nocapszy, but just an observation over the years that discussions of type using rudimentary dichotomies in general means that there remains a great deal of learning to do. Nocaps I hope this has not discouraged you from moving forward with your original intent for this thread, so back on point.
7. Originally Posted by "?"
MBTI Practitioner, Vicky Jo also reminds that whenever she see’s a conversation deteriorate to the point that J and P are the only letters being discussed, then she realizes that the discussion is not about type theory anymore, but a discussion of bias and stereotype. Vicky Jo says that clearly the participants don't know enough about type theory overall to keep the conversation going properly.
Well, that's because people, and especially MBTI Practitioners, don't understand what Perceiving and Judging mean.
8. Originally Posted by Jack Flak
Well, that's because people, and especially MBTI Practitioners, don't understand what Perceiving and Judging mean.
How about putting it this way, giving J/P so much attention for introverts is no difference that focusing on the auxiliary function Ti/Fi for extraverts. What would be the point? Ti/Fi is important but why would an EP type give it so much attention. That is exactly what you do when you allude to J/P, particularly as it pertains to introverted types.
9. Originally Posted by "?"
How about putting it this way, giving J/P so much attention for introverts is no difference that focusing on the auxiliary function Ti/Fi for extraverts. What would be the point? Ti/Fi is important but why would an EP type give it so much attention. That is exactly what you do when you allude to J/P, particularly as it pertains to introverted types.
Uhhh my point being that Introverted Ps shouldn't have a J function as their primary function. See sig.
10. Originally Posted by "?"
I absolutely agree with the latter portion that we may misunderstand the question(s) being asked or infer it to have a different connotation. However as for your former statement, I think it's safe to say that even if you are taking a test that merely results in dichotomy selections, you should still result in your dominant function (ie introversion/thinking=Ti). Furthermore if you consistently test the same way then it's safe to say that you are most likely that function.
Take a lesson boys and girls:
This is the kind of thinking that necessitates the gargantuan post I'm still working on.
If the testing is initially wrong, then consistency means it's consistently wrong.
An ITS ought to know that...
I agree that testing should result in the correct function, I can say with a great amount of confidence (being a victim myself) that it doesn't happen empirically.
I was originally tested as ISTP. I am not an ISTP.
The point dissonance was making was that if the questions are written in a way that the user thinks they understand, they will answer it, and (ideally) accurately so, to their understanding. However, if their assumption of the intended meaning differs from the actual intended meaning of only a single word, the question, given the boolean logic of the tests, will misrepresent the person. I emphasize the factor of idyllic violation, because people do and consistently observe themselves incorrectly.
Especially E__Ps, who, often without even knowing it, will lie or just make things up, even believing these things themselves.
Aside from that that, I find a great many people would rather have an answer, than the right answer, which almost invariably discourages further investigation. Apply this principle to introspection, and you'll have, with certainty, a lot of people who, even providing they do understand the question, will still not be able to answer accurately, because they have bad information.
After that, it may be a free-for-all even with the auxiliary function. When I hear people say that they're introverts but score higher on their auxiliary, I am always perplexed. If you're scoring higher on any extraverted function then you are most likely an extravert.
Ha!
So an INTJ who has 25% Si, 35% and 40% Te ought to be an ENTJ then?
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YELLA
New member
On a road trip of 252 miles in each direction, the average rate of travel going was 7mph more than the rate returning. Find the rate for each direction if the entire trip took 21 hours
tkhunny
Moderator
Staff member
Distance = Rate * Time
Let's see what you get.
Subhotosh Khan
Super Moderator
Staff member
YELLA said:
On a road trip of 252 miles in each direction, the average rate of travel going was 7mph more than the rate returning. Find the rate for each direction if the entire trip took 21 hours
Use the method shown at:
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https://www.coursehero.com/file/6743409/chapter-13-solution/ | 1,490,343,224,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218187744.59/warc/CC-MAIN-20170322212947-00599-ip-10-233-31-227.ec2.internal.warc.gz | 887,201,119 | 119,813 | chapter 13 solution
# chapter 13 solution - Basic 1. The portfolio weight of an...
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Basic 1. The portfolio weight of an asset is total investment in that asset divided by the total portfolio value. First, we will find the portfolio value, which is: Total value = 180(\$45) + 140(\$27) = \$11,880 The portfolio weight for each stock is: Weight A = 180(\$45)/\$11,880 = .6818 Weight B = 140(\$27)/\$11,880 = .3182 2. The expected return of a portfolio is the sum of the weight of each asset times the expected return of each asset. The total value of the portfolio is: Total value = \$2,950 + 3,700 = \$6,650 So, the expected return of this portfolio is: E(R p ) = (\$2,950/\$6,650)(0.11) + (\$3,700/\$6,650)(0.15) = .1323 or 13.23% 3. The expected return of a portfolio is the sum of the weight of each asset times the expected return of each asset. So, the expected return of the portfolio is: E(R p ) = .60(.09) + .25(.17) + .15(.13) = .1160 or 11.60%
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4. Here we are given the expected return of the portfolio and the expected return of each asset in the portfolio, and are asked to find the weight of each asset. We can use the equation for the expected return of a portfolio to solve this problem. Since the total weight of a portfolio must equal 1 (100%), the weight of Stock Y must be one minus the weight of Stock X. Mathematically speaking, this means: E(R p ) = .124 = .14w X + .105(1 – w X ) We can now solve this equation for the weight of Stock X as: .124 = .14w X + .105 – .105w X .019 = .035w X w X = 0.542857 So, the dollar amount invested in Stock X is the weight of Stock X times the total portfolio value, or: Investment in X = 0.542857(\$10,000) = \$5,428.57 And the dollar amount invested in Stock Y is: Investment in Y = (1 – 0.542857)(\$10,000) = \$4,574.43 5. The expected return of an asset is the sum of the probability of each return occurring times the probability of that return occurring. So, the expected return of the asset is: E(R) = .25(–.08) + .75(.21) = .1375 or 13.75% 6. The expected return of an asset is the sum of the probability of each return occurring times the probability of that return occurring. So, the expected return of the asset is: E(R) = .20(–.05) + .50(.12) + .30(.25) = .1250 or 12.50% 7. The expected return of an asset is the sum of the probability of each return occurring times the probability of that return occurring. So, the expected return of each stock asset is: E(R A ) = .15(.05) + .65(.08) + .20(.13) = .0855 or 8.55% E(R B ) = .15(–.17) + .65(.12) + .20(.29) = .1105 or 11.05% To calculate the standard deviation, we first need to calculate the variance. To find the variance, we find the squared deviations from the expected return. We then multiply each possible squared deviation by its probability, then add all of these up. The result is the variance. So, the variance and standard deviation of each stock is: σ A 2 =.15(.05 – .0855) 2 + .65(.08 – .0855) 2 + .20(.13 – .0855) 2 = .00060 σ A = (.00060) 1/2 = .0246 or 2.46%
σ B 2 =.15(–.17 – .1105) 2 + .65(.12 – .1105) 2 + .20(.29 – .1105) 2 = .01830
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## This note was uploaded on 01/29/2012 for the course MAN 4635 taught by Professor Q during the Spring '11 term at Metro State.
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chapter 13 solution - Basic 1. The portfolio weight of an...
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# We say a binary operation ∗ is associative when a ∗ (b ∗ c) = (a ∗ b) ∗ (a ∗c)
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## The correct option is B False We say a binary operation ∗ is associative if a ∗ (b ∗ c) = (a ∗ b) ∗ c. So the given statement is false.
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Join BYJU'S Learning Program | 191 | 616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 2, "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} | 2.5625 | 3 | CC-MAIN-2024-30 | latest | en | 0.842656 |
https://scitechdaily.com/simple-algebra-enables-faster-large-volume-covid-19-testing/ | 1,721,053,901,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514707.52/warc/CC-MAIN-20240715132531-20240715162531-00676.warc.gz | 445,653,502 | 23,180 | # Simple Algebra Enables Faster, Large-Volume COVID-19 Testing
A Cardiff University mathematician created a new method for processing large volumes of COVID-19 tests, potentially allowing more tests to be performed simultaneously and delivering quicker results.
Cardiff University mathematician proposes new technique to drastically increase the amount of testing for COVID-19.
A mathematician from Cardiff University has developed a new method for processing large volumes of COVID-19 tests which he believes could lead to significantly more tests being performed at once and results being returned much quicker.
Dr. Usama Kadri, from the University’s School of Mathematics, believes the new technique could allow many more patients to be tested using the same amount of test tubes and with a lower possibility of false negatives occurring.
Dr. Kadri’s technique, which has been published in the journal Health Systems, uses simple algebraic equations to identify positive samples in tests and takes advantage of a testing technique known as ‘pooling.’
Pooling involves grouping a large number of samples from different patients into one test tube and performing a single test on that tube.
If the tube is returned negative then you know that everybody from that group does not have the virus.
Pooling can be applied by laboratories to test more samples in a shorter space of time, and works well when the overall infection rate in a certain population is expected to be low.
If a tube is returned positive then each person within that group needs to be tested once again, this time individually, to determine who has the virus. In this instance, and particularly when it is known that infection rates in the population are high, the savings from the pooling technique in terms of time and cost become less significant.
However, Dr. Kadri’s new technique removes the need to perform a second round of tests once a batch is returned positive and can identify the individuals who have the virus using simple equations.
The technique works with a fixed number of individuals and test tubes, for example, 200 individuals and 10 test tubes, and starts by taking a fixed number of samples from a single individual, for example, 5, and distributing these into 5 of the 10 test tubes.
Another 5 samples are taken from the second individual and these are distributed into a different combination of 5 of the 10 tubes.
This is then repeated for each of the 200 individuals in the group so that no individual shares the same combination of tubes.
Each of the 10 test tubes is then sent for testing and any tube that returns negative indicates that all patients that have samples in that tube must be negative.
If only one individual has the virus, then the combinations of the tubes that return positive, which is unique to the individual, will directly indicate that individual.
However, if the number of positive tubes is larger than the number of samples from each individual, in this example 5, then there should be at least two individuals with the virus.
The individuals that have all of their test tubes return positive are then selected.
The method assumes that each individual that is positive should have the same quantity of virus in each tube, and that each of the individuals testing positive will have a unique quantity of virus in their sample which is different from the others.
From this, the method then assumes that there are exactly two individuals with the virus and, for every two suspected individuals, a computer is used to calculate any combination of virus quantity that would return the actual overall quantity of virus that was measured in the tests.
If the right combination is found then the selected two individuals have to be positive and no one else. Otherwise, the procedure is repeated but with an additional suspected individual, and so on until the right combination is found.
“Applying the proposed method allows testing many more patients using the same number of testing tubes, where all positives are identified with no false negatives, and no need for a second round of independent testing, with the effective testing time reduced drastically,” Dr. Kadri said.
So far, the method has been assessed using simulations of testing scenarios and Dr. Kadri acknowledges that lab testing will need to be carried out to increase confidence in the proposed method.
Moreover, for clinical use, additional factors need to be considered including sample types, viral load, prevalence, and inhibitor substances.
Reference: “Variation of quantified infection rates of mixed samples to enhance rapid testing during an epidemic” by Usama Kadri, 13 September 2020, Health Systems.
DOI: 10.1080/20476965.2020.1817801 | 908 | 4,754 | {"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-2024-30 | latest | en | 0.953674 |
https://socratic.org/questions/what-is-the-derivative-of-y-ln-x-2x-3-1-2#177444 | 1,643,423,560,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00430.warc.gz | 564,647,595 | 5,628 | # What is the derivative of y=ln[x/(2x+3)]^(1/2)?
$y ' = \frac{3}{2 x \left(2 x + 3\right)}$
$y = \ln {\left[\frac{x}{2 x + 3}\right]}^{\frac{1}{2}} = \frac{1}{2} \ln \left[\frac{x}{2 x + 3}\right]$
$y ' = \frac{1}{2} \cdot \frac{1}{\frac{x}{2 x + 3}} \cdot \frac{2 x + 3 - 2 x}{2 x + 3} ^ 2 = \frac{3}{2 x \left(2 x + 3\right)}$ | 169 | 330 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 3, "mathjax_inline_tex": 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} | 4.0625 | 4 | CC-MAIN-2022-05 | latest | en | 0.276623 |
https://math.stackexchange.com/questions/1740991/equivalence-relations-on-intersections | 1,623,797,853,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487621627.41/warc/CC-MAIN-20210615211046-20210616001046-00092.warc.gz | 343,775,748 | 37,845 | Equivalence relations on intersections
Let $R_1$ and $R_2$ be equivalence relations on a set $S$. Their intersection is on the relations considered as sets of ordered pairs. Identify the equivalence classes of the relation $R_1 \cap R_2$.
I don't know how to go about showing this with proof, any help would be appreciated!
• $[x]_{R_1\cap R_2}=\{y\in S\mid y(R_1\cap R_2)x\}$. Work this out. – drhab Apr 13 '16 at 16:59
If $E$ denotes an equivalence relation then the following statements are equivalent:
• $\langle x,y\rangle\in E$
• $y\in[x]_E$
Here $[x]_E$ denotes the equivalence class represented by $x$.
By definition of intersection:$$\langle x,y\rangle\in R_1\cap R_2\iff\langle x,y\rangle\in R_1\wedge\langle x,y\rangle\in R_2$$ Apparantly this statement can be translated into:$$y\in[x]_{R_1\cap R_2}\iff y\in[x]_{R_1}\wedge y\in[x]_{R_2}$$
Based on this find a relation between $[x]_{R_1\cap R_2}$, $[x]_{R_1}$ and $[x]_{R_2}$.
Maybe this will help:
This is a visual representation of two equivalence relations. Do you think you can translate what is going on here into the language of math?
• Did you paint this guy? – B. Pasternak Apr 13 '16 at 17:19 | 385 | 1,175 | {"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} | 3.71875 | 4 | CC-MAIN-2021-25 | latest | en | 0.831021 |
http://www.theguardian.com/commentisfree/2008/nov/22/health-bad-science-colds/print | 1,440,867,553,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440644064517.22/warc/CC-MAIN-20150827025424-00269-ip-10-171-96-226.ec2.internal.warc.gz | 747,141,880 | 23,711 | # When it comes to a cold, you might as well try goat entrails
• The Guardian,
I have a cold (and they're worse than you remember when you're well). Throughout the nation, homeopaths and self-declared nutrition therapists are celebrating. More importantly, I know that there is almost nothing I can do, except sit it out, and wait. Vitamin C will shave a few hours off it, at high doses, like 7g a day, which I can happily live without.
Although literally anything I try will appear, to me, to work: because unless I'm seroconverting with HIV (not that I'm prone to hypochondria) I will get better anyway. This is the natural history of the illness, and it's true with most things. When your back pain is at its worst and you visit your doctor - or your friendly local spoonbender - it's bound to get better, because these things come in cycles, or as statisticians say, they "regress to the mean". You can look at regression to the mean mathematically, if you like. On Bruce Forsyth's Play Your Cards Right, when Brucey puts a three on the board, the audience all shout: "Higher!" because they know the odds are that the next card is going to be higher than a three. "Do you want to go higher or lower than a jack? Higher?" "Lower!"
So I could take homeopathy. Or I could, equally stupidly, harass my GP for antibiotics, even though they are ineffective in treating a viral cold.
In one study, prescribing antibiotics rather than giving advice on self-management for sore throat resulted in an increased overall workload through repeat attendance. If a GP prescribed antibiotics for sore throat to 100 fewer patients each year, they calculated: 33 fewer would believe that antibiotics were effective, 25 fewer would intend to consult with the problem in the future, and 10 fewer would come back within the next year.
If you were an alternative therapist, or a drug salesman, you could turn those figures on their head and use them as a blueprint to drum up more trade: because we are all prone to see patterns where there is none, and more than that, to believing that our actions have results. This was demonstrated in a chilling experiment several decades ago. Subjects were recruited to play the role of a teacher trying to make a child arrive on time for school at 8.30am. They sat at a computer, on which it appeared that each day, for 15 consecutive days, a child would arrive at some time between 8.20 and 8.40.
Since this was a psychology experiment, the subjects were lied to: they did not know that the arrival times were entirely random, and predetermined before the experiment began.
Nevertheless, participants thoughtfully deployed punishments for lateness, and rewards for punctuality.
When they were asked at the end to rate their strategy, 70% concluded that reprimand was more effective than reward in producing punctuality from the child. It's a touching testament to their own beliefs about the world.
These people were convinced that their actions had an impact on the punctuality of the child, even though the arrival time was entirely random. The joy is, you have no way of knowing how many areas of your life this experiment might be relevant to. Now I'm going to dangle some goat entrails around my neck and get chanting. | 704 | 3,256 | {"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} | 2.625 | 3 | CC-MAIN-2015-35 | latest | en | 0.972054 |
https://www.advance-africa.com/KCSE-Past-Papers-2018-Physics-Paper-3.html | 1,611,242,381,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524858.74/warc/CC-MAIN-20210121132407-20210121162407-00279.warc.gz | 654,138,923 | 11,874 | # KCSE Past Papers 2018 Physics paper 3 (232/3)
## Physics - Paper 3 - November 2018 - 2.30 Hours
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Practical
2018 Physics paper 3
Question 1
You are provided with the following:
• Some water in a container
• A 10 ml measuring cylinder
• A piece of a glass rod
• A 10 g mass
• 5 paper clips
• A half metre rule
• A metre rule
• Two stands, two bosses and two clamps Three pieces of sewing thread
Proceed as follows:
(a) Pour 6 ml of the water into the measuring cylinder. Lower the glass rod into the water and determine the volume V of the glass rod.
V=............................cm3(1 mark)
Remove the glass rod from water.
(b) Using a stand and a piece of string, suspend the half metre rule at its centre of gravity C so that it balances horizontally with the scale facing you.
Using a second stand, clamp a metre rule vertically near one end of the half metre rule to note the height at which the half metre rule is horizontal.
Maintain this height throughout the experiment
Record the centimetre mark of the centre of gravity C.
C........................cm(1 mark)
(c) Using the string, suspend the 10 g mass on the half metre rule at a distance
d = 2 cm from C. The distance d = 2 cm should be maintained throughout the experiment.
Balance the half metre rule by suspending the glass rod using a string at a distance X from C.
Record the value of X
X=.........................cm(1 mark)
(d) Using the results in part (a) and (c) determine the;
(i) mass of the glass rod, (2 marks)
(ii) density of the glass rod. (2 marks)
(e) Remove the glass rod. Push the half metre rule through one paper clip and adjust the position of the clip to a point P where the half metre rule balances horizontally. See Figure 1.
(f) Record the centimetre mark for point P in Table 1.
(g) Repeat part (f) for the other number of clips shown in Table 1 and complete (the clips may be suspended by connecting them us a chain of the required number) (4 marks)
(h) Plot a graph of l/L(y axis) against the number of clips N.(4 marks)
(i) Determine the slope S of the graph.(3 marks)
(j) Determine K given that 1/L = 0.05 KN(2 marks)
Question 2
You are provided with the following:
(i) A voltmeter
(ii) A resistor labelled 10 1
(iii) A resistance wire mounted on a half metre rule labelled X
(iv) Two cells in a cell holder
(v) A switch
(vi) Eight connecting wires
(vii) A micrometer screw gauge
(viii) A resistor labelled 10 KB
(ix) A galvanometer
(x) A beaker containing a liquid labelled L
(xi) Two copper plates
(xii) A resistance wire labelled AB and mounted on a millimetre scale
(xiii) A jockey
(xiv) A vernier calliper Proceed as follows:
Part A
(a) Measure and record the diameter d of the resistance wire x
d= ...............mm
=.................m
(b) Set up the circuit as shown in Figure 2.
(i) Close the switch and record the potential difference V across the 10a resistor. (1 mark)
(ii) Open the switch. Determine the current I flowing in the circuit. (2 marks)
(c) (i) Now connect the voltmeter across wire X. Close the switch and record the potential difference V2 across wire X.
v1.............................
(ii) Determine the resistance R of wire X.(2 marks)
(iii) Determine K the resistance per metre of wire X.(1 marks)
(iv) Determine Q given that Q = n Kd2/4 (where d is in metres).(1 marks)
Part B
(d) (i) Using the vernier callipers measure and record the width W of one of the copper plates
W =............(1 mark)
(ii) Determine the area A of a 5 cm length of the copper plate
A =......................... .. cm2(1 mark)
(e) Using stands and clamps, hold the copper plates in the beaker such that both plates:
(i) reach the bottom of the beaker;
(ii) are parallel, vertical and facing each other;
(iii) are separated from each other by a distance S.
(f) Connect the copper plates to the circuit as shown in Figure 3.
(g) Set the separation distance between the copper plates S to 3 cm. Using the jockey tap wire AB at various points to obtain a point P at which the galvanometer does not show any deflection. Record the balance length L (from A to P) in Table 2.
(h) Repeat part (g) for other values of S shown in Table 2 and complete the table. (6 marks)
(i) Determine the average value of Z. (2 marks)
2018 Physics paper 3
Question 1
You are provided with the following:
• Some water in a container
• A 10 ml measuring cylinder
• A piece of a glass rod
• A 10 g mass
• 5 paper clips
• A half metre rule
• A metre rule
• Two stands, two bosses and two clamps Three pieces of sewing thread
Proceed as follows:
(a) Pour 6 ml of the water into the measuring cylinder. Lower the glass rod into the water and determine the volume V of the glass rod.
V=3.4 cmcm3± 0.2(1 mark)
Remove the glass rod from water.
(b) Using a stand and a piece of string, suspend the half metre rule at its centre of gravity C so that it balances horizontally with the scale facing you.
C = 25.3 cm ± 0.1
Using a second stand, clamp a metre rule vertically near one end of the half metre rule to note the height at which the half metre rule is horizontal.
Maintain this height throughout the experiment
Record the centimetre mark of the centre of gravity C.
CX = 2.2 cm ± 0.1 cm(1 mark)
(c) Using the string, suspend the 10 g mass on the half metre rule at a distance
d = 2 cm from C. The distance d = 2 cm should be maintained throughout the experiment.
Balance the half metre rule by suspending the glass rod using a string at a distance X from C.
Record the value of X
X=.........................cm(1 mark)
(d) Using the results in part (a) and (c) determine the;
(i) mass of the glass rod, (2 marks)
10 x 2 = mx 2.2
M =m/v 10 x 2/2.2
= 9.1g ± 1
(ii) density of the glass rod. (2 marks)
ƿ =m/ v
9.1/3.4
=2.68 gcm -3
(e) Remove the glass rod. Push the half metre rule through one paper clip and adjust the position of the clip to a point P where the half metre rule balances horizontally. See Figure 1.
(f) Record the centimetre mark for point P in Table 1.
(g) Repeat part (f) for the other number of clips shown in Table 1 and complete (the clips may be suspended by connecting them us a chain of the required number) (4 marks)
(h) Plot a graph of l/L(y axis) against the number of clips N.(4 marks)
(i) Determine the slope S of the graph.(3 marks)
(j) Determine K given that 1/L = 0.05 KN(2 marks)
Question 2
You are provided with the following:
(i) A voltmeter
(ii) A resistor labelled 10 1
(iii) A resistance wire mounted on a half metre rule labelled X
(iv) Two cells in a cell holder
(v) A switch
(vi) Eight connecting wires
(vii) A micrometer screw gauge
(viii) A resistor labelled 10 KB
(ix) A galvanometer
(x) A beaker containing a liquid labelled L
(xi) Two copper plates
(xii) A resistance wire labelled AB and mounted on a millimetre scale
(xiii) A jockey
(xiv) A vernier calliper Proceed as follows:
Part A
(a) Measure and record the diameter d of the resistance wire x
d=0.36mm۪.005
3.6X10-4M
(b) Set up the circuit as shown in Figure 2.
(i) Close the switch and record the potential difference V across the 10a resistor. (1 mark)
V1= 1.7 V ± 0.2
(ii) Open the switch. Determine the current I flowing in the circuit.(2 marks)
R=V/1
=1/0.17
=5.88 ☊
(c) (i) Now connect the voltmeter across wire X. Close the switch and record the potential difference V2 across wire X.
V2=1.0v۪.2
(ii) Determine the resistance R of wire X.(2 marks)
v1K=5.88/0.5
11.76☊m-1
(iii) Determine K the resistance per metre of wire X.(1 marks)
(iv) Determine Q given that Q = n Kd2/4 (where d is in metres).(1 marks)
Part B
(d) (i) Using the vernier callipers measure and record the width W of one of the copper plates
W =5.0cm± 0.10
(ii) Determine the area A of a 5 cm length of the copper plate
A =5X50 cm2
=25.00cm2
(e) Using stands and clamps, hold the copper plates in the beaker such that both plates:
(i) reach the bottom of the beaker;
(ii) are parallel, vertical and facing each other;
(iii) are separated from each other by a distance S.
(f) Connect the copper plates to the circuit as shown in Figure 3.
(g) Set the separation distance between the copper plates S to 3 cm. Using the jockey tap wire AB at various points to obtain a point P at which the galvanometer does not show any deflection. Record the balance length L (from A to P) in Table 2.
(h) Repeat part (g) for other values of S shown in Table 2 and complete the table. (6 marks)
(i) Determine the average value of Z. (2 marks)
Z average = 325.73+295.43+315.63/3
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Number 1188: mathematical and symbolic properties | Crazy Numbers
Discover a lot of information on the number 1188: properties, mathematical operations, how to write it, symbolism, numerology, representations and many other interesting things!
## Mathematical properties of 1188
Is 1188 a prime number? No
Is 1188 a perfect number? No
Number of divisors 24
List of dividers
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1, 2, 3, 4, 6, 9, 11, 12, 18, 22, 27, 33, 36, 44, 54, 66, 99, 108, 132, 198, 297, 396, 594, 1188
Sum of divisors 3360
Prime factorization 22 x 33 x 11
Prime factors
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2, 3, 11
## How to write / spell 1188 in letters?
In letters, the number 1188 is written as: One thousand one hundred and eighty-eight. And in other languages? how does it spell?
1188 in other languages
Write 1188 in english One thousand one hundred and eighty-eight
Write 1188 in french Mille cent quatre-vingt-huit
Write 1188 in spanish Mil ciento ochenta y ocho
Write 1188 in portuguese Mil cento oitenta e oito
## Decomposition of the number 1188
The number 1188 is composed of:
2 iterations of the number 1 : The number 1 (one) represents the uniqueness, the unique, a starting point, a beginning.... Find out more about the number 1
2 iterations of the number 8 : The number 8 (eight) represents power, ambition. It symbolizes balance, realization.... Find out more about the number 8
Other ways to write 1188
In letter One thousand one hundred and eighty-eight
In roman numeral
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MCLXXXVIII
In binary 10010100100
In octal 2244 | 737 | 2,430 | {"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-2024-33 | latest | en | 0.725995 |
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# Diffusion equation
The diffusion equation is a partial differential equation which describes density fluctuations in a material undergoing diffusion. It is also used to describe processes exhibiting diffusive-like behaviour, for instance the 'diffusion' of alleles in a population in population genetics.
The equation is usually written as:
$\frac{\partial\phi}{\partial t} = \nabla \cdot \bigg( D(\phi,\vec{r}) \, \nabla\phi(\vec{r},t) \bigg),$
where $\, \phi(\vec{r},t)$ is the density of the diffusing material at location $\vec{r}$ and time t and $\, D(\phi,\vec{r})$ is the collective diffusion coefficient for density φ at location $\vec{r}$; the nabla symbol $\, \nabla$ represents the vector differential operator del acting on the space coordinates. If the diffusion coefficient depends on the density then the equation is nonlinear, otherwise it is linear. If $\, D$ is constant, then the equation reduces to the following linear equation:
$\frac{\partial\phi}{\partial t} = D\nabla^2\phi(\vec{r},t),$
also called the heat equation. More generally, when D is a symmetric positive definite matrix, the equation describes anisotropic diffusion.
## Derivation
The diffusion equation can be derived in a straightforward way from the continuity equation, which states that a change in density in any part of the system is due to inflow and outflow of material into and out of that part of the system. Effectively, no material is created or destroyed:
$\frac{\partial\phi}{\partial t}+\nabla\cdot\vec{j}=0$,
where $\vec{j}$ is the flux of the diffusing material. The diffusion equation can be obtained easily from this when combined with the phenomenological Fick's first law, which assumes that the flux of the diffusing material in any part of the system is proportional to the local density gradient:
$\vec{j}=-D\,(\phi)\,\nabla\,\phi\,(\,\vec{r},t\,)$. | 467 | 1,906 | {"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": 11, "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} | 3.390625 | 3 | CC-MAIN-2024-33 | latest | en | 0.8521 |
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Question
# Find the coordinates of the intersection of the lines 2x−3y+1=0 and 2x+4y+5=0.
A
(1914, 47)
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B
(1914, 47)
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C
(1914, 47)
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D
(1914, 47)
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Solution
## The correct option is D (−1914, −47) 2x−3y+1=0 - - - - - (1) 2x+4y+5=0 - - - - - (2) Equation (1) can be written as 2x−3y=−1 - - - - - (3) Equation (2) can be written as 2x+4y=−5 - - - - - (4) (3) - (4) ⟹ −7y=4 ⟹ y=−47 Substituting y=−47 in equation (3), 2x−3(−47)=−1 ⟹2x+127=−1 ⟹2x=−1−127 ⟹x=−1914 ∴ The coordinates of the intersection of the lines 2x−3y+1=0 and 2x+4y+5=0 are (−1914, −47).
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https://www.jiskha.com/search?query=every+digit+in+8%2C999+is+greater+than+any+digit+in+24%2C005.explain+why+24%2C005+is+greater+than+8%2C999. | 1,632,384,549,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057417.92/warc/CC-MAIN-20210923074537-20210923104537-00513.warc.gz | 859,566,219 | 11,250 | # every digit in 8,999 is greater than any digit in 24,005.explain why 24,005 is greater than 8,999.
38,349 results
1. ## Mathematics
What is true of the number 55,555 A. Each digit has value less than the value of the digit time it's right and greater than the value of the digit itself. B. Each digit has a value 10 times the value of the digit to it's right and 10 times the value of the
2. ## mathematics
I am five digit number greater than 40,000 but less than 60,000.My ones digit is 3 more than my ten thousands digit.All my numbers are the same.The sum of my digit is 28.What number I am?
3. ## math
list all the 3 digit numbers that fit these clues. the hundreds digits is less than 3. the tens digit is less than 2. the ones digit is greater than 7
4. ## math
Form the greatest possible 5-digit number using the clues.All five digits are different. None of the five digit are 1.The digit in the ten thousands place is greater than 7.The sum of all five digit is 18.The greatest digit is equal to the sum of the other
5. ## Math
The greatest place value is hundred thousand. The digit in the tens place is 5.the remaining digit ate 2, 3, 8 and 7. One of the digit has a value of 30,000. One of the digit had a value of 800. The value of the digit in the thousands place is 10 times
6. ## math
use the clues to find the 7th digit mystery number. * it is between 7 and 8 million * it contains all of the digits greater than 1 and less than 8 * the hundred thousands digit is 3 * the ten thousands digit and the thousand digits are the same * the tens
7. ## Math
It is a 6 digit number the least digit is in the thousands place. The greatest digit is in the ones place. The digit in the tens place is 5 less than the digit in the ones place. The digit in the hundred thousands place is greater than the digit in the
8. ## maths
I am a three digit number. My tens digit is twice my ones digit, my hundreds digit is twice my tens digit. I am greater than 500. Who am i?
9. ## Mathe
The number has two digits. Both of the digits are even.the digit in the tens place is greater that the digit in the unit (or ones) place.the units (or ones digit) is not in the three times table. The tens digit is not double the ones digit. The sum of the
10. ## Algebra 1
How many two-digit numbers have a tens digit that is greater than the units digit? Explain.
11. ## Math
Use the clues to figure out the six-digit decimal number. 1. It is greater than 3,000 and less than 3,200 2. All the digits are different 3. The digit in the undredths place isa multiple of 5. 4. The product of 5 and th digit in the ones place is 0. 5. The
12. ## Math
It is a 7-digit number It has a digit 0. The greatest digit is in the hundred thousands place. The value of the digit 1 is 1,000,000. The digit 6 stands for 6,000. The value of the digit 5 is 5 ones. The digit 8 has a value greater than 700 but less than
13. ## math
I am a 4 digit number greater than 9,000 my omes digit is 3 and my hundreds digit is the smallest prime number greater than 5 the sum of my digits is 20. What number am i?
14. ## math
``My hundreds digit is 4 less than my tens digit. my ten thousands digit is 6 greater than my hundreds digit. my thousand digit is 5 less than my ten thousand digit and my ones digit is 6 greater than my hundred digit. if my tens digit is 7, what number am
15. ## Math
if the last digit of a weight measurement is equally likely to be any of the digits 0 through to 9.what is the probability that the last digit is greater than or equal to 5 ?
16. ## math
A four-digit number is selected at random. What is the probability that its ones place is greater than its tens place, its tens place is greater than its hundreds place, and its hundreds place is greater than its thousands place? Note that the first digit
17. ## Math
It is a 6 dagit number. The least digit is in the thousands place. The greatest digit is in the ones place. The digit in the tens place is 5 less than the digit in the ones place. The digit in the hundred thousands place is greater than the digit in the
18. ## Math Decimals
Jason has 4 tiles. Each tile has a number printed on it. The numbers are 2,3,6, and 8. A decimal number is formed using the tiles and the clues. Be a math detective and find the number. Clues: digit in tens place is the greatest number. digit in tenths is
19. ## math
I am five digit number greater than 40,000 but less than 70,000.My ones digit is more than my ten thousands digit.All my other digit are the same.The sum of my gigit is 19.What number am i?
20. ## Mathematics
The sum of the digits of a 3-digit number is three times the sum of the digits of a 2-digit number. When the 2-digit number is subtracted from the 3-digit number it gave another 2-digit number of which its first digit is three times the first digit of the
21. ## math
i am a number greater than 70000 but less than 90000. My ten digit is the same as my ones digit. My ten digit is the same as my ones digit. My ten thousand digit is four times my ones digits. Both my hundreds and thousands are half the ten thousand digit.
22. ## math
need help with math word problems. just don't understand. What is the number? you have a four digit number where the thousands digit is four greater than the hundreds digit.The tens digit is twice the thousands digit. The ones digit is half of the
23. ## algebra
It is a five digit whole number 1. It is a palindrome 2. It's hundreds digit is a multiple of 5 3. The product of its one digit and 8 is 12 greater than the square of its ones digit 4. The product of its thousands digit and 18 is 45 greater than the square
24. ## stonegate math
I m a 4- digit number. My ones digit is 2. My ten digit is 4 great than my ones digit. My hundreds and thousands digit are both 3 greater than my ones digit. What number am I
25. ## math
read the clues to find each number. it is a 7 digit number. it has a 0. the greatest digit is in the hundered thousands place. the value of the digit 1is i,000,000. the digit 6 stands for 6,000. the value of the digit 5 is 5 ones. the digit 8 has a value
26. ## math
find the number of two digit natural numbers such that the tens digit is greater than the units digit.
27. ## word problem
the different between the digit of a two digit number is 1, the number it self is 1 more than 5 times the sum of the digits, if the unit digit is greater than the tens digit, find the number.
28. ## math
I am having problems with this problem.In the mythical kingdom of Permutatus, wizards have special 2-digit license plats, The first digit is greater than 6, and the second digit is odd. The two digits cannot be the same, How many different license plates
29. ## math
My thousands digit is twice my hundreds digit. The sum of my tens and ones digit is my thousands digit. My hundreds digit is 2 more than my tens digit. My tens digit is greater than 1 and less than 3. What 4 digit number am I?
30. ## Mathematics
The sum of the digits of a 3-digit number is three times the sum of the digits of a 2-digit number. When the 2-digit number is subtracted from the 3-digit number it gave another 2-digit number of which its first digit is three times the first digit of the
31. ## Mathematics:help me please anyone
The sum of the digits of a 3-digit number is three times the sum of the digits of a 2-digit number. When the 2-digit number is subtracted from the 3-digit number it gave another 2-digit number of which its first digit is three times the first digit of the
32. ## singapore math
My 3rd grade son get this math problem. It states that John thinks of a 3-digit number. What is his number? Use the clues below to find John's number. Every digit is different. The sum of all thedigits is 19. The difference between the hundreds digit and
33. ## math
The number uses every number 0-9 the numbers are used only once the fourth digit is 4 the first digit in the billions place is 3 the 5 is next to the last digit the sixth digit is 7 the digit after 3 is 9 the digit before 5 is 2the third digit is half the
34. ## math
The unit digit of a 2-digit number is greater than the tens digit by 3. If the product of the digit is 18 find the number
35. ## How many of these number maths??
(a):how many digit number can be formed with the digit 2,2,2,3,3,4,5 (b):how many of these are greater than 3400000 (c):how many are greater than 3400000 and are divisible by 5 (d):how many are greater than 3400000 and are even ????
36. ## math
Why is 24,005 greater than 8,999?
37. ## Math
How many positive three-digit integers have the property that the tens digit is greater than or equal to the hundreds digit and the ones digit is greater than or equal to the tens digit? Please help :) thanks -Zach
38. ## Mathematics
The sum of the digits of a 3-digit number is three times the sum of the digits of a 2-digit number. When the 2-digit number is subtracted from the 3-digit number it gave another 2-digit number of which its first digit is three times the first digit of the
39. ## Math
My hundreds digit is 4 less than my tens digit. My ten thousand digit is 6 greater than my hundreds digit. My thousands digit is 5 less than my ten thousands digit and my ones digit is 6 greater than my hundreds digit. If my tens digit is 7, Question: What
40. ## Math
i am a four digit numeral number between 3000 and 4000.My ones digit is 3 greater than my thousands digit.The two other digits are the same.If the sum of my digit is 20,what number am i?
41. ## Math
If my clothes are like this the hundred digit is greater than 7 the tens is one more than the hundreds digit and the ones digit is less than 2 then what would my answer be
42. ## Mathematics
aminah thinks of a 4-digit number. what is the number? every digit is different. the sum of the thousands digit and the tens digit is 10. the sum of all the digits is 16. the difference between the hundreds digit and the ones digit is 6. the difference
43. ## math
How do you write the following digit:five thousandths.Does this right,0.005
44. ## FIND THE VALUE OF SQUARE
I AM A NUMBER GREATER THAN 99 AND LESS THAN 1,000. TWO OF MY DIGITS THAT AREN'T NEXT TO EACH OTHER ARE THE SAME. MY TENS DIGIT COULD NOT BE GREATER AND IT IS 1 MORE THAN MY HUNDREDS DIGIT. WHAT NUMBER AM I?
45. ## arithmetic
I am greater than 200 but less than 500. The sum of my digits is 18. the product of my tens place digit and my 100's place digit equal 20. What digit am I ???
46. ## math
how many 2 digit numbers can be written so that the second digit is greater than the digit before it ?
47. ## Math
I am 4-digit odd number. I am greater than 7000. The digit in my hundreds place is smaller than the digit in my tens place. What number can I be if I am made up of all the digits shown below? 6827
48. ## Math(Please check)
This is the same equation as before just with different numbers. y=yo + (vo sin Q) t - 1/2gt^2 This time the numbers are: 0 = 1.005 + (3.021 sin 30)t - 1/2(9.8)t^2 4.9t^2 - 1.5105 - 1.005 = 0 This is the part where I am stuck. Did I do this correctly? sin
49. ## Maths
The sum of my ones digit and tens digit is 10. My tens digit is greater than my ones digit I am a prime number. What number am I ? (Please answer before 5 pm)
50. ## Math
My tens digit is double the ones digit,Iam less than 50 and greater than 40...What number am I?
51. ## Math
Reasoning: What number am I? my tens digit is double my ones digit. I am less than 70 and greater than 60
52. ## math
every digit in 8,999 is greater than any digit in 24,005.explain why 24,005 is greater than 8,999.
53. ## Writing in Math
Every digit in 8,999 is greater ,than any digit in 24,005 is greater than 8,999.
54. ## 3 grade
explain why 24,005 is greater than 8,999.
55. ## math
explain why 24,005 is greater than 8,999
56. ## math
why is 24,005 greater than 8,999
57. ## math
It is between 7 and 8 million it contains all the digits greater then 1 and less then 8 the hundred thousands digit is 3 the ten thousands digit are the same their sum is the ones digit the tens digit is greater then the hundreds digit
58. ## math
i am a 6-digit number greater than 700,000.my hundred thousands digit is twice my ones digit,my thousands digit,and my hundreds digit.my remaining digits are the same-a prime number greater than 5
59. ## decimals
a number has four digits. the digit in the hundredth place is greater than the digit in the hundreths place of 4.361. the digit in the thousandths place is greater than the digit in the tenths place of 2.85. what are the possible four-digit numbers between
60. ## Mathematics some do something
The sum of the digits of a 3-digit number is three times the sum of the digits of a 2-digit number. When the 2-digit number is subtracted from the 3-digit number it gave another 2-digit number of which its first digit is three times the first digit of the
61. ## math
i have a six digit question my 100 thousands is one less than my ones digit,eight greater than my thousands digit,twice my tens digit and four times my hundreds digit,my ten thousand digit is zero,what number am i
62. ## math
I am 5 digit greater 60000, but less than 70000.My ones digit is less than my 10000 digit on my other digit are the same.The sum of digit is 23.What is my number
63. ## Math- Problem solving
A number has 4 digits. The digit in the hundredths place us greater than the digit in the hundredths place of 4.361. The digit in the tousandths place is greater than the digit in the tenths place of 2.85. What are the possible four-digit numbers between
64. ## algebra 2
a two-digit counting number has a value that is 13 greater than 3 times the sum of the digits. If the units digit is 1 greater than the tens digit, what is the number
65. ## math
i am a 5 digit palindromic number. My third digit is 5 more than my fifth digit , my fourth digit is 2 times greater than my first digit. The sum of my digit is 14 What number am i?
66. ## math
a three digit number has ten digit two greater than the units digit and the hundreds digit one greater than the tens digit. the sum of the tens and the hundreds is three times the units digit. what is the number
67. ## math
Im greater than 99 but less than 1000, two of my digits that are not next to each other are the same, my tenth digit can't be greater and is more than my hundreths digit. What number am I?
68. ## math
i'am a four digit numeral between 3000 and 4000.My ones digit is 2 greater than my thousands digit. the two digits are the same. If the sum of my digit is, 20, what number am i in roman numeral?
69. ## math
My hundredths digit is the product of 3 x 2. My tenths digit is the multiple of 3 that comes before the digit that is in the hundredths place. My thousandths digit could not be any greater. What number am I?
70. ## Math
What is the theoretical probability of a computer choosing a 2 digit number between 1 and 100 whose ones digit is greater than the tens digit?
71. ## berta jrhs
The sum of my digit is 0. My digt are not the same.i am an odd number.one of my digit is a multiple of 3,the other is not. I am greater than 50.the difference between my digit is less than 5. What am i?
72. ## Math
The sum of my tens and hundreds digit is my thousands digit. My hundreds digit is 1 less than my tens digit. My ones digit is greater than 7 and a multiple of 4. My tens digit is is 3 less than my ones digit. What is my number?
73. ## Math
I am a four digit number the tens digit is greater than the ones and hundreds digit the digits have a sum of 15
74. ## Math
My hundreds digit is 4 less than my tens my tens digit. My ten thousand digit is 6 greater than my hundreds digit. My thousands digit is 5 less than my ten thousands digit and my ones digit is 6 greater than my hundreds digit. If my tens digit is 7, what
75. ## maths
A number of two digit is such that four times the unit digit is five greater than trice the tense digit.when the digit are reserved,the number is increased by nine find the number.
76. ## 3rd math
okay Math genius! Help me out here. I am a number greater than 99 and less than 1,000. Two of my digits that aren't next to each other are the same. My tens digit could not be greater and it is 1 more than my hundreds digit. What number am I?
77. ## Algebra
Let P0 = \$1000.00. • k = 0.5% = _____0.005______ as a decimal. Compute and explain how to find the doubling time. Then, write the compound interest equation for this value of k and the 3 values for t. Number of years Equation New Value 1
78. ## maths I need answer
A 2-digit number has a greater tens digit than the ones digit. The difference between the tens and ones digit is 3. The sum of the two digits is 15. Find the 2-digit number.
79. ## word problem
the different between the digit of a two digit number is 1, the number it self is 1 more than 5 times the sum of its digits, if the unit digit is greater than the tens digit, find the number.
80. ## math
Determine how many 1000 digit numbers A have the following property: When any digit of A, aside from the first, is deleted to form a 999 digit number B, then B divides A.
81. ## Algebra 1
How many two-digit numbers have a tens digit that is greater than the units digit?
82. ## math
a 4 digit number my 2 nd digitis twice my 3 digit the sum of all my digits is 3 times greater than my last digit the productof my 3rd and 4th digits is 12 times greater than the ratio of my 2nd to 3rd whats my number?
83. ## math
i am a 2 digit number my ones digit is greater than than my tens digit my ones digit is 4 times my tens digit the sum of my digits is 10 what number am i
84. ## General Mathematics
Use a table of t-values to estimate the P-value for the specified one-mean t-test. Left-tailed test, n = 12, t = -3.412 Choose one below: P < 0.005 P > 0.005 0.005 < P < 0.01 0.01 < P < 0.025
85. ## math
I am a three-digit number the sum of my digits is eighteen Five letters are used to spell my second number The difference between my first two digits is one. I have fewer even digits than odd digits My second digit is greater than my first digit I am a
86. ## problem solving
three digit odd number divisible by 9 digits in consecutive order and digit in hundreds place is greater than the digit in the ones place?
87. ## math
I am five digit number greater than 60 000 but less than 70 000: my ones digit is 1 less than my ten thousands digit. All my digits are the same. The sum of my didits is 23.
88. ## Chemistry
If 25mL of 2.00M NAOH is added tpo 50mL of 0.10M HC2H3O2 (Ka=1.8*10^-5), what is the pH? moles HC2H3O2= c*v = .1 * .05L = .005 moles NaOH= c*v = .2 * .025L = .005 pH= Pka + log (B/A) = 4.75 + log (.005/.005) = 4.75 I'm not sure how to calculate the moles I
89. ## statistics
Use a table of t-values to estimate the P-value for the specified one-mean t-test. Left-tailed test, n=12, t= -3.412. a=p0.005. c= 0.005
90. ## math
this code has five digit 1,2,3,4,5. the first digit plus the second digit is equal to the third digit the second digit twice times the first digit the second digit is half the fourth digit and the fifth digit is the sum of the first and fourth digits. what
91. ## math
i am a digit greater than 20000, my first 2 & last digits are the same,the middle digits are the same, 1 digit is twice the other digit,i am an odd number
92. ## St.vincent
The sum of my tens and hundreds digit is greater than 7 and a multiple of 4. My tens digit is 3 less than my one's digit.what is my number?
93. ## Math - Algebra
What is the linear equation that will represent the second statement in the following problem: "The sum of the digits of a three-digit number is 12. If the hundreds digit is replaced by the tens digit, the tens digit by the units digit, and the units digit
94. ## math statistics and probability
how many 4 digit number can be formed from the digits 2,3,5,,6,7,8 if: a.)each digit be only used once b) how many are even? c) how many are odd? d) how many are greater than 25000?
95. ## Math
How many times greater is the value of the digit in 1 in 183,607 than the value of the digit in 1 in 362,105, can someone help
96. ## math (statistics)
how many 4 digit number can be formed from the digits 2,3,5,,6,7,8 if: a.)each digit be only used once b) how many are even? c) how many are odd? d) how many are greater than 25000?
97. ## Math
How many times greater is the value of the digit 5 in 583,608 than the value of the digit 5 in 362,501
98. ## algebra
Two factors, each with no digit greater than 5, have a product of 16,848. What is the largest possible 3-digit factor satisfying these conditions?
99. ## Math
It is a 5 digit no. ,palindrome,divisible by 4,10 digit is cube root of 1 digit,product of 100 digit and 1 digit is 54 ,sum of hundred and ones digit is 15.what is the number?
100. ## algebra
How many 5 digit number greater than 24000 can be formed if repetition of digit is allow. | 5,586 | 21,000 | {"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-2021-39 | latest | en | 0.861077 |
https://math-semester-final.wonderhowto.com/how-to/pythagorean-theorem-0147093/ | 1,723,134,978,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640736186.44/warc/CC-MAIN-20240808155812-20240808185812-00578.warc.gz | 302,181,872 | 27,349 | # How To: Pythagorean Theorem
## What Is Pythagorean Theorem?
Pythagorean Theorem is an equation discovered by a man named Pythagoras. It can be used to determine whether or not a triangle is a right triangle and find missing lengths of a triangle. The equation is:
## Parts of a Triangle
The hypotenuse is the longest side of the triangle and is opposite of the right triangle. There are also legs to a triangle.
Using the Pythagorean Theorem, 'a' and 'b' are the lengths of the two legs of a right triangle, while 'c' is the length of the hypotenuse.
## Is the Triangle a Right Triangle?
Let's say that in our example, we are told to determine whether the triangle is a right triangle or not. A picture of the triangle and its side lengths is given to us.
To figure out if the triangle is a right triangle or not, we need to use the Pythagorean Theorem. We use the equation a^2 +b^2 = c^2. Using the side lengths we are given, we know that a = 5, b = 12, and c = 13. Now all we have to do is complete the current equation which is 5^2 + 12^2 = 13^2. After solving the equation, our final answer is 169 = 169. Since a^2 + b^2 does equal c^2, we know that this triangle is a right triangle.
In the next example, we will also determine whether or not the triangle is a right triangle. We are given a picture of the triangle and its lengths.
From the picture, we know that a = 9, b = 12, and c = 14 and using the Pythagorean Theorem, we know that the equation that we need to solve is 9^2 + 12^2 = 14^2. Once it is solved, we know that 225 does not equal 196, meaning that this triangle is NOT a right triangle.
## Finding Missing Side Lengths
The Pythagorean Theorem can also be used to find missing side lengths. In this example, we can gather information given to us from the picture below.
We know that a = 12, b = x, and c = 13. We plug all of the numbers into a^2 + b^2 = c^2, the Pythagorean Theorem. We need to find out what b is equal to by solving the equation, 12^2 + b^2 = 13^2. After making calculations, we find out that 12^2 is equal to 144 and 13^2 is equal to 169. The current problem is now 144 + b^2 = 169. We need to get 'b' by itself, also known as isolating the variable. To do so, we subtract 144 - 144 and 169 - 144, since whatever you do on the side of the equal sign also has to be done on the other side of the equal sign. Now, b^2 = 25. How do we find out what 'b' is when we see that it's squared? Well, it's very simple actually. The opposite of squaring is finding the square root, so that is what we are going to do. We find the square root of b^2 and the square root of 25. Our final answer turns out to be b = 5. Since we know that b = x, we can plug in the number 5 for x.
## Real Life Example
Let's say you want to find the diagonal length of a television screen. The length is 4 inches while the width is 9 inches. We can find the diagonal length by using the Pythagorean Theorem. a^2 + b^2 = c^2. C is the unknown diagonal length. a = 4 and b = 9. The equation is 4^2 + 9^2 = c^2. We get 16 + 81 = c^2, which is then simplified to 97 = c^2. We find the square root of 97 and c^2, which is then 9.84885780179615 = c. We can round the number to the nearest hundreth. Our final answer is 9.85 = c. We now know that the diagonal length of the television is approximately 9.85.
## Practice Time
For more practice, play the following games
http://www.math-play.com/Pythagorean-Theorem-Game.html
http://www.kidsnumbers.com/pythagorean-theorem-game.php
http://www.math-play.com/Pythagorean-Theorem-Jeopardy/Pythagorean-Theorem-Jeopardy.html
## Citations
http://www.mathsisfun.com/pythagoras.html
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• Hot
• Latest | 1,080 | 3,878 | {"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} | 4.9375 | 5 | CC-MAIN-2024-33 | latest | en | 0.922956 |
https://www.programmingwithbasics.com/2015/11/write-program-to-check-primeness-of.html | 1,722,896,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640455981.23/warc/CC-MAIN-20240805211050-20240806001050-00744.warc.gz | 753,178,803 | 32,349 | # C++ Program To Check Primness of 1D Array
Problem:- Program To Check Elements of a 1-D Integer An Array For Primness. Means in array your task is to find the prime number in an array or C++ Program To Check Primness of an Array.
Check This:- Hacker rank solution for Strings, Classes, STL, Inheritance in C++.
Logic:- To Check primness of array we need to know that how to find the prime number then use a loop and run the program that's it we solve the problem. So Prime number is a number which only can be divided by the 1 or number itself, here are some prime number 2, 3, 5, 7, 11, 13, 17, 19. Important notice here 1 is not a prime number. Check Here C++ Program To Find Number Is Prime Or Not.
Also Check:- Geeksforgeeks solution for School, Basic, Easy, Medium, Hard in C++.
Solution:-
#include<iostream>
using namespace std;
int main()
{
int a[100],i,n,k,j,flag;
cout<<"Enter The Size of Array\n";
cin>>n;
cout<<"Enter The Element\n";
for(i=0;i<n;i++)
{
cin>>a[i];
}
cout<<"\n\n";
for(i=0;i<n;i++)
{
flag=0;
k=a[i]/2;
for(j=2;j<=k;j++)
{
if(a[i]%j==0)
{
flag=1;
break;
}
}
if(flag==1)
cout<<a[i]<<" is Not Prime\n";
else
cout<<a[i]<<" is Prime\n";
}
return 0;
}
Output:-
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Hi, I’m Ghanendra Yadav, SEO Expert, Professional Blogger, Programmer, and UI Developer. Get a Solution of More Than 500+ Programming Problems, and Practice All Programs in C, C++, and Java Languages. Get a Competitive Website Solution also Ie. Hackerrank Solutions and Geeksforgeeks Solutions. If You Are Interested to Learn a C Programming Language and You Don't Have Experience in Any Programming, You Should Start with a C Programming Language, Read: List of Format Specifiers in C. | 477 | 1,744 | {"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-2024-33 | latest | en | 0.708841 |
https://homeworkarea.com/solutions/analyse-the-outputs-and-outcomes-of-models-computer-based-and-non-computer-based-in-terms-of-their-managerial-implications/ | 1,618,163,712,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038064898.14/warc/CC-MAIN-20210411174053-20210411204053-00292.warc.gz | 406,175,172 | 7,985 | # Analyse the outputs and outcomes of models (comput
## Analyse the outputs and outcomes of models (computer based and non-computer based) in terms of their managerial implications.
The primary theme of the paper is Analyse the outputs and outcomes of models (computer based and non-computer based) in terms of their managerial implications. in which you are required to emphasize its aspects in detail. The cost of the paper starts from \$79 and it has been purchased and rated 4.9 points on the scale of 5 points by the students. To gain deeper insights into the paper and achieve fresh information, kindly contact our support.
Learning Outcomes for the module:
a.Formulate problems in terms of models to support decision-making, both from qualitative and quantitative perspectives.
b.Analyse the outputs and outcomes of models (computer based and non-computer based) in terms of their managerial implications.
1.Draw the cognitive map, without potential actions, to show the linkages between the five attributes listed in the paper, with “total inventory cost” at the top of the map. Explain the rationale for your cognitive map.
2.Discuss the advantages and disadvantages of generating weights by individual interviews (and combining the results) or by a group-based method.
3.Using your individual data set provided on Blackboard, calculate the scores for each of the SKUs in an EXCEL spread sheet, using the approach described in the paper, and the specific weights listed in Table 2. Put the scores in order, identifying the top 10%, the next 20% and the remaining 70%. Discuss how you would work with the managers in the company to “check for plausibility” (see case study page 212-213).
4.Extend the EXCEL spread sheet to calculate the overall weighted Symmetric Mean Absolute Percentage Error (SMAPE) for each of the three groups
Suppose forecasting method A has a lower overall unweighted SMAPE than method B, but a higher weighted SMAPE. Which method would you recommend and why?
5.The approach adopted in this paper assumes that managers are able to identify the relative importance of five criteria. What approach would you take if the managers were not able to do this? | 458 | 2,195 | {"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} | 3.28125 | 3 | CC-MAIN-2021-17 | latest | en | 0.924504 |
https://mathematica.stackexchange.com/questions/8994/standardform-on-graph-edges | 1,721,063,594,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00335.warc.gz | 331,859,996 | 42,646 | # StandardForm on Graph edges
The Graph command displays edge labels in TraditionalForm. I instead want to use StandardForm.
I tried
EdgeLabelStyle -> Directive[StandardForm, Background -> White], but of course it does not work.
So how to do it, acting on EdgeLabelStyle?
Edit: Comment to @Sjoerd examples Your Binomial example is somewhat misleading, since Binomial has an intrinsic TraditionalForm version. What we are seeing in the graph is a sort of hybrid of Standardform (for n and k) and TraditionalForm for the parenthesis (I think). This is the proof:
As you see, even though I used braces in the notation, I still get the parenthesis in the graph (from the internal TraditionalForm), but the n, k are indeed in StandardForm.
This shows that the "StandardForm" in EdgelabelStyle does something, but it is not enough. Perhaps it is possible to play with BaseStyle?
The same thing happens if you use MakeBoxes (without ToString).
The style names should be between quotes (this is stated in the Style doc page, under the 'More Information' section)
edges = {1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1};
GraphicsRow[{
Graph[edges, EdgeLabels -> Exp[I x/10],
EdgeLabelStyle -> Directive["StandardForm", Red, 30],
Graph[edges, EdgeLabels -> Exp[I x/10],
}]
EDIT
Based on your comments I tried to work with Notation and MakeBoxes.
Notation doesn't seem to cause problems:
MakeBoxes seem to work if you use ToString:
gplus /: MakeBoxes[gplus[x_, y_, n_], StandardForm] :=
RowBox[{MakeBoxes[x, StandardForm],
SubscriptBox["\[CirclePlus]", MakeBoxes[n, StandardForm]],
MakeBoxes[y, StandardForm]}]
edges = {1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1};
GraphicsRow[{
Graph[edges,
EdgeLabels -> ToString[gplus[a, b, c], StandardForm],
EdgeLabelStyle -> Directive["StandardForm", Red, 30],
Graph[edges,
EdgeLabels -> ToString[gplus[a, b, c], StandardForm],
}]
• This is very good!....but now there is another problem. For some reason, StandardForm is indeed displayed, but my custom notation (created with Notation and MakeBoxes[....StandardForm]) are no longer working. I also tried "Output" and "Text", to no avail. Any suggestions? Commented Aug 1, 2012 at 20:35
• @magma Tried something (see edits). If this doesn't work for you, you probably should be more specific in how you use Notation and MakeBoxes. Commented Aug 1, 2012 at 21:14
• thank you @Sjoerd. Please read my edited question. Commented Aug 2, 2012 at 2:07
You can use StandardForm to wrap the labels:
edges = {1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1};
GraphicsRow[{Graph[edges, EdgeLabels -> Exp[I x/10],
EdgeLabelStyle -> Directive[Red, 30], ImagePadding -> 15],
EdgeLabelStyle -> Directive[Red, 30], ImagePadding -> 15],
Graph[edges, EdgeLabels -> StandardForm[Exp[I x/10]],
EdgeLabelStyle -> Directive[Red, 30], ImagePadding -> 15]}]
• I tried that before asking here, but I really do not wish to mess up with the labels Commented Aug 2, 2012 at 2:06
I personally prefer using String for labels instead of evaluable expressions (in case they are messed up because of incautious assignments).
edges = {1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1};
Graph[edges, EdgeLabels -> {
1 \[UndirectedEdge] 2 -> Style[
ToString[HoldForm[
Integrate[Exp[(I*x)/10], {x, \[Alpha], \[Beta]}]
], StandardForm],
FontFamily -> "Times", Blue, 15, Bold],
2 \[UndirectedEdge] 3 -> "text",
3 \[UndirectedEdge] 1 -> ToString[HoldForm[
Integrate[Exp[(I*x)/10], {x, \[Alpha], \[Beta]}]
},
EdgeLabelStyle -> Directive[Red, 30], ImagePadding -> 15]
• This is not good for me Silvia. In my case case the label are "the message", so I want them to evaluate and stay untouched Commented Aug 1, 2012 at 19:58
• @magma In that case maybe you could remove the HoldForm wrapper, so the expression will be evaluated when generating the label content. Converting it to String is aimed at preventing the generated graph being modified by latter evaluation. eg. The graph is wrapped in a Dynamic, then a latter assignment, say x = 0, may cause Exp[I x/10] becoming 1. But if this behavior is indeed what you need, then using String will be inappropriate. Commented Aug 1, 2012 at 20:09
I think you can just use the FormatType option of Graph. First, your notation:
Then, the graph:
Graph[{1<->2, 2<->3, 3<->1}, EdgeLabels->Binomial[n,k], FormatType->StandardForm]
• Thank you Carl. This was a very old question of mine. I will check your answer tomorrow (if I can still find the problem :-)) Commented Oct 3, 2018 at 7:31 | 1,265 | 4,585 | {"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} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.838005 |
https://studylib.net/doc/25239650/sheet-3-machines | 1,558,430,545,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232256314.25/warc/CC-MAIN-20190521082340-20190521104340-00300.warc.gz | 651,341,211 | 40,443 | Uploaded by Muhamed Elsayed - Zalat
# sheet 3 machines
```Sheet 3
1) The secondary winding of a transformer has a terminal voltage of Vs(t) = 282.8 sin
377t V . The turns ratio of the transformer is 50:200 ( a = 0.25 ) . if the secondary
current of the transformer is Is(t) = 7.07 sin (377t – 36.87) A , what is the primary
current of this transformer ? what are its voltage regulation and efficiency ? the
impedances of this transformer referred to the primary side are :
Req = 0.05 Ω
Rc = 75 ohm
Xeq = 0.225 ohm
Xm = 20 ohm
2) A 20 KVA 8000/277 –V Distribution transformer has the following resistances and
reactances :
Rp = 32 ohm
Rs = 0.05 ohm
Xm = 30k ohm .
Xp = 45 ohm
Xs = 0.06 ohm
Rc = 250k ohm
The excitation branch impedances are given referred to the high voltage side of the
transformer .
(a) Find the equivalent circuit of this transformer referred to the high-voltage side.
(b) Find the per-unit equivalent circuit of this transformer.
(c) Assume that this transformer is supplying rated load at 277 V and 0.8 PF lagging.
What is this
transformer’s input voltage? What is its voltage regulation?
(d) What is the transformer’s efficiency under the conditions of part (c)?
``` | 351 | 1,190 | {"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-2019-22 | latest | en | 0.872898 |
https://www.jiskha.com/display.cgi?id=1511899655 | 1,531,909,400,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590127.2/warc/CC-MAIN-20180718095959-20180718115959-00473.warc.gz | 898,475,331 | 4,001 | # Physics
posted by ML
A solid ball of mass 3 kg, rolls down a hill that is 7 meters high. What is the rotational KE at the bottom of the hill?
1. Damon
loss of potential energy = m g h = 3*9.81* 7 = 206 Joules
= translational Ke + rotational Ke
= (1/2) m v^2 + (1/2)I omega^2
Note
I = (2/5) m r^2
v = omega r
so
206=(1/2)m v^2 + (1/2)(2/5)m r^2 v^2
412 = 1.4 *3 v^2
v = 9.9 m/s
then calculate (1/2) (2/5)(3) v^2
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https://developers.google.cn/maps/documentation/solar/calculate-costs-us | 1,708,716,969,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474445.77/warc/CC-MAIN-20240223185223-20240223215223-00599.warc.gz | 205,294,207 | 24,799 | # Calculate solar costs and savings (US only)
This document explains how the Solar API calculates the various values that it uses to recommend solar panel installations and to estimate the costs and cost savings for US addresses.
If you enter the address of a residence in a covered region of the US, the Solar API shows you the following estimates:
• How much sunlight the house receives annually
• How much space the roof has for a solar installation
• How much savings, in US dollars, the home can expect over the 20 year life of a solar system
• The average monthly electricity bill for homes in your area, which you can adjust for your home
• A recommended size, measured in kilowatts (kW), for a solar system on the house
Although the Solar API provides estimates for any structure that it has data for, the estimates that it provides are best suited for residences or small commercial structures. The Solar API recommends solar installation sizes that maximize savings without producing more energy in a year than a household can consume. The Solar API does not calculate values related to excess energy production.
The recommended installation sizes are limited to annual energy consumption for a number of reasons, but primarily because US households currently get little or no financial benefit from excess energy production. In US locations that have net metering, credits earned from excess energy production typically expire over time.
## Required values for financial analysis for non-US locations
From each `SolarPanelConfig` instance in the API response, you need two values to perform the financial analysis for that instance:
• `panelsCount`: The number of solar panels in an installation. You use this value in your calculation of the `installationSize`.
• `yearlyEnergyDcKwh`: How much sunlight energy a layout captures over the course of a year, in DC kWh, given a specific `panelsCount`. You use this value in your calculation of the annual solar energy AC production (`initialAcKwhPerYear`) of each `installationSize`.
Additionally, you need to gather location-specific values for the following variables that you will use in the calculations:
• billCostModel(): Your model for determining the cost, in local currency, paid by a household for using a given number of kWh. How much a utility charges for electricity can vary from day to day or hour to hour depending on things like demand, time of day, and how much electricity the household consumes. You might need to estimate an average cost.
• costIncreaseFactor: The Solar API uses 1.022 (2.2% annual increase) for US locations.
• dcToAcDerate: The efficiency at which an inverter converts the DC electricity that is produced by the solar panels to the AC electricity that is used in a household. The Solar API uses 85% for US locations.
• discountRate: The Solar API uses 1.04 (4% annual increase) for US locations.
• efficiencyDepreciationFactor: How much the efficiency of the solar panels declines each year. The Solar API uses 0.995 (0.5% annual decrease) for US locations.
• incentives: Include any monetary incentives to install solar panels given by government entities in your area.
• installationCostModel(): Your method for estimating the cost of installing solar in local currency for a given `installationSize`. The cost model would typically account for local labor and material costs for a given `installationSize`.
• installationLifeSpan: The expected lifespan of the solar installation. The Solar API uses 20 years. Adjust this value as needed for your area.
• kWhConsumptionModel(): Your model for determining how much energy a household consumes based on a monthly bill. In its simplest form, you would divide the bill by the average cost of a kWh in the household's location.
• monthlyBill: the average monthly electric bill for a subject household.
• monthlyKWhEnergyConsumption: An estimate of the average amount of electricity the household at a given location consumes in a month, measured in KWh.
With these values and the information provided by the API response, you can perform the calculations necessary to recommend the best `installationSize` for locations not covered by the Solar API.
## How it works
The average monthly electric bill is the key to the rest of the calculations.
The Solar API initially bases its calculations on a preselected monthly bill amount. If needed, you can select a different amount that more accurately reflects your own average monthly bill.
Knowing the amount of a monthly bill and the current cost of electricity in a given location, the Solar API can estimate the number of kilowatt hours (kWh) of electricity a household consumes each month. For the current cost of electricity around the US, and to determine kWhs from a monthly bill, the Solar API references databases maintained by Clean Power Research.
Using the number of kWh a household consumes, the usable area of a home's roof, and the solar potential of the home's location, the Solar API evaluates one or more possible solar installation sizes and recommends the size that provides the most savings.
The size of a solar panel installation is measured by its kW rating. The kW rating depends on the number of solar panels in the configuration and the power rating, measured in watts, of each panel.
The kW rating of an installation is not the same as the energy output of an installation, which is measured in kWh and is variable. The kWh output of an installation is dependent on factors like the following:
• The time of day
• The weather
• The orientation of the panel to the sun
• Any shadows cast on the panels by nearby objects
• The regional solar potential
• The age of the installation
The Solar API includes factors like regional solar potential and the age of the installation in its estimate of the annual energy production of a solar installation.
To determine the usable area of a roof and estimate the solar installation size it can support, the Solar API uses aerial imagery and advanced 3D modelling.
## Detailed explanation of the values and calculations
The following sections explain how the Solar API calculates the costs, savings, and sizes of solar installations for a given structure in the U.S.
The explanations of the calculations use terms to represent values in the calculations. For an explanation of the terms, see Definition of the terms used in our calculations.
### Annual household energy consumption
As mentioned earlier, the Solar API determines the monthly consumption of electricity based on the monthly bill amount and the cost of electricity where a household is located. After determining the monthly consumption of electricity of a household, we calculate annual energy consumption in KWh by using the following formula:
```annualKWhEnergyConsumption = monthlyKWhEnergyConsumption x 12
```
A household's energy consumption is assumed to remain the same year to year over the life of a solar installation. The Solar API assumes the life of a solar installation to be 20 years.
### Annual solar energy production
The Solar API estimates the annual energy production of a solar installation by considering factors like the intensity of sunlight, angle of sunlight, and number of hours of usable sunlight a region gets annually.
Solar installations produce direct current (DC) electricity, which has to be converted to alternating current (AC) electricity by an inverter before you can use it in your home. Some electricity is lost during the conversion process, and the efficiency of the inverter determines how much is lost.
The efficiency of the conversion process is referred to as the DC to AC derate. To account for the loss, the Solar API multiplies the annual output of the solar installation by a DC to AC derate of 0.85. The result is the annual production of AC electricity, as shown in the following formula:
```initialAcKwhPerYear = yearlyEnergyDcKwh x 0.85
```
The amount of energy an installation produces declines by about 0.5% each year over the life of the installation. To account for this, after the first year, the Solar API multiplies the annual AC output of an installation by 99.5%, or 0.995, each year over the estimated 20-year lifetime of the installation. This is illustrated in the following table.
Year Yearly solar energy production (kWh)
1 initialAcKwhPerYear
2 initialAcKwhPerYear x 0.995
: :
20 initialAcKwhPerYear x 0.99519
Because the solar panel efficiency decays at a constant rate, it is essentially a geometric series where a = initialAcKwhPerYear and r = efficiencyDepreciationFactor. We can use a geometric sum to calculate the `LifetimeProductionAcKwh`:
```LifetimeProductionAcKwh = (dcToAcDerate * initialAcKwhPerYear * (1 - pow(efficiencyDepreciationFactor, installationLifeSpan)) / (1 - efficiencyDepreciationFactor))
```
## The cost of electricity with solar If the size of an installation is limited by the roof size or other factors, the solar installation might produce less electricity than a household consumes. In these cases, the household will likely have to pay a utility for some amount of electricity each year, as shown in the following formula:
```annualKWhEnergyConsumption - initialAcKwhPerYear = annualUtilityEnergyRequired
```
To account for this cost, the Solar API applies a bill cost model to the estimated amount of electricity, in kWh, the household will need from a utility over the life of the solar installation. The following formula illustrates this calculation:
```annualUtilityBillEstimate = billCostModel(utilityEnergyRequired)
```
To account for the yearly increase in the cost of electricity, we apply a costIncreaseFactor of 2.2%, or 0.22, per year for US locations:
```costIncreaseFactor = 1 + 2.2% = 1.022
```
Due to inflation, we have to discount the value of the currency value in our estimates of future costs. To account for this, we apply a 4% discount rate to our model for US locations:
```discountRate = 1 + 4% = 1.04
```
The following table shows how each year's utility bill is calculated over the life of a solar installation. The remainingLifetimeUtilityBill is the sum total of the utility bills for each of the 20 years of the solar installation's lifetime.
Year Annual utility bill in current local currency value (USD) (annualUtilityBillEstimate)
1 billCostModel (yearlyKWhEnergyConsumption - initialAcKwhPerYear) = annualUtilityBillEstimateYear1
2 billCostModel (yearlyKWhEnergyConsumption - initialAcKwhPerYear x 0.995) x 1.022 / 1.04 = annualUtilityBillEstimateYear2
: :
20 billCostModel (yearlyKWhEnergyConsumption - initialAcKwhPerYear x 0.99519) x 1.02219 / 1.0419 = annualUtilityBillEstimateYear2
Total remainingLifetimeUtilityBill = annualUtilityBillEstimateYear1 + annualUtilityBillEstimateYear2 + …. + annualUtilityBillEstimateYear20
### The cost of electricity without solar
To calculate how much a household might save if they install solar, we also have to calculate how much the household might pay if they don't.
We again have to account for the increasing cost of electricity and inflation by applying the costIncreaseFactor of 1.022 and the discountRate of 1.04 to the calculation, as we did when we calculated the cost of electricity with solar.
The following table shows how each year's utility bill without solar is calculated over the life of a solar installation. The costOfElectricityWithoutSolar is the sum total of the utility bills over the same 20 year period that we used for the cost of electricity with solar.
Year Yearly utility bill (USD)
1 monthlyBill x 12
2 monthlyBill x 12 x 1.022 / 1.04
: :
20 monthlyBill x 12 x 1.02219 / 1.0419
Total Sum of all annual bills, which can also be expressed as costOfElectricityWithoutSolar = 204.35 x monthlyBill
### The cost of installing solar
The Solar API includes the cost of installing the recommended solar configuration in the estimates that it provides. To estimate the cost of an installation, the Solar API uses a localized installation cost model and the size of the installation.
```installationCost = InstallationCostModel (installationSize)
```
### Incentives
Government entities might provide incentives for installing solar. The incentives are often in the form of tax credits. Based on a household's location, the Solar API subtracts any incentives that are currently available to the household from the estimate of the total costs.
### The total cost with solar installation
The Solar API calculates the total 20-year cost of a solar configuration by using the following formula:
```totalCostWithSolar = installationCost + remainingLifetimeUtilityBill - incentives
```
### The total savings
The Solar API calculates the savings for the household using the following formula:
```savings = costOfElectricityWithoutSolar - totalCostWithSolar
```
The Solar API performs the above calculations for each possible installation size and then recommends the installation size that provides the maximum savings for the household. The amount of the estimated savings is returned with the recommendation.
[]
[] | 2,754 | 13,087 | {"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} | 2.59375 | 3 | CC-MAIN-2024-10 | latest | en | 0.888062 |
https://www.aqua-calc.com/calculate/weight-to-volume/substance/palladium-blank-monoxide | 1,679,982,747,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00639.warc.gz | 728,051,115 | 8,043 | ## palladium monoxide: convert weight to volume
### Volume of 100 grams of Palladium monoxide
centimeter³ 12.05 milliliter 12.05 foot³ 0 oil barrel 7.58 × 10-5 Imperial gallon 0 US cup 0.05 inch³ 0.74 US fluid ounce 0.41 liter 0.01 US gallon 0 meter³ 1.2 × 10-5 US pint 0.03 metric cup 0.05 US quart 0.01 metric tablespoon 0.8 US tablespoon 0.81 metric teaspoon 2.41 US teaspoon 2.44
### The entered weight of Palladium monoxide in various units of weight
carat 500 ounce 3.53 gram 100 pound 0.22 kilogram 0.1 tonne 0 milligram 100 000
#### How many moles in 100 grams of Palladium monoxide?
There are 816.86 millimoles in 100 grams of Palladium monoxide
#### Foods, Nutrients and Calories
NEAPOLITAN CHOCOLATE STRAWBERRY VANILLA ICE CREAM, NEAPOLITAN, UPC: 070253712395 weigh(s) 136 grams per metric cup or 4.6 ounces per US cup, and contain(s) 198 calories per 100 grams (≈3.53 ounces) [ weight to volume | volume to weight | price | density ]
179809 foods that contain Iron, Fe. List of these foods starting with the highest contents of Iron, Fe and the lowest contents of Iron, Fe, and Recommended Dietary Allowances (RDAs) for Iron
#### Gravels, Substances and Oils
CaribSea, Freshwater, Instant Aquarium, Tahitian Moon weighs 1 473.7 kg/m³ (92.00009 lb/ft³) with specific gravity of 1.4737 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Carbonic dichloride, liquid [COCl2] weighs 1 402.8 kg/m³ (87.57394 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-403B, liquid (R403B) with temperature in the range of -56.67°C (-70.006°F) to 48.89°C (120.002°F)
#### Weights and Measurements
The grain per cubic millimeter density measurement unit is used to measure volume in cubic millimeters in order to estimate weight or mass in grains
The electric resistance of a conductor determines the amount of current, that flows through the conductor for a given voltage at the ends of the conductor.
long tn/yd³ to µg/US gal conversion table, long tn/yd³ to µg/US gal unit converter or convert between all units of density measurement.
#### Calculators
Cubic equation calculator. Real and complex roots of cubic equations | 682 | 2,466 | {"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} | 2.578125 | 3 | CC-MAIN-2023-14 | latest | en | 0.683008 |
https://www.unibo.it/it/studiare/dottorati-master-specializzazioni-e-altra-formazione/insegnamenti/insegnamento/2019/423530 | 1,723,599,496,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641086966.85/warc/CC-MAIN-20240813235205-20240814025205-00133.warc.gz | 805,777,280 | 79,966 | # 85164 - NONPARAMETRIC STATISTICS
• Docente: Laura Anderlucci
• Crediti formativi: 6
• SSD: SECS-S/01
• Lingua di insegnamento: Inglese
• Modalità didattica: Convenzionale - Lezioni in presenza
• Campus: Bologna
• Corso: Laurea Magistrale in Statistical sciences (cod. 9222)
Valido anche per Laurea Magistrale in Statistical sciences (cod. 9222)
## Conoscenze e abilità da conseguire
By the end of the course the student knows the fundamentals of the most relevant nonparametric techniques for statistical inference. In particular, the student is able to solve hypothesis testing problems where the conditions for the traditional parametric inferential tools to be applied are not fulfilled; to build nonparametric density estimators or nonparametric estimators of the regression function; and to use these methods in an effective and coherent way.
## Contenuti
Introduction.
Hypothesis testing about the probability of an event, on a single sample, on two paired samples and on two independent samples; the case of more than two (dependent/independent) samples.
Hypothesis testing about a location measure, on one sample, on two paired samples and on two independent samples.
One-way and two-way analysis of variance by ranks.
Goodness-of-fit tests.
Measures of association and related hypothesis testing methods.
Nonparametric methods for probability density function estimation: histogram, naive, kernel, nearest neighbour estimators.
Nonparametric methods for regression function estimation: kernel smoother; loess.
Nonparametric methods for statistical learning.
For the topics introduced by the lecturer, tutorial applications will be carried out in computer laboratory using R software.
## Testi/Bibliografia
Beyond teaching material provided by the lecturers (and available at http://campus.unibo.it/) the following references are recommended as additional readings:
• M. Hollander, A. W. Douglas, E. Chicken, Nonparametric statistical methods, Wiley, New York, 2014, 3rd edition.
ISBN: 978-0-470-38737-5.
• Scott, David W. Multivariate density estimation: theory, practice, and visualization. John Wiley & Sons, 2015.
• T. Hastie, R. Tibshirani, and J. Friedman (2001) The Elements of Statistical Learning: data mining, inference and prediction. Springer Verlag.
Freely available at: https://web.stanford.edu/~hastie/Papers/ESLII.pdf
## Metodi didattici
Lectures. Tutorial applications in class and in computer laboratory.
## Modalità di verifica e valutazione dell'apprendimento
The exam aims to test the student's achievement of the following purposes:
• deep knowledge of the theoretical methods described and discussed;
• critical skill in choosing the most adequate inferential tool to solve a given problem and ability to interpret the corresponding results.
The exam consists of a mandatory written exam (110 minutes). Oral exam is on demand.
The written exam consists of exercises, some of which can be solved by using R software, and theoretical questions. During the exam, using lecture notes, books or electronic devices is forbidden.
The overall evaluation is expressed in marks out of 30. Having the oral exam, the evaluation obtained in the written exam can change by plus or minus 3 marks at most.
## Strumenti a supporto della didattica
Blackboard; PC; videoprojector; computer laboratory
Teaching material is available at http://iol.unibo.it/ (download is only allowed to the students that have the course in their study plan)
## Orario di ricevimento
Consulta il sito web di Laura Anderlucci | 792 | 3,541 | {"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-2024-33 | latest | en | 0.756651 |
https://www.physicsforums.com/threads/why-use-capacitors-to-store-energy.374950/ | 1,624,144,001,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649731.59/warc/CC-MAIN-20210619203250-20210619233250-00418.warc.gz | 849,559,070 | 16,173 | # Why use capacitors to store energy?
Why are capacitors used to give pules of very high potential difference (for e.g. x)?...what's the advantage?
The source charging the capacitors should have potential difference less than x if capacitors have an advantage here; for e.g. with a source x/2, we can getting a potential x...is this happening here?
Last edited:
I'm not sure I really understand your question, but maybe the capacitor's ability to release energy quickly is a part of the answer?
EDIT: In response to your edit, are you talking about a http://en.wikipedia.org/wiki/Charge_pump" [Broken]?
Last edited by a moderator:
I'm not sure I really understand your question, but maybe the capacitor's ability to release energy quickly is a part of the answer?
EDIT: In response to your edit, are you talking about a http://en.wikipedia.org/wiki/Charge_pump" [Broken]?
May be it's ability to release a very high amount of current is the answer. The source might just be inept of giving a high amount of current at that potential, thus capacitors are charged first with a resistance in series to lower the current rate. Then the high current output is taken from the capacitors.
Any other ways to step up the voltage using capacitors other than charge pumps?
Last edited by a moderator:
Born2bwire
Gold Member
Any actual source is going to diverge from the behavior of the ideal sources that we use in circuits. The first rough estimation of a more realistic source is to assume that there is an internal resistance. This would be modeled as having the source have a resistor in series (for a voltage source) or in parallel (for a current source I think) between the source and the output terminals. In this way, we see that power gets siphoned off from the source and expended into the internal resistance. It is thus easy to see that this will restrict high current draws on a voltage source for example. That is, if we place a low resistive load across the voltage source terminals, the effective resistance will be dominated by the internal resistance. So even as we place a smaller and smaller resistive load, the drawn current will change very little due to the comparatively larger internal resistance.
Another effect of the internal resistance is that it behaves as a dampener in a LC circuit. If I place a function generator that has an output impedance of 50 ohms onto a LC circuit, its behavior will be like an RLC circuit, where the R element comes from the output impedance/internal resistance of the source. This affects the supplied voltage/current relationships of the LC circuit. For example, it would shift the resonant time constant of the circuit from what it would have been with an ideal source. So the internal resistance can also affect the transient behavior of a circuit.
This is one reason why fast transient and high current draws can be better taken off of capacitors since they can be precharged with enough charge to drive the necessary fast current spikes. Capacitors also are simple filter elements. Not only can they provide current compensation when the source lacks the current for a sudden transient, they also can act as a simple low pass or high pass filter at the same time. In a simple power supply circuit that uses a transformer to rectifier to filter to voltage regulator, the filter stage is just a resistor and capacitor. Both elements work double duty as the capacitor and resistor first act as low pass filters to help smooth out the rectified AC wave from the rectification stage. In addition, the capacitor also helps by being a charge resevoir and the resistor controls the bias current that biases the zener diode that acts as the voltage regulator.
Ok...thanks for the explanation. Sort of like maximum power theory.
Why are capacitors used to give pulses of very high potential difference (for e.g. x)?...what's the advantage?
In automobile ignition systems up to about 1970, the energy was stored as current in the ignition coil, and when the coil current was suddenly interrupted, there was a 300-volt spike on the coil primary, leading to a 30,000 pulse on the coil secondary.
Bob S | 861 | 4,156 | {"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} | 3.21875 | 3 | CC-MAIN-2021-25 | latest | en | 0.955369 |
https://www.dummies.com/education/science/biology/how-to-simulate-error-propagation/ | 1,569,192,266,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514575751.84/warc/CC-MAIN-20190922221623-20190923003623-00221.warc.gz | 791,179,439 | 14,978 | How to Simulate Error Propagation - dummies
# How to Simulate Error Propagation
Probably the most general error-propagation technique is called Monte-Carlo analysis. You can use this technique to solve many difficult statistical problems. Calculating how SEs propagate through a formula for y as a function of x works like this:
1. Generate a random number from a normal distribution whose mean equals the value of x and whose standard deviation is the SE of x.
2. Plug the x value into the formula and save the resulting y value.
3. Repeat this step a large number of times.
The resulting set of y values will be your simulated sampling distribution for y.
4. Calculate the SD of the y values.
The SD of the simulated y values is your estimate of the SE of y. (Remember, the SE of a number is the SD of the sampling distribution for that number.)
You can perform these calculations very easily using the free program Statistics 101. With very little extra effort, this software can give you the confidence interval and even a histogram of the simulated areas. And simulation can easily and accurately handle non-normally distributed measurement errors.
Consider the example of estimating the SE of the area of a circle whose diameter is 2.3 cm, with a SE of 0.2 cm. The formula for the area of a circle, in terms of its diameter (d) is A = (π/4)r2
This problem can easily be solved by simulation, using the Statistics 101 program. The program (only four lines long) generates the output shown. The SE of the coin area from this simulation is about 0.72, in good agreement with the value obtained by the other methods.
Credit: Screenshot courtesy of John C. Pezzullo, PhD | 365 | 1,683 | {"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} | 3.296875 | 3 | CC-MAIN-2019-39 | longest | en | 0.88521 |
http://blog.stela.org.br/reception-school-mkexmyq/mathematics-for-economics-1-pdf-4b00cb | 1,652,891,147,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522284.20/warc/CC-MAIN-20220518151003-20220518181003-00479.warc.gz | 7,896,344 | 14,061 | # mathematics for economics 1 pdf
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Bayesian Logical Positivism
I just read Eliezer Yudkowsky's introduction to Bayes Theorem. Quite illuminating.
Eliezer writes:
"Previously, the most popular philosophy of science was probably Karl Popper's falsificationism - this is the old philosophy that the Bayesian revolution is currently dethroning. Karl Popper's idea that theories can be definitely falsified, but never definitely confirmed, is yet another special case of the Bayesian rules; if p(X|A) ~ 1 - if the theory makes a definite prediction - then observing ~X very strongly falsifies A. On the other hand, if p(X|A) ~ 1, and we observe X, this doesn't definitely confirm the theory; there might be some other condition B such that p(X|B) ~ 1, in which case observing X doesn't favor A over B. For observing X to definitely confirm A, we would have to know, not that p(X|A) ~ 1, but that p(X|~A) ~ 0, which is something that we can't know because we can't range over all possible alternative explanations."
Of course, this raises the possibility that we can reformulate logical positivism in Bayesian terms:
For every meaningful proposition Q there is some finitely executable experimental test E for which:
p(E|Q) <> p(~E|Q)
That is, for a proposition to be meaningful, there must be some repeatable experience that is more likely to occur if the proposition is true than if it is false.
Definitely food for thought.
I'm just beginning to read about something called postpositivism. At first glance, postpositivism this looks like another sad attempt to revive metaphysics by focusing on the weaknesses of prototypical logical positivism all-the-while ignoring the central principle of positivism. In other words, postpositivism seems to score highly on the waffle-o-meter... Stay tuned!
Monday, November 29, 2004
Robot Descendants
At the last WTA chapter meeting, I posed the following rhetorical question:
Which scenario is better: a) that humanity destroy itself in a thermonuclear fireball, or b) that we develop an AI that learns everything we know before supplanting us as the dominant (or only) life form on Earth?
Normally, I hate this kind of question. You know, questions like "would you rather lose an arm or a leg?" Of course, I don't want to lose either.
Still, my question about alternate futures serves to highlight an interesting point. I think it is the dream of most scientists, inventors and artists to produce work that will be appreciated long after they are dead. They don't care who is doing the appreciating. They don't produce a new cure for disease just for their family, their country or their race, they do it for humanity. But what is our individual connection to humanity? How much of an emotional connection do we really have with homo sapiens sapiens in particular? Not that much, I would argue. It's too abstract a concept.
Suppose that in 10,000 years, our descendants have brains that are twice as large as ours, have twenty times our strength and virtually unlimited lifespans. If you met such a creature today, would you call her human? More than human? And if our woman of the future says she appreciates your paintings, your novels and your contributions to science, would you be offended? And if her body were made of silicon and carbon nanotubes, would this change anything?
The bottom line is this. Any creature that we consider to be intelligent and virtuous is a good person, independent of its biology.
D'Lenn, Abe Sapien, Stitch, Arwen, Romanadvoratrelundar, Yoda, John Parker, Kalel, Data... if they were real, would we not consider each of their lives as important as any human life?
As an aside, I wonder how appreciation of science fiction correlates with tolerance for homosexuals. Surely, the sexual practices of fictional alien species must be more bizarre than homosexuality.
Sunday, November 28, 2004
And so it begins...
Taking a cue from George W. Bush, Maryland's Governor has banned certain representatives of the press from speaking with state officials because the journalists refuse to write what the Governor wants to read.
Control of the media is one of the hallmarks of fascism.
Read this list of fascist principles and see how many are being employed today:
1. Powerful and continuing expressions of nationalism.
2. Disdain for the importance of human rights.
3. Identification of enemies/scapegoats as a unifying cause.
4. The supremacy of the military/avid militarism.
5. Rampant sexism.
6. A controlled mass media.
7. Obsession with national security.
8. Religion and ruling elite tied together.
9. Power of corporations protected.
10. Power of labor suppressed or eliminated.
11. Disdain and suppression of intellectuals and the arts.
12. Obsession with crime and punishment.
13. Rampant cronyism and corruption.
14. Fraudulent elections.
Tuesday, November 23, 2004
The Principle of Verifiable Equivalence
The Principle of Verifiability says that the meaning of a proposition is its method of verification, or, as I (and Karl Popper) prefer, its method of falsification.
Here's an example. Suppose I say "the Lockheed SR-71 Blackbird is the fastest aircraft in the world." I could mean several different things. I could mean that it accelerates faster than any other airplane. I could mean that it flies in a straight line faster than any other plane. Or, I could mean that it climbs fastest, etc. I can tell you precisely what I mean by telling you the conditions of the experiment that falsifies my statement, e.g., "In level flight, no other airplane that can take off under its own power and is powered by air-breathing engines, can sustain a higher speed than the SR-71 over a 500 mile course."
Okay, that was an easy one.
Look at this proposition: "All Carbon 14 nuclei will eventually decay." Thanks to Nicolas for presenting me with this puzzler. At first glance, scientists will agree with this proposition. The problem is that, as written, we can never falsify it. We would have to wait for all time and observe all C14 nuclei in the universe decay before we were satisfied. This isn't even possible in principle, let alone in practice. This means that, as a logical positivist, I can assign no meaning to this proposition.
This is typical of the kind of challenge philosophers have put forward in opposition to logical positivism. By carefully wording accepted scientific knowledge, opponents of logical philosophy hope to find places where science and logical positivism cannot be reconciled. Poor them - it's a hopeless cause.
The actual law of radioactive decay for C14 says this: "the mean lifetime of C14 nuclei is 8,000 years." Of course, this proposition is perfectly falsifiable. Indeed, the law of radioactive decay does not say that a given nucleus must ever decay. It just says that it will last on average 8,000 years before decaying.
Equivalence
Given a proposition, I can generally reword that proposition in some equivalent form. Instead of saying "I am soaked," I could say "I am covered in water," or "I am drenched." There are a very large number of ways of saying the same thing, though many of these ways become more and more verbose. How do we know whether our alternately-worded propositions are equivalent or not?
Certainly, two propositions cannot be equivalent if an experiment will falsify one proposition, but not the other. That is, verification is one (and, in my opinion, the only) arbiter of equivalency. (Note: I include rigorous mathematical proof as a form of verification of mathematical propositions, so the principle applies to mathematics also.)
This might be called the "Principle of Verifiable Equivalence" (okay, so I'm not good at naming principles). It's weaker than the Principle of Verifiability for two reasons. First, it can be used to compare two similar propositions that don't have identical meaning, but which have overlapping meaning. If I say that "George is lying to me," I am also satisfying the condition that "George is being dishonest towards me." If I falsify the latter proposition (dishonesty), I must falsify the former proposition (lying).
Second, the Principle of Verifiable Equivalence is weaker because it does not say anything about propositions that cannot be verified or falsified. It merely says that any two such propositions can never be shown to be equivalent.
Directives
Thanks again to Nicolas for this proposition:
"Homosexuality is unnatural."
This type of proposition is confusing to people because they make the mistake of thinking that it is a proposition about the world. It isn't. Those who utter this proposition are not talking about social convention - they generally do not mean that "homosexuality is practiced by less than 10% of humans." If they did, their utterance would be no more controversial than saying "private aviation is practiced by less than 10% of humans."
What is controversial is what they are really saying, namely: "I consider homosexuality to be wrong." This isn't about the world, per se. It is about how the speaker feels about the issue.
Similarly, if I say "eating fast food is bad," I really mean either "I don't like fast food," or "do not eat fast food!" Of course, this proposition is either an expression of opinion/strategy, or else it is a directive. Neither one can be true or false at all.
So, before you get too deep trying to analyze the proposition "God is good," just remember that this proposition is, at best, not about the world, and, most likely, utter nonsense.
Tuesday, November 16, 2004
Incredibly, it's not just me
I knew I couldn't be the only one who sensed right-wing undertones in The Incredibles. Check out this piece in the New York Observer:
The simple message that President Bush managed to dumbly repeat until it seemed true for so many can find itself illustrated in diverse places because it sticks so easily. Team America and The Incredibles are different films that arrive at the same conclusion: At some point the Evil Ones must die, and at some point a special, chosen, brave and happy few will vanquish them; it’s up to the rest of us to sit by and trust them to take care of it, without questioning their methods. In The Incredibles, Mrs. Incredible—voiced by Holly Hunter—lectures her children pointedly: These people will kill you, she says, unless you use your superpowers.
I also found this nice post at the Turk's Head Review blog.
Monday, November 15, 2004
Drill Deeper
One of the big stories this week is the extent to which Saddam Hussein cheated the U.N. Oil for Food Program.
Apparently, Hussein was able to obtain influence among corrupt officials at the U.N. and in neighboring states by awarding them "oil allocations." The allocations could be sold to oil companies who were supposed to be the sole recipient of such allocations from the Iraqi government.
So, let's get this straight. The authoritarian regressives on the right highlight corruption at the U.N. as evidence that the sanctions weren't working. The Republicans then hold Democrats responsible for the failure of the embargo, citing misplaced trust in the U.N. See the problems here?
I'll spell it out.
First, you can't blame the reality-based community for the failure of sanctions WHEN THE SANCTIONS WORKED!!! Saddam was contained, and posed no threat. We had leverage right up until Bush traded Iraq's secular state for a fundamentalist, fanatical chaos.
Second, the corruption was obviously facilitated by your beloved oil companies! You know the ones. Like Halliburton, which may have paid \$180 million in bribes to Nigerian officials while Dick Cheney was still CEO. Don't try to tell me the CEO won't notice that little \$180 million line-item.
There's your real national security threat: big oil and its influence.
3D Politics
Last week, I figured out that Liberal is not the antonym of Conservative. Liberal is the antonym of Authoritarian. Simplisticaly, I thought that the opposite of Conservative would be Progressive.
However, the resulting 2D plot of political space failed to generate the depth of insight I was hoping for. The problem is that Conservative, the tendency to want to keep things the way they are/were, is loaded with too much historical baggage. Conservatives were the first to work to protect our environment; Nixon signed the Endangered Species Act into law in 1973. However, if go back in time to 19th century, the colonial push towards the West was marked by environmental disregard. Settlers killed millions of Buffalo without consideration for the ecological or economic damage they were doing. In other words, whether a position is conservative or not depends on how far back you want to go.
Obviously, using a continuum from progressive to conservative to regressive isn't an ideal analyical tool for mapping political positions. Still, referring to contemporary right-wingers as regressives is both apt and gratifying.
So what is it about the conservative/regressive worldview that is time-independent?
Pessimism
Certainly, there's the view on the right that problems of human affairs are fundamentally intractible at the social level. This is a pessimism that resonates with the religious: man is too deeply flawed to solve his own problems, so he must appeal to a divine power for help.
Power to the Powerful
Conservatism primarily conserves power among the already powerful. It's no coincidence that "conservatives" are identified with efforts to preserve racial segregation, regressive tax schemes, elimination of social benefits for the underprivileged (including affirmative action), and the transfer of public assets and public enterprise into corporate hands.
Nepotism
Who among the powerful can be trusted to maintain the status quo? Friends and relatives, of course. Consider the Bush administration's current priorities. No bid contracts in Iraq have benefited Halliburton (the company formerly led by the Vice President) and Bechtel with lucrative contracts. George Shultz, the former U.S. Secretary of State, and the man credited with the ascendency of George W. Bush to the Presidency of the United States, served on the board of Bechtel and was Executive Vice President of the company in the 1980's. The recent Medicare prescription drug benefit is a benefit not for seniors, but for HMO's and pharmaceutical companies. The new priority, social security privatization will benefit the brokerage industry by mandating that every citizen risk part of his earnings in brokerage accounts.
Foreign Policy for Profit
Until recently, conservatives supported foreign policy that was designed to benefit American multinational corporations, while protecting U.S. treasure with defensive use of military power, and offensive use of covert power. Republicans have generally favored economic engagement with corrupt and oppressive regimes when there are profits to be made. The interests of the downtrodden are of no sincere interest to them. Communism was a threat that merited special attention, for the communist worldview swore to eliminate private property and sweep away privilege.
Continuing in this tradition, George W. Bush saw an opportunity for profit in his invasion of Iraq. Unlike his predecessors, Bush has completely ignored the defense of treasure paradigm. He has permanently damaged the America's brand, exposed us to increased danger of terror attack, bogged down the military in an unwinnable war, and risked the entire economy in a gamble for Mid-East oil. However, his friends and family stand to make out like kings.
Fear
Unofficially, the view from the left is that the Republicans are a haven for racists and bigots. This could just be coincidence. Racists aren't going to admit to being racist, so they have to find a legitimate cover story. The Republican's just happen to have economic reasons for maintaining the status quo. Nonetheless, the right's opposition to the progression of civil rights is real enough.
Naturally, any party whose political platform is fundamentally based on pessimism will benefit in a climate of fear. That's why we are now engaged in a "war on terror." Conveniently, this war, inasmuch as it has been defined, is interminable.
So, let's return to our political geometry. We now have as axes: Liberal-Authoritarian, Optimistic-Pessimistic, and Fairness-Nepotism. There are eight combinations of these attributes, more if you include neutral positions.
Here are some of those combinations:
Political LocationPositions
Liberal-Optimistic-FairnessDemocratic idealists
Liberal-Optimistic-NepotismCorrupt Democrats
Liberal-Pessimistic-FairnessLibertarian idealists
Liberal-Pessimistic-NepotismLibertarian with pro-corporate agenda
Authoritarian-Optimistic-FairnessSocialist revolutionary idealists
Authoritarian-Optimistic-NepotismSoviet-style communists
Authoritarian-Pessimistic-FairnessReligious right-wing idealists
Neutral-Pessimistic-FairnessThe original American right
Authoritarian-Pessimistic-NepotismFascism
Neutral-Pessimistic-NepotismThe new American right
Seven Generals
Seven Generals explain why the Bush still doesn't get it:
The people in control in the Pentagon and the White House live in a fantasy world. They actually thought everyone would just line up and vote for a new democracy and you would have a sort of Denmark with oil. I blame Defense Secretary Donald Rumsfeld and the people behind him -- Deputy Defense Secretary Paul Wolfowitz and Undersecretary Douglas Feith. The vice president himself should probably be included; certainly his wife. These so-called neocons: These people have no real experience in life. They are utopian thinkers, idealists, very smart, and they have the courage of their convictions, so it makes them doubly dangerous.
-Gen. Merrill "Tony" McPeak
Air Force chief of staff, 1990-94
Movie Reviews
I'm working on another "deep" post, but for now, here's are three (short) movie reviews.
Saw the Incredibles last night. I rate it 6.5/10. Technically, the movie is well crafted. I thought the violence and suspense were a bit strong for young children, though none of the young kids at the movie theater cried during the show. The thing that spoiled the movie for me was the dialogue. They just had to go and throw in references to terrorism and tort reform. I dunno, maybe I'm getting overly sensitive in my old age. Also, Pixar's opening short, Bounding, was their lamest ever.
7/10. Nice, romantic film. Young Hugh Grant put in a good performance as Chopin. Don't know why, but I always find Bernadette Peters annoying. Loved Emma Thompson, as usual.
I really enjoyed this movie, even more than the original. 8.5/10. Bridget's hilarious friends played a less prominent role in the story :(, but the embarrassing moments weren't quite as embarrassing as in the first picture :). If you liked the first movie, you'll probably like the sequel.
Sunday, November 07, 2004
Liberal vs. Conservative: the false dichotomy
We tend to think of there being two competitors in the political ring...
On the left, wearing the blue shorts, friend to the arts and sciences, tree-hugger extraordinaire, progress is his middle name... the Liberal.
On the right, wearing the red trunks, friend to industry and millionaire alike, Mr. "Three Strikes and you're out!", God bless this contender... the Conservative.
However, this is a very poor, one-dimensional approximation to something that has at least two dimensions.
Dictionary.com defines conservative as: Favoring traditional views and values; tending to oppose change.
The opposite of conservative is progressive: Promoting or favoring progress toward better conditions or new policies, ideas, or methods
Neither conservatism nor progressiveness is a panacea. In extreme doses, both are dangerous. I suppose conservatives would say that change brings unpredictable results, including the potential loss of something that is good. Progressives might say that we never have the perfect solution, and that changes made with due deliberation have the potential to improve our lives. Fundamentally, conservatives are pessimists and progressives are optimists.
What is the opposite of liberalism? To be a liberal is to value freedom. The closest antonym for liberalism I have found so far is authoritarianism.
In the following table, I've tried to classify certain political positions within this two-dimensional political universe:
If you want to know who Bush is talking about when he refers to those who "hate freedom," it is the occupants of the bottom half of this diagram.
Now to political parties. Democrats are liberal progressives with a hint of environmental conservatism. Republicans used to be conservative in general, now they are squarely in are bottom right quadrant (a.k.a., conservative freedom-haters). Ralph Nader is probably in the bottom left corner. The libertarians live in the top right quadrant.
I feel sort of dim for taking so long to figure this out. I'm optimistic that eliminating the false liberal/conservative dichotomy in the minds of voters will lead to positive results. However, there are dark forces that don't want the dichotomy to go away.
So when Ann Coulter (may her name be cursed throughout the Local Group) spits her venom at us liberals, just remember that she's an authoritarian: she hates freedom.
Election Analysis
Dick Meyer at CBS has some interesting post-election analysis.
Meyer convincingly argues that moral values were not the most significant factor in this election. The way the question was asked in the exit polls, and the cultural definition of morality skews the question in favor of the Republicans.
First of all, Terrorism and Iraq were not grouped together as a category. Had they been, the Terrorism/Iraq category would have been the biggest single issue for voters at 34%.
Second, the question effectively reads something like: "Do you favor the Republican view of morality?" Not surprisingly, 80% of them voted for the Republicans.
Friday, November 05, 2004
Who doesn't want to be a liberal?
We live in a liberal democracy because human nature makes it impossible for us to reach anything but a broad consensus. Before we start throwing our freedoms in the trash, let's take time to ponder what we might lose if we surrender our liberal ideals.
Privacy
If we don't watch people all the time, we won't know what they're getting up to. They could be plotting a terrorist attack. They could be printing counterfeit money. They could be having sex in illegal positions! We had better install cameras in every room of every home, just in case. Big Brother is watching!
Our liberal philosophy is what gives us the right to privacy. We accept responsibility to use our privacy wisely, and grant others the right to do the same. In certain locations (like airports), privacy may be curtailed due to the immediacy of the threat to public safety. Libraries are not such locations.
Guns
The United States has liberal gun ownership laws. Not conservative gun ownership laws, liberal ones. We allow people the freedom to have guns, and trust that they will use them responsibly. Yes, a gun owner might kill someone accidentally or on purpose, but we should not take rights away from people unless we can show that those rights would serve purely anti-social purposes. For example, assault weapons are designed to kill people in large numbers. The only reason to justify the availability of assault weapons is to enable the populace to resist tyranny. The government owns tanks and attack helicopters; should we allow individuals to own those, too? I don't think that the tyranny argument justifies private ownership of military class weapons.
Hunting
You want to shoot a defenseless animal for sport. It's perfectly legal. Such courage, such guile. What a sense of power, what a thrill, what a man. Why don't you strangle your dog with your bare hands? That would be thrilling wouldn't it?
As you can tell, I don't think hunting for sport shows good character. If you hunt because you must, that's another thing entirely. Hey, I'm as happy as the next guy to eat burgers and chicken nuggets. I just don't think that the act of killing the weak and the innocent should be a source of pleasure. However, this is a liberal democracy, if you want to kill an animal for pleasure, you have the right to fire away! As long as hunters don't kill endangered species or turn their guns on people, I don't see an imperative to take away their rights.
Choice.
When does human life (as we know it) begin? We all agree that a viable fetus is a human being with human rights. However, choosing the moment of conception as the start of life is ARBITRARY!! Where will this insanity lead? Is it better to treat women as chattel - mere breeding machines that we value less than an undeveloped fetus? Is it better to kill a sentient woman than to kill an embryo? Should we force women to salvage their unfertilized eggs during menstruation so we can fertilize the eggs and implant them in any available woman? Should all women of child-bearing age be pregnant to ensure that no potential human lives are lost?
This is a liberal democracy, and women have the right to privacy, the right to self-defense, and have responsibility for their own lives. What a woman does with a non-viable embryo is between her and her doctor. We already restrict abortion to non-viable fetuses (except when birth would cost the life of the mother), so there's no threat to society from the practice of abortion.
Sexuality
There's a whole bunch of people whose sexual advances I would rather not receive. Only a subset of these people are men!
In my opinion, those who want to deny human rights to gays and lesbians (the right to marry, the right to work, the right to healthcare, the right to exist, etc.) are simply bigots. How do I know? I used to irrationally fear gays, too. Twenty years ago, I used to rationalize all sorts of stupid reasons why gays were bad people, or at least, doing bad things. When I finally met some gay people and learned that they're no different than the rest of us, my rationalizations melted away.
Decades ago, non-whites were seen as less than human, and all sorts of rationalizations were given for infringement of their rights. In time, that too passed.
Hey, bigots have the right to be bigots, but they don't have the right to make homosexuals (or anyone else) second-class citizens. There's no public safety hazard posed by gays and lesbians.
Conscience
I think organized religion is a form of mass delusion. For their part, superstitious people think my atheism is wicked. But as long as a man's delusion isn't a safety threat, let him be deluded. If you want to discriminate against me because I insist on seeing reality instead of fantasy, you had better be ready to be on the receiving end of the same discrimination. Once they've come after me, they'll come after you next. Yes, religion is yet another freedom guaranteed by liberal democracy.
Now, who doesn't want to be a liberal?
Virtue
Just blogging out loud for a bit...
Good is not absolute. Good is defined for each of us by the sum of our personal experiences. The "good" is a part of our brain that is co-activated when we have positive experiences: we see a beautiful scene, make contact with a loved one, or eat chocolate cake. Similarly, the "bad" is that part of our brain co-activated with negative experiences like physical pain, emotional loss and bad smells.
There's not necessarily a fixed, physical "goodness center" in the brain. Our neocortex is a highly generalized pattern matching machine; good and bad are just patterns we extract from experience.
The more distant an action is from simple pleasure or pain, the harder it is to assign a goodness rating to that action. Nonetheless, we can still say that filing away recipes in alphabetical order is good because it will reduce the pain of finding the recipes later.
Naturally, precisely what a man defines as good or bad is as unique as his life experiences.
Ethics
Ethics is a set of principles we adopt to facilitate living with others. Ethics are not absolute either. In free societies, promotion of the "common good" is a social objective. The common good is a sort of an average of the individual good over the entire group. A individual's personal goodness almost always differs from ethical goodness, even if only slightly.
Since a person is not an invariant machine, her individual experiences will change her perception of good and bad. Good and bad are not fixed at the individual level, so good and bad cannot be optimally fixed at a social level. Ethics must be adaptable.
Morality
I'll define morality as the set of principles an individual uses to determine right and wrong. Individual moral codes and ethical or legal codes frequently diverge.
Since moral code is defines by personal good, moral codes are not absolute either. Sorry.
Virtue
Following society's ethical rules to the letter would be considered virtuous behavior. Is that all there is to it? If a person follows the rules under duress, is that still virtue? I'm not so sure. In my view, it is the thought that counts, and a good act performed for the wrong reason isn't good, though it is preferable to a bad act.
An autonomous automobile that that never breaks traffic laws might also be considered virtuous by some. However, the car cannot visualize the consequences of breaking those laws, and consequently make the choice to obey them. The robot car is generally good for us, but it is not virtuous.
Finally, is it not a virtue to be able to see the good exception to the rule of law?
Wednesday, November 03, 2004
"Ignorance more frequently begets confidence than does knowledge: it is those who know little, not those who know much, who so positively assert that this or that problem will never be solved by science."
-Charles Darwin, The Descent of Man
Unless there's evidence of massive voter fraud on the part of the Republicans, it looks like George W. Bush has won re-election.
Before last night, I had spent quite a while wondering what I would do to protest a Bush victory. I really couldn't think of anything fitting.
After last night, I realized that this isn't really about America or politics anymore. It requires personal adaptation.
The Scales Fall
It is now apparent that more than 50% of American voters cannot make rational political decision. They vote based on their gut or based on wedge issues. They cannot see the big picture. To be fair, one can probably lump many voters on the left into this category, too. You know the ones I mean. The environmentalist who thinks that environmentalism trumps national security, or the pacifist who cannot justify war for any reason.
Yesterday, America re-elected the worst President in U.S. history. A man who has brought wide-scale corruption, environmental degradation, disastrous foreign policy and increased terrorism. What does a President have to do to get thrown out of office?
Under normal circumstances, Americans would fire the president when they feel an acute malaise. There has to be a noticeable drop in the standard of living due to unemployment or underemployment or due to high crime rates. Things are definitely worse than they were under Clinton, but that's no longer enough. Bush's perpetual "war on terrorism" is a political weapon that uses fear to keep people from voting the way they should. This means that terrorist attacks against America will actually help Bush win re-election, not hurt him. We shouldn't be surprised that the Bush administration pretends it's doing something abroad (e.g., the war in Iraq), while simultaneously weakening homeland security. The "American Death Spiral," if you will.
Things will get a lot worse before they get better.
The Stars Fall
Government funding for America's technological infrastructure is collapsing. Science funding is down, funding for education (including higher education) is down, and there is no government road map to keep America competitive.
The Bush administration's policy is to stack government science committees with representatives from industries that have something to lose if any progress is made. Drug companies to run the FDA, energy executives to set energy policy, polluters to govern the EPA, and right-wing ignoramuses to oversee stem cell research (or rather, to prevent it).
This is a perversion. Government exists to serve the needs of the people, and without it, people cannot compete with corporate influence. Government is supposed to do those things that corporations and private interests cannot or will not do.
Corporations will happily hire their PhD's in India or China, and build factories elsewhere. Why should they pay more to hire Americans? Americans have no special skills anymore.
America's star is sure to fall.
Personal Strategy
Politically, not much has changed. The Democratic political machine is just getting fired up, and it will be stronger in 2006 and stronger still in 2008. I shall redouble my efforts as a Democratic volunteer.
However, on a personal level, things need to change. I have always had faith in American innovation and American democracy. We had the best universities and the best scientific research. Our society was imperfect, but it was both liberal and wonderful. I always believed America could meet any challenge. I believed that the American dream would always be out there. I believed that an investment in America was an investment in myself.
I no longer believe this. I'm not saying that America is doomed to fall way behind the rest of the world. I just don't have any confidence that it can stay ahead. America's values are no longer much different from any other country. The United States is violent, ignorant, unjust, and now it tortures prisoners.
I can and I will work to change this, but I no longer fight for America's honor as if it were my own. America isn't me anymore.
I'm taking a card from the Republican playbook. In the Republican world, social Darwinism is the order of the day. They say that society must have its winners and its losers. Well then, I will be a winner. I will be more competitive than ever. I will fight for my own interests first, and my country's interests second. I will ask not what I can do for my country. I won't even ask what my country can do for me. I will ask what I can do for myself.
In practical terms what does this mean? I haven't had time to think things through in much detail, but there are a couple of areas that I can work on.
Financially, I need to get my house in order. I can't make financial decisions based on loyalty to America or to progressive agendas. I can't sit back and have faith that hard work and traditional virtues are enough. If I have to invest in overseas institutions or in oil companies, so be it.
As for homeland security, it doesn't really exist. The government is actually motivated to promote insecurity. I must learn to be responsible for my own security, and that means stocking up on food, water and medical supplies. It means ensuring that I am much more self-sufficient.
Darwin's Revenge
If you're a Republican reading this, you're probably celebrating right now. You're wondering why I wasn't some sort of survivalist all along. You have a point. I should be the master my own destiny, and not trust in the goodwill of others.
What you Republicans haven't figured out is that your strategy has destroyed America's reputation, and now it is destroying America. It rewards investment in the best solution for the individual, not the best solution for the country. It promotes disinvestment in the United States.
So, if it is social Darwinism you seek, it is social Darwinism you shall have. Just remember, in natural selection, only the fittest survive, and I am the fittest! | 7,557 | 35,884 | {"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} | 2.546875 | 3 | CC-MAIN-2018-22 | latest | en | 0.936557 |
http://www.statemaster.com/correlations/gov_200_tot_reg_vot-2004-election-total-registered-voters | 1,571,437,631,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986684854.67/warc/CC-MAIN-20191018204336-20191018231836-00438.warc.gz | 320,816,741 | 13,659 | FACTOID # 1: The total number of state executions in 2005 was 60: 19 in Texas and 41 elsewhere. The racial split was 19 Black and 41 White.
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Correlations > Government Statistics > 2004 Election > Total registered voters
DEFINITION: Total number of registered voters for the 2004 elections.
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A correlation is a statistical measure of similarity between at least two given sets of data. StateMaster's correlations compare two variables from our database and reveal statistical relationships between them. The percentages you see represent the strength (or likelihood) that a change in the topic variable is matched by a change in the listed variables below it. But remember: These correlations do not imply causation, that is, one does not necessarily cause the other. Also, not all variables contain all states, rather subsets of states matched together.
NOTES:
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• Only variable pairs where at least 15 states match for each have been considered.
• Strength is given by the correlation coefficient (R squared). It is the fraction of variation in Y that can be attributed to the variation in X. 100% signifies a perfect fit (R squared of 1). The top 50 such stats are displayed
SOURCE: U.S. Census Bureau, Current Population Survey, 2005. | 366 | 1,752 | {"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} | 2.546875 | 3 | CC-MAIN-2019-43 | latest | en | 0.869375 |
https://www.opengl.org/discussion_boards/archive/index.php/t-146451.html | 1,534,922,936,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221219692.98/warc/CC-MAIN-20180822065454-20180822085454-00053.warc.gz | 934,108,402 | 5,076 | PDA
View Full Version : How to generate Tangent & Binormal from a given Face?
Antorian
07-18-2003, 03:16 PM
I'm implementing a dot3 Bump mapping using 3 texture unit with Normal Cube Map.
So I've got a problem with Tangent & binormal Vectors Generation.
Tangent_List = new CVector3D[Num_Vertices];
Binormal_List = new CVector3D[Num_Vertices];
for(unsigned int j=0; j<Num_Faces*3; j+=3)
{
CVector3D vec1, vec2;
int i = FaceIndex_List[j];
int i1 = FaceIndex_List[j+1];
int i2 = FaceIndex_List[j+2];
vec1 = Vertex_List[i2] - Vertex_List[i];
vec2 = Vertex_List[i1] - Vertex_List[i];
float d_u1 = Mapping_List[i2].x - Mapping_List[i].x;
float d_u2 = Mapping_List[i1].y - Mapping_List[i].y;
Tangent_List[i] = (vec1 * d_u1) - (vec2 * d_u2);
Tangent_List[i].Normalize();
Tangent_List[i1] = Tangent_List[i];
Tangent_List[i2] = Tangent_List[i];
Binormal_List[i] = Normal_List[i].CrossProduct(Tangent_List[i]);
Binormal_List[i1] = Normal_List[i1].CrossProduct(Tangent_List[i1]);
Binormal_List[i2] = Normal_List[i2].CrossProduct(Tangent_List[i2]);
}
But when Rendering, the bumped object (sphere cylinder torus etc..) doesn't look weel.
Someplaces bump is inverted....
How must it be done?
Suggestions Are welcome.
Thanks.
jwatte
07-18-2003, 06:24 PM
You can get inverted bumps if the texture mapping is mirrored. Either use the +3 Knobbly Stick Of Artist Enlightenment, or a tool like NVMeshMender to patch it up. Or generate a texture atlas for your normal maps, rather than re-using texure coordinates.
Eric Lengyel
07-18-2003, 07:53 PM
This can be a difficult calculation if you're not thinking about the problem in the right way. See Section 6.8 of my book, Mathematics for 3D Game Programming and Computer Graphics (http://www.terathon.com/books/mathgames.html) for a good explanation of how to do this.
cseger
07-19-2003, 07:47 AM
I've tried a number of ways and I end up using the one in Eric's book. By far the best and robust of all the onces I tried.
Also remember to take UV discontinuitise, material, and smoothing groups into mind when averaging.
Also a re-orthogonalization of the basis is good after the averaging.
If I remember correctly, all this is mentioned in his book, highly recommended.
Chris | 630 | 2,223 | {"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} | 2.5625 | 3 | CC-MAIN-2018-34 | latest | en | 0.646575 |
https://www.geeksforgeeks.org/solve-the-equation-2x-32-3-4/ | 1,653,738,422,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663016373.86/warc/CC-MAIN-20220528093113-20220528123113-00744.warc.gz | 869,609,743 | 24,151 | # Solve the equation (2x – 3)2/3 = 4
• Last Updated : 30 Nov, 2021
Algebra is the branch of mathematics that includes numerals and variables with operators. The term that has constant value is known as numerals, it is represented by numbers. The term that does not have a constant value is known as the variable, its value is not fixed, it is represented by letters and symbols. Algebra is basically used to calculate the value of the unknowns. With the help of the information given in the question, we apply the operation and get the value of unknows.
Algebraic Expression
An algebraic expression is the representation of the mathematical statement into the mathematical form with the help of numerals, variables, and appropriate operators.
For example: ‘Three times a number is added to 12’ can be written as ‘3x + 12’. Here we do not know the value of the number so we suppose it as x, it can take any value. Plus sign separate the statement into two-part, we name it as terms. In 3x + 12, there are two terms.
So on the basis of the number of terms, the algebraic expression can be classified into the following types.
• Monomial: If the number of terms in the expression is one then it is known as a monomial. Example: 5x, 6y, etc
• Binomial: If the number of terms in the expression is two then it is known as binomial. For example: 5x+3, 12y-3, etc.
• Trinomial: If the number of terms in the expression is three then it is known as the trinomial expression. For example: 5x-6y+3z, 5q-6f+3x, etc.
• Polynomial: If the number of terms in an expression is one or more than one then it is termed as the polynomial.
Solution of an equation:
Earlier in this article, we said, variables can take any value. But when we compare an algebraic expression with numerals then these variables should have any fixed value. With the help mathematical operation, we can find the value of the variables. The solution of an equation is the value of numerals at which the given equation got satisfied. The left-hand side and right-hand side of the equation should have to equal.
### Solve the equation (2x – 3)2/3 = 4
Solution:
Step to solve the problem:
Step 1: To solve the exponent of an equation, do the inverse operation on both sides.
In the given problem, we can clearly see there is the cube root and square on the variable part so the inverse operation is the cube and square root respectively.
⇒ (2x – 3) = 4(3/2)
⇒ (2x – 3) = (43)(1/2)
⇒ (2x – 3) = (64)(1/2)
As we know that the square root of 64 is +8 and -8.
Step 2: Here we have two cases because the square root of 64 gives two values.
Case 1: When the square root is +8
⇒ 2x – 3 = 8
Case 2: When the square root is -8.
⇒ 2x – 3 = -8
Step 3: Transfer all the numerals on one side and all the variables on the other side of the equal sign. And get the value of the variables.
Case 1:
⇒ 2x – 3 = 8
⇒ 2x = 8 + 3
⇒ 2x = 11
⇒ x = 11/2
Case 2:
⇒ 2x – 3 = -8
⇒ 2x = -8 + 3
⇒ 2x = -5
⇒ x = -5/2
So the solutions of the given equation are x = 11/2 and x = -5/2.
### Similar Questions
Question 1: Find all the solutions to the equation: (x – 2)2/3 = 9.
Solution:
In the given problem, we can clearly see there is the cube root and square on the variable part so the inverse operation is the cube and square root respectively.
⇒ (x – 2) = 9(3/2)
⇒ (x – 2) = (93)(1/2)
⇒ (x – 2) = (729)(1/2)
So, the square root of 729 is -27 and +27. So we have two cases.
Case 1:
⇒ (x – 2) = 27
⇒ x = 27 + 2
⇒ x = 29
Case 2:
⇒ (x – 2) = -27
⇒ x = -27 + 2
⇒ x = -25
So the solution of the given equation is x = +29 and x = -25.
Question 2: Find all the solutions to the equation: (3x – 2)2/3 = 1.
Solution:
In the given problem, we can clearly see there is the cube root and square on the variable part so the inverse operation is the cube and square root respectively.
⇒ (3x – 2) = 1(3/2)
⇒ (3x – 2) = (13)(1/2)
⇒ (3x – 2) = 1(1/2)
So, the square root of 1 is +1 and -1. So we have two cases.
Case 1:
⇒ 3x – 2 = 1
⇒ 3x = 1 + 2
⇒ 3x = 3
⇒ x = 1
Case 2:
⇒ 3x – 2 = -1
⇒ 3x = -1 + 2
⇒ 3x = +1
⇒ x = 1/3
So the solution of the given question is x = 1/3 and x = 1.
Question 3: Find all the solutions to the equation: (7x + 4)2/3 = 1.
Solution:
In the given problem, we can clearly see there is the cube root and square on the variable part so the inverse operation is the cube and square root respectively.
⇒ (7x + 4) = 1(3/2)
⇒ (7x + 4) = (13)(1/2)
⇒ (7x + 4) = 1(1/2)
So, the square root of 1 is +1 and -1. So we have two cases.
Case 1:
⇒ 7x + 4 = 1
⇒ 7x = 1 – 4
⇒ 7x = -3
⇒ x = -3/7
Case 2:
⇒ 7x + 4 = -1
⇒ 7x = -1 -4
⇒ 7x = -5
⇒ x = -5/7
So the solution of the given question is x = -3/7 and x = -5/7.
My Personal Notes arrow_drop_up | 1,578 | 4,752 | {"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} | 4.78125 | 5 | CC-MAIN-2022-21 | latest | en | 0.933002 |
http://math.stackexchange.com/questions/231918/is-all-of-mathbbr-the-only-open-set-containing-mathbbq | 1,469,661,709,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827080.38/warc/CC-MAIN-20160723071027-00317-ip-10-185-27-174.ec2.internal.warc.gz | 151,130,236 | 17,972 | # Is all of $\mathbb{R}$ the only open set containing $\mathbb{Q}$?
Is all of $\mathbb{R}$ the only open set containing $\mathbb{Q}$?
False, take any irrational $p$ in $\mathbb{R}$. Then $\mathbb{Q} \subset \mathbb{R} \setminus \{p\}$ and $\mathbb{R} \setminus \{p\}$ is open.
Of course we can take any subset $A$ of $\mathbb{R} \setminus \mathbb{Q}$ that is closed in $\mathbb{R}$ and take $\mathbb{R} \setminus A$.
Is what I got right? or did I forget something?
What fact or theorem can I use?
-
What is R{p}?... – DonAntonio Nov 7 '12 at 4:31
Your reasoning is valid. – Austin Mohr Nov 7 '12 at 4:34
Maybe you meant to write $R - \{p\}$ instead of $R\{p\}$. If so I agree with your example. – coffeemath Nov 7 '12 at 4:34
Maximiliano wrote R\{p}. Markdown eats backslashes. – MJD Nov 7 '12 at 4:36
Yes. What you have written is correct. In fact, you can do much better in the sense that you can cover $\mathbb{Q}$ with open set with arbitrarily small length $\epsilon$, for instance $$\bigcup_{k=0}^{\infty} \left(q_k - \dfrac{\epsilon}{2^{k+2}},q_k + \dfrac{\epsilon}{2^{k+2}}\right)$$ – user17762 Nov 7 '12 at 4:58
Yes. What you have written is correct. In fact, you can do much better in the sense that you can cover $\mathbb{Q}$ with open set with arbitrarily small length $\epsilon$, for instance $$\mathbb{Q}_{\text{open cover}} = \bigcup_{k=0}^{\infty} \left(q_k - \dfrac{\epsilon}{2^{k+2}},q_k + \dfrac{\epsilon}{2^{k+2}} \right)$$ | 506 | 1,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2016-30 | latest | en | 0.811583 |
https://www.antiessays.com/free-essays/Lab-1-3-Bit-And-Byte-Structure-609102.html | 1,653,635,820,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662636717.74/warc/CC-MAIN-20220527050925-20220527080925-00455.warc.gz | 721,724,717 | 9,286 | # Lab 1.3: Bit and Byte Structure
399 Words2 Pages
Exercise 1.3.1 What is the decimal value of Byte 1 by itself? What is the decimal value of Byte 2 by itself? -6400 -233 Exercise 1.3.2 What is the decimal equivalent of the binary sequence in Figure 1- 12 (the combined sequence of Byte 1 and Byte 2 as a single decimal value)? How does this compare to the individual values of Byte 1 and Byte 2? -6633 -There is an increase of bits. From 8 to 16, which increase the decimal number. Exercise 1.3.3 Given a device with a storage capacity of 120 MB, how many bytes can be stored on this device? Show your calculations. -1024KB*120=122880 122880KB*1024=125829120B Exercise 1.3.4 Given a computer with a disk capacity of 16 GB and a word size of 32 bits, how many words can be stored on the disk? Show your calculations. -16*1024MB=16384MB -16384MB*1024KB=16777216KB -16777216KB*1024B=1.71798918E10 -1.7179891E10/32=536870912 words Exercise 1.3.5 Represent the binary value 110110 2 in hexadecimal. Show the steps of conversion that you used. -0011 0110 0011 represents 3 0110 represents 6 - 36 Exercise 1.3.6 Represent the hexadecimal value f6 16 in binary and decimal. Show the steps of conversion that you used. - The best way I feel to do this propose is to change it to binary first. - F is 1111 - 6 is 0110 - 11110110 in binary - Then do the decimal step , 246 Lab 1.3 Reviews Explain why it is important to know how many system words will fit in a primary storage device on a computer (such as the hard drive). -So that you know how much ad primary storage unit can hold. Explain why more information can be contained in multiple bytes joined together than in a single byte. -Multiple bytes has more ones and zero as bytes with holds more information a single byte only has a series of 8 bit which has less data then a series of multiple sets of 8bit. Explain why it is more | 522 | 1,878 | {"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} | 4.4375 | 4 | CC-MAIN-2022-21 | longest | en | 0.868906 |
https://www.sololearn.com/ru/Discuss/764731/how-the-answer-is-32-for-25 | 1,720,917,179,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514527.38/warc/CC-MAIN-20240714002551-20240714032551-00794.warc.gz | 861,730,967 | 179,726 | how the answer is 32 for 2**5? | Sololearn: Learn to code for FREE!
+ 2
# how the answer is 32 for 2**5?
5th Oct 2017, 12:04 AM
Su Jun Li
6 ответов
+ 3
exponentiation. 2**5 is 2 multiplied by its self 5 times. 2*2*2*2*2
5th Oct 2017, 12:13 AM
(DMT)
+ 3
2*2=4*2=8*2=16*2=32 ==2*2*2*2*2 2**5= 2^5 or 5 times product 2
5th Oct 2017, 12:15 AM
0
2**5 is nothing but 2 to the power of 5 which equals to 32.
6th Oct 2017, 7:29 AM
PINK CHOUDHURY
0
Lol
6th Oct 2017, 10:27 AM
Vale
0
5 is power of 2 which is 2X2X2X2X2 So answer is 32
9th Oct 2017, 9:02 PM
정원우
- 1
it means 2 raised to the power 5 OR 2*2*2*2*2 = 32 , got it?
8th Oct 2017, 3:46 PM
Kirtxn
Актуальное сегодня
HELP ME | 335 | 673 | {"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-2024-30 | latest | en | 0.855609 |
https://www.math-only-math.com/worksheet-on-rational-numbers.html | 1,721,129,752,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514745.49/warc/CC-MAIN-20240716111515-20240716141515-00248.warc.gz | 778,138,426 | 14,378 | Worksheet on Rational Numbers
Practice the questions given in the worksheet on rational numbers.
We know, numbers that can be expressed in the form x/y, where y is a non - zero integer and x is any integer are called rational numbers.
The questions are based on rational number, useful results on rational number, positive rational number and negative rational number.
1. Write down the numerator of each of the following rational numbers:
(i) (-7)/5
(ii) 15/(-4)
(iii) (-17)/(-21)
(iv) 8/9
(v) 5
2. Write down the denominator of each of the following rational numbers:
(i) (-4)/5
(ii) 11/(-34)
(iii) (-15)/(-82)
(iv) 15
(v) 0
3. Write down the rational number whose numerator is (-3) × 4, and whose denominator is (34 - 23) × (7 - 4).
4. Write the following rational numbers as integers:
17/1, (-23)/1, 35/1, (-77)/1, 91/1.
5. Write the following integers as rational numbers with denominator 1:
-19, 27, 71, -101.
6. Write down the rational number whose numerator is the smallest four digit number and denominator is the largest five digit number.
7. Separate positive and negative rational numbers from the following rational numbers:
(-5)/(-7), 12/(-5), 7/4, 13/(-9), 0, (-18)/(-7), (-95)/116, (-1)/(-9)
8. Which of the following rational numbers are positive?
(i) (-8)/7 (ii) 9/8 (iii) (-19)/(-13) (iv) (-21)/13
9. Which of the following rational numbers are negative?
(i) (-3)/7 (ii) (-5)/-8 (iii) 9/(-83) (iv) (-115)/-197
10. Which is the following statement are true or false?
(i) Every whole number is a rational number.
(ii) Every integer is a rational number.
(ii) 0 is a whole number but it is not a rational number.
Answers for the worksheet on rational numbers are given below to check the exact answers of the above questions on rational number.
1. (i) -7
(ii) 15
(iii) -17
(iv) 8
(v) 5
2. (i) 5
(ii) -34
(iii) -82
(iv) 1
(v) any non zero integer
3. (-12)/33
4. 17, -23, 35, -77, 91
5. (-19)/1, 27/1, 71/1, (-101)/1
6. 1000/99999
7. Positive rational numbers: (-5)/(-7), 7/4, (-18)/(-7), (-1)/(-9);
Negative rational numbers: 12/(-5), 13/(-9), (-95)/116
8. (ii) 9/8
(iii) (-19)/(-13)
9. (i) (-3)/7
(iii) 9/(-83)
10. (i) true
(ii) true
(ii) false
Rational Numbers - Worksheets
Worksheet on Rational Numbers
Worksheet on Lowest form of a Rational Number
Worksheet on Standard form of a Rational Number
Worksheet on Equality of Rational Numbers
Worksheet on Comparison of Rational Numbers
Worksheet on Representation of Rational Number on a Number Line
Worksheet on Properties of Addition of Rational Numbers
Worksheet on Rational Expressions Involving Sum and Difference
Worksheet on Multiplication of Rational Number
Worksheet on Division of Rational Numbers
Worksheet on Finding Rational Numbers between Two Rational Numbers
Worksheet on Word Problems on Rational Numbers
Objective Questions on Rational Numbers
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Rational Numbers
What is Rational Numbers?
Is Every Rational Number a Natural Number?
Is Zero a Rational Number?
Is Every Rational Number an Integer?
Is Every Rational Number a Fraction?
Positive Rational Number
Negative Rational Number
Equivalent Rational Numbers
Equivalent form of Rational Numbers
Rational Number in Different Forms
Properties of Rational Numbers
Lowest form of a Rational Number
Standard form of a Rational Number
Equality of Rational Numbers using Standard Form
Equality of Rational Numbers with Common Denominator
Equality of Rational Numbers using Cross Multiplication
Comparison of Rational Numbers
Rational Numbers in Ascending Order
Representation of Rational Numbers on the Number Line
Addition of Rational Number with Same Denominator
Addition of Rational Number with Different Denominator
Properties of Addition of Rational Numbers
Subtraction of Rational Number with Different Denominator
Subtraction of Rational Numbers
Properties of Subtraction of Rational Numbers
Simplify Rational Expressions Involving the Sum or Difference
Multiplication of Rational Numbers
Properties of Multiplication of Rational Numbers
Rational Expressions Involving Addition, Subtraction and Multiplication
Division of Rational Numbers
Rational Expressions Involving Division
Properties of Division of Rational Numbers
To Find Rational Numbers | 1,447 | 5,889 | {"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} | 4.03125 | 4 | CC-MAIN-2024-30 | latest | en | 0.781396 |
https://www.neetprep.com/questions/55-Physics/684-Mechanical-Properties-Solids?courseId=8&testId=1061963-Past-Year--onward--NTA-Papers-MCQs | 1,722,872,924,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00308.warc.gz | 723,259,795 | 54,058 | The maximum elongation of a steel wire of $$1~\text m$$ length if the elastic limit of steel and its Young's modulus, respectively, are $$8 × 10^8 ~\text{N m}^{-2 }$$ and $$2 × 10^{11} ~\text{N m}^{-2},$$ is:
1. $$0.4~\text{mm}$$
2. $$40~\text{mm}$$
3. $$8~\text{mm}$$
4. $$4~\text{mm}$$
Subtopic: Young's modulus |
69%
From NCERT
NEET - 2024
Hints
Let a wire be suspended from the ceiling (rigid support) and stretched by a weight $$W$$ attached at its free end. The longitudinal stress at any point of cross-sectional area $$A$$ of the wire is:
1. zero
2. $$\dfrac{2W}{A}$$
3. $$\dfrac{W}{A}$$
4. $$\dfrac{W}{2A}$$
Subtopic: Stress - Strain |
65%
From NCERT
NEET - 2023
Hints
The amount of elastic potential energy per unit volume (in SI unit) of a steel wire of length $$100$$ cm to stretch it by $$1$$ mm is: (Given: Young's modulus of the wire = $$2.0\times 10^{11}$$ Nm-2)
1 $$10^{11}$$ 2 $$10^{17}$$ 3 $$10^{7}$$ 4 $$10^{5}$$
Subtopic: Potential energy of wire |
62%
From NCERT
NEET - 2023
Hints
Given below are two statements:
Assertion (A): The stretching of a spring is determined by the shear modulus of the material of the spring. Reason (R): A coil spring of copper has more tensile strength than a steel spring of the same dimensions.
1 Both (A) and (R) are True and (R) is the correct explanation of (A). 2 Both (A) and (R) are True but (R) is not the correct explanation of (A). 3 (A) is False but (R) is True. 4 (A) is True but (R) is False.
Subtopic: Shear and bulk modulus |
From NCERT
NEET - 2022
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
A wire of length $$L$$, area of cross section $$A$$ is hanging from a fixed support. The length of the wire changes to $$\mathrm{L}_1$$when mass $$M$$ is suspended from its free end. The expression for Young's modulus is:
1 $$\frac{{Mg(L}_1-{L)}}{{AL}}$$ 2 $$\frac{{MgL}}{{AL}_1}$$ 3 $$\frac{{MgL}}{{A(L}_1-{L})}$$ 4 $$\frac{{MgL}_1}{{AL}}$$
Subtopic: Young's modulus |
78%
From NCERT
NEET - 2020
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NEET 2025 - Target Batch
Hints
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NEET 2025 - Target Batch
When a block of mass $$M$$ is suspended by a long wire of length $$L$$, the length of the wire becomes ($$L+l$$). The elastic potential energy stored in the extended wire is:
1 $${1}/{2}~MgL$$ 2 $$Mgl$$ 3 $$MgL$$ 4 $${1}/{2}~Mgl$$
Subtopic: Potential energy of wire |
69%
From NCERT
NEET - 2019
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Hints
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The stress-strain curves are drawn for two different materials $$X$$ and $$Y.$$ It is observed that the ultimate strength point and the fracture point are close to each other for material $$X$$ but are far apart for material $$Y.$$ We can say that the materials $$X$$ and $$Y$$ are likely to be (respectively):
1 ductile and brittle 2 brittle and ductile 3 brittle and plastic 4 plastic and ductile
Subtopic: Stress - Strain Curve |
81%
From NCERT
NEET - 2019
To view explanation, please take trial in the course.
NEET 2025 - Target Batch
Hints | 1,061 | 3,269 | {"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} | 3.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.658796 |
https://en.wikipedia.org/wiki/Trivial_hash_function | 1,580,170,868,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251737572.61/warc/CC-MAIN-20200127235617-20200128025617-00267.warc.gz | 429,509,252 | 11,032 | # Index mapping
(Redirected from Trivial hash function)
Index mapping (or direct addressing, or a trivial hash function) in computer science describes using an array, in which each position corresponds to a key in the universe of possible values.[1] The technique is most effective when the universe of keys is reasonably small, such that allocating an array with one position for every possible key is affordable. Its effectiveness comes from the fact that an arbitrary position in an array can be examined in constant time.
## Applicable arrays
There are many practical examples of data whose valid values are restricted within a small range. A trivial hash function is a suitable choice when such data needs to act as a lookup key. Some examples include:
• month in the year (1–12)
• day in the month (1–31)
• day of the week (1–7)
• human age (0–130) – e.g. lifecover actuary tables, fixed term mortgage
• ASCII characters (0–127), encompassing common mathematical operator symbols, digits, punctuation marks and English language alphabet
## Examples
Using a trivial hash function, in a non-iterative table lookup, can eliminate conditional testing and branching completely, reducing the instruction path length of a computer program.
### Avoid branching
Roger Sayle gives an example[2] of eliminating a multiway branch caused by a switch statement:
```inline bool HasOnly30Days(int m)
{
switch (m) {
case 4: // April
case 6: // June
case 9: // September
case 11: // November
return true;
default:
return false;
}
}
```
Which can be replaced with a table lookup:
```inline bool HasOnly30Days(int m)
{
static const bool T[] = { 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0 };
return T[m-1];
}
``` | 414 | 1,704 | {"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-2020-05 | latest | en | 0.809925 |
https://algorithm.zone/blogs/codeforces-round-572-div2-a-e-problem-solution.html | 1,675,598,025,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500251.38/warc/CC-MAIN-20230205094841-20230205124841-00820.warc.gz | 110,153,617 | 6,223 | # Codeforces Round #572 Div2 A~E problem solution
Site link: Codeforces Round #572 Div2
# gossip
It was a great fight, with 148 in the end However, this is mainly a conclusion question, so I won't do much explanation D2 is more immortal, so you can fill it out later
# A. Keanu Reeves
Given a binary string s, defining a string is awesome if and only if the number of zeros and 1s in the string is different Now we want to divide this s into several continuous substrings, and each substring is awesome. At least we should divide it into several segments and output the specific scheme
Data range:
$$1 \leq |s| \leq 100$$
## thinking
Because the data range is too small, it can be messed up Because each substring must be continuous, an obvious greedy idea is to look back from the beginning to see where the right endpoint of the longest substring can be found. We can find that this is correct, because the combination of two shorter schemes must correspond to a scheme with the longest plus one section
## code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
ios::sync_with_stdio(0);cin.tie(0);
int n;cin >> n;
string s;cin >> s;
vector<string> p;
for(int i = 0;i < n;++i)
{
if(s[i] == '#') continue;
int m = i,z = 0,o = 0;
for(int j = i;j < n;++j)
{
if(s[j] == '0') ++o;
else ++z;
if(z != o) m = j;
}
string res;
for(int j = i;j <= m;++j)
{
res += s[j];
s[j] = '#';
}
p.push_back(res);
}
cout << p.size() << endl;
for(auto& v : p) cout << v << " ";
return 0;
}
# B. Number Circle
Given a sequence a of \ (n \) elements, it is required that each element on the whole ring satisfies that the sum of two adjacent elements is strictly greater than him It is required to construct a specific scheme or explain that there is no solution
## thinking
One of the most direct ideas is to sort and output directly But obviously this question can't be done like this. For example, the data: 5 2 3 5 7 If you sort directly, you will think that 7 is illegal, because the left and right elements are 2 and 5 respectively But there is obviously another construction scheme: 732335 is legal The problem is that the smallest and the second largest may be the same as the largest. In other words, the construction scheme of this problem should be an evenly distributed sum, rather than a direct ranking sum from small to large
Therefore, the idea is relatively clear. Since it is necessary to distribute evenly, sort first, and then distinguish the elements according to parity. Finally, judge whether there is a solution and output it
## code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+7;
int a[N];
int main()
{
int n;scanf("%d",&n);
for(int i = 1;i <= n;++i) scanf("%d",&a[i]);
sort(a + 1,a + n + 1);
vector<int> res;
for(int i = 1;i <= n;i += 2) res.push_back(a[i]);
reverse(res.begin(),res.end());
// for(auto& v : res) cout << v << " ";cout << endl;
for(int i = 2;i <= n;i += 2) res.push_back(a[i]);
for(int i = 1;i <= n;++i) a[i] = res[i - 1];
// for(int i = 1;i <= n;++i) cout << a[i] << " ";cout << endl;
int ok = 1;
if(a[n] + a[2] <= a[1]) ok = 0;
if(a[n - 1] + a[1] <= a[n]) ok = 0;
for(int i = 2;i <= n - 1;++i)
if(a[i - 1] + a[i + 1] <= a[i])
ok = 0;
if(!ok) puts("NO");
else
{
puts("YES");
for(int i = 1;i <= n;++i) printf("%d ",a[i]);
}
return 0;
}
# C. Candies!
Given a sequence a, make sure its length is a power of 2 An operation is now specified. For a subsequence whose interval length is a power of 2, every two consecutive adjacent elements are divided into a group, and then the sum of two elements in each group is calculated. If a sum is greater than 10, a candy is added, and the sum \ (\% 10 \) is the new value, and the process is repeated until there is only one element in the whole sequence Give \ (q \) queries. Each query contains an interval. Ask this interval and you can finally get several sweets
Data range:
$$1 \leq n \leq 10^5$$
$$1 \leq s_i \leq 9$$
$$1 \leq q \leq 10^5$$
## thinking
Because the data range of this question is too strong, it can be estimated that the processing complexity of each query is either \ (O(1) \) or \ (O(logn) \) Since it is the sum of the whole interval, it is not difficult to think that it should be the sum of the whole interval. The contribution of the whole interval is that the whole interval contains several \ (10 \) Just divide it
Why is he right It can be understood perceptually: if the sum is less than 10, then this part will inevitably pass to the next. If it exceeds 10, then the whole 10 will be removed and then go to the next step. No matter what the situation is, there will not be a situation where the combination of both sides exceeds 10, but after taking the mold respectively, it is only 10, that is, the situation of contradiction
## code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+7;
int a[N];
int main()
{
int n;scanf("%d",&n);
for(int i = 1;i <= n;++i) scanf("%d",&a[i]);
for(int i = 1;i <= n;++i) a[i] += a[i - 1];
int q;scanf("%d",&q);
while(q--)
{
int l,r;scanf("%d%d",&l,&r);
printf("%d\n",(a[r] - a[l - 1]) / 10);
}
return 0;
}
# D1. Add on a Tree
Given a rootless tree with \ (n \) points, two leaf nodes can be found at a time. Add or subtract a real number from all edges on the simple path between the two leaf nodes Q: for this tree, is it possible for any situation to occur in a limited number of operations, that is, there is no situation that cannot occur in a limited number of operations at all
Data range:
$$2 \leq n \leq 10^5$$
## thinking
At first glance, this topic is very awesome However, after simulating the example, it can be found that for a point with degree 2, if one side is 1 and the other side is 0, the problem will not be solved, because the two sides must pass through together, and it is impossible for one side to be different from the other side However, this conclusion cannot be generalized. For example, it is possible to have a point with a degree of \ (4 \), and a point is connected with four points. In this case, there is a solution Continue to guess that this should be the conclusion of the whole topic Just check it directly
## code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5+7;
int deg[N];
int main()
{
int n;scanf("%d",&n);
for(int i = 1;i < n;++i)
{
int u,v;scanf("%d%d",&u,&v);
++deg[u];++deg[v];
}
int ok = 1;
for(int i = 1;i <= n;++i)
{
if(deg[i] == 2)
{
ok = 0;
break;
}
}
if(!ok) puts("NO");
else puts("YES");
return 0;
}
D2 is too awesome. Write it later
# E. Count Pairs
Given a sequence a with a length of \ (n \), find the number that satisfies \ (1 \ Leq I < J \ Leq n \) and \ ((a_i+a_j)*(a_i^2+a_j^2)\equiv k (mod p) \), where \ (k \) is a positive integer and \ (P \) is a prime number
Data range:
$$2 \leq n \leq 3*10^5$$
$$2 \leq p \leq 10^9$$
$$0 \leq k \leq n - 1$$
$$0 \leq a_i \leq p - 1$$
## thinking
This form is sum of squares on one side and sum on the other, which is a little ugly There are several directions you can try, such as cubic and expansion. After trying some, you find that they are not very good The correct idea is to multiply both sides by a \ ((a_i - a_j) \) so that the left is a square difference and a square sum, and then continue to reduce the form to obtain this formula: \ (a_i^4-a_j^4\equiv k(a_i-a_j) (mod p) \), continue to pull the right to the left, and find that this expression is equivalent to \ (p| a_i ^ 4-a_j ^ 4-K (a_i-a_j) \), and continue to get a centralized form: \ (p| a_i (a| I ^ 3-K) - A_ J (a_j ^ 3-K) \) is then equivalent to defining a \ (c_i = a_i(a_i^3-k) \), and requires that for each \ (c_i \), how many \ (c_j \) before him satisfy the congruence with its module \ (P \) Use \ (map \) to make statistics
Pay attention to the form of subtraction and prevent negative numbers from taking remainder You can easily get through this problem by typing a fast power and fast multiplication
## code
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 3e5+7;
ll a[N];
ll qmul(ll a, ll b, ll P)
{
ll L = a * (b >> 25LL) % P * (1LL << 25) % P;
ll R = a * (b & ((1LL << 25) - 1)) % P;
return (L + R) % P;
}
ll qpow(ll a,ll b,ll p)
{
ll res = 1 % p;
while(b)
{
if(b & 1) res = qmul(res,a,p);
a = qmul(a,a,p);
b >>= 1;
}
return res;
}
int main()
{
ios::sync_with_stdio(0);cin.tie(0);
ll n,p,k;cin >> n >> p >> k;
map<ll,int> tb;
int res = 0;
for(int i = 1;i <= n;++i) cin >> a[i];
for(int i = 1;i <= n;++i)
{
ll r = ((qpow(a[i],3,p) - k) % p + p) % p;
r = qmul(r,a[i],p);
if(tb.count(r)) res += tb[r];
++tb[r];
}
cout << res;
return 0;
}
Tags: CodeForces
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Chapter 13:
Applications of Hadamard Matrices in Communication Systems
Abstract
Modern communication systems and digital signal processing (signal modeling), image compression and image encoding,3 and digital signal processing systems are heavily reliant on statistical techniques to recover information in the presence of noise and interference. One of the mathematical structures used to achieve this goal is the Hadamard matrix. Historically, Plotkin18 first showed the error-correcting capabilities of codes generated from Hadamard matrices. Later, Bose and Shrikhande found the connection between Hadamard matrices and symmetrical block code designs. In this chapter, we will discuss some of these applications in error-control coding and in CDMAs.
13.1 Hadamard Matrices and Communication Systems
13.1.1 Hadamard matrices and error-correction codes
The storage and transmission of digital data lies at the heart of modern computers and communications systems. When a message is transmitted, it has the potential to become scrambled by noise. The goal of this section is to provide a brief introduction to the basic definitions, goals, and constructions in coding theory. We describe some of the classical algebraic constructions of error-correcting codes, including the Hadamard codes. The Hadamard codes are relatively easy to decode; they are the first large class of codes to correct more than a single error. A Hadamard code was used in the Mariner and Voyager space probes to encode information transmitted back to the Earth when the probes visited Mars and the outer planets of the solar system from 1969 to 1976. Mariner 9 was a space shuttle whose mission was to fly to Mars and transmit pictures back to Earth. Fig. 13.1 is one of the pictures transmitted by 9. With Mariner 5, six-bit pixels were encoded using 32-bit long Hadamard code that could correct up to seven errors.
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CHAPTER 13
30 PAGES
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A265484 Numbers n such that 36^n - 6^n - 1 is prime. 1
1, 2, 3, 4, 24, 63, 81, 92, 165, 232, 236, 591, 669, 1343, 2973, 3630, 5600, 7019, 17169, 31945, 52172 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS For n = 1, 2, 3, 4, 24 the corresponding primes are 29, 1259, 46439, 1678319, 22452257707354557235348829785471057919. a(22) > 10^5. - Robert Price, Jan 06 2020 LINKS Table of n, a(n) for n=1..21. EXAMPLE 4 is in the sequence because 36^4 - 6^4 - 1 = 1678319 is prime. MATHEMATICA Select[Range[1000], PrimeQ[36^# - 6^# - 1] &] PROG (Magma) [n: n in [0..300] | IsPrime(36^n-6^n-1)]; (PARI) for(n=1, 1e3, if(ispseudoprime(36^n - 6^n - 1), print1(n, ", "))) \\ Altug Alkan, Dec 12 2015 CROSSREFS Cf. similar sequences listed in A265481. Sequence in context: A233344 A329566 A329532 * A000336 A287433 A235041 Adjacent sequences: A265481 A265482 A265483 * A265485 A265486 A265487 KEYWORD nonn,more AUTHOR Vincenzo Librandi, Dec 12 2015 EXTENSIONS a(14)-a(18) from Ray Chandler, Sep 25 2019 a(18) corrected and a(19)-a(21) added by Robert Price, Jan 06 2020 STATUS approved
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Last modified June 23 05:53 EDT 2024. Contains 373629 sequences. (Running on oeis4.) | 586 | 1,584 | {"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-2024-26 | latest | en | 0.645297 |
http://parishkaram.org.in/83289_density-of-20-mm-stone-aggregate.html | 1,547,654,575,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583657555.87/warc/CC-MAIN-20190116154927-20190116180927-00574.warc.gz | 177,440,883 | 5,618 | # density of 20 mm stone aggregate
Density of coarse aggregate 20mm - AnswersRaw materials that are a component in concrete, aggregates are inert granular materials such as sand, gravel or crushed stone. Coarse aggregates are particles.density of 20 mm stone aggregate,DENSITY OF 20 MM STONE AGGREGATESep 26, 2016 . More Details : .pakistancrushers/contact.php Aggregate Importers Aggregate Buyers - I am looking to buy High Density Iron Ore.
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DENSITY OF AGGREGATE 20MM AND 10MM COMPACTOct 24, 2016 . Process Crusher,density of 20 mm stone aggregate - ore crusher . what is the density of crushed stone aggregate 10mm 20mm 40mm; density.density of 20 mm stone aggregate,Density of Coarse Aggregate - GharExpertThe normal weight of stone aggregate in loose state is= 1400 to 1600 kg per cum, . Percentage 40mm. Passing 20mm. (by weight) 16mm. for nominal 12.5mm.
John Frank
### How to calculate cement aggregate and sand in KG by knowing the .
All the other answers provided here gives you a detailed calculation steps and theoretical . Bulk density of coarse aggregate (20mm) is approximately 1560kg/m3. So we will . How do I calculate stone in 1 cubic metre with 1:6 cement mortar?
### Aggregates in Concrete
Crushed Stone or Manufactured Mineral Aggregate . 2.9, Bulk Density (of Bulk Unit Weight) - 1520 to 1680 kg/m3 (95 to . ▫Size: <4.75 mm; >75 μm (0.003 in.) .. Concrete Technology. 20. B. Aggregate Grading (distribution of particles of.
### IS 383 (1970): Specification for Coarse and Fine Aggregates From .
2.1.2 Crushed Stone Sand - Fine aggregate produced . 2.2 Coarse Aggregate -- Aggregate most of which is retained on 4*75-mm .. 20 mm. 16mm. 12.5 mm. 10 mm. 4.75 mm. 236mm. TABLE 2 COARSE AGGREGATES .. b) Bulk density,.
### PROPERTIES OF AGGREGATES - The Constructor
1. AGGREGATES: BULK DENSITY, SPECIFIC GRAVITY AND VOIDS BULK DENSITY River sand Fine 1. . Broken stone, graded. 25 mm maximum size. 46 . 5.0. 80. Moist sand. 2.5. 40. Moist gravel or crushed rock. 1.25 – 2.5. 20 – 40.
### Density of coarse aggregate 20mm - Answers
Raw materials that are a component in concrete, aggregates are inert granular materials such as sand, gravel or crushed stone. Coarse aggregates are particles.
### Density of Coarse Aggregate - GharExpert
The normal weight of stone aggregate in loose state is= 1400 to 1600 kg per cum, . Percentage 40mm. Passing 20mm. (by weight) 16mm. for nominal 12.5mm.
### What is bulk density of cement, aggregate, sand? - Quora
The bulk density is the mass of the material related to a specific volume and for cement is . Bulk Density of coarse aggregate 20mm 1560 Kg/M3 HOW TO TEST AGGREGATE FOR BULK DENSITY AND VOIDS? 83.1k Views · View Upvotes · Answer.
### How to calculate cement aggregate and sand in KG by knowing the .
All the other answers provided here gives you a detailed calculation steps and theoretical . Bulk density of coarse aggregate (20mm) is approximately 1560kg/m3. So we will . How do I calculate stone in 1 cubic metre with 1:6 cement mortar?
### Aggregates in Concrete
Crushed Stone or Manufactured Mineral Aggregate . 2.9, Bulk Density (of Bulk Unit Weight) - 1520 to 1680 kg/m3 (95 to . ▫Size: <4.75 mm; >75 μm (0.003 in.) .. Concrete Technology. 20. B. Aggregate Grading (distribution of particles of.
### IS 383 (1970): Specification for Coarse and Fine Aggregates From .
2.1.2 Crushed Stone Sand - Fine aggregate produced . 2.2 Coarse Aggregate -- Aggregate most of which is retained on 4*75-mm .. 20 mm. 16mm. 12.5 mm. 10 mm. 4.75 mm. 236mm. TABLE 2 COARSE AGGREGATES .. b) Bulk density,.
### PROPERTIES OF AGGREGATES - The Constructor
1. AGGREGATES: BULK DENSITY, SPECIFIC GRAVITY AND VOIDS BULK DENSITY River sand Fine 1. . Broken stone, graded. 25 mm maximum size. 46 . 5.0. 80. Moist sand. 2.5. 40. Moist gravel or crushed rock. 1.25 – 2.5. 20 – 40.
### How we calculate of Sand, cement and aggregate of M20 ratio or .
Feb 16, 2012 . M20 (1 cement :1.5 sand :3 stone/brick aggregate). . Once the W/C ratio is determined use 186 kg water (for 20mm CA) to find out the qty. of.
density of 20 mm stone aggregate,
### Pumice, the natural lightweight aggregate
Pumice stone is a natural lightweight aggregate which is formed by the . TO FINEST. COARSE : 16-32 mm . DRY LOOSE BULK DENSITY. From 250 kg/m3 . Lightweight concrete for floors C20-C30 1,300 kg/m³ - 1,700 kg/m³. Lightweight.
### Experimental Study on Achievement of Density and Strength in Light .
Floating concrete, Pumice stone, Aluminum powder, Fly ash, Density, Compressive strength saw dust . 392 17.422. Aggregate – Pumice Stones – 10 to 20 mm.
density of 20 mm stone aggregate,
### cbr value of sandy subgrade blended with coarse aggregate
to 115.83% and 31.30% to 151.94% for 10mm and 20mm coarse aggregates respectively at OMC. In soaked . discrete natural fiber in sand at a density of 16.60. KN/m3 there is . lime & stone dust on CBR value of black cotton soil with the.
### Aggregate - Afrisam
Concrete aggregates include a wide range of stone sizes within target gradings as . 6-mm Flats .. The apparent relative density of aggregate generally varies .. 0 - 20. 0 - 50. 85 - 100. 100. 9,5. 0 - 5. 0 - 5. 0 - 25. 0 - 55. 85 - 100. 100. 6,7.
### Physiochemical characterization of coarse aggregates in Qatar for .
Jul 27, 2013 . 15–20 mm (also known as 20 mm), Aggregates in concrete and asphalt .. Relative density, Oven dry, 2.86, 2.85, 2.69, 2.68, 2.46, 2.60, 2.49, 2.36 ... comparison to imported lime stone and gabbro aggregates; thereby failing.
### Farm structures . - Ch3 Building materials: Concrete - FAO
Normal values for density of aggregate (sand and stone) are 2600 to 2700 kg/ m3 and . aggregate must be able to pass between the reinforcement bars, 20mm. | 1,667 | 5,767 | {"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-2019-04 | longest | en | 0.760628 |
https://www.r-bloggers.com/2011/03/r-tutorial-series-anova-pairwise-comparison-methods/ | 1,726,558,506,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651750.27/warc/CC-MAIN-20240917072424-20240917102424-00032.warc.gz | 871,986,707 | 25,564 | Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
When we have a statistically significant effect in ANOVA and an independent variable of more than two levels, we typically want to make follow-up comparisons. There are numerous methods for making pairwise comparisons and this tutorial will demonstrate how to execute several different techniques in R.
### Tutorial Files
Before we begin, you may want to download the sample data (.csv) used in this tutorial. Be sure to right-click and save the file to your R working directory. This dataset contains a hypothetical sample of 30 participants who are divided into three stress reduction treatment groups (mental, physical, and medical). The values are represented on a scale that ranges from 1 to 5. This dataset can be conceptualized as a comparison between three stress treatment programs, one using mental methods, one using physical training, and one using medication. The values represent how effective the treatment programs were at reducing participant’s stress levels, with higher numbers indicating higher effectiveness.
### Beginning Steps
To begin, we need to read our dataset into R and store its contents in a variable.
1. > #read the dataset into an R variable using the read.csv(file) function
3. > #display the data
4. > dataPairwiseComparisons
The first ten rows of our dataset
### Omnibus ANOVA
For the purposes of this tutorial, we will assume that the omnibus ANOVA has already been conducted and that the main effect for treatment was statistically significant. For details on this process, see the One-Way ANOVA with Pairwise Comparisons tutorial, which uses the same dataset.
### Means
Let’s also look at the means of our treatment groups. Here, we will use the tapply() function, along with the following arguments, to generate a table of means.
• X: the data
• INDEX: a list() of factor variables
• FUN: the function to be applied
1. > #use tapply(X, INDEX, FUN) to generate a table displaying each treatment group mean
2. > tapply(X = dataPairwiseComparisons\$StressReduction, INDEX = list(dataPairwiseComparisons\$Treatment), FUN = mean)
The treatment group means
### Pairwise Comparisons
We will cover five major techniques for controlling Type I error when making pairwise comparisons. These methods are no adjustment, Bonferroni’s adjustment, Holm’s adjustment, Fisher’s LSD, and Tukey’s HSD. All of these techniques will be demonstrated on our sample dataset, although the decision as to which to use in a given situation is left up to the reader.
#### pairwise.t.test()
Our first three methods will make use of the pairwise.t.test() function, which has the following major arguments.
• x: the dependent variable
• g: the independent variable
• p.adj: the p-value adjustment method used to control for the family-wise Type I error rate across the comparisons; one of “none”, “bonferroni”, “holm”, “hochberg”, “hommel”, “BH”, or “BY”
Using p.adj = “none” in the pairwise.t.test() function makes no correction for the Type I error rate across the pairwise tests. This technique can be useful for employing methods that are not already built into R functions, such as the Shaffer/Modified Shaffer, which use different alpha level divisors based on the number of levels composing the independent variable. The console results will contain no adjustment, but the researcher can manually consider the statistical significance of the p-values under his or her desired alpha level.
1. > #use pairwise.t.test(x, g, p.adj) to test the pairwise comparisons between the treatment group means
3. > pairwise.t.test(dataPairwiseComparisons\$StressReduction, dataPairwiseComparisons\$Treatment, p.adj = “none”)
Pairwise comparisons of treatment group means with no adjustment
With no adjustment, the mental-medical and physical-medical comparisons are statistically significant, whereas the mental-physical comparison is not. This suggests that both the mental and physical treatments are superior to the medical treatment, but that there is insufficient statistical support to distinguish between the mental and physical treatments.
The Bonferroni adjustment simply divides the Type I error rate (.05) by the number of tests (in this case, three). Hence, this method is often considered overly conservative. The Bonferroni adjustment can be made using p.adj = “bonferroni” in the pairwise.t.test() function.
2. > pairwise.t.test(dataPairwiseComparisons\$StressReduction, dataPairwiseComparisons\$Treatment, p.adj = “bonferroni”)
Pairwise comparisons of treatment group means using Bonferroni adjustment
Using the Bonferroni adjustment, only the mental-medical comparison is statistically significant. This suggests that the mental treatment is superior to the medical treatment, but that there is insufficient statistical support to distinguish between the mental and physical treatments and the physical and medical treatments. Notice that these results are more conservative than with no adjustment.
The Holm adjustment sequentially compares the lowest p-value with a Type I error rate that is reduced for each consecutive test. In our case, this means that our first p-value is tested at the .05/3 level (.017), second at the .05/2 level (.025), and third at the .05/1 level (.05). This method is generally considered superior to the Bonferroni adjustment and can be employed using p.adj = “holm” in the pairwise.t.test() function.
2. > pairwise.t.test(dataPairwiseComparisons\$StressReduction, dataPairwiseComparisons\$Treatment, p.adj = “holm”)
Pairwise comparisons of treatment group means using Holm adjustment
Using the Holm procedure, our results are practically (but not mathematically) identical to using no adjustment.
#### LSD Method
The Fisher Least Significant Difference (LSD) method essentially does not correct for the Type I error rate for multiple comparisons and is generally not recommended relative to other options. However, should the need arise to employ this method, one should seek out the LSD.test() function in the agricolae package, which has the following major arguments.
• y: the dependent variable
• trt: the independent variable
• DFerror: the degrees of freedom error
• MSerror: the mean squared error
Note that the DFerror and MSerror can be found in the omnibus ANOVA table.
1. > #load the agricolae package (install first, if necessary)
2. > library(agricolae)
3. #LSD method
4. #use LSD.test(y, trt, DFerror, MSerror) to test the pairwise comparisons between the treatment group means
5. > LSD.test(dataPairwiseComparisons\$StressReduction, dataPairwiseComparisons\$Treatment, 30.5, 1.13)
Pairwise comparisons of treatment group means using LSD method
Using the LSD method, our results are practically (but not mathematically) identical to using no adjustment or the Holm procedure.
#### HSD Method
The Tukey Honest Significant Difference (HSD) method controls for the Type I error rate across multiple comparisons and is generally considered an acceptable technique. This method can be executed using the TukeyHSD(x) function, where x is a linear model object created using the aov(formula, data) function. Note that in this application, the aov(formula, data) function is identical to the lm(formula, data) that we are already familiar with from linear regression.
1. > #HSD method
2. > #use TukeyHSD(x), in tandem with aov(formula, data), to test the pairwise comparisons between the treatment group means
3. TukeyHSD(aov(StressReduction ~ Treatment, dataPairwiseComparisons))
Pairwise comparisons of treatment group means using HSD method
Using the HSD method, our results are practically (but not mathematically) identical to using the Bonferroni, Holm, or LSD methods.
### Complete Pairwise Comparisons Example
To see a complete example of how various pairwise comparison techniques can be applied in R, please download the ANOVA pairwise comparisons example (.txt) file. | 1,733 | 7,963 | {"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} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.885495 |
https://www.kilomegabyte.com/11-kib-to-gibit | 1,716,614,682,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058773.28/warc/CC-MAIN-20240525035213-20240525065213-00473.warc.gz | 737,125,747 | 5,838 | #### Data Units Calculator
###### Kibibyte to Gibibit
Online data storage unit conversion calculator:
From:
To:
The smallest unit of measurement used for measuring data is a bit. A single bit can have a value of either zero(0) or one(1). It may contain a binary value (such as True/False or On/Off or 1/0) and nothing more. Therefore, a byte, or eight bits, is used as the fundamental unit of measurement for data storage. A byte can store 256 different values, which is sufficient to represent standard ASCII table, such as all numbers, letters and control symbols.
Since most files contain thousands of bytes, file sizes are often measured in kilobytes. Larger files, such as images, videos, and audio files, contain millions of bytes and therefore are measured in megabytes. Modern storage devices can store thousands of these files, which is why storage capacity is typically measured in gigabytes or even terabytes.
# 11 kib to gibit result:
11 (eleven) kibibyte(s) is equal 0.00008392333984375 (zero point zero × 4 eight trillion three hundred and ninety-two billion three hundred and thirty-three million nine hundred and eighty-four thousand three hundred and seventy-five) gibibit(s)
#### What is kibibyte?
The kibibyte is a multiple of the unit byte for quantities of digital information. The binary prefix kibi means 2^10, or 1024; therefore, 1 kibibyte is 1024 bytes. The unit symbol for the kibibyte is KiB.
#### What is gibibit?
The gibibit is a multiple of the bit, a unit of information, prefixed by the standards-based multiplier gibi (symbol Gi), a binary prefix meaning 2^30. The unit symbol of the gibibit is Gibit. 1 gibibit = 2^30 bits = 1073741824 bits = 1024 mebibits
#### How calculate kib. to gibit.?
1 Kibibyte is equal to 0.00000762939453125 Gibibit (zero point zero × 5 seven hundred and sixty-two billion nine hundred and thirty-nine million four hundred and fifty-three thousand one hundred and twenty-five gibit)
1 Gibibit is equal to 131072 Kibibyte (one hundred and thirty-one thousand and seventy-two kib)
1 Kibibyte is equal to 8192.00000000000000 bits (eight thousand one hundred and ninety-two point zero × 14 zero bits)
1 Gibibit is equal to 1073741824 bits (one billion seventy-three million seven hundred and forty-one thousand eight hundred and twenty-four bits)
11 Kibibyte is equal to 90112 Bit (ninety thousand one hundred and twelve bit)
Gibibit is greater than Kibibyte
Multiplication factor is 131072.
1 / 131072 = 0.00000762939453125.
11 / 131072 = 0.00008392333984375.
Maybe you mean Kilobyte?
11 Kibibyte is equal to 11.264 Kilobyte (eleven point two hundred and sixty-four kb) convert to kb
### Powers of 2
kib gibit (Gibibit) Description
1 kib 0.00000762939453125 gibit 1 kibibyte (one) is equal to 0.00000762939453125 gibibit (zero point zero × 5 seven hundred and sixty-two billion nine hundred and thirty-nine million four hundred and fifty-three thousand one hundred and twenty-five)
2 kib 0.0000152587890625 gibit 2 kibibyte (two) is equal to 0.0000152587890625 gibibit (zero point zero × 4 one hundred and fifty-two billion five hundred and eighty-seven million eight hundred and ninety thousand six hundred and twenty-five)
4 kib 0.000030517578125 gibit 4 kibibyte (four) is equal to 0.000030517578125 gibibit (zero point zero × 4 thirty billion five hundred and seventeen million five hundred and seventy-eight thousand one hundred and twenty-five)
8 kib 0.00006103515625 gibit 8 kibibyte (eight) is equal to 0.00006103515625 gibibit (zero point zero × 4 six billion one hundred and three million five hundred and fifteen thousand six hundred and twenty-five)
16 kib 0.0001220703125 gibit 16 kibibyte (sixteen) is equal to 0.0001220703125 gibibit (zero point zero × 3 one billion two hundred and twenty million seven hundred and three thousand one hundred and twenty-five)
32 kib 0.000244140625 gibit 32 kibibyte (thirty-two) is equal to 0.000244140625 gibibit (zero point zero × 3 two hundred and forty-four million one hundred and forty thousand six hundred and twenty-five)
64 kib 0.00048828125 gibit 64 kibibyte (sixty-four) is equal to 0.00048828125 gibibit (zero point zero × 3 forty-eight million eight hundred and twenty-eight thousand one hundred and twenty-five)
128 kib 0.0009765625 gibit 128 kibibyte (one hundred and twenty-eight) is equal to 0.0009765625 gibibit (zero point zero × 3 nine million seven hundred and sixty-five thousand six hundred and twenty-five)
256 kib 0.001953125 gibit 256 kibibyte (two hundred and fifty-six) is equal to 0.001953125 gibibit (zero point zero × 2 one million nine hundred and fifty-three thousand one hundred and twenty-five)
512 kib 0.00390625 gibit 512 kibibyte (five hundred and twelve) is equal to 0.00390625 gibibit (zero point zero × 2 three hundred and ninety thousand six hundred and twenty-five)
1024 kib 0.0078125 gibit 1024 kibibyte (one thousand and twenty-four) is equal to 0.0078125 gibibit (zero point zero × 2 seventy-eight thousand one hundred and twenty-five)
2048 kib 0.015625 gibit 2048 kibibyte (two thousand and forty-eight) is equal to 0.015625 gibibit (zero point zero × 1 fifteen thousand six hundred and twenty-five)
4096 kib 0.03125 gibit 4096 kibibyte (four thousand and ninety-six) is equal to 0.03125 gibibit (zero point zero × 1 three thousand one hundred and twenty-five)
8192 kib 0.0625 gibit 8192 kibibyte (eight thousand one hundred and ninety-two) is equal to 0.0625 gibibit (zero point zero × 1 six hundred and twenty-five)
### Convert Kibibyte to other units
kib System Description
11 kib 90112 bit 11 kibibyte (eleven) is equal to 90112 bit (ninety thousand one hundred and twelve)
11 kib 11264 b 11 kibibyte (eleven) is equal to 11264 byte (eleven thousand two hundred and sixty-four)
11 kib 11.264 kb 11 kibibyte (eleven) is equal to 11.264 kilobyte (eleven point two hundred and sixty-four)
11 kib 0.011264 mb 11 kibibyte (eleven) is equal to 0.011264 megabyte (zero point zero × 1 eleven thousand two hundred and sixty-four)
11 kib 0.000011264 gb 11 kibibyte (eleven) is equal to 0.000011264 gigabyte (zero point zero × 4 eleven thousand two hundred and sixty-four)
11 kib 0.000000011264 tb 11 kibibyte (eleven) is equal to 0.000000011264 terabyte (zero point zero × 7 eleven thousand two hundred and sixty-four)
11 kib 0.0107421875 mib 11 kibibyte (eleven) is equal to 0.0107421875 mebibyte (zero point zero × 1 one hundred and seven million four hundred and twenty-one thousand eight hundred and seventy-five)
11 kib 0.00001049041748046875 gib 11 kibibyte (eleven) is equal to 0.00001049041748046875 gibibyte (zero point zero × 4 one quadrillion forty-nine trillion forty-one billion seven hundred and forty-eight million forty-six thousand eight hundred and seventy-five)
11 kib 0.0000000102445483207702636719 tib 11 kibibyte (eleven) is equal to 0.0000000102445483207702636719 tebibyte
11 kib 90.112 kbit 11 kibibyte (eleven) is equal to 90.112 kilobit (ninety point one hundred and twelve)
11 kib 0.090112 mbit 11 kibibyte (eleven) is equal to 0.090112 megabit (zero point zero × 1 ninety thousand one hundred and twelve)
11 kib 0.000090112 gbit 11 kibibyte (eleven) is equal to 0.000090112 gigabit (zero point zero × 4 ninety thousand one hundred and twelve)
11 kib 0.000000090112 tbit 11 kibibyte (eleven) is equal to 0.000000090112 terabit (zero point zero × 7 ninety thousand one hundred and twelve)
11 kib 88 kibit 11 kibibyte (eleven) is equal to 88 kibibit (eighty-eight)
11 kib 0.0859375 mibit 11 kibibyte (eleven) is equal to 0.0859375 mebibit (zero point zero × 1 eight hundred and fifty-nine thousand three hundred and seventy-five)
11 kib 0.00008392333984375 gibit 11 kibibyte (eleven) is equal to 0.00008392333984375 gibibit (zero point zero × 4 eight trillion three hundred and ninety-two billion three hundred and thirty-three million nine hundred and eighty-four thousand three hundred and seventy-five)
11 kib 0.000000081956386566162109375 tibit 11 kibibyte (eleven) is equal to 0.000000081956386566162109375 tebibit | 2,339 | 8,014 | {"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-2024-22 | latest | en | 0.864766 |
https://number.academy/3013296 | 1,718,925,055,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862006.96/warc/CC-MAIN-20240620204310-20240620234310-00892.warc.gz | 357,581,643 | 12,100 | # Number 3013296 facts
The even number 3,013,296 is spelled 🔊, and written in words: three million, thirteen thousand, two hundred and ninety-six, approximately 3.0 million. The ordinal number 3013296th is said 🔊 and written as: three million, thirteen thousand, two hundred and ninety-sixth. The meaning of the number 3013296 in Maths: Is it Prime? Factorization and prime factors tree. The square root and cube root of 3013296. What is 3013296 in computer science, numerology, codes and images, writing and naming in other languages
## What is 3,013,296 in other units
The decimal (Arabic) number 3013296 converted to a Roman number is (M)(M)(M)(X)MMMCCXCVI. Roman and decimal number conversions.
#### Time conversion
(hours, minutes, seconds, days, weeks)
3013296 seconds equals to 1 month, 6 days, 21 hours, 1 minute, 36 seconds
3013296 minutes equals to 6 years, 2 months, 2 weeks, 6 days, 13 hours, 36 minutes
### Codes and images of the number 3013296
Number 3013296 morse code: ...-- ----- .---- ...-- ..--- ----. -....
Sign language for number 3013296:
Number 3013296 in braille:
QR code Bar code, type 39
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#### Is Prime?
The number 3013296 is not a prime number.
#### Factorization and factors (dividers)
The prime factors of 3013296 are 2 * 2 * 2 * 2 * 3 * 11 * 13 * 439
The factors of 3013296 are
1, 2, 3, 4, 6, 8, 11, 12, 13, 16, 22, 24, 26, 33, 39, 44, 48, 52, 66, 78, 88, 104, 132, 143, 156, 176, 208, 264, 286, 312, 429, 439, 528, 572, 624, 858, 878, 1144, 1317, 1716, 1756, 2288, 2634, 3432, 3512, 4829, 5268, 5707, 6864, 7024, 9658, 3013296. show more factors ...
Total factors 80.
Sum of factors 9166080 (6152784).
#### Powers
The second power of 30132962 is 9.079.952.783.616.
The third power of 30132963 is 27.360.585.403.058.958.336.
#### Roots
The square root √3013296 is 1735,884789.
The cube root of 33013296 is 144,437711.
#### Logarithms
The natural logarithm of No. ln 3013296 = loge 3013296 = 14,918545.
The logarithm to base 10 of No. log10 3013296 = 6,479042.
The Napierian logarithm of No. log1/e 3013296 = -14,918545.
### Trigonometric functions
The cosine of 3013296 is 0,957439.
The sine of 3013296 is -0,288637.
The tangent of 3013296 is -0,301467.
## Number 3013296 in Computer Science
Code typeCode value
3013296 Number of bytes2.9MB
Unix timeUnix time 3013296 is equal to Wednesday Feb. 4, 1970, 9:01:36 p.m. GMT
IPv4, IPv6Number 3013296 internet address in dotted format v4 0.45.250.176, v6 ::2d:fab0
3013296 Decimal = 1011011111101010110000 Binary
3013296 Decimal = 12200002110120 Ternary
3013296 Decimal = 13375260 Octal
3013296 Decimal = 2DFAB0 Hexadecimal (0x2dfab0 hex)
3013296 BASE64MzAxMzI5Ng==
3013296 MD542240e76462e5b9b01bc94baa939653a
3013296 SHA1d7ace7e368b03af986113918619e888274e2ce2b
3013296 SHA224c8c3987b5cf0d87c57da1d1ed29fecf2cc3100427e2de120ee3c5d86
3013296 SHA256aa7f64ae4b23b5fc2eec645fb9fd9b0a8e0efa357be31771672f52e500444ef0
3013296 SHA384d09f4abaf3dde81b88c662f1bfd44a14fc13de10d4a060c3e9782aec48e7927cccd609a872d3352511f6f2c03d0b1a4d
More SHA codes related to the number 3013296 ...
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## Numerology 3013296
### Character frequency in the number 3013296
Character (importance) frequency for numerology.
Character: Frequency: 3 2 0 1 1 1 2 1 9 1 6 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 3013296, the numbers 3+0+1+3+2+9+6 = 2+4 = 6 are added and the meaning of the number 6 is sought.
## № 3,013,296 in other languages
How to say or write the number three million, thirteen thousand, two hundred and ninety-six in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 3.013.296) tres millones trece mil doscientos noventa y seis German: 🔊 (Nummer 3.013.296) drei Millionen dreizehntausendzweihundertsechsundneunzig French: 🔊 (nombre 3 013 296) trois millions treize mille deux cent quatre-vingt-seize Portuguese: 🔊 (número 3 013 296) três milhões e treze mil, duzentos e noventa e seis Hindi: 🔊 (संख्या 3 013 296) तीस लाख, तेरह हज़ार, दो सौ, छियानवे Chinese: 🔊 (数 3 013 296) 三百零一万三千二百九十六 Arabian: 🔊 (عدد 3,013,296) ثلاثة ملايين و ثلاثة عشر ألفاً و مئتان و ستة و تسعون Czech: 🔊 (číslo 3 013 296) tři miliony třináct tisíc dvěstě devadesát šest Korean: 🔊 (번호 3,013,296) 삼백일만 삼천이백구십육 Danish: 🔊 (nummer 3 013 296) tre millioner trettentusinde og tohundrede og seksoghalvfems Hebrew: (מספר 3,013,296) שלושה מיליון ושלושה עשר אלף מאתיים תשעים ושש Dutch: 🔊 (nummer 3 013 296) drie miljoen dertienduizendtweehonderdzesennegentig Japanese: 🔊 (数 3,013,296) 三百一万三千二百九十六 Indonesian: 🔊 (jumlah 3.013.296) tiga juta tiga belas ribu dua ratus sembilan puluh enam Italian: 🔊 (numero 3 013 296) tre milioni e tredicimiladuecentonovantasei Norwegian: 🔊 (nummer 3 013 296) tre million tretten tusen to hundre og nittiseks Polish: 🔊 (liczba 3 013 296) trzy miliony trzynaście tysięcy dwieście dziewięćdziesiąt sześć Russian: 🔊 (номер 3 013 296) три миллиона тринадцать тысяч двести девяносто шесть Turkish: 🔊 (numara 3,013,296) üçmilyononüçbinikiyüzdoksanaltı Thai: 🔊 (จำนวน 3 013 296) สามล้านหนึ่งหมื่นสามพันสองร้อยเก้าสิบหก Ukrainian: 🔊 (номер 3 013 296) три мільйони тринадцять тисяч двісті дев'яносто шість Vietnamese: 🔊 (con số 3.013.296) ba triệu mười ba nghìn hai trăm chín mươi sáu Other languages ...
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If you know something interesting about the number 3013296 or any other natural number (positive integer), please write to us here or on Facebook.
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The content of the comments is the opinion of the users and not of number.academy. It is not allowed to pour comments contrary to the laws, insulting, illegal or harmful to third parties. Number.academy reserves the right to remove or not publish any inappropriate comment. It also reserves the right to publish a comment on another topic. Privacy Policy. | 2,232 | 6,237 | {"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-2024-26 | latest | en | 0.789671 |
https://www.studentlance.com/solution/the-table-depicts-a-two-way-anova-in-which-gender-has-two-groups-male-and-female-marital-status | 1,508,805,050,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187827662.87/warc/CC-MAIN-20171023235958-20171024015958-00631.warc.gz | 997,264,734 | 5,869 | # The table depicts a two-way ANOVA in which gender has two groups (male and female), marital status - 23378
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The table depicts a two-way ANOVA in which gender has two groups (male and female), marital status has three groups (married, single never married, divorced), and the means refer to happiness scores (n = 100):
1. What is/are the independent variable(s)? What is/are the dependent variable(s)?
2. What would be an appropriate null hypothesis? Alternate hypothesis?
3. What are the degrees of freedom for 1) gender, 2) marital status, 3) interaction between gender and marital status, and 4) error or within variance?
4. Calculate the mean square for 1) gender, 2) marital status, 3) interaction between gender and marital status, and 4) error or within variance.
5. Calculate the F ratio for 1) gender, 2) marital status, and 3) interaction between gender and marital status.
6. Identify the criterion Fs at alpha = .05 for 1) gender, 2) marital status, and 3) interaction between gender and marital status.
7. If alpha is set at .05, what conclusions can you make?
Source Sum of Squares (degrees of freedom [df]) Mean Square Fobt. Fcrit. Gender 68.15 ? ? ? ? Marital Status 127.37 ? ? ? ? Gender * Marital Status (A x B) 41.90 ? ? ? ? Error (Within) 864.82 ? ? NA NA Total 1102.24 99 NA NA NA
Please Note: The table that you see in the assignment has been slightly modified from the one presented in the module notes since it is beyond the scope of this unit to have students calculate p values. Instead you are asked to calculate the F value and compare it to the critical F value to determine whether the test is significant or not.
Solution Description
Previously, thank you for purchasing my tutorial. I try to
Attachments | 498 | 1,964 | {"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} | 3.296875 | 3 | CC-MAIN-2017-43 | longest | en | 0.863399 |
http://pietro-corna.com/health/how-to-square-a-building-345.php | 1,560,749,408,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998376.42/warc/CC-MAIN-20190617043021-20190617065021-00389.warc.gz | 141,226,912 | 5,806 | How to square a building 345
Using the Pythagorean theorem we would find: Craig Sweeney said this on November 6, 2016 at 3: For a large room, use 6-8-10 or 9-12-15 feet or meters. While your room doesn't need to be perfectly square, it's best to get corners as close to 90 degrees as possible. Measure four units along the other side. This technique simply requires that the carpenter create a triangle in the corner of the lines that are to be square 90 degrees to each other. In addition, precisely how may possibly most people keep up some sort of distance learning?
If it's less than 5 units, your corner is less than 90 degrees; if it's greater than 5 units, your corner angle is too large. Share yours! Pegs D and A form the line perpendicular to the base line and the angle between the line CD and the base line is a right angle see Fig.
Try 30-40-50 centimeters if using the metric system. Adding string lines on bigger project are useful with property marking for say if laying a pad of a house.
How to Square Joists for a Deck
When all sides of the tape are stretched, a triangle with lengths of 3 m, 4 m and 5 m is formed see Fig. Awesome picture!
IB Ian Burnby Apr 27, 2016. Just multiply the length and width together to get the square footage. Craig Siefkas said this on August 13, 2013 at 5: I am a cabinet maker and 20 years ago we wood use this regularly, today with all the computer assist software we are getting LAZY. The triangle must have one side leg that is 3 feet long, a second side that is 4 feet long and a third side that is 5 feet long.
When using the 3-4-5 method for squaring corners, if your last measurement [the third side] connecting the two legs measuring [5 foot side] is off and not square you will need to make adjustments. On his website, Rob uses his knowledge and experience to help and educate on best practices in the remodeling industry.
Peg A is not on the base line. Edit Related wikiHows. Not Helpful 21 Helpful 22.
Pro Tips: How To Square a Large Project Using the 3-4-5 Rule
The operator then rotates the instrument until the image of pole A can be seen. This circle crosses the base line twice see Fig. Arlie said this on April 5, 2013 at 5: Of course any lengths could be used to create the right angle for construction — as long as they were correct when applied to the Pythagorean theorem.
What's the best way to check both corners to see if my deck is square? | 591 | 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} | 4 | 4 | CC-MAIN-2019-26 | latest | en | 0.942622 |
https://math.stackexchange.com/questions/580521/generalizing-int-01-frac-arctan-sqrtx2-2-sqrtx2-2 | 1,721,590,769,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517768.40/warc/CC-MAIN-20240721182625-20240721212625-00541.warc.gz | 324,555,843 | 40,845 | # Generalizing $\int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{\operatorname dx}{x^{2}+1} = \frac{5\pi^{2}}{96}$
The following integral
\begin{align*} \int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{dx}{x^{2}+1} = \frac{5\pi^{2}}{96} \tag{1} \end{align*}
is called the Ahmed's integral and became famous since its first discovery in 2002. Fascinated by this unbelievable closed form, I have been trying to generalize this result for many years, though not successful so far.
But suddenly it came to me that some degree of generalization may be possible. My conjecture is as follows: Define the (generalized) Ahmed integral of parameter $p$, $q$ and $r$ by
\begin{align*} A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1}}{q \sqrt{p^{2}x^{2} + 1} } \, \frac{pqr \, dx}{(r^{2} + 1)p^{2} x^{2} + 1}. \end{align*}
Now suppose that $p q r = 1$, and define its complementary parameters as
\begin{align*} \tilde{p} = r \sqrt{\smash{q}^{2} + 1}, \quad \tilde{q} = p \sqrt{\smash{r}^{2} + 1}, \quad \text{and} \quad \tilde{r} = q \sqrt{\smash{p}^{2} + 1}, \tag{2} \end{align*}
Then my guess is that
\begin{align*} A(p, q, r) = \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} (1 / \tilde{p} ) - \arctan^{2} ( \tilde{q} ) - \arctan^{2} ( \tilde{r} ) \right\}. \end{align*}
Plugging the values $(p, q, r) = (1/\sqrt{2}, \sqrt{2}, 1)$, the corresponding complementary parameters become $(\tilde{p}, \tilde{q}, \tilde{r}) = (\sqrt{3}, 1, \sqrt{3})$. Then for these choices, the original Ahmed's integral $\text{(1)}$ is retrieved:
\begin{align*} \int_{0}^{1} \frac{\arctan \sqrt{x^{2} + 2} }{\sqrt{x^{2} + 2} } \, \frac{dx}{x^{2} + 1} &= \frac{\pi^{2}}{8} + \frac{1}{2} \left\{ \arctan^{2} \frac{1}{\sqrt{3}} - \arctan^{2} 1 - \arctan^{2} \sqrt{3} \right\} \\ &= \frac{5\pi^{2}}{96}. \end{align*}
In fact, I have a more generalized conjecture involving dilogarithms depending on complementary parameters. But since this specialized version is sufficiently daunting, I won't deal with it here.
Unfortunately, proving this relation is not successful so far. I just heuristically calculated and made some ansatz to reach this form. Can you help me improve the situation by proving this or providing references to some known results?
EDIT. I finally succeeded in proving a general formula: let $k = pqr$ and complementary parameters as in $\text{(2)}$. Then whenever $k \leq 1$, we have
\begin{align*} A(p, q, r) &= 2\chi_{2}(k) - k \arctan (\tilde{p}) \arctan \left( \frac{k}{\tilde{p}} \right) \\ &\quad + \frac{k}{2} \int_{0}^{1} \frac{1}{1-k^{2}x^{2}} \log\left( \frac{1+\tilde{p}^{2}x^{2}}{1+\tilde{p}^{2}} \times \frac{1+\tilde{q}^{2}x^{2}}{1+\tilde{q}^{2}} \times \frac{1+\tilde{r}^{2}x^{2}}{1+\tilde{r}^{2}} \right) \, dx. \end{align*}
Then the proposed conjecture follows as a corollary. I'm planning to gather materials related to the Ahmed's integrals and put into a combined one. You can find an ongoing proof of this formula here.
• Sorry for the off-topic question: I have looked through some of your very beautiful answers here, and it seems you are developing cool new methods. Are you publishing those methods/generalisations in some math-journals, or are they on Math.SE only? Commented Oct 5, 2014 at 17:34
• @NicoDean, Thank you! I think that most of the techniques are already well established, so I am only posting them on Math.SE. Commented Oct 5, 2014 at 22:02
• This is a great work! I've just met Ahmed's integral (and it's solution) in the book "Inside interesting integrals...", I had no idea it's possible to generalize it to such extent Commented Mar 21, 2016 at 8:10
• This might be one of the more beautiful things I've seen, but I'm going to have to read over that proof 3 or 4 times to fully appreciate this. Commented Apr 12, 2016 at 16:09
• Thank you so much for all these responses. My plan is to polish the result before I make any edit on this question, but unfortunately, I am recently so busy with my own study to improve it. And moreover on this, I suspect that my computation is essentially a disguise of Lobachevski's volume formula for an orthoscheme (given the relationship between the generalized Ahmed's integral and the Coxeter's integral). I would like to clarify this linkage before I make any update. Commented May 25, 2016 at 23:04
Below is a simpler, more direct proof of the proposed conjecture. First, establish
\begin{align*} K(a)=\int_{0}^{1} \frac{\ln\frac{a+x^2}{a+1}}{1-x^2} dx =&\int_0^1 \int_{\infty}^a\frac1{(y+1)(y+x^2)}dy\ dx\\ =&\int_\infty^a \frac{\text{arccot} \sqrt {y}}{\sqrt{y}(y+1)}dy= -\text{arccot}^2 \sqrt {a} \end{align*} Then, for $$r=\frac1{pq}$$ \begin{align*} & \int_{0}^{1} \frac{1}{(p^{2} + \frac1{q^2})x^{2} + 1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1}}{q \sqrt{p^{2}x^{2} + 1} } \ dx\\ =& \int_0^1 \frac{q^2}{(p^{2}q^2 + 1)x^{2} + q^2} \int_0^x \frac{x}{x^2+q^2(p^2x^2+1)y^2}dy\ dx\\ =&\int_0^1 \int_y^1\frac{q^2x}{[(p^2q^2+1)x^2+q^2][(1+p^2q^2y^2)x^2+q^2y^2]}dx\ dy\\ =&\ \frac12 \int_0^1 \frac1{1-y^2}\ln\frac{[1+(p^2+1)q^2y^2][q^2+(1+p^2q^2)y^2]}{y^2[1+q^2+p^2q^2y^2] [1+q^2 +p^2q^2]} \ dy\\ =&\ \frac12\int_0^1 \frac1{1-y^2}\left[-\ln y^2 -\ln \frac{1+q^2+p^2q^2y^2}{1+q^2 +p^2q^2}\right.\\ &\hspace{25mm}\left.+ \ln \frac{1+(1+p^2)q^2y^2}{1+q^2 +p^2q^2} + \ln \frac{q^2+(1+p^2q^2)y^2}{1+q^2 +p^2q^2}\right] dy\\ =&\ \frac12\left[-K(0)-K\left(\frac{q^2+1}{p^2q^2} \right) + K\left(\frac{1}{p^2q^2+q^2} \right)+ K\left(\frac{q^2}{p^2q^2+1} \right) \right]\\ =&\ \frac{\pi^2}8+\frac12\bigg[ \arctan^2\frac{pq}{\sqrt{q^2+1}}-\arctan^2 q\sqrt{p^2+1} -\arctan^2{\frac{ \sqrt{p^2q^2+1}}q}\bigg] \end{align*}
• Genius +1, thank you for sharing. Commented Jan 27 at 15:45
Too long for comment ,Define the generalized Ahmed integral $$I\left( p,q,r \right) = pqr \int_0^1{\frac{\arctan \left( q\sqrt{p^2u^2+1} \right)}{q\sqrt{p^2u^2+1}}\frac{1}{\left( p^2\left( r^2+1 \right) u^2+1 \right)}}\,du$$
$$1 \text{, when } p,q,r>0, \, pqr\le 1 \text{, then: }$$ \begin{aligned} &I\left( p,q,r \right) = pqr\int_0^1{\frac{\arctan \left( q\sqrt{p^2u^2+1} \right)}{q\sqrt{p^2u^2+1}}\frac{1}{\left( p^2\left( r^2+1 \right) u^2+1 \right)}}\,du \\ &=A\left( p,q,r \right) -\arctan \left( \frac{pq}{\sqrt{q^2+1}} \right) \arctan \left( r\sqrt{q^2+1} \right) \\ &A\left( p,q,r \right) =2\sum_{k=0}^{\infty}{\frac{\left( pqr \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}+ \\ &\frac{pqr}{2}\int_0^1{\frac{\ln \left( \frac{1+\left( r\sqrt{q^2+1} \right) ^2x^2}{1+\left( r\sqrt{q^2+1} \right) ^2} \right) +\ln \left( \frac{1+\left( p\sqrt{r^2+1} \right) ^2x^2}{1+\left( p\sqrt{r^2+1} \right) ^2} \right) +\ln \left( \frac{1+\left( q\sqrt{p^2+1} \right) ^2x^2}{1+\left( q\sqrt{p^2+1} \right) ^2} \right)}{1-\left( pqr \right) ^2x^2}dx} \end{aligned}
$$2 \text{, for any } p,q,r>0, \text{ then: }$$ \begin{aligned} &I\left( p,q,r \right) = pqr\int_0^1{\frac{\arctan \left( q\sqrt{p^2u^2+1} \right)}{q\sqrt{p^2u^2+1}}\frac{1}{\left( p^2\left( r^2+1 \right) u^2+1 \right)}}\,du \\ &=\hat{A}\left( p,q,r \right) -\arctan \left( \frac{pq}{\sqrt{q^2+1}} \right) \arctan \left( r\sqrt{q^2+1} \right) \\ &\hat{A}\left( p,q,r \right) =\frac{\pi ^2}{4}+2\sum_{k=0}^{\infty}{\frac{\left( \frac{pqr-1}{pqr+1} \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}+ \\ &\frac{pqr}{2}\int_0^1{\frac{\ln \left( \frac{1+\left( r\sqrt{q^2+1} \right) ^2x^2}{1+\left( r\sqrt{q^2+1} \right) ^2}\frac{1+\left( p\sqrt{r^2+1} \right) ^2x^2}{1+\left( p\sqrt{r^2+1} \right) ^2}\frac{1+\left( q\sqrt{p^2+1} \right) ^2x^2}{1+\left( q\sqrt{p^2+1} \right) ^2} \right) +4\ln \left( pqr \right)}{1-\left( pqr \right) ^2x^2}dx} \end{aligned}
For this problem, note that
$$I\left( p,q,r \right) =\int_0^1{\frac{\arctan \left( q\sqrt{p^2u^2+1} \right)}{q\sqrt{p^2u^2+1}}\frac{1}{\left( p^2\left( r^2+1 \right) u^2+1 \right)}du}=\int_0^1{\frac{\arctan \left( \frac{2\sqrt{2u^2+1}}{\sqrt{5}\left( u^2+1 \right)} \right)}{\sqrt{2u^2+1}\left( 3u^2+1 \right)}du}, \\ \text{using the addition formula for } \arctan \left( x \right): \arctan \left( \frac{2\sqrt{2u^2+1}}{\sqrt{5}\left( u^2+1 \right)} \right) \\ =\arctan \left( \sqrt{5}\sqrt{2u^2+1} \right) -\arctan \left( \frac{\sqrt{2u^2+1}}{\sqrt{5}} \right) \text{; so } I=\int_0^1{\frac{\arctan \left( \sqrt{5}\sqrt{2u^2+1} \right) -\arctan \left( \frac{\sqrt{2u^2+1}}{\sqrt{5}} \right)}{\sqrt{2u^2+1}\left( 3u^2+1 \right)}du} \\ =I\left( \sqrt{2},\sqrt{5},\frac{1}{\sqrt{2}} \right) -I\left( \sqrt{2},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{2}} \right)$$
where $$I\left( \sqrt{2},\sqrt{5},\frac{1}{\sqrt{2}} \right) = 2\sum_{k=0}^{\infty}{\frac{\left( \frac{\sqrt{5}-1}{\sqrt{5}+1} \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}+ \\ \int_0^1{\frac{\sqrt{5}\left( \ln \left( \frac{\left( \frac{\sqrt{\sqrt{5}^2+1}x}{\sqrt{2}} \right) ^2+1}{\left( \frac{\sqrt{\sqrt{5}^2+1}}{\sqrt{2}} \right) ^2+1}\frac{\left( \frac{\sqrt{\sqrt{2}^2+1}x}{\frac{1}{\sqrt{5}}} \right) ^2+1}{\left( \frac{\sqrt{\sqrt{2}^2+1}}{\frac{1}{\sqrt{5}}} \right) ^2+1}\frac{\left( \sqrt{2}\sqrt{\left( \frac{1}{\sqrt{2}} \right) ^2+1}x \right) ^2+1}{\left( \sqrt{2}\sqrt{\left( \frac{1}{\sqrt{2}} \right) ^2+1} \right) ^2+1} \right) +4\ln \left( \sqrt{5} \right) \right)}{2\left( 1-\left( \sqrt{5}x \right) ^2 \right)}}\,dx+ \\ \frac{\pi ^2}{4}-\arctan \left( \frac{\sqrt{\sqrt{5}^2+1}}{\sqrt{2}} \right) \arctan \left( \frac{\sqrt{2}\sqrt{5}}{\sqrt{\sqrt{5}^2+1}} \right) \\ =\frac{2\left( \frac{1}{2}\sqrt{5}\ln\left(1+\sqrt{2}\right) -\frac{1}{2}\sqrt{5}\ln\left(1-\sqrt{2}\right) \right)}{\sqrt{5}}+ \\ \int_0^1{\frac{\sqrt{5}\left( 2\ln \left( 3x^2+1 \right) +\ln \left( \frac{25}{256}\left( 15x^2+1 \right) \right) \right)}{2\left( 1-\left( \sqrt{5}x \right) ^2 \right)}}\,dx \\ +\frac{1}{12}\pi \left( 3\pi -4\arctan \left( \sqrt{\frac{5}{3}} \right) \right);$$
Similarly, we get $$I\left( \sqrt{2},\frac{1}{\sqrt{5}},\frac{1}{\sqrt{2}} \right) =2\sum_{k=0}^{\infty}{\frac{\left( \frac{1}{\sqrt{5}} \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}+ \\ \int_0^1{\frac{2\ln \left( \frac{5}{8}\left( \frac{3x^2}{5}+1 \right) \right) +\ln \left( \frac{1}{4}\left( 3x^2+1 \right) \right)}{2\sqrt{5}\left( 1-\frac{x^2}{5} \right)}}\,dx-\arctan \left( \frac{\sqrt{2}}{\sqrt{5}\sqrt{\left( \frac{1}{\sqrt{5}} \right) ^2+1}} \right) \arctan \left( \frac{\sqrt{\left( \frac{1}{\sqrt{5}} \right) ^2+1}}{\sqrt{2}} \right) \\ =\frac{2\left( \frac{1}{2}\sqrt{5}\ln\left(1+\sqrt{\frac{5}{2}}\right) -\frac{1}{2}\sqrt{5}\ln\left(1-\sqrt{\frac{5}{2}}\right) \right)}{\sqrt{5}}+ \\ \int_0^1{\frac{2\ln \left( \frac{5}{8}\left( \frac{3x^2}{5}+1 \right) \right) +\ln \left( \frac{1}{4}\left( 3x^2+1 \right) \right)}{2\sqrt{5}\left( 1-\frac{x^2}{5} \right)}}\,dx-\frac{1}{6}\pi \tan ^{-1}\left( \sqrt{\frac{3}{5}} \right);$$
Then the integral is given by $$\int_0^1{\frac{2\ln \left( \frac{5}{8}\left( \frac{3x^2}{5}+1 \right) \right) +\ln \left( \frac{1}{4}\left( 3x^2+1 \right) \right)}{2\sqrt{5}\left( 1-\frac{x^2}{5} \right)}dx}, \\ \int_0^1{\frac{\sqrt{5}\left( 2\ln \left( 3x^2+1 \right) +\ln \left( \frac{25}{256}\left( 15x^2+1 \right) \right) \right)}{2\left( 1-\left( \sqrt{5}x \right) ^2 \right)}}\,dx$$ an be computed accordingly
Regarding Sangchul Lee's paper "Some properties on generalized Ahmed's integral," two corrections are made (as amended in this document)
(1) Formula (1,3) $$\arctan \left( \frac{pq}{\sqrt{q^2+1}} \right) \arctan \left( r\sqrt{p^2+1} \right) should be corrected to \arctan \left( \frac{pq}{\sqrt{q^2+1}} \right) \arctan \left( r\sqrt{q^2+1} \right)$$
(2) In formula (1,4), $$\chi_2(z)=\sum_{k=0}^{\infty}{\frac{\left(z \right) ^{2k+1}}{\left( 2k+1 \right) ^2}}$$, the subscript should start from 0 | 5,228 | 11,612 | {"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": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2024-30 | latest | en | 0.700591 |
https://topgradeessaywritings.com/calculate-the-heat-of-reaction-ahyo-of-the-reaction-3mnos-302g-mn304s-assume-that-ah-is-independent-of-t/ | 1,713,632,418,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817670.11/warc/CC-MAIN-20240420153103-20240420183103-00594.warc.gz | 515,460,807 | 10,079 | # Calculate the heat of reaction AHyO of the reaction 3MnO(s) + 302(g) = Mn304(s) Assume that AH,” is independent of T.
Calculate the heat of reaction AHyO of the reaction 3MnO(s) + 302(g) = Mn304(s) Assume that AH,” is independent of T..
1.The equilibrium constant for the ammonia synthesis reaction is 775 at 25°C based on 1 atm ideal gas standard states. The enthalpy change associated with the reaction, or simply the heat of reaction, AH,”, is -45.9kJ. Assuming that AH,” is independent of temperature, estimate the equilibrium constant for the reaction at 45°C.
2.Equilibrium of the system containing MnO(s), h4n304(s) and O&) was examined. Both MnO and Mn304 were pure and stable states. It was found that equilibrium partial pressures of oxygen were PO, = 1.4xlO”atm at l,OOO°C PO, = 2.8~1O-~atm at l,lOO°C Calculate the heat of reaction AHyO of the reaction 3MnO(s) + 302(g) = Mn304(s) Assume that AH,” is independent of T.
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Calculate the heat of reaction AHyO of the reaction 3MnO(s) + 302(g) = Mn304(s) Assume that AH,” is independent of T.
Posted in Uncategorized | 363 | 1,279 | {"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} | 2.921875 | 3 | CC-MAIN-2024-18 | latest | en | 0.94017 |
https://www.jiskha.com/questions/14690/ok-i-figured-out-part-a-but-i-am-having-trouble-with-b-part-a-was-zero-a-hypothetical | 1,620,488,766,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.94/warc/CC-MAIN-20210508151721-20210508181721-00019.warc.gz | 864,608,311 | 5,740 | # Physics/Math
Ok...I figured out part a but I am having trouble with b. part a was:
Zero, a hypothetical planet, has a mass of 1.0x10^23 kg, a radius of 3.0x10^6 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface.
(a) If the probe is launched with an initial kinetic energy of 5.0x10^7 J, what will be its kinetic energy when it is 4.0x10^6 m from the center of Zero?
>>>>>>>>>>>>>>>>>>
so I did KE + U = constant,
V = sqrt(2GM/R)
V = sqrt(2(6.67e-11)(1.0e23)/(3.0e6))
V = 2108.7
(1/2) mV^2 - GMm/R = constant
(.5(10)(2108.7)^2 - ((6.67e-11)(1.0e23)(10)/(4.0e6))
111165392.3 - 16675000 = 94490392.3 = 9.4e7=constant
9.4e7-5e7=4.4e7 J which is the right answer...but for b:
>>>>>>>>>>>>>>>>>>>>>>
If the probe is to achieve a maximum distance of 8.0x10^6 m from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?
->I am not sure what to do. I recalculated the constant using the radius of 8e6 m and got 2.78e6 J but then I am not sure what to use for U. U=mgd so I tried U=(10)(9.8)(3e6) and subtracted that from 2.78e6 J but that is wrong. What am I missing?
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July 26, 2013, 00:18 PimpleFoam-Crashed #1 Member Amin Join Date: May 2013 Posts: 76 Rep Power: 6 Hello In order to simulate a laminar and incompressible flow past a cube i tried to use icoFoam, but the i did know how to stabilize this solver, because of the high courant number. That is why i am trying to run the simulation now with pimpleFoam, but unlikely the solver is crashing too. I think because to time-step is very low... Do you know how to stabilize icoFoam and how to fix the problem with pinmpleFoam? It would be great if someone could help me solve the problem : Here is the output of OF (simulation with pimpleFoam. OF 2.2.0): Code: ```Courant Number mean: 0.00905642 max: 10.7784 deltaT = 1.3457e-101 --> FOAM Warning : From function Time::operator++() in file db/Time/Time.C at line 1029 Increased the timePrecision from 261 to 262 to distinguish between timeNames at time 12.8518 Time = 12.851786558825548212325884378515183925628662109375 DILUPBiCG: Solving for Ux, Initial residual = 0.425712, Final residual = 2.75527e-06, No Iterations 5 DILUPBiCG: Solving for Uy, Initial residual = 0.285945, Final residual = 5.18152e-07, No Iterations 5 DICPCG: Solving for p, Initial residual = 0.657594, Final residual = 0.014474, No Iterations 2 time step continuity errors : sum local = 4.37835e-05, global = -2.86105e-05, cumulative = -0.00487001 DICPCG: Solving for p, Initial residual = 0.592278, Final residual = 8.3816e-07, No Iterations 163 time step continuity errors : sum local = 3.37411e-09, global = 6.57181e-11, cumulative = -0.00487001 ExecutionTime = 174.2 s ClockTime = 201 s Courant Number mean: 0.00538199 max: 5.49296 deltaT = 1.22493e-101 --> FOAM Warning : From function Time::operator++() in file db/Time/Time.C at line 1029 Increased the timePrecision from 262 to 263 to distinguish between timeNames at time 12.8518 Time = 12.851786558825548212325884378515183925628662109375 DILUPBiCG: Solving for Ux, Initial residual = 0.319248, Final residual = 5.97266e-06, No Iterations 5 DILUPBiCG: Solving for Uy, Initial residual = 0.284562, Final residual = 5.43627e-06, No Iterations 5 DICPCG: Solving for p, Initial residual = 0.428836, Final residual = 0.0167637, No Iterations 2 time step continuity errors : sum local = 0.00011502, global = 6.84904e-05, cumulative = -0.00480151 DICPCG: Solving for p, Initial residual = 0.306846, Final residual = 8.56454e-07, No Iterations 153 time step continuity errors : sum local = 1.01177e-08, global = 5.813e-11, cumulative = -0.00480151 ExecutionTime = 174.63 s ClockTime = 201 s Courant Number mean: 0.00810236 max: 7.03961 deltaT = 8.70027e-102 --> FOAM Warning : From function Time::operator++() in file db/Time/Time.C at line 1029 Increased the timePrecision from 263 to 264 to distinguish between timeNames at time 12.8518 Time = 12.851786558825548212325884378515183925628662109375 DILUPBiCG: Solving for Ux, Initial residual = 0.456665, Final residual = 6.35141e-07, No Iterations 8 DILUPBiCG: Solving for Uy, Initial residual = 0.375937, Final residual = 3.13479e-06, No Iterations 8 DICPCG: Solving for p, Initial residual = 0.489683, Final residual = 0.0239404, No Iterations 26 time step continuity errors : sum local = 0.000190772, global = -4.11878e-05, cumulative = -0.0048427 DICPCG: Solving for p, Initial residual = 0.449154, Final residual = 7.10247e-07, No Iterations 175 time step continuity errors : sum local = 1.31591e-08, global = -4.67166e-11, cumulative = -0.0048427 ExecutionTime = 175.14 s ClockTime = 202 s Courant Number mean: 0.0375239 max: 28.0087 deltaT = 1.55314e-102 --> FOAM Warning : From function Time::operator++() in file db/Time/Time.C at line 1029 Increased the timePrecision from 264 to 265 to distinguish between timeNames at time 12.8518 Time = 12.851786558825548212325884378515183925628662109375 #0 Foam::error::printStack(Foam::Ostream&) in "/opt/openfoam220/platforms/linux64GccDPOpt/lib/libOpenFOAM.so" #1 Foam::sigFpe::sigHandler(int) in "/opt/openfoam220/platforms/linux64GccDPOpt/lib/libOpenFOAM.so" #2 in "/lib/x86_64-linux-gnu/libc.so.6" #3 double Foam::sumProd(Foam::UList const&, Foam::UList const&) in "/opt/openfoam220/platforms/linux64GccDPOpt/lib/libOpenFOAM.so" #4 Foam::PBiCG::solve(Foam::Field&, Foam::Field const&, unsigned char) const in "/opt/openfoam220/platforms/linux64GccDPOpt/lib/libOpenFOAM.so" #5 in "/opt/openfoam220/platforms/linux64GccDPOpt/bin/pimpleFoam" #6 in "/opt/openfoam220/platforms/linux64GccDPOpt/bin/pimpleFoam" #7 in "/opt/openfoam220/platforms/linux64GccDPOpt/bin/pimpleFoam" #8 in "/opt/openfoam220/platforms/linux64GccDPOpt/bin/pimpleFoam" #9 in "/opt/openfoam220/platforms/linux64GccDPOpt/bin/pimpleFoam" #10 __libc_start_main in "/lib/x86_64-linux-gnu/libc.so.6" #11 in "/opt/openfoam220/platforms/linux64GccDPOpt/bin/pimpleFoam" Floating point exception (core dumped)``` Thx
July 26, 2013, 03:45 #2 Senior Member Alexey Matveichev Join Date: Aug 2011 Location: Nancy, France Posts: 1,736 Rep Power: 29 There are lots of reasons your solution is diverging (bad mesh, wrong IC, wrong BC etc). Can you please post you case files?
July 26, 2013, 03:50 #3 Member Amin Join Date: May 2013 Posts: 76 Rep Power: 6
July 26, 2013, 04:53
#4
Senior Member
Alexey Matveichev
Join Date: Aug 2011
Location: Nancy, France
Posts: 1,736
Rep Power: 29
Well...
To make the case run it was enough to increase nNonOrthogonalCorrectors from 0 to 1.
But results are still rather strange so you can also increase nOuterCorrectors and nCorrectors. But better switch to residualControl for exit condition from PIMPLE loop.
This is just the correction of nNonOrthogonalCorrectors (fvSolution)
Code:
```PIMPLE
{
nOuterCorrectors 1;
nCorrectors 2;
nNonOrthogonalCorrectors 1;
pRefCell 1001;
pRefValue 0;
}```
Velocity field is on velocity-field-1.png. You've switched off turbulence but still you get instabilities in the flow.
If we now switch on residualControl (fvSolution), reduce maxCo from 5 to 1 (in controlDict) and rerun the case we'll get results presented on velocity-field-2.png. And they seems to be more promising. Here's the fragment of fvSolution:
Code:
```PIMPLE
{
nOuterCorrectors 100;
nCorrectors 3;
nNonOrthogonalCorrectors 2;
pRefCell 1001;
pRefValue 0;
residualControl
{
U
{
tolerance 1e-2;
relTol 0;
}
p
{
tolerance 1e-2;
relTol 0;
}
}
}```
Also I've changed residuals for individual solvers:
Code:
``` p
{
solver PCG;
preconditioner DIC;
tolerance 1e-06;
relTol 1e-05;
}
pFinal
{
solver PCG;
preconditioner DIC;
tolerance 1e-08;
relTol 0;
}```
Attached Images
velocity-field-1.png (30.6 KB, 56 views) velocity-field-2.png (22.8 KB, 50 views)
July 26, 2013, 05:25
#5
Senior Member
Alexey Matveichev
Join Date: Aug 2011
Location: Nancy, France
Posts: 1,736
Rep Power: 29
Was not exactly right about flow being laminar. Velocity field at 30 s is on attached file. But still flow has this "stable" von Carman vortex street
Attached Images
velocity-field-3.png (61.2 KB, 43 views)
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https://mathequalslove.blogspot.com/2016/06/evaluating-functions-puzzles-open.html | 1,550,513,955,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247487624.32/warc/CC-MAIN-20190218175932-20190218201932-00126.warc.gz | 619,193,071 | 34,651 | Math = Love: Evaluating Functions Puzzles - Open Middle Style
## Saturday, June 4, 2016
### Evaluating Functions Puzzles - Open Middle Style
After making this activity to practice classifying relations as a function or not a function, I was eager to make a similar activity to practice a different skill. I settled on creating a practice activity for evaluating functions and better understanding function notation.
I wanted to make this activity use the same integer cards (-4 to +4) that the previous activity used to save me time cutting and laminating.
I grabbed a notepad, wrote a few functions, and fiddled around with the numbers until they worked.
This is what I came up with:
Place the integers between -4 and +4, inclusive, in the missing boxes to make each function evaluate properly. Each number may only be used once.
I printed off the puzzle and gave it to my husband/lovely beta tester to give it a shot. After trying what felt like a zillion combinations, he came up with a solution. I took a look at it and realized it was a slightly different solution than I had in mind when I originally created the problem. For one of the functions, he used the same three integers I had; however, he had the integers placed in different locations within the puzzle.
We discussed the problem and decided we didn't know how long our students would stick with it. The overall idea was good, but it seemed like there was just too much guessing and checking needed. The presence of a square root symbol did mean there could only be a 0, 1, or 4 in that slot. But there were no other similar clues throughout the puzzle.
I dropped the idea for a few days until picking it back up today. I stared at the puzzle for a while before deciding maybe it would be more logically solve-able if I provided the solver with some of the input and output values.
This led to a different version. My functions were longer and more complex, so I had to change the paper orientation. That was annoying.
Instructions:
Place the integers between -4 and +4, inclusive, in the missing boxes to make each function evaluate properly. Each number may only be used once.
My husband found a solution to this puzzle in about five minutes. Given that he's one of the mathiest people I know, I'm expecting it would take students at least 15 minutes to solve the puzzle.
One exciting thing was that the solution I intended when making the puzzle and the solution Shaun came up with were totally different! For one of the functions, we used the same three integers, but we placed them in different places. For the other two functions, we used completely different sets of integers to solve each. This makes me wonder if there are even more solutions to this puzzle?
I'm not as thrilled with these as I am with the function/not a function puzzle. I just feel like these puzzles are missing something that I can't put my finger on.
If you're interested in using either of these puzzles in your classroom or trying them yourself, I've uploaded them here as editable Publisher files and non-editable PDF files.
Each file comes with a puzzle sheet and a sheet of integer squares to cut out.
1. Hi Sarah, I actually like the first version of your problem better but I agree that is a bit challenging. I only found 6 possible solutions. You could change the functions a to allow more possible solutions. If you remove the squared in the g(x) function you can increase it to 10 possible solutions and if you change it to g(x) = __x-5 you can increase it to 18 possible solutions. Anyway, just my 2 cents worth. Thanks for the great ideas.
1. Thanks for the feedback. I like the change you suggested!
2. I did a problem like this for function notation, but called it "drag and drop" to mirror the question type of our state's EOC. You have inspired me to write a blog entry about it. Check it out http://mathdyal.blogspot.com/2016/06/function-notation-drag-and-drop-open.html
My problem is easier than yours but it may be a good way to scaffold up to your problem. Keep the good ideas coming! :)
1. Awesome! Thanks for sharing!
3. Thank you for sharing!
It takes time to create any activity and I appreciate your time and effort to put it out there to share.
4. I did this activity today with my kids. THey LOVED it. LOVED. Great way to review function notation. There was excellent "number talk" happening. Thanks for sharing
1. Yay! Looking forward to trying it with my kids now! | 1,005 | 4,482 | {"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} | 4.3125 | 4 | CC-MAIN-2019-09 | latest | en | 0.972192 |
https://math.stackexchange.com/questions/1699625/you-are-dealt-five-cards-from-a-standard-deck-what-is-the-probability-that-you | 1,721,371,606,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.83/warc/CC-MAIN-20240719043706-20240719073706-00345.warc.gz | 350,321,002 | 41,493 | # You are dealt five cards from a standard deck. What is the probability that you'll have at most three kings in your hand?
You are dealt five cards from a standard and shuffled deck of playing cards. Note that a standard deck has 52 cards and four of those are kings. What is the probability that you'll have at most three kings in your hand?
I know that the answer is $\frac{54144}{54145}$ from the answer key, and I know that the sample space is ${^{52}\mathrm C_5}$. What I don't get is how to find the event.
Do I just add the combination for each number of kings together (${^5\mathrm C_2} + {^5\mathrm C_3}$)?
Or do I need to multiply as well to account for the other cards in the 5-card hand?
• Formatting tips here.
– Em.
Commented Mar 16, 2016 at 4:07
• Also, consider giving a check mark.
– Em.
Commented Mar 16, 2016 at 4:10
The probability that you will have at most 3 kings is the probability that you will have less than 4.
$$\mathsf P(K\leq 3) = 1 -\mathsf P(K=4)$$
The probability that you will have exactly all four kings is the count of ways to select 4 kings and 1 other card divided by the count of ways to select any 5 cards.
$$\mathsf P(K=4)~=~\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$
Put it together.
$$\mathsf P(K\leq 3) ~=~ 1 -\dfrac{{^{4}\mathrm C_{4}}\cdot{^{(52-4)}\mathrm C_{(5-4)}}}{^{52}\mathrm C_5}$$
• This does not agree with the other answers when calculated. It should be a $\binom{48}{1}$ instead of a $\binom{48}{4}$, as you stated in your explanation. Commented Mar 16, 2016 at 4:25
• Typo,.... corrected. Thank you. Commented Mar 16, 2016 at 4:46
• @PaddlingGhost Ah, I see what you meant. Good eye. I thought I read a one, my mistake.
– Em.
Commented Mar 16, 2016 at 4:54
The number of ways to select $5$ cards that have $0,1,2,3$ kings are: $\binom{48}{5}, \binom{4}{1}\binom{48}{4}, \binom{4}{2}\binom{48}{3}, \binom{4}{3}\binom{48}{2}$ respectively. Thus the probability of at most $3$ kings is: $Pr(x \leq 3) =\dfrac{\binom{48}{5}+\binom{4}{1}\binom{48}{4}+\binom{4}{2}\binom{48}{3}+\binom{4}{3}\binom{48}{2}}{\binom{52}{5}}$
At most 3 kings is the same thing as not 4 kings.
$P(4$ kings)$= 48 / {52\choose 5} = \frac{1}{54145}$
• This is wrong. Please review the other responses.
– Em.
Commented Mar 16, 2016 at 4:04
• @probablyme I believe you are mistaken. The Graham Kemp's answer has an error. Commented Mar 16, 2016 at 4:22
• You're right, but it's easy to misread $P(4 \text{ kings})=\frac{1}{54145}$ as saying that the final answer should be $\frac{1}{54145}$. Commented Mar 16, 2016 at 5:05
• @Kevin I agree with you. And not just because of your awesome name Commented Mar 16, 2016 at 14:31
• @Kevin Since the final answer was known in advance, I do not think misreading is so likely. Commented Mar 16, 2016 at 19:08
Split it into disjoint events, and then add up their probabilities:
The probability of exactly $\color\red0$ kings is:
$$\frac{\binom{4}{\color\red0}\cdot\binom{52-4}{5-\color\red0}}{\binom{52}{5}}$$
The probability of exactly $\color\red1$ king is:
$$\frac{\binom{4}{\color\red1}\cdot\binom{52-4}{5-\color\red1}}{\binom{52}{5}}$$
The probability of exactly $\color\red2$ kings is:
$$\frac{\binom{4}{\color\red2}\cdot\binom{52-4}{5-\color\red2}}{\binom{52}{5}}$$
The probability of exactly $\color\red3$ kings is:
$$\frac{\binom{4}{\color\red3}\cdot\binom{52-4}{5-\color\red3}}{\binom{52}{5}}$$
Hence the overall probability is:
$$\sum\limits_{n=0}^{3}\frac{\binom{4}{n}\cdot\binom{52-4}{5-n}}{\binom{52}{5}}=\frac{54144}{54145}$$
Yet another way: there are 52 cards in the deck, 4 of which are kings. The probability the first card is a king is 4/52= 1/13. There are then 51 cards left, 3 of them kings. The probability the second card is a king is 3/51= 1/17. There are then 50 cards left, 2 of them kings. The probability the third card is a king is 2/50= 1/25. There are then 49 cards left, 1 of them a king. The probability the fourth card is a king is 1/49. There are no longer any kings in the deck so the probability the fifth card is not a king is 1.
The probability of getting "four kings and one non-king" in that order is (1/13)(1/17)(1/25)(1/49). There are 5 ways to order "four kings and one non- king" (the non-king being in any of the 5 places) so the probability of "four kings and one non-king" in any order is 5(1/13)(1/17)(1/25)(1/49).
Finally, the probability of get "at most three kings in a five card hand" is 1- 5(1/13)(1/17)(1/25)(1/49).
Since you need to find probability $P(\leq 3~\text{kings})$, you can find $P(>3~\text{kings})$ and subtract it from $1$.
$P(>3~\text{kings}) \iff P(4~\text{kings})$, which is $48/52C5 = 1/54145$
Here, we take $48$, since out of $5$ cards four are kings, and fifth can be any one out of the remaining $48$ cards.
So your final answer is $1 - (1/54145)$ which is $54144/54145$.
• Please read this tutorial on how to typeset mathematics on this site. Commented Mar 16, 2016 at 10:27 | 1,715 | 4,990 | {"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} | 4.46875 | 4 | CC-MAIN-2024-30 | latest | en | 0.918436 |
https://oneclass.com/textbook-notes/ca/ubc/elec-comp-engr/eece-320/282527-320s14hw1spdf.en.html | 1,521,262,506,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257644271.19/warc/CC-MAIN-20180317035630-20180317055630-00642.warc.gz | 677,824,906 | 42,097 | Textbook Notes (362,730)
EECE 320 (1)
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# 320S14HW1s.pdf
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School
University of British Columbia
Department
Electrical and Computer Engineering
Course
EECE 320
Professor
All
Semester
Spring
Description
Homework 1 – Solutions Spring 2014 – EECE 320 – UBC 1. (Proofs.) (a) Show that, for two real numbers a and b, a+b can be less than max(a;b). (Using predicates and quantifiers, we can state this as: Show that 9(a;b) 2 R2 a+b < max(a;b). R is the set of real numbers.) Solution. Consider a = 6;b = ▯3. Then a+b = 6▯3 = 3 and thus a+b < max(6;▯3)=6. (This is a proof by example, which is a somewhat easy proof.) (b) Prove by induction (on n ▯ 0, n is a non-negative integer): !2 n 3 n å (i) = å i : i=0 i=0 Solution. We want to prove ! n n 2 P(n) : (i) = i : i=0 i=0 0 3 ▯ 0 ▯2 Base case: n = 0åi=0i = 0 = å i=0 : Thus the proposition is true for the base case, P(0). Induction hypothesis: Assume that P(1);P(2);:::;P(k) are all true. Induction step: We want to show that P(k+1) is true. ! k+1 k+1 2 P(k+1) : (i) = i : å å i=0 i=0 Therefore, to establish P(k+1) we need to show that: ! ! k k 2 k (k+1) + (i)3 = i +2 i (k+1)+(k+1) 2 i=0 i=0 i=0 By our induction hypothesis, P(k) is true. So, we need to show that: ! 3 k 2 (k+1) = 2 å i (k+1)+(k+1) i=0 Using the identity k i = k(k+1)=2, we need to show that åi=0 (k+1)3 = 2k(k+1) (k+1)+(k+1) 2 2 2 2 = k(k+1) +(k+1) = (k+1) (k+1) = (k+1) : ▯ 1 (c) Prove that for all integers n ▯ 1: 1 ▯ 2: åk=1 k Solution. We can show the stronger result that n P(n) : 1 ▯ 2▯ :1 k=1k2 n Base case: P(1) : 1 ▯ 2▯1 = 1. The base case holds. Induction hypothesis: Let P(2);P(3);:::;P(m) be true. Induction step: We would like to establish P(m+1). m+1 1 1 P(m+1) : å ▯ 2▯ : k=1k2 m+1 Using P(m), we would like to show that 1 1 1 (m+1) 2+2▯ m ▯ 2▯ m+1 : Rearranging terms, we need to show that 1 1 (m+1) 2▯ m(m+1) ; and in turn we need to show that 1 1 m+1 ▯ m : This is true for m ▯ 1: ▯ (d) Prove that we cannot divide a sheet of paper into more than n(n+1)=2+1 regions by drawing n straight lines on the sheet. Solution. We proceed by induction. Base Case: If there are no lines then the plane is divided into 1=0(0+1)=2+1 regions, as desired. Induction Hypothesis: Suppose that any set of n lines divide the plane into at most n(n+1)=2+1 regions. Induction Step: Let X be some set of n+1 lines. Let l be an arbitrary line in X, and letY =X ▯flg (the rest of X). Let A and B be the part of the sheet of paper on the left and right halves of l. By the inductive hypothesis, Y divides the plane into at most n(n+1)=2+1. Let us call these regions1R 2R ;:::. Observe that l can divide each Riinto at most two sub-regions, R i A and R \iB. Moreover, unless l intersects i , one of these regions will be empty. Thus the number of new regions created by drawing l is at most the number of regions that l intersects. Between any two regions that l intersects, there is at least one line which l intersects. Moreover, l intersects each line in Y at most once (since any two lines intersect at most once), and there are n lines in Y. Thus the number of new regions is at most n+1. Thus X divides the piece of paper into at most n(n+1)=2+1+(n+1) = (n+1)(n+2)=2+1 regions, as desired. ▯ 2. (Proofs about games.) Adam and Eve are playing a game with a pile of 2013 cards that proceeds as follows. Each player can, when it is their turn, remove cards (at least one card and no more than three cards) from the pile. If, after a player’s turn, no cards remain then the player who removed the last card(s) is the winner. I
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So we can recommend you notes for your school. | 1,355 | 3,871 | {"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-2018-13 | latest | en | 0.84221 |
https://discuss.codechef.com/t/abgame-editorial/21024 | 1,686,058,187,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00247.warc.gz | 250,526,979 | 10,046 | ABGAME - EDITORIAL
Setter: Praveen Dhinwa
Tester: Teja Vardhan Reddy
Editorialist: Taranpreet Singh
Easy
PREREQUISITES:
Game Theory, Nim’s Game and Observations.
PROBLEM:
Two players A and B plays a game on a string consisting of characters ‘.’, ‘A’ and ‘B’. Each occurrence of character ‘A’ and ‘B’ is assigned direction from left to right alternatively, starting from the right direction. ‘.’ represent empty spot.
Player A can move any character marked ‘A’ in the predetermined direction any number of steps as long as not jumping over any of character ‘A’ and ‘B’, while player B can move any character marked ‘B’ in the predetermined direction in the same manner.
Players make move alternatively, with player A playing first. Determine who will win, if both of them play optimally and the player making the last move wins.
For this problem, I am counting only A and B as characters, for a simpler explanation.
QUICK EXPLANATION
• Except for the last character if the number of characters present is odd, all characters are assigned directions such that in pairs of two, Characters are moving toward each other. ie. First and second are moving toward each other, Third and Fourth characters are moving toward each other and so on. If there are an odd number of characters, the last character is moving toward the right end.
• Let reserve moves represent the number of moves a player can perform independently of other moves. More the reserve moves, better for that person. Since he wants to play the last move.
• If any pair of adjacent characters moving toward each other belongs to the same player, that player can move either of his characters toward the other in maximum d moves, if d is the number of empty cells between both characters. So, That player’s reserve moves increases by d. In case of the last character where the last character is moving toward the right border, that player has an additional d reserve moves, if there are d empty cells between last character and right border.
• If the pair of adjacent characters are different and there are d empty cells between both characters, It can be reduced to the game of Nim, with sizes of piles being the number of cells between every pair of adjacent numbers, where the losing player has to use its reserve moves first.
• The final game is the game of Nim with a finite number of pass moves for both players.
• If Number of pass moves differ, The player having the higher number of pass moves will always win. Otherwise, It is simply a Game of Nim, with the winner of Nim game being the winner of the Game.
EXPLANATION
This is not a single game, but a combination of Nim’s game combined with a simple turn game.
Let us see how the directions first. Consider example A…A…A…B…B…B. The first character is assigned direction right, second is assigned left, third is assigned right and so on. This way. Directions are assigned as R…L…R…L…R…L.
It is easy to notice that the first and second characters are moving together, third and fourth character are moving toward each other and so on. So, Every pair of characters can be treated separately.
Let us define the concept of reserve moves for a player as the number of moves a player can perform independently of the moves performed by the opponent.
Playing a simple game for understanding
Player A can perform X moves, while player B can perform Y moves, independent of each others’ moves. If player A is first to move, player A wins if X > Y. But if player B is second to move, player A wins if X \geq Y. It is not easy to prove how.
The first move matters.
Coming back to original problem
When both characters of the pair are same, both characters are controlled by the same player, and cannot be affected by opponent’s moves. This player can make minimum 1 and maximum d moves using these pair of characters only. Since a player wants to play as many moves as possible, it is optimal to spend d moves on this pairs. This will add to reserve moves available to the player.
Similarly, If there are an odd number of characters available, the last character is assigned right direction and it will move toward the right border. This will add d moves to reserve moves of player, to whom the last character belongs.
Now comes the tricky part.
When both characters of the pair are different. Either of the players can move any number of steps at a time, moving only one character in one move until there is no empty cell left in between. Notice that Either player will run out of moves if and only if there are no such pairs, with empty cells in between. Hence, in the final state, all such pairs will compress, removing all empty cells between the characters.
Let us consider an example A…B.A…B
Winning player is A with the strategy to move his character at position 6 to position 7, and then in subsequent moves, moving his character in the same way as B moves.
Let’s try to reframe the problem as Number of empty cells between first such pair is 2, while the number of empty cells between the second pair is 3. Playing in turns with A moving first, Each player at a move is allowed to reduce exactly one of the numbers by any non-zero number. The player unable to make a move loses. Find the winner if both players play optimally.
Does the above rings a bell?
Give it a try!!
The above statement is exactly the Nim’s game. (Explaining Nim’s Game later).
Hence, the problem is just the Nim’s game, with both players having a given number of pass moves (possibly different number of moves). In case initially the Nim’s game is in a winning position, A will Make move so as to give B a losing position in Nim’s game. In case the given game was already in a losing position, A will use a reserve move. In either way, Now player B and A, in turns, use their pass moves till pass moves are exhausted for a player, and is forced to make a move in losing position in Nim, thus losing the game.
Another way of writing the winning idea is, If both players have an unequal number of reserve moves, the player with more reserve moves wins, otherwise the winner of Nim’s game wins. It is not hard to follow, once you get the idea of the above solution.
I had seen a post asking details on Game Theory but couldn’t reply, so, here we go!
I will not be going in too much detail, will be using informal language, just explaining the ideas with brief proof. Details you can find on various blogs, for example, here and here.
The Nim’s game at the core is the problem with N piles of stones containing a non-zero number of stones. Both players, in their turns, are allowed to remove any positive number of stones from exactly one pile. The player unable to make a move loses.
The solution is based on the idea of xor-sum.
Let us see the final state. In this state, none of the pile has any stones left. In terms of xor-sum, the xor-sum of the number of stones in each pile is 0. The previous state must have included one pile containing say x stones, having x as xor-sum and is a winning state.
We know, that in game theory, The states visited from last to first is something like losing state, winning state, losing state alternatively and so on. So, the second last state visited must be losing state for the player to move.
If it was like, that before second last move, there was only one pile having more than x stones and losing player removed stones to leave x stones, Then this is not an optimal strategy as the losing player could have removed the whole pile, winning the game.
So, in second last move, there should be exactly two piles. Suppose there were y stones in that pile. Suppose x \neq y. Even now, the player making the current move can win, by removing stones in such a way to get the same number of stones in both piles, thus again winning the game.
Thus, for second last state to be losing, the required condition becomes x == y.
Similar reasoning escalates to more piles. (Equality changes into zero xor-sum).
The reasoning is simplified in terms of xor-sum of piles. See, All losing states have xor-sum of stones count in piles 0. The reasoning is, that If some piles have xor-sum 0, then in next turn, it will definitely change into some non-zero value.
And, If the piles have xor-sum non-zero, it is always possible to remove stones so as to make xor-sum 0.
This way, a parallel is drawn between Nim’s game and xor-sum, and we get the result as we know today.
That’s all for today. Feel free to share anything useful related to Game theory.
Time Complexity
Time complexity is O(|S|) where |S| is the length of string and we need to iterate over it only once.
AUTHOR’S AND TESTER’S SOLUTIONS:
Feel free to Share your approach, If it differs. Suggestions are always welcomed.
7 Likes
``````<copy>
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``````Woah!! This long editorial, scary even after solving this problem.
``````
``````</copy>
``````
For those who are clueless. FCTR - Editorial - #2 by taran_1407 - editorial - CodeChef Discuss
Looks like someone went through a change of mind @taran_1407 ?
3 Likes
Nice solution.
CodeChef seems to be obsessed with using Nim everywhere this year xD.
i did the same but it gave WA:-
Hello @saurav0001, I am unable to see your code.
1 Like
Video solution with explanation: https://youtu.be/Gp90sUHZ6Y0
3 Likes
Another way to look at it is to compare the total number of each player’s moves if they play optimally. Player A wins iff he has more moves available. First, count the number of “pass moves” for each player. If the Nim game is initially in a winning position it gives one extra move to player A - add 1 to his move count, otherwise both players have equal numbers of moves in the Nim game.
@vipin1407 please check this: k6pI0F - Online C++0x Compiler & Debugging Tool - Ideone.com
@saurav0001 what you are doing in your submission is wrong. You need to study more about nim games and how to determine the winner in case of nim games (xor of the piles’ sizes etc.).
Can someone plz explain why we are making pairs and make them move towards each other only? I think i’m missing something. Plz help!
I think it was the first question on NIM’s game- rest game theory questions were brute forcing optimal solutions?
…XDD
1 Like
I certainly remember what i said at that time. And, it was true
This one went long because of additional explanation of nim’s game i provided.
This one actually used Nim’s game. Rest were just game theory ones, not exactly Nim’s games.
Congrats, its true in this case as well now
``````This one went long because of additional explanation of nim's game i provided.
``````
Mine went long because I wanted a nice formatting to explain maths
You can say that for game theory…
Codechef showed me how bad I am at game theory…XD
Or at only logical questions which has nothing about data structure…
2 Likes
Short and sweet explaination…
Can be called “quick explaination”…
Nice one…
Because every character at odd position (out of non-empty characters) is moving toward right, while other characters are moving toward left.
See, .A…B…B…B First A moves toward right, First B toward left, Second B toward right and third A toward left. So, they end up moving toward each other. | 2,519 | 11,202 | {"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} | 3.8125 | 4 | CC-MAIN-2023-23 | latest | en | 0.940109 |
http://www.markedbyteachers.com/gcse/science/investigation-into-cars-going-down-slopes.html | 1,526,825,388,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794863570.21/warc/CC-MAIN-20180520131723-20180520151723-00305.warc.gz | 408,657,859 | 19,381 | • Join over 1.2 million students every month
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# Investigation into cars going down slopes
Extracts from this document...
Introduction
## Investigation into cars going down slopes
1) Variables
1. The material and surface of the slope
2. Gradient of the slope (height)
3. Type of car – weight, height and the size of the car etc.
4. Whether you give the car a push as you release it
5. Height of the car up the slope.
2) The variable, that I am going to investigate, is the height of the car up the slope.
3) Prediction
### My prediction of what I will think will happen is that as the car is placed higher up the slope, then this will give the car more potential energy. Then as the car is released it will have morekinetic energy allowing the car to go faster & further along the run-out. Therefore I predict that the height will be proportional to the velocity squared (speed).
4)The scientificreason for my prediction
### B C
• The velocity at A will be 0 as the car isn’t moving but the car however has very high potential energy at A.
Middle
(This can be written as ½ mv )
So now the whole of the cars energy =kinetic energy + potential energy
At A on the diagram the potential energy (mgh) is equal to the kinetic energy (½ mv ) at B.
The equation for this is:
mgh = ½ mvTherefore the two Ms will cancel out and leave the new equation as h = v - As you can see here h is proportional to v
2g
It is also correct to say that the maximum velocity of the car will be at B.
The average velocity is calculated as: DISTANCE
#### TIME
-So from this equation 2x this will simply give the Maximum velocity and again squaring this value will give you the maximum velocity squared
5) Method for doing my investigation
#### A
B C
Set up experiment is set up exactly as it is in my prediction.
1. Firstly, place the car with the front wheels on the line you are starting from.
Conclusion
height and up the side of the graph is the Velocity squared. From knowing that the height is proportional to the velocity squared the graph should look something like this-
If there are any anomalous results then these should be repeated again.
##### At the start of our investigation the car we used was giving us unreliable results and so we decided to do the whole investigation from the beginning but with a different car and also with a different person doing the timing .We definitely made the right decision as we then got better results. From these results we found that the line of best fit on the graph showed a positive correlation, which is what we wanted. However the results weren’t quite right so we re-tested them and found we got better, more reliable results that we could then work from. From the new results we plotted them on a graph and found that what we got was perfect and agreed with our prediction at the start.
This student written piece of work is one of many that can be found in our GCSE Forces and Motion section.
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• Ideas and feedback to | 1,108 | 4,830 | {"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} | 4.25 | 4 | CC-MAIN-2018-22 | latest | en | 0.891907 |
https://www.numbersaplenty.com/3334004333 | 1,696,139,711,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510781.66/warc/CC-MAIN-20231001041719-20231001071719-00072.warc.gz | 969,157,817 | 3,366 | Search a number
3334004333 = 111072832629
BaseRepresentation
bin1100011010111000…
…1101111001101101
322121100111222010222
43012232031321231
523312001114313
61310455141125
7145422366152
oct30656157155
98540458128
103334004333
111460a5a890
127906747a5
13411959b3a
14238b0c229
15147a6d208
hexc6b8de6d
3334004333 has 8 divisors (see below), whose sum is σ = 3671088480. Its totient is φ = 3002585680.
The previous prime is 3334004299. The next prime is 3334004371.
It is a happy number.
3334004333 is nontrivially palindromic in base 10.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 3334004333 - 218 = 3333742189 is a prime.
It is a super-2 number, since 2×33340043332 = 22231169784925549778, which contains 22 as substring.
It is a Duffinian number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (3334004393) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 1415138 + ... + 1417491.
It is an arithmetic number, because the mean of its divisors is an integer number (458886060).
Almost surely, 23334004333 is an apocalyptic number.
It is an amenable number.
3334004333 is a deficient number, since it is larger than the sum of its proper divisors (337084147).
3334004333 is a wasteful number, since it uses less digits than its factorization.
3334004333 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 2832747.
The product of its (nonzero) digits is 11664, while the sum is 26.
The square root of 3334004333 is about 57740.8376541248. The cubic root of 3334004333 is about 1493.9018094971.
It can be divided in two parts, 333400 and 4333, that added together give a palindrome (337733).
The spelling of 3334004333 in words is "three billion, three hundred thirty-four million, four thousand, three hundred thirty-three". | 635 | 2,105 | {"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-2023-40 | latest | en | 0.853393 |
http://experiment-ufa.ru/how_to_write_2306_in_roman_numerals | 1,529,726,576,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864940.31/warc/CC-MAIN-20180623035301-20180623055301-00479.warc.gz | 106,912,690 | 6,084 | # How to write 2306 in roman numerals
## Write 2306 in roman numerals.
If it's not what You are looking for, type in into the box below your integer to convert it into roman numerals.
## 2306 in roman numerals:
When you convert 2306 into roman numerals you get
MMCCCVI
In summary 2306 in roman numerals is MMCCCVI
Now when you know how to write 2306 in roman numerals you can use it whenever you want.
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http://planetmath.org/SemimartingaleTopology | 1,521,517,520,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647280.40/warc/CC-MAIN-20180320033158-20180320053158-00627.warc.gz | 230,664,760 | 5,285 | # semimartingale topology
Let $(\Omega,\mathcal{F},(\mathcal{F}_{t})_{t\in\mathbb{R}_{+}}),\mathbb{P})$ be a filtered probability space and $(X^{n}_{t})$, $(X_{t})$ be cadlag adapted processes. Then, $X^{n}$ is said to converge to $X$ in the semimartingale topology if $X^{n}_{0}\rightarrow X_{0}$ in probability and
$\int_{0}^{t}\xi^{n}\,dX^{n}-\int_{0}^{t}\xi^{n}\,dX\rightarrow 0$
in probability as $n\rightarrow\infty$, for every $t>0$ and sequence of simple predictable processes $|\xi^{n}|\leq 1$.
This topology occurs with stochastic calculus where, according to the dominated convergence theorem (http://planetmath.org/DominatedConvergenceForStochasticIntegration), stochastic integrals converge in the semimartingale topology. Furthermore, stochastic integration with respect to any locally bounded (http://planetmath.org/LocalPropertiesOfProcesses) predictable process $\xi$ is continuous under the semimartingale topology. That is, if $X^{n}$ are semimartingales converging to $X$ then $\int\xi\,dX^{n}$ converges to $\int\xi\,dX$, a fact which does not hold under weaker topologies such as ucp convergence.
Also, for cadlag martingales, $L^{1}$ convergence implies semimartingale convergence.
It can be shown that semimartingale convergence implies ucp convergence. Consequently, $X^{n}$ converges to $X$ in the semimartingale topology if and only if
$X^{n}_{0}-X_{0}+\int\xi^{n}\,dX^{n}-\int\xi^{n}\,dX\xrightarrow{\rm ucp}0$
for all sequences of simple predictable processes $|\xi^{n}|\leq 1$.
The topology is described by a metric as follows. First, let $D^{\rm ucp}(X-Y)$ be a metric defining the ucp topology. For example,
$D^{\rm ucp}(X)=\sum_{n=1}^{\infty}2^{-n}\mathbb{E}\left[\min\left(1,\sup_{t
Then, a metric $D^{\rm s}(X-Y)$ for semimartingale convergence is given by
$D^{\rm s}(X)=\sup\left\{D^{\rm ucp}(X_{0}+\xi\cdot X):|\xi|\leq 1\textrm{ is % simple previsible}\right\}$
($\xi\cdot X$ denotes the integral $\int\xi\,dX$). This is a proper metric under identification of processes with almost surely equivalent sample paths, otherwise it is a pseudometric.
If $\lambda_{n}\not=0$ is a sequence of real numbers converging to zero and $X$ is a cadlag adapted process then $\lambda_{n}X\rightarrow 0$ in the semimartingale topology if and only if
$\lambda_{n}\int_{0}^{t}\xi^{n}\,dX\rightarrow 0$
in probability, for every $t>0$ and simple predictable processes $|\xi^{n}|\leq 1$. By the sequential characterization of boundedness (http://planetmath.org/SequentialCharacterizationOfBoundedness), this is equivalent to the statement that
$\left\{\int_{0}^{t}\xi\,dX:|\xi|\leq 1\textrm{ is simple predictable}\right\}$
is bounded in probability for every $t>0$. So, $\lambda_{n}X\rightarrow 0$ in the semimartingale topology if and only if $X$ is a semimartingale. It follows that semimartingale convergence only becomes a vector topology (http://planetmath.org/TopologicalVectorSpace) when restricted to the space of semimartingales. Then, it can be shown that the set of semimartingales is a complete topological vector space (http://planetmath.org/CompletenessOfSemimartingaleConvergence).
Title semimartingale topology SemimartingaleTopology 2013-03-22 18:40:41 2013-03-22 18:40:41 gel (22282) gel (22282) 6 gel (22282) Definition msc 60G48 msc 60G07 msc 60H05 semimartingale convergence UcpConvergence UcpConvergenceOfProcesses | 1,044 | 3,380 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 36, "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} | 2.796875 | 3 | CC-MAIN-2018-13 | latest | en | 0.676275 |
https://book.gateoverflow.in/tag/numerical-ability | 1,620,528,798,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988953.13/warc/CC-MAIN-20210509002206-20210509032206-00194.warc.gz | 171,910,197 | 18,569 | # Recent questions tagged numerical-ability
1
In how many different ways can the letters of the word ‘MATHEMATICS’ be arranged so that the vowels always come together? $10080$ $4989600$ $120960$ None of the options
2
If $\log_{x}y=100$ and $\log_{2}x=10$, then the value of $y$ is : $2^{10}$ $2^{100}$ $2^{1000}$ $2^{10000}$
1 vote
3
A college cricket team with $11$ players consists of $4$ batsman, $3$ all-rounders, $3$ bowlers and $1$ wicket keeper. $3$ players are selected randomly. Find the probability that the selection contains a batsman, a bowler and an all-rounder. $\dfrac{12}{60} \\$ $\dfrac{13}{25} \\$ $\dfrac{12}{55} \\$ $\dfrac{104}{165}$
4
Ali is thrice as good as workman as Birju and therefore is able to finish a job in $60$ days less than Birju. If both Ali and Birju, work together, they can do it in: $20$ days $22.5$ days $25$ days $30$ days
1 vote
5
Rishi got two halls in his house painted. It cost him $₹6,000$ and $₹6,100$ respectively to paint the $4$ walls of his $2$ square halls, of the same height. If the length of one hall exceeds the length of the other by $1$ m and the cost of painting is $₹5$ per sq.m, what is the height of the two halls? $3$ m $5$ m $7.5$ m $10$ m
6
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8
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9
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12
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13
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14
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15
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16
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1 vote
17
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18
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19
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20
If on May $1$, $60$% of the permanent workers and $40$% of the temporary workers employed by $M/s$ Mega Corp Limited were women, how many of the workers employed by $M/s$ Mega Corp Limited at that time were women? $200$ $120$ $410$ $260$
21
Based on the information given above, the minimum number of faculty members involved in both training and teaching , but not in research is: $1$ $3$ $4$ $5$
The completion of one cycle of the network results in one bottle ready to be sold in the market . The project involves a total of $800$ bottles. What is the average cost of the entire project? Rs $74400$ Rs $372000$ Rs $15000$ Rs $18500$
$A$ can complete half work in $7$ days $B$ can do $\dfrac{1}{3}$ of work in $14$ days .while $C$ can complete the $20\%$ of the remaining work in $\dfrac{28}{5}$ days. In how many days $A,B,C$ will complete the work together ?
A plane left half an hour than the scheduled time and in order to reach its destination $1500 \hspace{0.1cm} km$ away in time, it had to increase its speed by $33.33\%$ over its usual speed. find its increased speed? $250 \hspace{0.1cm} kmph$ $500 \hspace{0.1cm} kmph$ $750 \hspace{0.1cm} kmph$ none the answer given is $750 \hspace{0.1cm} kmph$ but i m getting $250$ which is correct?? | 1,769 | 5,618 | {"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} | 3.4375 | 3 | CC-MAIN-2021-21 | latest | en | 0.909185 |
https://studylib.net/doc/9714108/mechanics-glossary | 1,716,878,529,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00514.warc.gz | 464,909,496 | 11,913 | # MECHANICS GLOSSARY
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``` No Brain Too Small SCIENCE
AS 90940
Mechanics Glossary
acceleration
change in speed over time, whether speeding up or slowing down
air resistance
friction acting on an object moving through air
average speed
the total distance travelled divided by the time required to travel that distance
balanced forces
when two forces combine to cancel each other out
conservation
(energy) cannot be created or destroyed
constant speed
speed is not increasing or decreasing but remain consistent over time
constant
unchanging
contact area
amount of area shared by two objects e.g. ice skates have sharp edges, and thus a
small area in contact with the ice
deceleration
decrease in the speed over time of an object
distance
how far an object has travelled (usually horizontal)
drag
for an object moving through air, friction is called "air resistance" or "drag"
EK
kinetic energy; the kind of energy possessed by moving objects
EP
gravitational potential energy; energy possessed by objects that have been lifted up
force
a push or pull that can change an object’s movement
free fall
falling motion caused by gravity alone
friction
friction is an opposing force; it acts in the opposite direction to a force which is
applied to an object
g
acceleration caused by the earth’s mass (approx. 10 ms-2)
gradient
slope of a graph (rise/run)
gravity
the force of attraction between any two objects; the Earth is very big and so has a
large gravity pulling everything down towards it
height
distance (vertical)
instantaneous
at an instant of time
mass
amount of matter in an object or substance
mechanical energy
kinetic or potential (or heat) energy
net force
the single force that represents all the forces acting on a body; overall force
power
the rate that energy is changed from one type to another; power is the energy
changed divided by the time it takes for the energy to change
pressure
a measure of the force exerted on a certain area of surface; force per unit area
reaction
An equal and opposite force exerted by a body against a force acting upon it
relationship
information that can be used to link two things together e.g. force and acceleration
No Brain Too Small SCIENCE
resultant force
a stationary object remains stationary if the sum of the forces acting upon it resultant force - is zero. A moving object with a zero resultant force keeps moving at
the same speed and in the same direction. If the resultant force acting on an object
is not zero, a stationary object begins to accelerate in the same direction as the
resultant force.
speed
how fast an object is travelling; units are distance time-1, e.g. ms-1
stationary
stopped, not moving or “at rest”
thrust
to push or drive with force
unbalanced forces
forces that are NOT balanced; situation leading to acceleration or deceleration of an
object
velocity
speed (Note: velocity is really the measurement of the rate and direction of motion
but at Level 1 Science we treat speed and velocity as the same thing)
weight (force)
the force on an object caused by gravity
work
energy required to make an object move
Extra notes:
No Brain Too Small SCIENCE
``` | 746 | 3,183 | {"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-2024-22 | latest | en | 0.908259 |
https://masterlyassignments.com/math-3/ | 1,675,067,164,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499804.60/warc/CC-MAIN-20230130070411-20230130100411-00668.warc.gz | 396,637,267 | 15,885 | math
math. Tracks of a Killer
Introduction:
The body of famous pop music producer Johnathan Wallace was found in his bathtub. It is our hypothesis that an intruder surprised the victim and drowned him. The only clue at the crime scene was a set of muddy footprints leading from a nearby window to the bathroom and back again. The footprints were smeared, so their exact size could not be determined, and the soles of the shoes had no pattern. It will be difficult to match the footprints to any particular pair of shoes.
You will use data collected from a sample group of people to determine whether there is a correlation between height and either shoe length or stride length. If there is, you can then use the predictive equation to predict the height of the killer based on the muddy footprint evidence at the crime scene.
Learning objectives:
• Evaluate statistical claims using eight guidelines and an understanding of how statistical studies are created
Before you begin:
• Review the key principles of correlation and causation in unit 5E of your textbook.
• Download the file “Online QR Activity – Tracks of a Killer – Data file.xls” which provides you the data on the sample of people’s heights, shoe lengths, and stride lengths. Save it as “LastName-Tracks of a Killer activity spreadsheet.xls.” You will be submitting this saved spreadsheet through Canvas when it’s complete.
• Locate the resources for “Interpreting correlation coefficients,” “Exploring correlation with Excel,” and “Graphing with Excel” on our Canvas site which you will use to help you complete this Activity.
Procedure:
• Use the data in the “Online QR Activity – Tracks of a Killer – Data file.xls” spreadsheet, to calculate two correlation coefficients in Excel, one for Shoe Length vs. Height and one for Stride Length vs. Height.
• Create two scatter plots in Excel, one for each of the relations listed above. You should insert the trendline, equation for the trendline, and correlation coefficient () on each graph. Don’t forget to label your axes and include a title for each graph.
• Answer the Analysis questions below in a Word Document which you should save as Save it as “LastName-Tracks of a Killer analysis.” The analysis should be a minimum of 1 page (at least 250-300 words), typed double-spaced in Times New Roman font.
• Submit the completed and saved Excel budget spreadsheet and the analysis Word Document, through this assignment in Canvas.
Crime scene and suspect data
Three suspects were questioned immediately following the murder:
• Penelope Paige, pop star, 5’4”/green eyes/blonde hair. Possible motive: She is suing Wallace over the failure of her last album.
• Rex Chapman, rock guitarist, 5’8”/brown eyes/brown hair. Possible motive: He accused Wallace of stealing profits from his hit single “Walk It Off.”
• Dirty Dawg, rapper, 6’0”/brown eyes/black hair. Possible motive: He wants out of a record contract with Wallace.
Here is the evidence from the crime scene:
Analysis:
1. Based on the data, do you think there is a linear relationship between Stride Length and Height? Explain your answer and incorporate your interpretation of the correlation coefficient and trendline.
2. Based on the data, do you think there is a linear relationship between Shoe Length and Height? Explain your answer and incorporate your interpretation of the correlation coefficient and trendline.
3. Do you think it is possible to infer a person’s height from his or her shoe size? Explain your answer.
4. Do you think it is possible to infer a person’s height from his or her stride length? Explain your answer.
5. Use the relationship between Stride Length and Height to determine the approximate heights of people with the following stride lengths. You should include in your solutions work that shows the unit conversions necessary.
1. 75 m
2. 45 m
3. 50 m
6. Suppose you measure the stride length of a set of footprints, and you predict that the person who made the footprints is 1.75 m tall. Later, you find out that the person who made the footprints is actually only 1.52 m tall. Give two possible reasons why your prediction was incorrect, and thoroughly explain your reasons.
7. Use the relationships that you calculated and determine which of the three suspects most likely left the footprints to and from Jonathan Wallace’s bathroom. Show all your calculations. Hint: Use the axis from your graph to determine which variables are the x and y in your trendline equation.
Excellent Proficient Satisfactory Needs Improvement Not Evident Graphs Points: 30 Both scatterplots have all 6 required elements present (for a total of 12 required elements). Points: 25 Up to 2 required elements are missing from the scatterplots. Points: 20 Up to 3 required elements are missing and/or the graphs are not scatterplots Points: 18 More than 4 required elements are missing and/or the graphs are not scatterplots Points: 0 Graphs were not submitted Analysis questions Points: 60 All 7 analysis questions were answered thoroughly and completely Points: 53 Most analysis questions were answered thoroughly and completely, but 1-2 were not. Points: 47 Some analysis questions were answered thoroughly and completely, but 3-4 were not. Points: 40 Some analysis questions were answered thoroughly and completely, nut 5-6 were not, OR all analysis questions were answered, but not thoroughly. Points: 0 (0.00%) Not included Writing Points: 10 Writing exhibited complex sentence structure, and was free of spelling and grammatical errors Points: 9 Writing exhibited simple sentence structure, and was free of spelling and grammatical errors Points: 8 Writing exhibited simple sentence structure, and had less than 3 spelling and/or grammatical errors Points: 7 Writing exhibited simple sentence structure, and had 4-5 spelling and/or grammatical errors OR writing had incomplete sentences and no spelling/grammatical errors. Points: 0 Writing exhibited poor sentence structure, and had many spelling and/or grammatical errors 100 87 75 65
Adapted from: Texas Instruments. (2010). Forensics Case 1 – Tracks of a Killer: Using footprints to estimate height. Retrieved 2010, from Education technology – Classroom Activities: http://education.ti.com/calculators/downloads/US/Activities/Detail?id=6369
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https://www.assignmentessays.com/suppose-v-w-%E2%88%88-cn-an-exercise-similar-to-the-one-leading/ | 1,580,219,958,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251778272.69/warc/CC-MAIN-20200128122813-20200128152813-00141.warc.gz | 748,325,678 | 12,551 | # Suppose v, w ∈ Cn . An exercise similar to the one leading
August 30, 2017
Question
Assignment 6–Algebra I
Due at the beginning of tutorial Nov. 16
1. Suppose v, w ∈ Cn . An exercise similar to the one leading up to the
Cauchy-Schwarz inequality yields the equation
v+w
2
= v
2
+ 2Re( v, w ) + w 2 ,
where Re(a + bi) = a for a, b ∈ R. Take this equation for granted. It is
also not that difficult to prove that Re(a + bi) ≤ a2 ≤ |a + bi| for every
a, b ∈ R. Using these two facts, prove the triangle inequality which states
/4
v+w ≤ v + w ,
(Hint: Use the Cauchy-Schwarz inequality to prove that the square of the
LHS is less than or equal to the square of the RHS.)
2. Let
2
−1
−4
2
v1 =
0 , v2 = 1
−4
2
−i
, v3 = 3 .
1
i
The set {v1 , v3 , v3 } is a linearly independent set in C4 (do not prove this).
Compute an orthonormal set {u1 , u2 , u3 } from it using the Gram-Schmidt
procedure. Begin the procedure with the vector v1 .
/4
3. Let {v1 , . . . , vp } and {u1 , . . . , up } be as in Theorem GSP in the text.
(a) Prove by strong induction on 1 ≤ k ≤ p that uk ∈ span{v1 , . . . vk }
for every 1 ≤ k ≤ p. (Hint: Have a look at the link on strong induction on the course website. The difference between strong induction
and just plain induction here is that you will make the induction
assumption of uj ∈ span{v1 , . . . , vj } for all 1 ≤ j ≤ k)
(b) Prove that uk = 0 for any 1 ≤ k ≤ p. (Hint: Look at the proof of
Theorem GSP. You will need part (a).)
SUGGESTED EXERCISES
(see next page)
• Prove the first two facts in exercise 1.
• Compute an orthonormal set from {u1 , u2 , u3 } in exercise 2.
• All reading questions in section MO and exercises T13-T18 in section MO.
• All reading questions and exercises in section MM.
• Prove Theorem SLSLC in section LC by using Theorem SLEMM and what
1
/4
/4
• Choose a set of two linearly independent vectors in C3 (why is this easy?)
and apply the Gram-Schmidt procedure to the set. Normalize the resulting
orthogonal set to obtain an orthonormal set.
• Suppose {v1 , . . . , vk } is an orthonormal set in Cn and a1 , . . . , ak ∈ C.
Prove that
2
k
aj vj
k
|aj |2
=
j=1
j=1
• Give an explicit example of an orthogonal set {v1 , . . . , vk } in C2 and scalars
a1 , . . . , ak ∈ C for which the previous equation fails. (Prove both orthogonality and the failure of the equation. Hint: Take k = 2.)
• Suppose {v1 , v2 } ⊂ Cn is linearly independent. Use the Gram-Schmidt
procedure to prove that v2 = u1 + u2 where u1 is some vector orthogonal
to v1 and u2 ∈ span{v1 }.
• Prove that v2 = u1 +u2 as in the previous exercise, also when {v1 , v2 } ⊂ Cn
is linearly dependent.
• Prove that v =
n
j=1
v, ej ej . This proves that Cn = span{e1 , . . . en }.
• Use the previous exercise to prove that v ∈ Cn is orthogonal to every
vector in Cn if and only if v = 0.
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Positive SSL | 969 | 2,931 | {"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} | 3.90625 | 4 | CC-MAIN-2020-05 | latest | en | 0.846297 |
http://levendeurdegoyaves.com/weather/greedy-method-algorithm-pdf.php | 1,620,721,101,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243991904.6/warc/CC-MAIN-20210511060441-20210511090441-00264.warc.gz | 31,592,906 | 7,014 | # Greedy method algorithm pdf
12.02.2021 | By Takinos | Filed in: Weather.
26/09/ · The Pilot method, randomized greedy, and iterated greedy are all useful and easy to implement. Each of the three generates a num ber of feasible solutions. The carousel greedy algorithm . 1 Greedy algorithms Today and in the next lecture we are going to discuss greedy algorithms. \Greedy" in this context means \always doing the locally optimal thing". E.g., a greedy algorithm for driving to some destination might be one that at each intersection always takes the street heading most closely in the direction of the destination. In general, this approach might or might not nd an. Thus after the greedy algorithm added its kth activity to S, the (k + 1)st activity from S* would still belong to U. But the greedy algorithm ended after k activities, so U must have been empty. We have reached a contradiction, so our assumption must have been wrong. Thus the greedy algorithm must be optimal.
# Greedy method algorithm pdf
June Cutting-plane method Reduced gradient Frank—Wolfe Subgradient method. National Institute of Standards and Technology NIST. Mathematics portal. It is important, however, to note that the greedy algorithm can be used as a selection algorithm to prioritize options within a search, or branch-and-bound algorithm. See our User Agreement and Privacy Policy.Interval Partitioning: Greedy Algorithm Greedy algorithm. Consider lectures in increasing order of start time: assign lecture to any compatible classroom. 17 Sort intervals by starting time so that 1 2 . ←0 // # classrooms used so far for ←1to if lecture is compatible with some classroom then. Module 3 The Greedy Method. It Is the most straight forward technique. It can be applied to a wide variety of problems. Most of these problems have n inputs and require is to obtain a subset that satisfies these constraints. Any subset that satisfies these constraints is called a feasible solution. We need to find a feasible solution that either maximizes or minimizes a given objective function. 26/09/ · The Pilot method, randomized greedy, and iterated greedy are all useful and easy to implement. Each of the three generates a num ber of feasible solutions. The carousel greedy algorithm . The greedy method is a general algorithm design paradigm, built on the following elements: n configurations: different choices, collections, or values to find n objective function: a score assigned to configurations, which we want to either maximize or minimize It works . Thus after the greedy algorithm added its kth activity to S, the (k + 1)st activity from S* would still belong to U. But the greedy algorithm ended after k activities, so U must have been empty. We have reached a contradiction, so our assumption must have been wrong. Thus the greedy algorithm must be optimal. 1 Greedy algorithms Today and in the next lecture we are going to discuss greedy algorithms. \Greedy" in this context means \always doing the locally optimal thing". E.g., a greedy algorithm for driving to some destination might be one that at each intersection always takes the street heading most closely in the direction of the destination. In general, this approach might or might not nd an. 5 GREEDY ALGORITHMS 5 Greedy Algorithms The second algorithmic strategy we are going to consider is greedy algorithms. In lay-man’s terms, the greedy method is a simple technique: build up the solution piece by piece, picking whatever piece looks best at the time. This is not meant to be precise, and sometimes, it can take some cleverness to gure out what the greedy algorithm really is. But. Lecture 9: Greedy Algorithms version of September 28b, A greedy algorithm always makes the choice that looks best at the moment and adds it to the current partial solution. Greedy algorithms don’t always yield optimal solutions, but when they do, they’re usually the simplest and most efficient algorithms available. Interval Scheduling Interval scheduling. Job starts at and finishes at. 07/04/ · Greedy algorithm 1. Short Explanation, Caisar Oentoro 2. What is Greedy Algorithm? In the hard words: A greedy algorithm is an algorithm that follows the problem solving heuristics of making the locally optimal choice at each stage with the hope of finding a global optimum. class so far, take it! See Figure. for a visualization of the resulting greedy schedule. We can write the greedy algorithm somewhat more formally as shown in in Figure.. (Hopefully the first line is understandable.) After the initial sort, the algorithm is a simple linear-time loop, so the entire algorithm runs in O(nlogn) time.
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Greedy Method - Design \u0026 Algorithms - Lec-38 - Bhanu Priya, time: 13:05
Tags: Car action magazine pdf, Tuba euphonium quartet pdf, 1 Greedy algorithms Today and in the next lecture we are going to discuss greedy algorithms. \Greedy" in this context means \always doing the locally optimal thing". E.g., a greedy algorithm for driving to some destination might be one that at each intersection always takes the street heading most closely in the direction of the destination. In general, this approach might or might not nd an. Greedy algorithms build up a solution piece by piece, always choosing the next piece that offers the most obvious and immediate benet. Although such an approach can be disastrous for some computational tasks, there are many for which it is optimal. Our rst example is that of minimum spanning trees. Minimum spanning trees Suppose you are asked to network a collection of computers by . Lecture 9: Greedy Algorithms version of September 28b, A greedy algorithm always makes the choice that looks best at the moment and adds it to the current partial solution. Greedy algorithms don’t always yield optimal solutions, but when they do, they’re usually the simplest and most efficient algorithms available. Interval Scheduling Interval scheduling. Job starts at and finishes at. Greedy Method ˜ Objective: ˜General approach: • Given a set of n inputs. • Find a subset, called feasible solution, of the n inputs subject to some constraints, and satisfying a given objective function. • If the objective function is maximized or minimized, the feasible solution is optimal. • It is a locally optimal method. ˜Algorithm: F Step 1: Choose an input from the input set. Interval Partitioning: Greedy Algorithm Greedy algorithm. Consider lectures in increasing order of start time: assign lecture to any compatible classroom. 17 Sort intervals by starting time so that 1 2 . ←0 // # classrooms used so far for ←1to if lecture is compatible with some classroom then.Finally, not every greedy algorithm is associated with a matroid, but ma-troids do give an easy way to construct greedy algorithms for many problems. Relevant Readings • Kleinberg and Tardos, Algorithm Design, Chapter 4 (Greedy Algo-rithms). This book has an excellent treatment of greedy algorithms. It. Thus after the greedy algorithm added its kth activity to S, the (k + 1)st activity from S* would still belong to U. But the greedy algorithm ended after k activities, so U must have been empty. We have reached a contradiction, so our assumption must have been wrong. Thus the greedy algorithm must be optimal. 07/04/ · Greedy algorithm 1. Short Explanation, Caisar Oentoro 2. What is Greedy Algorithm? In the hard words: A greedy algorithm is an algorithm that follows the problem solving heuristics of making the locally optimal choice at each stage with the hope of finding a global optimum. 26/09/ · The Pilot method, randomized greedy, and iterated greedy are all useful and easy to implement. Each of the three generates a num ber of feasible solutions. The carousel greedy algorithm . Interval Partitioning: Greedy Algorithm Greedy algorithm. Consider lectures in increasing order of start time: assign lecture to any compatible classroom. 17 Sort intervals by starting time so that 1 2 . ←0 // # classrooms used so far for ←1to if lecture is compatible with some classroom then. Greedy Method ˜ Objective: ˜General approach: • Given a set of n inputs. • Find a subset, called feasible solution, of the n inputs subject to some constraints, and satisfying a given objective function. • If the objective function is maximized or minimized, the feasible solution is optimal. • It is a locally optimal method. ˜Algorithm: F Step 1: Choose an input from the input set. The greedy method is a general algorithm design paradigm, built on the following elements: n configurations: different choices, collections, or values to find n objective function: a score assigned to configurations, which we want to either maximize or minimize It works . 5 GREEDY ALGORITHMS 5 Greedy Algorithms The second algorithmic strategy we are going to consider is greedy algorithms. In lay-man’s terms, the greedy method is a simple technique: build up the solution piece by piece, picking whatever piece looks best at the time. This is not meant to be precise, and sometimes, it can take some cleverness to gure out what the greedy algorithm really is. But. Module 3 The Greedy Method. It Is the most straight forward technique. It can be applied to a wide variety of problems. Most of these problems have n inputs and require is to obtain a subset that satisfies these constraints. Any subset that satisfies these constraints is called a feasible solution. We need to find a feasible solution that either maximizes or minimizes a given objective function. AGREEDYALGORITHMFORALIGNINGDNASEQUENCES L U U+1 L’ U’ i j k k–1 k–2 FIG levendeurdegoyaves.comteboxesonantidiagonalk¡1designatepointswhereS(i,j.
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### 2 comments on “Greedy method algorithm pdf”
1. Zulkishakar says:
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It is the valuable information | 2,141 | 9,917 | {"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} | 3.625 | 4 | CC-MAIN-2021-21 | latest | en | 0.890742 |
https://www.onlinemath4all.com/derivative-of-e-to-the-power-of-square-root-x.html | 1,716,197,178,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058254.21/warc/CC-MAIN-20240520080523-20240520110523-00229.warc.gz | 843,735,054 | 14,082 | # Derivative of e to the Power of Square Root x
We know the derivative of ex, which is ex.
(ex)' = ex
We can find the derivative of ex using chain rule.
If y = ex, find ᵈʸ⁄d.
Let t = √x.
Then, we have
y = et
By chain rule,
Substitute y = et and t = √x.
Substitute t = √x.
Therefore,
Kindly mail your feedback to v4formath@gmail.com
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The Mean Value Theorem Worksheet | 184 | 615 | {"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-2024-22 | latest | en | 0.797579 |
https://homework.cpm.org/category/CON_FOUND/textbook/caac/chapter/13/lesson/13.RE3-S/problem/2-153 | 1,576,150,145,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543252.46/warc/CC-MAIN-20191212102302-20191212130302-00181.warc.gz | 391,163,947 | 15,507 | ### Home > CAAC > Chapter 13 > Lesson 13.RE3-S > Problem2-153
2-153.
Find the formula for t(n) for the arithmetic sequence in which t(15) = 10 and t(63) = 106. Homework Help ✎
Visualize the sequences as two points on a graph, (15, 10) and (63, 106).
Finding the common difference is the same as finding the constant rate of change of a line, the slope.
Use the slope or common difference and one of the points, say (15, 10) to substitute into t(n) = mn + b. And solve: 10 = 2(15) + b
t(n) = 2n − 20 | 160 | 504 | {"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-2019-51 | longest | en | 0.88436 |
http://www.thescienceforum.com/physics/1477-heat-flow-grad-t-grad-q.html | 1,674,769,326,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764494826.88/warc/CC-MAIN-20230126210844-20230127000844-00189.warc.gz | 87,757,502 | 11,375 | 1. I was thinking about analogies between the fields of electric force, pressure force in liquids and heat flow and am wondering whether the heat flow is grad T or grad Q.
Ive come to the conclusion that it must be grad T for the following reason:
Consider two objects of the same mass and form, at the same temperature, but with one object having a higher thermal capacity than the other, so both have the same temperature, but one has more thermal energy than the other.
If I put the two objects into contact there will be no change in temperature, so no flow of heat.
If I put the two objects together when they have different temperatures there will be heat flow.
By this reasoning I come to the conclusion that heat flow = grad T.
Im interested in any comments because I fuck up physics alot.
2.
3. Temperature is a measure of the average energy of the parts of the systems (be those parts molecules, atoms, nuclei, or electrons). The statistical definition of temperature is T = 1/(dS/dU) This definition leads to all the properties of temperature we know and like - although that may not be readily apparent.
Heat flows from high to low temperature. This being because the enthropy of a system must increase or be maintained under all processes (2LoT). Suppose I have two temperature reservoirs with T1 > T2 and an amount of energy dQ moves through heat exchange (with the T1 -> T2 direction being the positive one).
The enthropy change for the entire system is dS = dQ/T2 - dQ/T1
Since dS >= 0 the heat must flow from T2 to T1, T2 needs not have higher total thermal energy than T1 (which it could have if it had greater mass or a higher heat capacity, for instance).
Temperature is an intesive quantity: If you place two heat reservoirs of the same temperature in contact, the new reservoir remains the same temperature as the two individual reservoirs. OTOH, two identical heat reservoirs put together will have twice the thermal energy of one or the other - since energy is an extensive quantity.
Heat is energy that gets redistributed through thermal contact (the technical definition is that heat changes the energy distribution - as opposed to work, which moves energy by shifting the entire distribution up or down the energy axis). Note that 'heat' denotes only the energy that's actually transferred. | 494 | 2,330 | {"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} | 3.5625 | 4 | CC-MAIN-2023-06 | latest | en | 0.936844 |
http://slideplayer.com/slide/3269059/ | 1,506,070,466,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818688926.38/warc/CC-MAIN-20170922074554-20170922094554-00662.warc.gz | 306,297,635 | 20,624 | # Fun with Vectors. Definition A vector is a quantity that has both magnitude and direction Examples?
## Presentation on theme: "Fun with Vectors. Definition A vector is a quantity that has both magnitude and direction Examples?"— Presentation transcript:
Fun with Vectors
Definition A vector is a quantity that has both magnitude and direction Examples?
Represented by an arrow A (initial point) B (terminal point) v, v, or AB
If two vectors, u and v, have the same length and direction, we say they are equivalent u v
a b Vector addition: a + b
a b a+ba+b
a b a+ba+b
Scalar Multiplication a
a 2a2a -a -½ a
Subtraction b a
b a -b
Subtraction b a -b a+(-b)
Subtraction b a -b a+(-b)
Subtraction b a -b a+(-b) If a and b share the same initial point, the vector a-b is the vector from the terminal point of b to the terminal point of a
Let’s put these on a coordinate system We can describe a vector by putting its initial point at the origin. We denote this as a= where (a 1,a 2 ) represent the terminal point
Graphically x y a= (a 1,a 2 ) y x z v= a b c (a,b,c)
Given two points A=(x1,y1) and B=(x2,y2), The vector v = AB is given by v = …or in 3-space, v =
Graphically A=(-1,2) B=(2,3) A B v = = v
Recall, a vector has direction and length Definition: The magnitude of a vector v = is given by
Properties of Vectors Suppose a, band c are vectors, c and d are scalars 1.a+b=b+a 2.a+(b+c)=(a+b)+c 3.a+0=a 4.a+(-a)=0 5.c(a+b)=ca+cb 6.(c+d)a=ca+da 7.(cd)a=c(da) 8.1a=a
Standard Basis Vectors Definition: vectors with length 1 are called unit vectors
Example: We can express vectors in terms of this basis a = a = 2i -4j+6k Q. How do we find a unit vector in the same direction as a? A. Scale a by its magnitude
Example a =
12.3 The Dot Product Motivation: Work = Force* Distance Box F D Fx Fy
Box D F Fx Fy To find the work done in moving the box, we want the part of F in the direction of the distance
One interpretation of the dot product Where is the angle between F and D
A more useful definition You can show these two definitions are equal by considering the following triangle and applying the law of cosines! See page 808 for details y x z b a-b a Think, what is |a| 2 ?
Example a=, b= Find a. b and the angle between a and b
The Dot Product If a = and b= then The dot product of a and b is a NUMB3R given by
The Dot Product a and b are orthogonal if and only if the dot product of a and b is 0 Other Remarks: a b
Properties of the dot product Suppose a, b, and c are vectors and c is a scalar 1.a. a=|a| 2 2.a. b=b. a 3.a. (b+c) = (a. b)+(a. c) 4.(ca). b=c(a. b)=a. (cb) 5.0. a=0
Yet another use of the dot product: Projections a. b=|a| |b| cos( ) Think of our work example: this is ‘how much’ of b is in the direction of a b a |b| cos( )
We call this quantity the scalar projection of b on a Think of it this way: The scalar projection is the length of the shadow of b cast upon a by a light directly above a
Q. How do we get the vector in the direction of a with length comp a b? A.We need to multiply the unit vector in the direction of a by comp a b. We call this the vector projection of b onto a
Examples/Practice!
Key Points Vector algebra: addition, subtraction, scalar multiplication Geometric interpretation Unit vectors The dot product and the angle between vectors Projections (algebraic and geometric)
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Similar presentations | 992 | 3,497 | {"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} | 4.4375 | 4 | CC-MAIN-2017-39 | longest | en | 0.867085 |
https://www.adidasshoesoutletwholesale.com/definition-of-volume-science/ | 1,624,194,297,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487662882.61/warc/CC-MAIN-20210620114611-20210620144611-00326.warc.gz | 560,010,798 | 11,030 | A quantity of mail. In different phrases quantity is a measure of the scale of an object similar to top and width are methods to explain measurement.
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### An instance of quantity is how a lot water a glass can maintain.
Definition of quantity science. That is written as 1 cm³. Children will be capable of observe the variations and similarities among the many containers after they pour the water and examine the outcomes. Quantity refers back to the quantity of area the item takes up.
The diploma of sound depth or audibility. By conference the phrase quantity implies a three-dimensional context the place. A mass of boulders quantity applies to an combination with out form or define and able to flowing or fluctuating.
The quantity of area measured in cubic items that an object or substance occupies. In laptop knowledge storage a quantity or logical drive is a single accessible storage space with a single file system usually although not essentially resident on a single partition of a tough disk. The amount of a mountain is way bigger than the amount of a rock for example.
Frequent items used to specific quantity embody liters cubic meters gallons milliliters teaspoons and ounces although many different items exist. Fullness or amount of tone. If the size of the room is 50 meters the width is 30 meters and the peak is 25 meters then the amount of the room is.
A cubic cm block takes up 1 cubic cm. What’s the definition of Quantity. We measure quantity in cubic items.
The amount of gross sales. The quantity of area an object takes up. An amazing quantity of water.
The state of understanding. In science in addition to in our day-to-day lives quantity is taken into account the measure of a three-dimensional area whether or not its a substance within one thing or enclosed inside one thing. The amount of a sound is how loud or quiet the sound is.
The amount of an object is a measure of the quantity of area occupied by that object and isn’t to be confused with mass. Quantity is the quantity of area inside a strong determine. A division of systematized information as an object of research the science of theology.
The density of pure water is outlined to be 1 gram per milliliter. Quantity 50 m 30 m 25 m 375 m 3. What device do you utilize to measure the amount of liquids.
You possibly can work out the amount of a form by multiplying top width depth. Density mass quantity. Quantity Definition Quantity is the three-dimensional area occupied by a substance or enclosed by a floor.
Sounds are vibrations that journey via the air. Data as distinguished from ignorance or misunderstanding. The reciprocal of density Vm is its particular quantity.
Bulk implies an combination that’s impressively massive heavy or quite a few. The size is the longest distance between the objects extremities. An instance of quantity is how loud the radio is.
Though a quantity may be completely different from a bodily disk drive it could actually nonetheless be accessed with an working programs logical interface. Density tends to be greater for solids than for liquids that are in flip extra dense than gases. A nail hit onerous with a hammer will make a robust vibration which implies it’s going to make a.
Quantity is the amount of three-dimensional area occupied by a liquid strong or fuel. The most typical equation to search out quantity is mass divided by density. A mass or amount particularly a big amount of one thing.
Quantity is the extent at which one thing is heard or the quantity of area that one thing takes up. One thing reminiscent of a sport or method which may be studied or realized like systematized information have it all the way down to a science. In scientific phrases quantity is three-dimensional area occupied by a fuel liquid or strong.
Quantity size width top. If the item is hole. To show up the amount on a radio.
The darkened bulk of the skyscrapers mass suggests an combination made by piling collectively issues of the identical sort. Bulk mass quantity imply the combination that kinds a physique or unit. Scientists measure quantity in cubic items reminiscent of liters cubic meters gallons and ounces.
Learn on to be taught extra about what quantity is how its measured and different vital key chemistry phrases. These figures are three-d which means the determine has three components together with size width and top. Children will be taught that quantity is the area a substance or object takes up or that may be enclosed in a container like those we now have above.
Quantity is the quantity of area a 3D form takes up.
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Pin On Faculty | 1,187 | 6,330 | {"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-2021-25 | latest | en | 0.902986 |
https://metanumbers.com/255285299999999 | 1,628,221,480,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152112.54/warc/CC-MAIN-20210806020121-20210806050121-00110.warc.gz | 390,842,074 | 11,227 | ## 255285299999999
255,285,299,999,999 (two hundred fifty-five trillion two hundred eighty-five billion two hundred ninety-nine million nine hundred ninety-nine thousand nine hundred ninety-nine) is an odd fifteen-digits composite number following 255285299999998 and preceding 255285300000000. In scientific notation, it is written as 2.55285299999999 × 1014. The sum of its digits is 101. It has a total of 2 prime factors and 4 positive divisors. There are 254,797,182,791,064 positive integers (up to 255285299999999) that are relatively prime to 255285299999999.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 15
• Sum of Digits 101
• Digital Root 2
## Name
Short name 255 trillion 285 billion 299 million 999 thousand 999 two hundred fifty-five trillion two hundred eighty-five billion two hundred ninety-nine million nine hundred ninety-nine thousand nine hundred ninety-nine
## Notation
Scientific notation 2.55285299999999 × 1014 255.285299999999 × 1012
## Prime Factorization of 255285299999999
Prime Factorization 523 × 488117208413
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 255285299999999 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) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 255,285,299,999,999 is 523 × 488117208413. Since it has a total of 2 prime factors, 255,285,299,999,999 is a composite number.
## Divisors of 255285299999999
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 2.55773e+14 Sum of all the positive divisors of n s(n) 4.88117e+11 Sum of the proper positive divisors of n A(n) 6.39434e+13 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 1.59777e+07 Returns the nth root of the product of n divisors H(n) 3.99237 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 255,285,299,999,999 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 255,285,299,999,999) is 255,773,417,208,936, the average is 63,943,354,302,234.
## Other Arithmetic Functions (n = 255285299999999)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 254797182791064 Total number of positive integers not greater than n that are coprime to n λ(n) 127398591395532 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 7941062709433 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 254,797,182,791,064 positive integers (less than 255,285,299,999,999) that are coprime with 255,285,299,999,999. And there are approximately 7,941,062,709,433 prime numbers less than or equal to 255,285,299,999,999.
## Divisibility of 255285299999999
m n mod m 2 3 4 5 6 7 8 9 1 2 3 4 5 3 7 2
255,285,299,999,999 is not divisible by any number less than or equal to 9.
## Classification of 255285299999999
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (255285299999999)
Base System Value
2 Binary 111010000010111000111101100111110111010011111111
3 Ternary 1020110220000222111012002211002
4 Quaternary 322002320331213313103333
5 Quinary 231430043243244444444
6 Senary 2302540240420320515
8 Octal 7202707547672377
10 Decimal 255285299999999
12 Duodecimal 2477000254313b
20 Vigesimal 14ic1cg4jjjj
36 Base36 2ihog4ck5b
## Basic calculations (n = 255285299999999)
### Multiplication
n×i
n×2 510570599999998 765855899999997 1021141199999996 1276426499999995
### Division
ni
n⁄2 1.27643e+14 8.50951e+13 6.38213e+13 5.10571e+13
### Exponentiation
ni
n2 65170584396089489429400000001 16637092188730958965246811730765855899999999 4247205070527822841638533175773115506376538978858800000001 1084249020591212165269474905514350885446810893064156039101276426499999999
### Nth Root
i√n
2√n 1.59777e+07 63436.9 3997.21 761.036
## 255285299999999 as geometric shapes
### Circle
Diameter 5.10571e+14 1.604e+15 2.04739e+29
### Sphere
Volume 6.96893e+43 8.18958e+29 1.604e+15
### Square
Length = n
Perimeter 1.02114e+15 6.51706e+28 3.61028e+14
### Cube
Length = n
Surface area 3.91024e+29 1.66371e+43 4.42167e+14
### Equilateral Triangle
Length = n
Perimeter 7.65856e+14 2.82197e+28 2.21084e+14
### Triangular Pyramid
Length = n
Surface area 1.12879e+29 1.9607e+42 2.0844e+14
## Cryptographic Hash Functions
md5 f9cf96990a3d8a5b1d4c0418d6b42c4a 58015195309f9f89b414c8ec8c7a84e76117ce26 f1bfdb868623ec7df044be09b8757212bcd96798751c06961b836573dd35eaea 62acea7caebe2e19b9d68a4abf8600839613ba90298ecf2c5bfbdca956739d7dc866945a0b5b25f35c002ef3778006704042ec713b66a699d278c5cc1a3b9212 487f23f21c9871695e104954f7d4510370c7ddcf | 1,815 | 5,205 | {"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} | 3.40625 | 3 | CC-MAIN-2021-31 | latest | en | 0.762864 |
https://www.jobilize.com/trigonometry/course/13-3-geometric-sequences-sequences-probability-and-counting-by-opensta?qcr=www.quizover.com&page=3 | 1,610,753,627,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703497681.4/warc/CC-MAIN-20210115224908-20210116014908-00790.warc.gz | 851,489,082 | 24,830 | # 13.3 Geometric sequences (Page 4/6)
Page 4 / 6
## Verbal
What is a geometric sequence?
A sequence in which the ratio between any two consecutive terms is constant.
How is the common ratio of a geometric sequence found?
What is the procedure for determining whether a sequence is geometric?
Divide each term in a sequence by the preceding term. If the resulting quotients are equal, then the sequence is geometric.
What is the difference between an arithmetic sequence and a geometric sequence?
Describe how exponential functions and geometric sequences are similar. How are they different?
Both geometric sequences and exponential functions have a constant ratio. However, their domains are not the same. Exponential functions are defined for all real numbers, and geometric sequences are defined only for positive integers. Another difference is that the base of a geometric sequence (the common ratio) can be negative, but the base of an exponential function must be positive.
## Algebraic
For the following exercises, find the common ratio for the geometric sequence.
$1,3,9,27,81,...$
$-0.125,0.25,-0.5,1,-2,...$
The common ratio is $-2$
$-2,-\frac{1}{2},-\frac{1}{8},-\frac{1}{32},-\frac{1}{128},...$
For the following exercises, determine whether the sequence is geometric. If so, find the common ratio.
$-6,-12,-24,-48,-96,...$
The sequence is geometric. The common ratio is 2.
$5,5.2,5.4,5.6,5.8,...$
$-1,\frac{1}{2},-\frac{1}{4},\frac{1}{8},-\frac{1}{16},...$
The sequence is geometric. The common ratio is $-\frac{1}{2}.$
$6,8,11,15,20,...$
$0.8,4,20,100,500,...$
The sequence is geometric. The common ratio is $5.$
For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio.
$\begin{array}{cc}{a}_{1}=8,& r=0.3\end{array}$
$\begin{array}{cc}{a}_{1}=5,& r=\frac{1}{5}\end{array}$
$5,1,\frac{1}{5},\frac{1}{25},\frac{1}{125}$
For the following exercises, write the first five terms of the geometric sequence, given any two terms.
$\begin{array}{cc}{a}_{7}=64,& {a}_{10}\end{array}=512$
$\begin{array}{cc}{a}_{6}=25,& {a}_{8}\end{array}=6.25$
$800,400,200,100,50$
For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio.
The first term is $2,$ and the common ratio is $3.$ Find the 5 th term.
The first term is 16 and the common ratio is $-\frac{1}{3}.$ Find the 4 th term.
${a}_{4}=-\frac{16}{27}$
For the following exercises, find the specified term for the geometric sequence, given the first four terms.
${a}_{n}=\left\{-1,2,-4,8,...\right\}.$ Find ${a}_{12}.$
${a}_{n}=\left\{-2,\frac{2}{3},-\frac{2}{9},\frac{2}{27},...\right\}.$ Find ${a}_{7}.$
${a}_{7}=-\frac{2}{729}$
For the following exercises, write the first five terms of the geometric sequence.
$\begin{array}{cc}{a}_{1}=-486,& {a}_{n}=-\frac{1}{3}\end{array}{a}_{n-1}$
$\begin{array}{cc}{a}_{1}=7,& {a}_{n}=0.2{a}_{n-1}\end{array}$
$7,1.4,0.28,0.056,0.0112$
For the following exercises, write a recursive formula for each geometric sequence.
${a}_{n}=\left\{-1,5,-25,125,...\right\}$
${a}_{n}=\left\{-32,-16,-8,-4,...\right\}$
$\begin{array}{cc}a{}_{1}=-32,& {a}_{n}=\frac{1}{2}{a}_{n-1}\end{array}$
${a}_{n}=\left\{14,56,224,896,...\right\}$
${a}_{n}=\left\{10,-3,0.9,-0.27,...\right\}$
$\begin{array}{cc}{a}_{1}=10,& {a}_{n}=-0.3{a}_{n-1}\end{array}$
${a}_{n}=\left\{0.61,1.83,5.49,16.47,...\right\}$
${a}_{n}=\left\{\frac{3}{5},\frac{1}{10},\frac{1}{60},\frac{1}{360},...\right\}$
$\begin{array}{cc}{a}_{1}=\frac{3}{5},& {a}_{n}=\frac{1}{6}{a}_{n-1}\end{array}$
${a}_{n}=\left\{-2,\frac{4}{3},-\frac{8}{9},\frac{16}{27},...\right\}$
${a}_{n}=\left\{\frac{1}{512},-\frac{1}{128},\frac{1}{32},-\frac{1}{8},...\right\}$
${a}_{1}=\frac{1}{512},{a}_{n}=-4{a}_{n-1}$
For the following exercises, write the first five terms of the geometric sequence.
${a}_{n}=-4\cdot {5}^{n-1}$
${a}_{n}=12\cdot {\left(-\frac{1}{2}\right)}^{n-1}$
$12,-6,3,-\frac{3}{2},\frac{3}{4}$
For the following exercises, write an explicit formula for each geometric sequence.
${a}_{n}=\left\{-2,-4,-8,-16,...\right\}$
${a}_{n}=\left\{1,3,9,27,...\right\}$
${a}_{n}={3}^{n-1}$
${a}_{n}=\left\{-4,-12,-36,-108,...\right\}$
${a}_{n}=\left\{0.8,-4,20,-100,...\right\}$
${a}_{n}=0.8\cdot {\left(-5\right)}^{n-1}$
${a}_{n}=\left\{-1.25,-5,-20,-80,...\right\}$
${a}_{n}=\left\{-1,-\frac{4}{5},-\frac{16}{25},-\frac{64}{125},...\right\}$
${a}_{n}=-{\left(\frac{4}{5}\right)}^{n-1}$
${a}_{n}=\left\{2,\frac{1}{3},\frac{1}{18},\frac{1}{108},...\right\}$
${a}_{n}=\left\{3,-1,\frac{1}{3},-\frac{1}{9},...\right\}$
${a}_{n}=3\cdot {\left(-\frac{1}{3}\right)}^{n-1}$
For the following exercises, find the specified term for the geometric sequence given.
Let ${a}_{1}=4,$ ${a}_{n}=-3{a}_{n-1}.$ Find ${a}_{8}.$
Let ${a}_{n}=-{\left(-\frac{1}{3}\right)}^{n-1}.$ Find ${a}_{12}.$
${a}_{12}=\frac{1}{177,147}$
For the following exercises, find the number of terms in the given finite geometric sequence.
${a}_{n}=\left\{-1,3,-9,...,2187\right\}$
${a}_{n}=\left\{2,1,\frac{1}{2},...,\frac{1}{1024}\right\}$
There are $12$ terms in the sequence.
## Graphical
For the following exercises, determine whether the graph shown represents a geometric sequence.
The graph does not represent a geometric sequence.
For the following exercises, use the information provided to graph the first five terms of the geometric sequence.
$\begin{array}{cc}{a}_{1}=1,& r=\frac{1}{2}\end{array}$
$\begin{array}{cc}{a}_{1}=3,& {a}_{n}=2{a}_{n-1}\end{array}$
${a}_{n}=27\cdot {0.3}^{n-1}$
## Extensions
Use recursive formulas to give two examples of geometric sequences whose 3 rd terms are $\text{\hspace{0.17em}}200.$
Answers will vary. Examples: ${\begin{array}{cc}{a}_{1}=800,& {a}_{n}=0.5a\end{array}}_{n-1}$ and ${\begin{array}{cc}{a}_{1}=12.5,& {a}_{n}=4a\end{array}}_{n-1}$
Use explicit formulas to give two examples of geometric sequences whose 7 th terms are $1024.$
Find the 5 th term of the geometric sequence $\left\{b,4b,16b,...\right\}.$
${a}_{5}=256b$
Find the 7 th term of the geometric sequence $\left\{64a\left(-b\right),32a\left(-3b\right),16a\left(-9b\right),...\right\}.$
At which term does the sequence exceed $100?$
The sequence exceeds $100$ at the 14 th term, ${a}_{14}\approx 107.$
At which term does the sequence begin to have integer values?
For which term does the geometric sequence ${a}_{{}_{n}}=-36{\left(\frac{2}{3}\right)}^{n-1}$ first have a non-integer value?
${a}_{4}=-\frac{32}{3}\text{\hspace{0.17em}}$ is the first non-integer value
Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10 th term.
Use the explicit formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8 th term.
Answers will vary. Example: Explicit formula with a decimal common ratio: ${a}_{n}=400\cdot {0.5}^{n-1};$ First 4 terms: $\begin{array}{cc}400,200,100,50;& {a}_{8}=3.125\end{array}$
Is it possible for a sequence to be both arithmetic and geometric? If so, give an example.
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
2x²-6x+1=0
Ife
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
y2=4ax= y=4ax/2. y=2ax
akash
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai | 3,292 | 9,717 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 82, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 4.75 | 5 | CC-MAIN-2021-04 | latest | en | 0.889724 |
https://amaanswers.com/what-percent-is-20-out-of-7000 | 1,660,220,093,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571284.54/warc/CC-MAIN-20220811103305-20220811133305-00639.warc.gz | 131,016,614 | 4,848 | # What percent is 20 out of 7000?
Lenore Alonso asked, updated on May 19th, 2022; Topic: percent
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Steps to solve "what percent is 20 of 7000?" If you are using a calculator, simply enter 20รท7000ร100 which will give you 0.29 as the answer.
Even, how do you find 10% value?
There are two steps to calculating a 10 percent discount:
• Step 1 is to convert your percentage to a decimal, the formula for which is 10 / 100 = 0.1. So 10 percent as a decimal is 0.1.
• Step 2 is to multiply your original price by your decimal.
• Hereof, what is the 10% in money? While 10 percent of any amount is the amount multiplied by 0.1, an easier way to calculate 10 percent is to divide the amount by 10. So, 10 percent of \$18.40, divided by 10, equates to \$1.84.
Come what may, what percent is 10 out of 1000000?
Nearby Results
10% ofResult
1000000100000
1000000.01100000.001
1000000.02100000.002
1000000.03100000.003
What percent is 1600 out of 3000?
Percentage Calculator: 1600 is what percent of 3000? = 53.33. | 352 | 1,047 | {"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-2022-33 | latest | en | 0.902972 |
https://highonessays.com/2022/11/12/use-factoring-to-solve-x2-x-20/ | 1,669,621,308,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710488.2/warc/CC-MAIN-20221128070816-20221128100816-00782.warc.gz | 349,739,390 | 24,445 | Use FACTORING to solve : x2 + x – 20 = 0
Use the QUADRATIC FORMULA to solve: 46 on p. 636
In this discussion, you will solve quadratic equations by two main methods: factoring and using the quadratic formula. Read the following instructions in order and view the MAT222 Week 4 Discussion Example to complete this discussion. Please complete the following problems according to your assigned number. (Instructors will assign each student their number.)
• For the factoring problem, be sure you show all steps to the factoring and solving. Show a check of your solutions back into the original equation.
• For the quadratic formula problem, be sure that you use readable notation while you are working the computational steps. Refer to the Inserting Math Symbols handout for guidance with formatting.
• Present your final solutions as decimal approximations carried out to the third decimal place. Due to the nature of these solutions, no check is required.
• Incorporate the following four math vocabulary words into your discussion. Use bold font to emphasize the words in your writing. Do not write definitions for the words; use them appropriately in sentences describing your math work. | 241 | 1,192 | {"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-2022-49 | latest | en | 0.895611 |
https://www.engineersadda.co/tag/nic-quiz | 1,604,159,251,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107919459.92/warc/CC-MAIN-20201031151830-20201031181830-00305.warc.gz | 679,350,889 | 22,051 | # Quiz Electronics Engineering 28 July 2020
Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 28/07/2020
Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes
Q1. The most appropriate matching for the following pairs:
COLUMN-I COLUMN-II
Z: Auto decrement addressing 3. Constants
Options:
(a) X-3, Y-2, Z-1
(b) X-2, Y-3, Z-1
(c) X-3, Y-1, Z-2
(d) X-2, Y-1, Z-3
Q2. Consider the following statements regarding a plane wave propagating through free space: The direction of field is
1. ‘E’ is perpendicular to the direction of propagation
2. ‘H’ is perpendicular to the direction of propagation
3. ‘E’ is perpendicular to the direction of field ‘H’
Which of these statements are correct?
(a) 1 and 2
(b) 2 and 3
(c) 1 and 3
(d) 1, 2 and 3
Q3. A wire having resistance 100 Ω is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is
(a) 100 Ω
(b) 50 Ω
(c) Zero
(d) ∞
Q4. The Nyquist sampling frequency (in Hz) of a signal given by
6 × 104sin𝑐2(400𝑡)∗106sin𝑐3(100𝑡) is
(a) 200
(b) 300
(c) 1500
(d) 1000
Q5. The total circuit resistance of the given circuit is
(a) 440 Ω
(b) 110Ω
(c) Zero
(d) 1100 Ω
Q6. How many 128 x 8 bit RAMs are required to design 32K x 32 bit RAM
(a) 512
(b) 1024
(c) 128
(d) 32
Q7. Match List-I(Diode) with List-II(Application) and select the correct answer using codes given below the lists:
List-I
A. Varactor Diode
B. Tunnel Diode
C. Photo Diode
D. Zener Diode
List-II
1. to charge auxiliary storage batteries
2. Reference voltage
3. High frequency tuning circuit
4. High frequency switching circuit
Codes:
A B C D
(a) 2 1 4 3
(b) 3 1 4 2
(c) 2 4 1 3
(d) 3 4 1 2
Q8. What is the power of periodic non-sinusoidal voltage and currents?
(a) The sum of the root mean square power of the sinusoidal components including the fundamental
(b) The sum of the average powers of the harmonics excluding the fundamental
(c) The sum of the average powers of the sinusoidal components including the fundamental
(d) The average power of the fundamental components alone.
Q9. Which of the following is a digital device
(a) Regulator of a fan
(b) Microphone
(c) Resistance of a material
(d) Light switch
Q10. Electric field inside a hallow metallic charged sphere is
(a) Zero
(b) Decreasing towards centre
(c) Increasing toward centre
(d) None of these
SOLUTIONS
S1. Ans.(b)
Sol.
In Indirect addressing mode, the instruction points indirectly to the address of memory location where the data is stored or to be stored.
In Immediate addressing mode, the data is directly used in instruction itself.
In Autodecrement addressing mode the value in the base register is decremented by the size of the data item which is to be accessed. This step is done before determining the effective address within a loop.
S2. Ans.(d)
S10. Ans.(b)
Sol. Effective resistance between two points on any diameter of the circle is 50Ω.
S4. Ans.(b)
Sol. The given signal is 6 × 104sin𝑐2(400𝑡)∗106sin𝑐3(100𝑡)
Let two functions be
f1(t) = 104sin𝑐2(400𝑡)
f2(t) = 106sin𝑐3(100𝑡)
We know that Convolution in time domain is multiplication in frequency domain
Bandwidth of F1(ω) = 2 x 400 rad/sec = 800 rad/sec or 400 Hz [for sin𝑐2]
Bandwidth of F2(ω) = 3 x 100 rad/sec = 300 rad/sec or 150 Hz [for sin𝑐3]
The F1(ω).F2(ω) will have bandwidth of 150Hz only because in case of product of two spectrums we consider the minimum frequency.
So, Sampling frequency = 2 x 150 Hz = 300 Hz
S5. Ans.(a)
Sol. We know that,
P=V^2/R
⇒ P = V^2/P
⇒ R_1=(200)^2/1000=4Ω
⇒ R_2=(200)^2/100=400 Ω
So, the total resistance = R1 + R2 = 40 + 400 = 440 Ω.
S6. Ans.(b)
Sol.
Number of RAM’s = (32K x 32 bits)/(128 x 8 bits) = 210 = 1024
S7. Ans.(d)
Sol.
Zener Diode can be used as a Voltage regulator or as a reference voltage.
Varactor diode is widely used in high frequency tuning circuits like VCO and frequency modulators.
A Tunnel diode is a heavily doped p-n junction diode. It is used as a high frequency fast switching device.
Photo diode is used as an energy converter. It absorbs photons and convert them into electric current. It is used to charge auxiliary storage batteries.
S8. Ans.(a)
S9. Ans. (b)
Sol. An electric operating equipment have switch with two condition either ON or OFF, which is similar to that of digital switch conditions of 1 i.e., ON and 0 i.e., OFF. Rest have Fuzzy logic.
S10. Ans.(a)
Sol. Inside a hallow metallic charged sphere, electric field is always zero.
# Quiz Electronics Engineering 10 July 2020
Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 10/07/2020
Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes
Q1. Default port for HTTP:
(a) 23
(b) 25
(c) 21
(d) 80
Q2. The binary equivalent of the decimal number 42.75 is
(a) 101010.110
(b) 100110.101
(c) 101010.101
(d) 100110.110
Q3. In DMA transfer scheme, the transfer scheme other than burst mode is
(a) cycle technique
(b) stealing technique
(c) cycle bypass technique
(d) cycle stealing technique
Q4. A symbol table of length 152 is possessing 25 entries at any instant. What is occupation density?
(a) 8.06
(b) 127
(c) 0.164
(d) 6.08
(a) A portion of memory is wasted because the core occupied by the assembler is unavailable to the object program.
(b) It is necessary to retranslate the user’s program deck every time it is run.
(c) It is difficult to handle multiple segments especially if the source programs are in different language and to produce orderly modular programs.
(d) All of the above
Q6. When we run a program in HTML coding _____ is used as backend and ____ works as front-end.
(c) MS-Word – Internet Explorer
(d) Both (1) and (2)
Q7. In JSP, the classes that allow primitive types to be accessed as objects are known as________
(a) Primitive classes
(b) Object classes
(c) Boxing classes
(d) Wrapper classes.
Q8. What are the ways of dealing with deadlock?
(d) All of the above
Q9.Which of the following is not a Data Definition Language Statement?
(a) DELETE
(b) CREATE
(c) DROP
(d) ALTER
Q10. What is the use of super keyword in Java?
(a) to refer to current class object
(b) to refer to immediate parent class of a class
(c) to refer to immediate child class of a class
(d) to refer to static member of parent class
SOLUTIONS
S1. Ans.(d)
Sol. HTTP means Hyper Text Transfer Protocol. This protocol is used to transfer data over the World Wide Web. HTTP use port number 80.
S2. Ans.(a)
Sol. Decimal number = 42.75
For left side of decimal, divide 42 by 2
Division Quotient Remainder
42/2 21 0
21/2 10 1
10/2 5 0
5/2 2 1
2/2 1 0
1/2 0 1
So, (42)10 = (101010)2
Similarly, For right side of decimal, multiply 0.75 by 2
0.75 x 2 = 1.5
0.5 x 2 = 1.0
0.0 x 2 = 0.0
Hence, the binary equivalent is 101010.110
S3. Ans.(d)
Sol. Cycle Stealing and burst transfer mode are other methods for DMA. Cycle stealing mode is a method of accessing computer memory (RAM) or bus without interfering with the CPU. In this mode, the DMA steals cycle from the processor in order to transfer the byte.
S4. Ans.(c)
Sol. Given:
Number of entries at any instant = 25
Length of symbol table = 152
We know that, Occupation density = (Number of entries )/(length of symbol table)
= 25/152 = 0. 164
S5. Ans.(b)
Sol. In Compile and go systems, the compilation or link steps are not separated from program execution. In the intermediate forms the program are not saved to the file system but generally kept in primary memory. It is a programming language processor.
S6. Ans.(a)
S7. Ans.(d)
Sol. Wrapper classes provide a way to use primitive data types as objects. Wrapper classes in java provides a mechanism to convert primitive into object and object into primitive. It can be used to store the same value as of a primitive type variable but the instances/objects of wrapper classes themselves are Non-Primitive.
S8. Ans-(d)
S9. Ans.(a)
Sol. Data Definition Language in short DDL deals with database schemas and descriptions, of how data should reside in the database. DDL commands are:- CREATE, ALTER, DROP, TRUNCATE, RENAME.
S10. Ans.(b)
Sol. The super keyword in java is a reference variable which refers to the parent class objects. Super keywords in Java is used to refer immediate parent class instance variables or methods or constructor.
# Quiz Electronics Engineering 5 June 2020
Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Date: 05/06/2020
Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes
Q1. The response of an initially relaxed linear constant parameter network to a unit impulse applied a t = 0 is 4e^(-2t)u(t). The response of this network to a unit step function will be
(a) 2[1 – e^(-2t)]u(t)
(b) 4[ e^(-t)-e^(-2t)]u(t)
(c) sin2t
(d) [1 – 4e^(-4t)]u(t)
L1 Difficulty 3
QTags Signals and systems
QCreator Vikram Kumar
Q2. A message signal band limited to 5 kHz is sampled at the minimum rate as dictated by the sampling theorem. The number of quantization levels is 64. If samples are encoded in binary form, then transmission rate is
(a) 50 kbps
(b) 60 kbps
(c) 32 kbps
(d) 10 kbps
L1 Difficulty 4
QTags Communication Engineering
QCreator Vikram Kumar
Q3. When a computer is first turned on or restarted, a special type of absolute loader is executed called
L1 Difficulty 2
QTags Operating System
QCreator Vikram Kumar
Q4. The Transaction Control Language
(a) refers to data using physical address.
(b) is used to define the physical characteristics of each record.
(c) used to control transactional processing
(d) None of the above
L1 Difficulty 3
QTags DBMS
QCreator Vikram Kumar
Q5. Pre-emphasis in FM system involves
(a) compression of the modulating signal
(b) expansion of the modulating signal
(c) amplification of lower frequency components of the modulating signal
(d) amplification of higher frequency components of the modulating signal
L1 Difficulty 3
QTags Communication Engineering
QCreator Vikram Kumar
Q6. In an Entity-Relationship (E-R) diagram which diagram links attributes to entity and entity with other relationship?
(a) Double ellipse
(b) Dashed ellipse
(c) Double lines
(d) lines
L1 Difficulty 3
QTags DBMS
QCreator Vikram Kumar
Q7. Pseudo-instructions are
(a) assembler derivative
(b) instruction in any program that have no corresponding machine code instruction
(c) instruction in any program whose presence or absence will not change the output for any input
(d) None of the above
L1 Difficulty 4
QTags Operating System
QCreator Vikram Kumar
Q8. The number of columns in table is referred to as
(a) Tuple
(b) Attribute
(c) Degree
(d) Entity
L1 Difficulty 2
QTags DBMS
QCreator Vikram Kumar
Q9. A B-tree of order 4 is built from scratch by 10 successive insertions. What is the maximum number of node splitting operations that may take place?
(a) 5
(b) 4
(c) 3
(d) 6
L1 Difficulty 3
QTags Data Structure
QCreator Vikram Kumar
Q10. A blank EPROM has:
(a) All bits set to logical 0
(b) All bits set to logical 1
(c) Half the total number of bits set to 0 and remaining half to logical 1
(d) Either A. or B.
L1 Difficulty 2
QTags Computer Organization & Microprocessor
QCreator Vikram Kumar
SOLUTIONS
S1. Ans.(a)
Sol. Given:
h(t) = 4e^(-2t)u(t)
input is a unit step function
We know that,
Y(s) = H(s).X(s)
Converting both h(t) and x(t) in Laplace transform, we get
⇒ H(s) = 4/(s + 2)
⇒ X(s) = 1/s
Now, Y(s) = 4/(s(s + 2))
By partial fraction, we get
⇒ Y(s) = 2/s – 2/(s + 2)
Converting it again to time domain form, we get
y(t) = 2[1 – e^(-2t)]u(t)
S2. Ans.(b)
Sol. Given:
Message signal bandwidth(f_m) = 5 kHz
Quantization level(L) = 64
Let number of bits be ‘n’.
Here, L = 2^n
⇒ 64 = 2^n
So, n = 6
Now, f_s = 2f_m
⇒ f_s = 10 kHz
Transmission rates is
N = nf_s
So, N = 60 kbps
S3. Ans.(a)
Sol. Bootstrap loader is a small executable code which is permanently stored in a ROM chip. It is also called Initial Program Loader. It is automatically executed by the processor, when a computer is first turned on or restarted.
S4. Ans.(c)
Sol. TCL(Transaction Control Language) used to control transactional processing in database. Examples: COMMIT, ROLLBACK, SAVEPONT, SET TRANSACTION.
S5. Ans.(d)
Sol. Higher frequency components of modulating signal have smaller magnitude so there is a chance of losing of those signal due to masking of those signals by noise. So, It is necessary to provide Pre-emphasis of high frequency component so that they have enough power to reach at the destination source.
S6. Ans.(d)
Sol.
Lines links attributes to entity and entity sets with other relationship.
Double Lines involves in total participation of an entity in a relationship set.
S7. Ans.(a)
Sol.
Pseudo-instructions are assembly language instruction which is translated by assembler into one or more machine language instruction. Pseudo-instruction often start with a dot to distinguish them from machine instructions. Pseudo-ops can make the assembly of the program dependent on parameters input by a programmer.
Assembler directives are instructions that direct the assembler to set the program or register address during assembly or something else.
S8. Ans.(c)
Sol. Number of columns in the table are called Degree.
S9. Ans.(a)
5
S10. Ans.(b)
All bits set to logical 1
# Quiz Electronics Engineering
Exam: NIC
Topic: Miscellaneous
Each Question carries 1 Mark
Negative Marking: 1/4
Time: 10 Minutes
Q1. A file is preferable to an array of structures because
• memory space will not be wasted
• file lives even after the program that is created terminates
• there are many system tools to manipulate files.
• All of the above
L1 Difficulty 3
QTags Data Structure
QCreator Vikram Kumar
Q2. Which of the following traversal techniques lists the nodes of a binary search tree in ascending order?
• In-order
• Post-order
• Pre-order
• None of these
L1 Difficulty 2
QTags Computer
QCreator Vikram Kumar
Q3. Which of the following is generally a benefit of normalization?
• Insertion anomalies are avoided
• Selection anomalies are avoided
• Number of tables are reduced
• Performance is improved
L1 Difficulty 3
QTags DBMS
QCreator Vikram Kumar
Q4. Time sharing provides
• Disk management
• Concurrent execution
• File system management
• All of the above
L1 Difficulty 3
QTags Operating System
QCreator Vikram Kumar
Q5. Match COLUMN-I with COLUMN-II and select the correct answer using the options given below:
COLUMNN-1 COLUMNN-II
1. 2421 1.Reflecting code
2. 8421 2. Non-weighted code
3. Gray Code 3. Sequential code
Options:
A B C
• 3 1 2
• 1 3 2
• 1 2 3
• 3 2 1
L1 Difficulty 3
QTags Digital Electronics
QCreator Vikram Kumar
Q6. An instruction cycle refers to
• Fetching an instruction
• Clock speed
• Executing an instruction
• Fetching, decoding and executing an instruction
L1 Difficulty 2
QTags Computer Organization & Microprocessor
QCreator Vikram Kumar
Q7. Pick the machine independent phase(s) of the compiler
• intermediate code generation
• Syntax analysis
• Lexical analysis
• All of the above
L1 Difficulty 4
QTags Compiler Design
QCreator Vikram Kumar
Q8. Communication via circuit switching involves three phases which are
• Circuit establishment, data compression, circuit disconnect
• Circuit establishment, data transfer, circuit disconnect
• Data transfer, data compression, circuit disconnect
• Circuit establishment, data compression.
L1 Difficulty 3
QTags Networking
QCreator Vikram Kumar
Q9. The development is supposed to proceed linearly through the phase in
• Spiral model
• Waterfall model
• Prototyping model
• None of the above
L1 Difficulty 3
QTags Software Engineering
QCreator Vikram Kumar
Q10. A PAM signal can be detected by using
• an integrator
• a band pass filter
• a high pass filter
L1 Difficulty 4
QTags Communication Engineering
QCreator Vikram Kumar
SOLUTIONS
S1. Ans(d)
Sol: Array is used during programming. It runs till the program runs and stops when program terminates. But file lives even after program is terminated. We can bring changes in file with the help of different system tools.
S2. Ans(a)
Sol: In order traversal of binary search tree list the nodes in ascending order.
S3. Ans.(d)
Sol: Normalization is an advantageous prospect in DBMS because
• A smaller database can be maintained as normalization eliminates the duplicate data.
• Better performance is ensured. As databases become lesser in size, the passes through the data becomes faster and shorter thereby improving response time and speed.
• Narrower tables are possible
• Fewer indexes per table ensures faster maintenance tasks
S4. Ans.(b)
Sol:
Time-sharing is a technique in which many people located at various terminals uses a particular computer system at the same time. Processor’s time is shared among multiple users at the same instant in time-sharing.
• quick response.
• prevents duplication of software.
• reduces CPU idle time.
• less reliable.
• less secure.
• affects integrity of user program and data
• issues related to data communication.
S5. Ans.(b)
Sol:
• A code is reflective if code for one element is complement for the code for another element. Ex- 2421,5211, XS3. 8421 is not reflective.
• A code is sequential if two subsequent codes, seen as numbers in binary representation, differ by one. Ex- 8421, Excess-3 but 5211 and 2421 are not.
• Non weighted codes are not positionally weighted. Ex- Gray code.
S6. Ans.(d)
Sol: An Instruction Cycle is the basic process of a computer. It is the process by which a computer retrieves a program instruction from its memory, determines what actions the instruction is asking to take and carries out those actions. It is also known as fetch decode-execute cycle.
S7. Ans.(d)
Sol: Machine independent phases of compiler are
• Lexical analysis
• Syntax analysis
• Intermediate code generation
S8. Ans.(b)
Sol: The three phases of Circuit Switching involves:-
• connection establishment
• data transfer and
• circuit disconnect
S9. (b)
S10. Ans.(a)
Sol: An integrator helps in detecting a PAM signal. It can also be detected by a Low Pass Filter. | 4,972 | 18,295 | {"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} | 2.71875 | 3 | CC-MAIN-2020-45 | longest | en | 0.740992 |
http://www.mathworks.com/help/finance/rsindex.html?nocookie=true | 1,441,448,434,000,000,000 | text/html | crawl-data/CC-MAIN-2015-35/segments/1440646242843.97/warc/CC-MAIN-20150827033042-00295-ip-10-171-96-226.ec2.internal.warc.gz | 564,745,430 | 9,534 | rsindex
Relative Strength Index (RSI)
Syntax
```rsi = rsindex(closep, nperiods)
rsits = rsindex(tsobj, nperiods)
rsits = rsindex(tsobj, nperiods, 'ParameterName', ParameterValue, ...)
```
Arguments
`closep` Vector of closing prices. `nperiods` (Optional) Number of periods. Default = `14`. `tsobj` Financial time series object.
Description
`rsi = rsindex(closep, nperiods)` calculates the Relative Strength Index (RSI) from the closing price vector `closep`.
`rsits = rsindex(tsobj, nperiods)` calculates the RSI from the closing price series in the financial time series object `tsobj`. The object `tsobj` must contain at least the series `Close`, representing the closing prices. The output `rsits` is a financial time series object whose dates are the same as `tsobj` and whose data series name is `RSI`.
```rsits = rsindex(tsobj, nperiods, 'ParameterName', ParameterValue, ...)``` accepts a parameter name/parameter value pair as input. This pair specifies the name for the required data series if it is different from the expected default name. The valid parameter name is
`CloseName`: closing prices series name
The parameter value is the string that represents the valid parameter name.
1. The relative strength factor is calculated by dividing the average of the gains by the average of the losses within a specified time period:
```RS = (average gains)/(average losses)```.
2. The first value of `RSI`, `RISI(1)`, is set as `NaN` to preserve the dimensions of `CLOSEP`.
Examples
collapse all
Calculate the Relative Strength Index (RSI)
This example shows how to calculate the RSI for Disney stock and plot the results.
```load disney.mat dis_RSI = rsindex(dis); plot(dis_RSI) title('Relative Strength Index for Disney') ```
References
Murphy, John J., Technical Analysis of the Futures Market, New York Institute of Finance, 1986, pp. 295–302.
Get trial now | 461 | 1,888 | {"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-2015-35 | latest | en | 0.663133 |
http://www.cinterviews.com/2009/11/difference-between-linear-search-and.html | 1,511,535,630,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934808254.76/warc/CC-MAIN-20171124142303-20171124162303-00574.warc.gz | 359,743,946 | 16,901 | ### difference between linear search and binary search
Answer:The main difference b/n Linear search and Binary search are given below:
Linear search: Linear search starts to search first records of the list and going on until it finds the matched record or to the end of list.In Linear Linked list their is no need to sort the file in a particular order.
Binary search: Their is three steps you have to follow when you using Binary search.
1.It starts with to select the middle item of the list.When the key item is less than the selected item than Binary search take the item between the current item and the start element of the list.Otherwise, if key item is greater than the selected item than Binary search take the item between the current item and the last element of the list.Benefit of its is that after taking one comparison it delete the half list.
2.If key item is less than the current item than we only take half list who is less than the key.Otherwise,If key item is greater than the current item than we only take half list who is greater than the key.And start in again.
3.We repeat step2 till we not reached our desired key. | 235 | 1,145 | {"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} | 3.09375 | 3 | CC-MAIN-2017-47 | longest | en | 0.912628 |
http://mathoverflow.net/questions/1142/is-very-ampleness-of-a-divisor-on-a-curve-determined-entirely-by-degree-and-genus | 1,467,242,988,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783397864.87/warc/CC-MAIN-20160624154957-00034-ip-10-164-35-72.ec2.internal.warc.gz | 201,156,382 | 17,429 | # Is very ampleness of a divisor on a curve determined entirely by degree and genus?
Edit: Apparently the answer is "no", so what is an example of two curves of genus g, and a divisor of degree d on each, such that one is very ample and the other is not?
Question as originally stated:
Suppose X is a complete nonsingular curve (smooth proper integral scheme of dimension 1 over C) and D ∈ DivX.
I have heard that very ampleness of D is determined entirely by its degree and the genus of X. How can this be done explicitly?
Thanks!
-
Let $C_1$ be a hyperelliptic curve of genus $g \geq 3$ (example: $y^7 = x^2 + 1$ for $g = 3$), and $C_2$ be a non-hyperelliptic curve of the same genus $g$ (for example, the Klein quartic with $g = 3$ again: I'll use it in the form $y^7 = x^2(x-1)$).
Then let $K_1$, $K_2$ be the canonical divisors of $C_1$, $C_2$ respectively (In the example above: $K_1 = 4[\infty_1]$ and $K_2 = 4[\infty_2]$, where $\infty_1, \infty_2$ are the "points at infinity" on the curves above: note that the projective curves corresponding to the equations above both have singularities at infinity, but in both cases these singularities can be resolved to give a single point at infinity on each curve.) Then $K_1$ is not very ample but $K_2$ is: this is a standard application of Riemann-Roch (see also Hartshorne, Chapter IV, Proposition 5.2.) The reason that $K_1$ is not very ample is that the map $C_1 \rightarrow P^2$ associated to $K_1$ is a double cover of a rational curve (a conic, in fact) in $P^2$, rather than an embedding.
-
Not entirely by its degree; see Hartshorne Chapter 4, section 3 (Prop. 3.1 and corollary 3.2) and section 5.
-
Hart. Prop 5.2 says the canonical divisor K on a curve C of genus g > 1 is very ample iff C is not hyperelliptic. – David Zureick-Brown Oct 19 '09 at 4:28
I'm late to the game, but I would like to point out that the answer is systematically no. One class of examples. Suppose g>2 for simplicity. In that case any general line bundle of degree 2g is very ample and special ones are not. This can be seen by using the criteria that a line bundle is very ample iff for any effective divisor $D$ of degree 2, $h^0(L(-D) = h^0(L)-2$. One checks using R-R that this holds iff $L$ is not of the form $L= K_C(D)$ where $D$ is an effective divisor of degree 2. Line bundles of the form $L= K_C(D)$ are a 2 dimensional subset of the g dimensional (Picard) variety. This can be expanded upon.
-
The other way to construct examples is just by considering curves in a given projective space. For example, a plane curve $C$ of degree $d$ has genus $g = \frac{(d-1)(d-2)}2$. One can, of course, rephrase this as saying that $\mathcal{O}_C(1)$ is a very ample divisor of degree $d$ on $C$. On the other hand, as $d < g$, a general divisor of degree $d$ on $C$ does not even have a single global section.
-
I am maybe misunderstanding, but I think that ypur example is fulfilled by a couple of curves of gwnus bigger than 3, one hyperelliptic and the other not. If you consider the canonical divisor, it is very ample on the non-hyperelliptic whereas it factors through the hyperelliptic pencil on the hyperelliptic curve.
-
This seems to be a duplicate of Alison Miller's accepted answer. – user5117 Feb 13 '14 at 15:40
ha! I was answering from my cell phone and I did not notice it... – IMeasy Feb 14 '14 at 7:43 | 979 | 3,378 | {"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} | 3.421875 | 3 | CC-MAIN-2016-26 | latest | en | 0.942115 |
http://openstudy.com/updates/4ecdeb76e4b04e045aeffaef | 1,448,417,635,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398444139.37/warc/CC-MAIN-20151124205404-00249-ip-10-71-132-137.ec2.internal.warc.gz | 181,579,649 | 13,089 | neverforgetvivistee 4 years ago Solving rational equations: (2x+2) (4x^2-16) (5x-5) ------- - ---------- = -------------- (3x-12) (3x^2-24x+48) (3x^2-24x+48) I have the answer already, but I don't know how it came to be I try to do it by factoring the 3x^2-24x+48 's and get 3(x^2-8+16) then I multiply that by everything, as with the other denominators, and so on but then I get a huge, incorrect answer Someone please help even if it's one sentence. I want to get this problem done by tonight
1. rld613
Firstly you can factor out a 3
2. rld613
and then it becomes 3(x-4)^2
3. rld613
and 3x-12 can also be factored
4. rld613
3(x-4)
5. rld613
So from what I see in order to get the same denomanator you must multiply the denomanator and numerator by (x-4)
6. rld613
Did u get that so far
7. neverforgetvivistee
yes and i'm fixing it on my paper... stupid factoring mistakes Thanks!
8. rld613
ok gnite
9. neverforgetvivistee
after this problem goodnight
10. neverforgetvivistee
Do you multiply the numerator of the first fraction by 3(x-4)^4? This is where i messed up the most I was struggling with factoring because I'm half asleep if you were wondering why I took so long by the way...
11. rld613
nope you multiply the numerator and the denomanator of the first term by (x-4)
12. neverforgetvivistee
13. neverforgetvivistee
because in the second fraction the denominator I get is 3(x-4)
14. rld613
well we already have a 3 in the denomanator of the first term
15. rld613
First term denomanator: 3(x-4) second term denomanator: 3(x-4) (x-4)
16. rld613
so all your missing in the first term is the (x-4)
17. neverforgetvivistee
how did you get rid of the 3?
18. rld613
I factored it
19. rld613
factor (3x-12) for me
20. neverforgetvivistee
3(x-4)
21. rld613
it is 3(x-4)
22. rld613
ya
23. rld613
so what doesn't make sense
24. neverforgetvivistee
when you're supposed to multiply everything by the denominator
25. rld613
nope we have two denomanators and we need them to be the same
26. rld613
so in order ofor 3(x-4) to be equal to 3(x-4)^2
27. rld613
we would have to multiply the first term by a (x-4)
28. rld613
so 3(x-4)*(x-4)=3(x-4)^2
29. rld613
Is that clear?
30. neverforgetvivistee
yes sort of , do we do that to the numerator too?
31. rld613
yup forsure
32. neverforgetvivistee
ooooooooooooooooooooooooooooohhhhhhhhhhhhhhhhhhhhhhhhhhhhhh ok I see I see
33. rld613
YAY i am so excited
34. rld613
I try to explain concepts but sometimes the person on the other end can't understand
35. neverforgetvivistee
THANK YOU OK I ACTUALLY SEE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! THANK YOU OK LISTEN TO ME RIGHT NOW LOOK AT YOURSELF IN THE MIRROR AND SAY HELL YEAH IM AWESOME IM SO GOOD AT MATH THANK YOU OKAY BECAUSE I WAS REALLY STRUGGLING WITH THAT AND NOW I CAN FINALLY SLEEP
36. neverforgetvivistee
NOW IM GOING TO FINISH THIS PROBLEM AND GET THE RIGHT ANSWER!!!!!!!!!!
37. rld613
Oh thanks for the boost I needed that one!!! LOL
38. neverforgetvivistee
lol
39. neverforgetvivistee
someone give rld613 some more medals
40. neverforgetvivistee
i'm going to solve this problem right now afk
41. rld613
I am really not into the medal thing
42. rld613
I just enjoy doing math
43. rld613
its like past 2 am and I am still doing math equations!!! I shld be long in bed
44. rld613
K now that we finished i am going to bed
45. neverforgetvivistee
rld are you still there? | 1,099 | 3,452 | {"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} | 3.90625 | 4 | CC-MAIN-2015-48 | longest | en | 0.878603 |
https://www.kidsacademy.mobi/printable-worksheets/online/activities/age-5-6/knowledge-check/counting/ | 1,716,667,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058834.56/warc/CC-MAIN-20240525192227-20240525222227-00349.warc.gz | 721,578,843 | 52,916 | # Counting worksheets activities for Ages 5-6
Step into the world of numbers with our captivating Counting Worksheets Activities! Designed to engage young learners, these worksheets are brimming with fun and educational tasks that transform counting from a mundane task into an exciting adventure. Ideal for preschool and kindergarten students, our Counting Worksheets Activities not only help in mastering the basics of numbers but also in enhancing fine motor skills and cognitive development. From vibrant images to count to intriguing puzzles that require number identification, these activities are crafted to stimulate curiosity and foster a love for mathematics. Start your child's numerical journey today with our engaging Counting Worksheets Activities!
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## Number Stories One More – Assessment 2 Worksheet
Tracing is a great activity for kids. They can count and trace numbers, recognize animals, and practice drawing on dotted lines. It's entertaining and educational, helping children learn valuable counting skills.
Number Stories One More – Assessment 2 Worksheet
Worksheet
## One Less: Assessment 2 Worksheet
Farmers are vital to our community. They raise animals and crops which provide us with food. Kids can learn a lot by visiting a farm—ask them to name the animals and crops they saw! Farmer Bill in this worksheet needs hay. Point to the haystacks, and ask your child to circle the one with 1 less.
One Less: Assessment 2 Worksheet
Worksheet
Learning Skills
Counting worksheets activities are an indispensable tool in the educational toolkit for young learners. These activities are not just about learning numbers; they're about building a strong foundation for mathematical skills that children will use throughout their lives. Here's why counting worksheets activities are so useful:
Firstly, counting worksheets activities introduce children to the concept of numbers in a structured and engaging way. They offer a clear, step-by-step approach to understanding what numbers represent, making the abstract concept of quantity more concrete for young minds. By associating numbers with physical objects or pictures, children can visually grasp the idea of counting, making the learning process both effective and enjoyable.
Moreover, these activities enhance number recognition and sequencing. Through repetitive practice, children become familiar with the order of numbers, which is crucial for their future mathematical understanding. This repetition solidifies their ability to recognize numbers not only in isolation but also in different contexts, further reinforcing their numerical fluency.
Counting worksheets activities also promote the development of fine motor skills. As children write numbers, circle items to count them, or draw lines to match numbers with quantities, they refine their hand-eye coordination and pencil control. This not only aids in their counting ability but also prepares them for more complex tasks such as writing.
Furthermore, these activities are adaptable to each child's learning pace and level. They can be easily customized to challenge more advanced learners or to provide additional support for those who need it. This flexibility ensures that all children can benefit from them, regardless of their starting point.
Lastly, counting worksheets activities foster a sense of achievement and confidence in young learners. By completing these worksheets, children can see tangible progress in their abilities, encouraging them to embrace more challenging mathematical concepts with enthusiasm and confidence.
In essence, counting worksheets activities are a fundamental stepping stone in early childhood education. They not only teach children about numbers but also build critical thinking, fine motor skills, and a positive attitude toward learning. | 682 | 3,881 | {"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-2024-22 | latest | en | 0.921179 |
http://cexams.com/Competitive-Exams/iq-quiz-07 | 1,531,945,698,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676590329.25/warc/CC-MAIN-20180718193656-20180718213656-00429.warc.gz | 72,723,327 | 3,868 | 1. A positive number when decreased by 4 is equal to 21 times reciprocal of the number. The number is :
2. In a 500 m race, the ratio of the speeds of two contestants A and B is 3 : 4. A has a start of 140 m. Then, A wins by :
3. If Borbs are better than Fribs, and Luns are worse than Jirts, Luns must be better than Fribs if Luns are better than Borbs.
4. One-fourth of one-third of a number is 15. What will be 40% of that number?
5. The number which when added to 10 times itself gives 264, is :
6. If + means /, - means x, / means + and x means - ,then 36 x 12 + 4 / 6 + 2 - 3 = ?
7. A number exceeds its four-seventh by 18. What is the number ?
8. A WINDOW always has
9. A train traveling at a speed of 90 kmph, crosses a pole in 10 seconds. The length of the train is :
10. If 16% of 40% of a number is 8, the number is :
Competitive Exams
1. IQ Quiz - 16
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6. IQ Quiz - 11
7. IQ Quiz - 10
8. IQ Quiz - 09
9. IQ Quiz - 08
10. IQ Quiz - 07
11. IQ Quiz - 06
12. IQ Quiz - 05
13. IQ Quiz - 04
14. IQ Quiz - 03
15. IQ Quiz - 02
16. IQ Quiz - 01
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Forgive yourself
When you violate someone's trust, you may feel so regretful that you have a hard time forgiving yourself for the violation. While a repentant heart is an essential part of making up with the person you betrayed, you also need to accept and learn to forgive yourself after you put the effort into making amends.Remember that no one is perfect. Whether your error in judgment was minor or major, it goes to show that you are only human. Accept your failure, and try to push forward into the future.By clinging to thoughts of past failure, you risk devaluing yourself. Once you begin to have such thoughts, it could zap your motivation for self improvement. | 587 | 1,981 | {"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} | 2.796875 | 3 | CC-MAIN-2018-30 | latest | en | 0.918208 |
https://www.fxsolver.com/browse/?like=1568&p=10 | 1,597,286,593,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738950.61/warc/CC-MAIN-20200813014639-20200813044639-00064.warc.gz | 683,349,812 | 42,188 | '
# Search results
Found 779 matches
Critical Damping Coefficient
A harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force, proportional to the displacement. If a ... more
Dolbear's Law - in degrees Celsius
Dolbear’s law states the relationship between the air temperature and the rate at which crickets chirp. It was formulated by Amos ... more
Dolbear's Law - in degrees Fahrenheit
Dolbear’s law states the relationship between the air temperature and the rate at which crickets chirp. It was formulated by Amos ... more
Future value is the value of an asset at a specific date. It measures the nominal future sum of money that a given sum of money is “worth” at a ... more
Spring constant
Hooke’s law is a principle of physics that states that the force F needed to extend or compress a spring by some distance X is proportional to that ... more
Absolute Magnitude of a Star - with distance modulus
Absolute magnitude is the measure of a celestial object’s intrinsic brightness. It is the hypothetical apparent magnitude of an object at a standard ... more
Absolute Magnitude of a Star - with parallax
Absolute magnitude is the measure of a celestial object’s intrinsic brightness. It is the hypothetical apparent magnitude of an object at a standard ... more
Wien's displacement law
Wien’s displacement law states that the black body radiation curve for different temperature peaks at a wavelength that is inversely proportional to ... more
Absolute Magnitude of a Star - with luminosity distance
Absolute magnitude is the measure of a celestial object’s intrinsic brightness. It is the hypothetical apparent magnitude of an object at a standard ... more
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The Gompertz–Makeham law states that the human death rate is the sum of an age-independent component (the Makeham term, named after William Makeham) and an ... more
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Category | 429 | 2,053 | {"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-2020-34 | latest | en | 0.892048 |
http://www.aorinevo.com/stevens/ma-123-calculus-iia/sandbox/ | 1,516,418,070,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888878.44/warc/CC-MAIN-20180120023744-20180120043744-00059.warc.gz | 379,150,767 | 21,562 | # MA 123: The SandBox
Use this page to ask anything related to MA 123. If you are registered and logged in, your posts will display your user name. If you would like to post anonymously, make sure you are logged out. Anyone is allowed to post.
## 47 thoughts on “MA 123: The SandBox”
1. Anonymous says:
Hello Aori,
I have a question with domain of multivariables $$\ln(1-x^2-y^2).$$ Is the inequality $$1-x^2-y^2 > 0$$ correct? Also, what does the graph of $$1-x^2-y^2 > 0$$ look like? Is it a circle with radius 1 centered at origin?
Thanks.
1. The domain of a logarithmic function is the set of all inputs for which the argument to the logarithmic function is positive. That is, the domain of $$\ln u$$ is the set of all $$u > 0.$$ Consequently, the domain of the multivariable function $$\ln(1-x^2-y^2)$$ is all real ordered pairs $$(x,y)$$ such that $$1-x^2-y^2 > 0.$$ We can represent the domain in set builder notation as
$$\left \lbrace (x,y) \; | \; 1-x^2 – y^2 >0 \right \rbrace.$$
We can also represent the domain graphically.
• First graph the corresponding equation $$1-x^2-y^2=0$$. This will be the graph of a circle centered at the origin of radius one. Make sure to sketch this circle using a dashed curve, since the inequality is strict.
• The graph obtained thus far partitions your coordinate plane into two regions. In each of the regions choose an ordered pair that clearly lies within that region.
• Test the inequality for each of these ordered pairs. If the inequality is satisfied, shade the region. Otherwise, do not shade.
The shaded portions and solid curves of the graph represent the domain of the function $$\ln(1-x^2-y^2).$$
2. Anonymous says:
Can you explain how to find the area of the parallelogram in Workshop 6 MA 123 and explain question 2 in the worksheet too please.
1. There isn’t a question in workshop 6 asking for the area of a parallelogram. Also, solutions to workshop 6 are available on moodle. Did you mean homework 4?
2. Anonymous says:
I am sorry. It asks for the area of Q. It is workshop 6 from feb 26,2013( MA 123). I need help with both questions 1 and 2.
3. P1. The four points $$A = (1,-2,1), B = (3,1,2), C= (0,2,1),$$ and $$D = (-2,-1,0)$$ form the corners of a quadrilateral $$Q$$.
1. Show that $$Q$$ is a parallelogram that is not a rectangle.
2. Find the area of $$Q$$.
Let
\begin{aligned} \vec{AB} &= \langle 3,1,2 \rangle – \langle 1,-2,1\rangle = \langle 2,3,1\rangle \\ \vec{CD} &= \langle -2,-1,0 \rangle – \langle 0,2,1 \rangle = \langle -2,-3,-1\rangle \\ \vec{BC} &= \langle 0,2,1 \rangle – \langle 3,1,2 \rangle = \langle -3,1,-1\rangle \\ \vec{AD} &= \langle -2,-1,0 \rangle – \langle 1,-2,1 \rangle = \langle -3,1,-1\rangle \\ \end{aligned}
Then
• $$Q$$ is a parallelogram if $$\vec{AB} \parallel \vec{CD}$$ and $$\vec{AD} \parallel \vec{BC}$$. That is, if $$\vec{AB} \times \vec{CD} = \vec{0}$$ and $$\vec{AD} \times \vec{BC} = \vec{0}$$. Since these cross products do equal the zero vector, we conclude that $$Q$$ is a parallelogram.
• $$Q$$ is not a rectangle if $$\vec{AB} \cdot \vec{AD} \not = 0$$. $$\vec{AB} \cdot \vec{AD} = -4 \not = 0.$$ So, $$Q$$ is not a rectangle.
4. A ball rests on a straight track aligned in the direction $$\langle 2, 1, 0 \rangle.$$ The ball will remain stationary unless it experiences a force of at least $$10$$ N along the direction of the track. If a wind is blowing in the direction $$\langle -1,-2, 6 \rangle,$$ what is the magnitude of the wind’s force required to move the ball, and in which direction will the ball move?
Let $$\vec{a} = \langle 2,1,0 \rangle$$ and $$\vec{b} = \langle -1,-2, 6 \rangle.$$ The amount of wind blowing in the direction of the track, $$\vec{a}$$, is given by
$$\text{proj}_{\vec{a}} \vec{b} = \frac{\vec{a} \cdot \vec{b}}{\vec{a} \cdot \vec{a}} \vec{a} = \left \langle \frac{-8}{5}, \frac{-4}{5}, 0 \right \rangle$$
We need that the magnitude of this vector be 10 N. That is, $$|c\text{proj}_{\vec{a}} \vec{b}| = \frac{4c}{\sqrt{5}} = 10.$$ So, $$c = \sqrt{5}/2.$$
The ball will move in the direction $$\vec{a} + c\text{proj}_{\vec{a}} \vec{b} = \left \langle 2-\frac{4}{\sqrt{5}},1-\frac{2}{\sqrt{5}},0 \right \rangle.$$ This is the same direction as $$\langle 2,1,0 \rangle.$$
5. Anonymous says:
Hi,
For the problem with the wind, you answered c = sqrt 5/2. Is this the wind force ( final answer for part a ) or do we have to work from there.
6. The force necessary to move the ball is $$c$$ times $$proj_{\vec{a}} \vec{b}.$$
3. Anonymous says:
Hello,
I need help with homework 3, question 1. I know that the series is ln(1+x) and I also did the ratio test, but I am not sure about how to use those values to find x. Basically, I got like convergence to -1 but they gave us big values and ask us to find x. How to do it?
1. Here is a similar problem.
For $$k = 1,-10$$, determine whether or not there exists a number $$x$$ such that
$$1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots = k$$
If $$x$$ exists, find it.
Note that
$$1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots = e^x$$
for all $$x$$ in the interval $$(-\infty, \infty)$$. Thus, we are looking for solutions to
$$e^x = k$$
for $$k = 1$$ and $$k = -1$$.
When $$k = 1$$, $$e^x = 1 \Rightarrow x = 0$$. When $$k = -1$$, $$e^x = -1$$ which has no solutions.
Be aware interval on which the series is allowed to be replaced by the corresponding elementary function, plays an important role. In the homework, the series is allowed to be be replaced by the elementary function $$\ln(1+x)$$ for $$x$$ in the interval $$(-1,1]$$. Therefore, any solutions you obtain after replacing the series with its corresponding elementary function must be also lie within this interval. If not, then it is not a solution to the original equality. If so, then it is a solution to the original equality.
2. Anonymous says:
Thank You. Can you explain the paragraph you typed in detail please.
3. Here is another example,
For $$k = -1, -1/2, 5000$$ determine whether or not there exists a number $$x$$ such that
$$1 + 2(x-1) + 2^2(x-1)^2 + 2^3(x-1)^3 + \cdots = k$$
If $$x$$, exists find it.
Note that
$$1 + 2(x-1) + 2^2(x-1)^2 + 2^3(x-1)^3 + \cdots = \frac{1}{1-2x}$$
for all $$x$$ in the interval $$(1/2,3/2)$$. The interval of convergence can be determined by applying the ratio test and testing the endpoints.
Thus, we are looking for solutions to
$$\frac{1}{1-2x} = k$$
for $$k = -1, -1/2, 5000$$.
If $$k = -1$$, then $$1/(1-2x) = -1 \Rightarrow x = 1$$. As $$x = 1$$ is in the interval of convergence, $$x = 1$$ solves the original equation where $$k = -1$$.
If $$k = -1/2$$, then $$1/(1-2x) = -1/2 \Rightarrow x = 3/2$$. As $$x = 3/2$$ is not in the interval of convergence, $$x = 3/2$$ does not solve the original equation where $$k = -1/2$$.
If $$k = 5000$$, then $$1/(1-2x) = 5000 \Rightarrow x = .4999$$. As $$x = .4999$$ is not in the interval of convergence, $$x = .4999$$ does not solve the original equation where $$k = 5000$$.
4. Anonymous says:
I have a WebAssign problem that I’m unsure why I am getting the wrong answer. We haven’t learned how to find the length of a curve yet, but I looked in the book to figure out how to solve it. My problem reads:
Find the length of the curve.
$$r(t)= \left \langle 4t,t^2, \frac{1}{6}t^3 \right \rangle, 0\leq t\leq 4$$
I took the derivative, squared it, square rooted that, and took the integral from 0 to 4 and got 128/3. I’m not exactly sure what I’m doing wrong.Thanks!
1. Aori Nevo says:
The steps you’ve outlined are the right ones. A bit of terminology is wrong, but the idea is correct. In particular, you mentioned taking the derivative and squaring it. What you mean to say is that you added the square of each component of $$\mathbf{r}'(t)$$ and then took the square root. I’ve worked out the solution to the problem and got an answer different from 128/3. There is probably a computational error somewhere.
5. Anonymous says:
Hello Aori, I don’t know how to approach #2 in Homework 5.
A ball rests on a straight track aligned in the direction (2,1,0). The ball will remain stationary unless it experiences a force of at least 10 N along the direction of the track. If a wind is blowing in the direction (-1,-2,6), what is the magnitude of the wind’s force required to move the ball, and in which direction will the ball move?
Can you give me a sort of hint on how to go about solving this?
1. Aori Nevo says:
Let $$B = \langle 2,1,0 \rangle$$ and $$W = \langle -1, -2, 6 \rangle$$. The effect of $$W$$ on $$B$$ is determined by
$$\text{comp}_\mathbf{B} \mathbf{W}$$
However, just computing this for the given values of $$W$$ and $$B$$ isn’t enough. We need to factor in that the wind can blow harder. This can be accomplished by scaling the vector $$W$$ by some nonzero constant, say $$c$$.
Working with the vectors $$cW$$ and $$B$$ one can find the appropriate value of $$c$$ and consequently the vector $$cW$$. Once this is done, the magnitude of the force will be $$\left| cW \right|$$. The direction can be computed using the formula
$$\text{proj}_{\mathbf{B}} c\mathbf{W}$$
6. Anonymous says:
Hi, I was wondering if you can help me with this problem. How can I find the
$$\sum_{n=1}^{\infty} n(n+1) (1/2)^{n-1}$$
from problem 4 on Homework 3?
1. Aori Nevo says:
Observe that,
\begin{aligned} \frac{d^2}{dx^2}\left[ \frac{1}{1-x}\right] = \frac{2}{(1-x)^3} \end{aligned}
and that,
\begin{aligned} \frac{d^2}{dx^2}\left[ \frac{1}{1-x}\right] & = \frac{d^2}{dx^2}\left[ \sum_{n=0}^{\infty} x^n \right] \\ & = \frac{d}{dx}\left[ \sum_{n=1}^{\infty} \frac{d}{dx}[x^n] \right] \\ & = \frac{d}{dx}\left[ \sum_{n=1}^{\infty}n x^{n-1} \right] \\ & = \sum_{n=2}^{\infty}\frac{d}{dx}[n x^{n-1}] \\ & = \sum_{n=2}^{\infty}n(n-1) x^{n-2} \\ & = \sum_{n=1}^{\infty}(n+1)n x^{n-1} \end{aligned}
So that
$$\frac{2}{(1-x)^3} = \sum_{n=1}^{\infty}(n+1)n x^{n-1}$$
which holds true for $$|x|<1$$.
Substituting in $$x = 1/2$$ we obtain the series we want to find the sum of on the right and the sum of that series on the left.
Some things to note in the above calculuations:
• When pushing the derivative through the summation we increment the index by one provided the first term is constant. Otherwise, leave the index alone.
• In deriving the series representation for $$2/(1-x)^3$$, we change the index in the last equality. You can make the substitution $$k-1=n-2$$ and, after completing the substitution, return the series to the original variable by replacing $$k$$ with $$n$$.
• The interval of convergence follows from the fact that the interval of convergence for $$1/(1-x)$$ is $$(-1,1)$$. Not that its necessary to know if the endpoints are included, since the value of $$x$$ we are using is well within the interval.
7. Anonymous says:
Hello Aori, I’m a bit confused with getting the next term, such as in the ratio test or estimating sums if its an alternating series.
For example:
$$\sum_{n=0}^{\infty} \frac{x^{2n}}{n^{2n+1}}$$
If I were to add 1 to n would it be
$$\sum_{n=0}^{\infty} \frac{x^{2(n+1)}}{n^{2(n+1)+1}}$$
or
$$\sum_{n=0}^{\infty} \frac{x^{2n+1}}{n^{2n+2}}$$
Thank you.
1. Aori Nevo says:
First note that the $$n$$-th term in the sum
$$\sum_{n=1}^{\infty} \frac{x^{2n}}{n^{2n+1}}$$
is denoted $$a_n$$ and is defined by
$$a_n = \frac{x^{2n}}{n^{2n+1}}$$
(By the way, the sum here must start at $$n=1$$ since $$n=0$$ makes the denominator zero).
Also note that when you are looking at the $$n$$-th term you must drop the summation symbol.
Now,
\begin{aligned} a_{n+1} &= \frac{x^{2(n+1)}}{n^{2(n+1)+1}}\\ & = \frac{x^{2n+2}}{(n+1)^{2n+2}} \end{aligned}
You simply replace where ever you see $$n$$ by $$n+1$$ and simplify to obtain the form of the $$n+1$$-th term.
However, these computations are done inline within the ratio test starting with
$$\lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|$$
Here is how it would look:
\require{cancel} \begin{aligned} \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right| &= \lim_{n \rightarrow \infty} \left| \frac{\frac{x^{2n+2}}{(n+1)^{2n+2}}}{\frac{x^{2n}}{n^{2n+1}}}\right| \\ & = \lim_{n \rightarrow \infty}\left| \frac{x^{2n+2}}{(n+1)^{2n+2}} \cdot \frac{n^{2n+1}}{x^{2n}} \right| \\ & = \lim_{n \rightarrow \infty}\left| \frac{\bcancel{x^{2n}}x^2}{(n+1)^{2n+2}} \cdot \frac{n^{2n+1}}{\bcancel{x^{2n}}} \right| \\ & = x^2 \cdot \lim_{n \rightarrow \infty} \left| \frac{n^{2n+1}}{(n+1)^{2n+2}} \right| \\ & = x^2 \cdot \lim_{n \rightarrow \infty} \frac{1}{n} = x^2 \cdot 0 = 0 \end{aligned}
Since the limit of the ratio is zero regardless of the value of $$x$$, the series converges for all values of $$x$$. That is the radius of convergence is $$\infty$$ and the interval of convergence is $$(-\infty, \infty)$$.
As for using the Alternating Series Estimation Theorem, please choose an example and I will go through a detailed solution. The question you’re asking is too broad.
8. Anonymous says:
Hello. I have a question regarding this WebAssign problem
$$\cos(x) \approx 1 – \frac{x^2}{2} + \frac{x^4}{24},\; (|error| < 0.05 )$$
It is asking for the range of values of x for which the given approximation is accurate to within the stated error. I get the the range of values of (-1.047,1.047), but that is not the correct answer.
1. Aori Nevo says:
Here is an example: Use the Alternating Series Estimation Theorem to estimate the range of values of x for which the given approximation is accurate to within the stated error.
$$\arctan(x) \approx x – \frac{x^3}{3} + \frac{x^5}{5}, \; (|error| < 0.0005 )$$
The term that plays the role of $$b_{n+1}$$ in the Alternating Series Estimation Theorem is the term in the Maclaurin series for $$\arctan x$$ after $$\frac{x^5}{5}$$ in absolute value, which is $$\left| \frac{x^7}{7} \right|$$.
This term bounds the error and so it suffices to set
$$\left| \frac{x^7}{7} \right| < 0.0005$$ which has as its solution set the interval $$(-\sqrt[7]{.0035}, \sqrt[7]{.0035})$$. Note: I tried this problem on WebAssign; it seems to have issues with rounding. If three decimal places doesn't work, do four.
9. Anonymous says:
Hello, I have a question about the maclaurin series of: $$f(x) = (1+x)^n = \sum_{n=0}^\infty\binom{k}{n} x^n$$
What exactly does the array of k and n mean? How does it expand to many terms?
Thank you.
1. Aori Nevo says:
By definition,
$${ k \choose n } = \frac{ k(k-1)(k-2) \cdots (k-n + 1)}{n!}$$
10. Anonymous says:
Hello Aori,
I have a question about the binomial series.
I understand that if $$f(x) = (1+x)^{k}$$,
$$\frac{d^{n}}{dx^n} = kPn * (1+x)^{k-n}$$,
and therefore the McLauren series for $$f(x) = \sum_{n=0}^{\infty }[ kCn * x^n ].$$
However, I don’t understand how to compute $$kCn$$ when k<n.
From what I understand, $$kCn = \frac{kPn}{ n!} = \frac{k!}{ n! (k-n)! }$$
For example, if $$k = 2$$, and $$n = 5$$,
$$kCn = 2C5 = \frac{2!}{5! (-3)!}$$
Or when k is a negative number, such as $$k = -2$$,
$$kCn = \frac{(-2)!}{ n! (-2-n)!}$$
I looked up negative factorials on wikipedia, but it said that factorials are not defined for negative integers, even when the factorial function is extended by the gamma function.
If this is the case, how does the binomial series work?
1. Aori Nevo says:
The problem seems to be with the identity
$$\frac{kPn}{n!} = \frac{k!}{n! (k-n)!}.$$
For your first example, note that
$$2 C5 = \frac{2(2-1)(2-2)(2-3)(2-4)}{5!} = 0$$
For your second example, with $$n = 5$$ note that
\begin{aligned} (-2) C5 &= \frac{-2(-2-1)(-2-2)(-2-3)(-2-4)(-2-4)}{5!} \\ & = \frac{(-2)(-3)(-4)(-5)(-6)}{5!} \end{aligned}
which equals $$-\frac{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}{5!}$$.
When $$k$$ is not a non-negative integer we compute, $$kCn$$ using the identity
$$kCn = \frac{kPn}{n!}$$ | 5,272 | 15,721 | {"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": 2, "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-2018-05 | latest | en | 0.897636 |
http://research.stlouisfed.org/fred2/graph/?chart_type=line&s[1][id]=PLOINVAOA624NUPN&s[1][transformation]=cch | 1,419,591,431,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447548791.91/warc/CC-MAIN-20141224185908-00037-ip-10-231-17-201.ec2.internal.warc.gz | 84,348,998 | 18,173 | # Graph: Price Level of Investment for Angola
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(a) Price Level of Investment for Angola, PPP of Investment over Exchange Rate, Not Seasonally Adjusted (PLOINVAOA624NUPN)
Price Level of GDP is the PPP over GDP divided by the exchange rate times 100. The PPP of GDP or any component is the national currency value divided by the real value in international dollars. The PPP and the exchange rate are both expressed as national currency units per US dollar.The value of price level of GDP for the United States is made equal to 100. Price Levels of the components Consumption, Investment, and Government are derived in the same way as the price level of GDP. While the U.S. = 100 over GDP, this is not true for the component shares. The purchasing power parity in domestic currency per \$US for GDP or any component, may be obtained by dividing the price level by 100 and multiplying by the Exchange Rate. More information is available at http://pwt.econ.upenn.edu/Documentation/append61.pdf.
For proper citation, see http://pwt.econ.upenn.edu/php_site/pwt_index.php
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Price Level of Investment for Angola
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A093031 Create a histogram of the digits used so far in the sequence. Add to the last element of the sequence the least used digit on the histogram. If two or more digits have the same score in the histogram, add the smallest. 0 is to be interpreted as 10 and so comes after 9. So if 0 is the digit to add, add 10 instead. Start the sequence with 0. 1
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OFFSET 0,3 LINKS CROSSREFS Sequence in context: A033037 A158333 A062505 * A143452 A193414 A138217 Adjacent sequences: A093028 A093029 A093030 * A093032 A093033 A093034 KEYWORD nonn,base AUTHOR Eric Angelini, May 07 2004 STATUS approved
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ConceptRelationship Between Zeroes and Coefficients of a Polynomial
#### Question
Verify that the numbers given along side of the cubic polynomials are their zeroes. Also verify the relationship between the zeroes and the coefficients.
2x^3 + x^2 – 5x + 2 ; 1/2, 1, – 2
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#### Reference Material
Solution for concept: Relationship Between Zeroes and Coefficients of a Polynomial. For the course 8th-10th CBSE
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# Logarithmic Depth Buffer
by Brano Kemen on 08/12/09 02:43:00 pm
The following blog post, unless otherwise noted, was written by a member of Gamasutra’s community.
The thoughts and opinions expressed are those of the writer and not Gamasutra or its parent company.
I assume pretty much every 3D programmer runs into Z-buffer issues sooner or later. Especially when doing planetary rendering; the distant stuff can be a thousand kilometers away but you still would like to see fine details right in front of the camera.
Previously I have dealt with the problem by splitting the depth range in two and using the first part for near stuff and another for distant stuff. The boundary was floating, somewhere around 5km - quad-tree tiles up to certain level were using the distant part, and the more detailed tiles that by law of LOD are occurring nearer the camera used the other part.
Most of the time this worked. But in one case it failed miserably - when a more detailed tile appeared behind a less detailed one.
I was thinking about the ways to fix it, grumbling why we can't have a Z-buffer with better distribution, when it occurred to me that maybe we can.
Steve Baker's document explains common problems with Z-buffer. In short, the depth values are proportional to the reciprocal of Z. This gives amounts of precision near the camera but little off in the distance. Common method is then to move your near clip plane further away, which helps but also brings its own problems, mainly that .. the near clip plane is too far
A much better Z-value distribution is a logarithmic one. It also plays nicely with LOD used in large scale terrain rendering.
Using the following equation to modify depth value after it's been transformed by the projection matrix:
```z = log(C*z + 1) / log(C*Far + 1) * w
```
Where C is constant that determines the resolution near the camera, and the multiplication by w undoes in advance the implicit division by w later in the pipeline.
Resolution at distance x, for given C and n bits of Z-buffer resolution can be computed as
``` log(C*Far + 1)
Res = ----------------
2^n * C/(C*x+1)
```
So for example for a far plane at 10,000 km and 24-bit Z-buffer this gives the following resolutions:
``` 1m 10m 100m 1km 10km 100km 1Mm 10Mm
------------------------------------------------------------
C=1 1.9e-6 1.1e-5 9.7e-5 0.001 0.01 0.096 0.96 9.6 [m]
C=0.001 0.0005 0.0005 0.0006 0.001 0.006 0.055 0.549 5.49 [m]
```
Along with the better utilization of z-value space it also (almost) gets us rid of the near clip plane.
And here comes the result.
Looking into the nose while keeping eye on distant mountains ..
10 thousand kilometers, no near Z clipping and no Z-fighting! HOORAY!
### More details
The C basically changes the resolution near the camera; I used C=1 for the screenshots, having theoretical resolution 1.9e-6m. However, the resolution near the camera cannot be utilized fully as long as the geometry isn't finely tessellated too, because the depth is interpolated linearly and not logarithmically. On models such as the guy on the screenshots it is perfectly fine to put camera on his nose, but with models with long stripes with vertices few meters apart the bugs from the interpolation can be visible. We will be dealing with it by requiring certain minimum tessellation.
Also I think I've read somewhere that some forthcoming generation of hardware will support different modes of interpolation too.
So yes, modifying C changes the resolution near the camera, setting it to a value that gives the largest acceptable resolution may be desirable to achieve more linear distribution in the near range and thus minimizing the interpolation problem.
Near clip plane can be put arbitrarily 'near' but not zero because of the 1/w division. I have put it to 0.0001m. This is using standard perspective projection setup.
### Negative Z artifact fix
Ysaneya suggested a fix for the artifacts occurring with thin or huge triangles when Z goes behind the camera, by writing the correct Z-value at the pixel shader level. This disables fast-Z mode but he found the performance hit to be negligible.
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Senior Technical Artist | 1,153 | 4,846 | {"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-2019-04 | latest | en | 0.967417 |
https://hub.jmonkeyengine.org/t/bug-with-barycentric-calculation-in-ray/27375 | 1,685,935,685,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224650620.66/warc/CC-MAIN-20230605021141-20230605051141-00546.warc.gz | 340,485,047 | 10,470 | # [BUG] with Barycentric calculation in Ray
In all definitions I can find of Barycentric coordinates they:
• Can not be negative numbers
• Always sum to 1
If you take a look below where w1 & w2 are calculated, you’ll see a third number w0 that is commented out. This looks to be the Barycentric coordinates that are calculated in Ray.class. Even leaving w0 uncommented and only taking a look at w1 & w2, I’m thinking these are wrong, because they do indeed return negative numbers. As far as I can tell, the output always sums to one… but not in a nonsensical manor, like (for instance):
(w0, w1, w2)
(-11.157536, 11.193249, 0.96428657)
You’ll also notice in the comments above w1, w2 this:
// these weights can be used to determine
// interpolated values, such as texture coord.
// eg. texcoord s,t at intersection point:
// s = w0s0 + w1s1 + w2s2;
// t = w0
t0 + w1t1 + w2t2;
Has anyone actually tried doing this? because it WILL NOT produce even remotely close results in its current form.
Here is the method from Ray.class in question:
[java]
private boolean intersects(Vector3f v0, Vector3f v1, Vector3f v2,
Vector3f store, boolean doPlanar, boolean quad) {
TempVars vars = TempVars.get();
`````` Vector3f tempVa = vars.vect1,
tempVb = vars.vect2,
tempVc = vars.vect3,
tempVd = vars.vect4;
Vector3f diff = origin.subtract(v0, tempVa);
Vector3f edge1 = v1.subtract(v0, tempVb);
Vector3f edge2 = v2.subtract(v0, tempVc);
Vector3f norm = edge1.cross(edge2, tempVd);
float dirDotNorm = direction.dot(norm);
float sign;
if (dirDotNorm > FastMath.FLT_EPSILON) {
sign = 1;
} else if (dirDotNorm < -FastMath.FLT_EPSILON) {
sign = -1f;
dirDotNorm = -dirDotNorm;
} else {
// ray and triangle/quad are parallel
vars.release();
return false;
}
float dirDotDiffxEdge2 = sign * direction.dot(diff.cross(edge2, edge2));
if (dirDotDiffxEdge2 >= 0.0f) {
float dirDotEdge1xDiff = sign
* direction.dot(edge1.crossLocal(diff));
if (dirDotEdge1xDiff >= 0.0f) {
if (!quad ? dirDotDiffxEdge2 + dirDotEdge1xDiff <= dirDotNorm : dirDotEdge1xDiff <= dirDotNorm) {
float diffDotNorm = -sign * diff.dot(norm);
if (diffDotNorm >= 0.0f) {
// this method always returns
vars.release();
// ray intersects triangle
// if storage vector is null, just return true,
if (store == null) {
return true;
}
// else fill in.
float inv = 1f / dirDotNorm;
float t = diffDotNorm * inv;
if (!doPlanar) {
direction.y * t, direction.z * t);
} else {
// these weights can be used to determine
// interpolated values, such as texture coord.
// eg. texcoord s,t at intersection point:
// s = w0*s0 + w1*s1 + w2*s2;
// t = w0*t0 + w1*t1 + w2*t2;
float w1 = dirDotDiffxEdge2 * inv;
float w2 = dirDotEdge1xDiff * inv;
//float w0 = 1.0f - w1 - w2;
store.set(t, w1, w2);
}
return true;
}
}
}
}
vars.release();
return false;
}
``````
[/java]
I guess I should mention, when calling a local copy of this to examine the numbers:
I am setting doPlanar to true and quad to false, as well as passing in the origin and direction for the original ray, plus the v1, 2 and 3 from the closest contact.
Do you get true or false as a result in these cases?
…though I guess nothing is stored for the false branches anyway.
Sorry had a terribly busy day yesterday. This is a positive collision that fires off the second check and the second check (calling the local function) returns a positive collision as well. Both return the same contact point
I thought, from Momoko_Fan 's description, this was calculated prior to the returned contact point, but it doesn’t seem to be.
EDIT: Check that… that’s only if doPlanar is false. Otherwise it returns t, w1, w2.
I thought it may be useful to know how I am testing this:
New Project
Rotate the default cube 45, 45, 0
Add RawInputListener and use the picking tutorial code in the onMouseMotion method.
Nothing more than this
After doing a bit of research for other threads relating to resolving texture coordinates from ray casting, it seems this has been a repetitive question since early 2010 with no resolve on any of the threads. I would think that if this wasn’t an issue, that at least one of them would have had a positive resolution. This doesn’t mean that this is proof positive that the issue exists, just a bit strange that in 3+ years no one has successfully gotten this to work.
EDIT: Most of these threads revolved around the “Render Nifty to Texture” example and being able to interact with the rendered results.
Ok… here is a comparison of JME’s Barycentric coords as apposed to two other methods… they were a quick conversion, not taking into account creating new vectors, etc. Here are the results (you’ll notice JME’s differ from the other two)
[java]
JME: (0.008120418, 0.47548336, 0.5163962)
BC1: (0.49115217, 0.1251316, 0.38371623)
BC2: (0.49115217, 0.1251316, 0.38371623)
JME: (0.008120418, 0.47548336, 0.5163962)
BC1: (0.56668615, -0.40701914, 0.840333)
BC2: (0.56668615, -0.40701914, 0.840333)
JME: (0.008120418, 0.47548336, 0.5163962)
BC1: (0.5166718, -0.35700476, 0.840333)
BC2: (0.5166718, -0.35700476, 0.840333)
JME: (0.008120418, 0.47548336, 0.5163962)
BC1: (0.46665692, -0.3069899, 0.840333)
BC2: (0.46665692, -0.3069899, 0.840333)
JME: (0.008120418, 0.47548336, 0.5163962)
BC1: (0.27750123, 0.07378864, 0.64871013)
BC2: (0.27750123, 0.07378864, 0.64871013)
[/java]
And here is one of the methods… the other is a direct swipe of Direct3D’s D3DIntersectTriangle
[java]
public Vector3f getBarycentricCoords(Triangle tri, Vector3f cp) {
Vector3f ret = new Vector3f();
Vector3f v1 = new Vector3f(tri.get2()).subtractLocal(tri.get1());
Vector3f v2 = new Vector3f(tri.get3()).subtractLocal(tri.get1());
Vector3f v3 = new Vector3f(cp).subtractLocal(tri.get1());
float d00 = v1.dot(v1);
float d01 = v1.dot(v2);
float d11 = v2.dot(v2);
float d20 = v3.dot(v1);
float d21 = v3.dot(v2);
float denom = d00 * d11 - d01 * d01;
float v = (d11 * d20 - d01 * d21) / denom;
float w = (d00 * d21 - d01 * d20) / denom;
float u = 1.0f - v - w;
ret.set(u,v,w);
return ret;
}
[/java]
What’s oddest to me is that the results are always the same for every run of the JME one… like you are passing the same value in every time. Where as the others change for each run. Are you actually passing different parameters each time?
@pspeed
I didn’t notice that… I’m passing in the exact same vectors to each routine. If you want, I can print those first to verify and post the results.
I’ve also tried pulling the vectors directly out of the float buffer to verify that they are correct. All seems right there
EDIT: This is happening from mouse move event… so the input vectors would be rather close in the example above. Though, I don’t see anything in the code that resembles any sort of rounding.
@pspeed
Ok here are the results, I also printed out the check on the verts from both the collision result and pulling them from the float buffer using the contact triangle index:
[java]
Triangle vert 1: (-1.0, 1.0, -1.0)
Buffer vert 1: (-1.0, 1.0, -1.0)
Triangle vert 2: (-1.0, -1.0, -1.0)
Buffer vert 2: (-1.0, -1.0, -1.0)
Triangle vert 3: (-1.0, -1.0, 1.0)
Buffer vert 3: (-1.0, -1.0, 1.0)
Calling methods using: (-1.0, 1.0, -1.0), (-1.0, -1.0, -1.0), (-1.0, -1.0, 1.0)
JME: (0.66071385, 0.18589167, 0.15339448)
BC1: (0.68471265, -0.45522714, 0.7705145)
BC2: (0.68471265, -0.45522714, 0.7705145)
Triangle vert 1: (-1.0, 1.0, 1.0)
Buffer vert 1: (-1.0, 1.0, 1.0)
Triangle vert 2: (-1.0, 1.0, -1.0)
Buffer vert 2: (-1.0, 1.0, -1.0)
Triangle vert 3: (-1.0, -1.0, 1.0)
Buffer vert 3: (-1.0, -1.0, 1.0)
Calling methods using: (-1.0, 1.0, 1.0), (-1.0, 1.0, -1.0), (-1.0, -1.0, 1.0)
JME: (0.66071385, 0.18589167, 0.15339448)
BC1: (0.32707047, 0.0737896, 0.5991399)
BC2: (0.32707047, 0.0737896, 0.5991399)
Triangle vert 1: (-1.0, 1.0, 1.0)
Buffer vert 1: (-1.0, 1.0, 1.0)
Triangle vert 2: (-1.0, 1.0, -1.0)
Buffer vert 2: (-1.0, 1.0, -1.0)
Triangle vert 3: (-1.0, -1.0, 1.0)
Buffer vert 3: (-1.0, -1.0, 1.0)
Calling methods using: (-1.0, 1.0, 1.0), (-1.0, 1.0, -1.0), (-1.0, -1.0, 1.0)
JME: (0.66071385, 0.18589167, 0.15339448)
BC1: (0.27596706, 0.10795689, 0.61607605)
BC2: (0.27596706, 0.10795689, 0.61607605)
Triangle vert 1: (-1.0, 1.0, 1.0)
Buffer vert 1: (-1.0, 1.0, 1.0)
Triangle vert 2: (-1.0, 1.0, -1.0)
Buffer vert 2: (-1.0, 1.0, -1.0)
Triangle vert 3: (-1.0, -1.0, 1.0)
Buffer vert 3: (-1.0, -1.0, 1.0)
Calling methods using: (-1.0, 1.0, 1.0), (-1.0, 1.0, -1.0), (-1.0, -1.0, 1.0)
JME: (0.66071385, 0.18589167, 0.15339448)
BC1: (0.4296639, -0.04390526, 0.61424136)
BC2: (0.4296639, -0.04390526, 0.61424136)
[/java]
EDIT: JME & D3D methods both use the origin and direction from the ray, the other uses the contact point from the original collision.
…what is the ray in each of these cases?
Well… what I figure out thus far is, the reason the numbers returned by intersects appear to be the same is, it’s rare that it meets the conditions set up in the method to calculate the barycentric coords. And since the method requires you to pass in the store, it’s just print out the last successful run.
@pspeed said: ...what is the ray in each of these cases?
The ray origin and direction are being set using the code example from the picking test example. Let me grab it…
[java]
click2d.set(getInputManager().getCursorPosition());
tempV2.set(click2d);
click3d.set(getCamera().getWorldCoordinates(tempV2, 0f));
pickDir.set(getCamera().getWorldCoordinates(tempV2, 1f).subtractLocal(click3d).normalizeLocal());
pickRay.setOrigin(click3d);
pickRay.setDirection(pickDir);
rayResults.clear();
[/java]
@t0neg0d said: Well... what I figure out thus far is, the reason the numbers returned by intersects appear to be the same is, it's rare that it meets the conditions set up in the method to calculate the barycentric coords. And since the method requires you to pass in the store, it's just print out the last successful run.
…which means it isn’t really intersecting, right?.. which means… What is the Ray you are passing in?
If the ray is in world coordinates then I think the triangle would need to also be in world coordinates, etc…
@t0neg0d said: The ray origin and direction are being set using the code example from the picking test example. Let me grab it...
[java]
click2d.set(getInputManager().getCursorPosition());
tempV2.set(click2d);
click3d.set(getCamera().getWorldCoordinates(tempV2, 0f));
pickDir.set(getCamera().getWorldCoordinates(tempV2, 1f).subtractLocal(click3d).normalizeLocal());
pickRay.setOrigin(click3d);
pickRay.setDirection(pickDir);
rayResults.clear();
[/java]
It might be quicker if you show us the rest of the code, too… like the code where you transform the Ray into mesh space or transform the triangle into world space or whatever.
Oh… when I force the calc anyways (which should always be valid, since the collision has already happened), I get all sorts of whacked out numbers like:
[java]
Calling methods using: (-1.0, 1.0, 1.0), (-1.0, 1.0, -1.0), (-1.0, -1.0, 1.0)
JME: (-87.069176, 87.069176, 1.0000037)
BC1: (0.610375, -0.158741, 0.548366)
BC2: (0.610375, -0.158741, 0.548366)
(-61969.805, 0.0)
[/java]
@pspeed said: It might be quicker if you show us the rest of the code, too... like the code where you transform the Ray into mesh space or transform the triangle into world space or whatever.
Um… I feeling like a real fucking tool atm. I had commented out the transforms on the verts and forgot to uncomment them.
[java]
geom.getWorldTransform().transformVector(tri.get1(), v1);
geom.getWorldTransform().transformVector(tri.get2(), v2);
geom.getWorldTransform().transformVector(tri.get3(), v3);
[/java]
Guess what works properly now? Feel free to smack me at any point >.<
Glad you figured it out. This is why keyhole debugging is so frustrating on this end.
1 Like
@pspeed said: Glad you figured it out. This is why keyhole debugging is so frustrating on this end.
Sorry to have taken up anyones time with this. I had been bouncing around so much trying to figure out what I had done wrong while I was calling the wrong method in Ray, that I forgot to undo what I had undone after narrowing down the proper method (thinks to you in the first place). Hah… I should just pay you to write my code for me. lol | 4,018 | 12,209 | {"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} | 2.90625 | 3 | CC-MAIN-2023-23 | latest | en | 0.751125 |
https://www.rhayden.us/managerial-economics/figure-133.html | 1,579,441,230,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00005.warc.gz | 1,066,704,991 | 6,921 | ## Figure 133
Monopoly Price Regulation: Optimal Price/Output Decision Making
Monopoly regulation imposes a price ceiling at P2 just sufficient to provide a fair return (area PnAECn) on investment. Under regulation, price falls from P1 to P2 and output expands from Q1 to Q2.
Price and cost
Price and cost
Quantity per time period
TC = \$3,750,000 + \$70Q + 0.00002Q2 MC = ATC/AQ = \$70 + \$0.00004Q
where cost is expressed in dollars.
To find the profit-maximizing level of output, demand and marginal revenue curves for annual service must be derived. This will give all revenue and cost relations a common annual basis. The demand curve for annual service is 12 times monthly demand:
Total and marginal revenue curves for this annual demand curve are
TR = \$270Q - \$0.00048Q2 MR = ATR/AQ = \$270 - \$0.00096Q
The profit-maximizing level of output is found by setting MC = MR (where Mn = 0) and solving for Q:
MC = MR \$70 + \$0.00004Q = \$270 - \$0.00096Q \$0.001 Q = \$200 Q = 200,000
The monthly service price is
This price/output combination generates annual total profits of n = \$270Q - \$0.00048Q2 - \$3,750,000 - \$70Q - \$0.00002Q2 = -\$0.0005Q2 + \$200Q - \$3,750,000 = -\$0.0005(200,0002) + \$200(200,000) - \$3,750,000 = \$16,250,000
If the company has \$125 million invested in plant and equipment, the annual rate of return on investment is v u t ^ 4. \$16,250,000 ni_ 1Q0, Return on Investment = = 0.13, or 13%
\$125,000,000
Now assume that the State Public Utility Commission decides that a 12 percent rate of return is fair given the level of risk taken and conditions in the financial markets. With a 12 percent rate of return on total assets, Malibu Beach would earn business profits of n = Allowed Return X Total Assets = 0.12 X \$125,000,000 = \$15,000,000
To determine the level of output that would generate this level of total profits, total profit must be set equal to \$15 million:
n = TR - TC \$15,000,000 = -\$0.0005Q2 + \$200Q - \$3,750,000
This implies that
which is a function of the form aQ2 + bQ - c = 0. Solving for the roots of this equation provides the target output level. We use the quadratic equation as follows:
_ -200 ± V 2002 - 4(-0.0005)(18,750,000) 2(-0.0005)
= 150,000 or 250,000 | 668 | 2,251 | {"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} | 3.328125 | 3 | CC-MAIN-2020-05 | latest | en | 0.882099 |
https://homework.zookal.com/questions-and-answers/given-that-f--x---x-2--641188948 | 1,621,155,919,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992516.56/warc/CC-MAIN-20210516075201-20210516105201-00297.warc.gz | 326,450,324 | 25,889 | 1. Math
2. Precalculus
3. given that f x x 2 ...
# Question: given that f x x 2 ...
###### Question details
Given that f ( x ) = x 2 − 5 x and g ( x ) = x − 2 , calculate (a) ( f ∘ g ) ( x ) = (b) ( g ∘ f ) ( x ) = (c) ( f ∘ f ) ( x ) = (d) ( g ∘ g ) ( x ) = | 114 | 261 | {"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-2021-21 | latest | en | 0.657163 |
http://mathhelpforum.com/algebra/165837-log-system-equations-print.html | 1,527,192,560,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866772.91/warc/CC-MAIN-20180524190036-20180524210036-00354.warc.gz | 183,948,296 | 3,018 | # Log System of equations
• Dec 9th 2010, 03:20 PM
Gerard
Log System of equations
I'm needing a little assistance with this problem, haven't any trouble until this.
A village of 1000 inhabitants increases at a rate of 10% per year. A neighbouring village of 2000 inhabitants decreases at a rate of 5% per year. After how many years will these two villages have the same population?
So I wrote them as two exponential functions and made them equal to each other. Not sure this is how I should do it but it feels right.
\$\displaystyle 1000(1.10)^x=2000(0.95)^x\$
Then by definition \$\displaystyle C^x=y <==> log(y)=x\$
Turned them into this
\$\displaystyle log(1.1)1000=log(0.95)2000\$
then I know it can be turned into this.
\$\displaystyle log1000/log1.1=log2000/log0.95\$
And now I'm stuck, help!
• Dec 9th 2010, 03:26 PM
pickslides
Quote:
Originally Posted by Gerard
So I wrote them as two exponential functions and made them equal to each other. Not sure this is how I should do it but it feels right.
\$\displaystyle 1000(1.10)^x=2000(0.95)^x\$
To make things easier, divide both sides by 1000.
What next?
• Dec 9th 2010, 03:33 PM
Gerard
Quote:
Originally Posted by pickslides
To make things easier, divide both sides by 1000.
What next?
\$\displaystyle (1.1)^x=2(0.95)^x\$
Then using the definition...
\$\displaystyle log1/log1.1 = log2/log0.95\$
?
• Dec 10th 2010, 02:01 AM
HallsofIvy
Using what definition? Taking the logarithm of both sides (any base) of
\$\displaystyle (1.1)^x= 2(0.95)^x\$ you get
x log(1.1)= x log(.95)+ log 2 | 494 | 1,553 | {"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-2018-22 | latest | en | 0.930024 |
https://www.r-bloggers.com/treating-your-data-the-old-school-vs-tidyverse-modern-tools/ | 1,586,327,968,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371810617.95/warc/CC-MAIN-20200408041431-20200408071931-00306.warc.gz | 1,078,723,919 | 54,574 | # Treating your data: The old school vs tidyverse modern tools
August 14, 2017
By
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By Gabriel Vasconcelos
When I first started using R there was no such thing as the tidyverse. Although some of the tidyverse packages were available independently, I learned to treat my data mostly using brute force combining pieces of information I had from several sources. It is very interesting to compare this old school programming with the tidyverse writing using the magrittr package. Even if you want to stay old school, tidyverse is here to stay and it is the first tool taught in many data science courses based on R.
My objective is to show a very simple example comparing the two ways of writing. There are several ways to do what I am going to propose here, but I think this example is enough to capture the main differences between old school codes and magrittr plus tidyverse. Magrittr is not new, but It seems to me that it is more popular now because of tidyverse.
## To the example
I am going to generate a very simple data where we have two variables indexed by letters. My objective is to sum the two variables only in the values corresponding to vowels.
```set.seed(123)
M = 1000
db1 = data.frame(id = sample(letters, 1000, replace = TRUE), v1 = rnorm(1000), v2 = rnorm(1000))
vowels=c("a", "e", "i", "o", "u")
```
```## id v1 v2
## 1 h -0.60189285 -0.8209867
## 2 u -0.99369859 -0.3072572
## 3 k 1.02678506 -0.9020980
## 4 w 0.75106130 0.6270687
## 5 y -1.50916654 1.1203550
## 6 b -0.09514745 2.1272136
```
The first strategy (old school) to solve this problem is to use aggregate and then some manipulation. First I aggregate the variables to have the sum of each letter, then I select the vowels and use `colsums` to have the final result.
```ag1 = aggregate( . ~ id, data = db1, FUN = sum)
ag1 = ag1[ag1\$id %in% vowels, ]
ag1 = colSums(ag1[, -1])
ag1
```
```## v1 v2
## 26.656837 6.644839
```
The second strategy (tidyverse) uses functions from the dplyr package and the foward-pipe operator (%>%) from the magrittr. The foward-pipe allows us to do many operations in a single shot to get the final result. We do not need to create these auxiliary objects like I did in the previous example. The first two lines do precisely the same as the aggregate. The `group_by` defines the variable used to create the groups and the summarize tells R the grouping function. In the third line I select only the lines corresponding to vowels and the last summarize sums each variable. As you can see, the results are the same. This approach generated an object type called tibble, which is a special type of data frame from the tidyverse with some different features like not using factors for strings.
```library(tidyverse)
ag2 = group_by(db1, id) %>%
summarise(v1 = sum(v1), v2 = sum(v2)) %>%
filter(id %in% vowels) %>%
summarize(v1 = sum(v1), v2 = sum(v2))
ag2
```
```## # A tibble: 1 x 2
## v1 v2
##
## 1 26.65684 6.644839
```
### The same thing using merge
Suppose that we want to do the same thing as the previous example but now we are dealing with two data frames: the one from the previous example and a second data frame of characteristics that will tell us which letters are vowels.
```aux = rep("consonant",length(letters))
aux[which(letters %in% vowels)] = "vowel"
db2 = data.frame(id = letters, type = aux)
```
```## id type
## 1 a vowel
## 2 b consonant
## 3 c consonant
## 4 d consonant
## 5 e vowel
## 6 f consonant
```
The first approach uses merge to combine the two data frames and then sum the observations that have `id==vowel`.
```merge1 = merge(db1, db2, by = "id")
```
```## id v1 v2 type
## 1 a -0.73657823 1.1903106 vowel
## 2 a 0.07987382 -1.1058145 vowel
## 3 a -1.20086933 0.4859824 vowel
## 4 a 0.32040231 -0.6196151 vowel
## 5 a -0.69493683 -1.0387278 vowel
## 6 a 0.15735335 1.6165776 vowel
```
```merge1 = colSums(merge1[merge1[,4] == "vowel", 2:3])
merge1
```
```## v1 v2
## 26.656837 6.644839
```
The second approach uses the function inner_join from the dplyr package, then it filters the vowels observations and uses summarize to sum the vowels observations.
```merge2 = inner_join(db1, db2, by = "id") %>%
filter(type == "vowel") %>%
summarise(v1 = sum(v1), v2 = sum(v2))
merge2
```
```## v1 v2
## 1 26.65684 6.644839
```
As you can see, the two ways of writing are very different. Naturally, there is some cost to change from the old school to the tidyverse codes. However, the second makes your code easier to read, it is part of the tidyverse philosophy to write codes that can be read by humans. For example, something like this:
```x = 1:10
sum(log(sqrt(x)))
```
```## [1] 7.552206
```
becomes something like this if you use the foward-pipe:
```x %>% sqrt() %>% log() %>% sum()
```
```## [1] 7.552206
```
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. | 1,564 | 5,280 | {"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-2020-16 | longest | en | 0.88788 |
https://www.unitconverter.dev/volume/usliquidpint/cubicmeter/0.01/ | 1,723,513,364,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641054522.78/warc/CC-MAIN-20240813012759-20240813042759-00400.warc.gz | 801,680,396 | 18,817 | # 🧮 Volume
## US Liquid Pint to Cubic Meter
The volume conversion of 0.01 us liquid pint is ~4.7326076668244E-6 cubic meter.
US Liquid Pint
to
Cubic Meter
US Liquid Pint Cubic Meter
0.01 ~4.7326076668244E-6
0.05 ~2.3663038334122E-5
0.1 ~4.7326076668244E-5
0.25 ~0.00011831519167061
1 ~0.00047326076668244
5 ~0.0023663038334122
10 ~0.0047326076668244
20 ~0.0094652153336488
50 ~0.023663038334122
100 ~0.047326076668244
### Volume
Volume is the quantity of three-dimensional space enclosed by a closed surface, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains. Volume is often quantified numerically using the SI derived unit, the cubic metre. The volume of a container is generally understood to be the capacity of the container; i. e., the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces. Three dimensional mathematical shapes are also assigned volumes. Volumes of some simple shapes, such as regular, straight-edged, and circular shapes can be easily calculated using arithmetic formulas. Volumes of complicated shapes can be calculated with integral calculus if a formula exists for the shape's boundary. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in the three-dimensional space. | 374 | 1,387 | {"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} | 2.71875 | 3 | CC-MAIN-2024-33 | latest | en | 0.814392 |
https://www.irishtimes.com/news/science/a-life-saving-whirligig-1.2999474 | 1,713,567,242,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817455.17/warc/CC-MAIN-20240419203449-20240419233449-00509.warc.gz | 723,787,499 | 88,060 | # A life-saving whirligig
## Research shows how the complex dynamics of a simple toy can be exploited for valuable global health applications
Modern science is big: the gravitational wave detector (Ligo) cost more than a billion dollars, and the large hadron collider (LHC) in Geneva took decades to build and cost almost five billion euros. It may seem that scientific advances require enormous financial investment. So, it is refreshing to read in Nature Biomedical Engineering (Vol 1, Article 9) about the development of an ultra-cheap hand-operated centrifuge that costs only a few cents to manufacture.
The centrifuge was inspired by a simple spinning toy called a whirligig. This consists of a circular disc such as a button, with a loop of twine threaded through it. Some readers may recall making these toys from tin-lids and wool. As the string loop is pulled and relaxed, it induces successive rapid winding and unwinding phases with alternate clockwise and anticlockwise spinning of the disc. What is surprising is that, if the toy is correctly operated, spin-rates of many thousands of revolutions per minute (rpm) can be achieved.
The designers of the centrifuge, from the Department of Bioengineering, Stanford University, call their invention a "paperfuge", as the disc is made from stiff paper card. Two discs, a loop of string or nylon and two wooden or plastic handles are all that is required to make it.
Commercial centrifuges are heavy and expensive and need electric power. Egg-beaters and salad-spinners have been tried without much success as spin-rates are far below the levels required for practical use. The paperfuge is capable of spinning much more rapidly. In trials, the smallest disc, of radius 5mm, reached peak angular speeds of 125,000 rpm. The disc most suitable for applications, with a 50mm radius, reached more than 20,000 rpm, comparable to commercial centrifuges.
#### Mathematical model
The designers developed a comprehensive mathematical model of the paperfuge to analyse its performance. Newton’s law of motion states that the rate of change of angular momentum is equal to the total torque or turning effect. There are three turning effects, due to the manual input force applied by the operator, the twisted state of the string and the air-drag on the disc. These torques were parameterised, that is, approximated by bulk formulae using measurements from several experiments. Then the theoretical motion of the disc was calculated from the equation of motion. Comparison of theoretical results and experimental data showed that the mathematical model simulated the motion faithfully.
In practical use, small capillaries of whole blood are mounted radially on the paper disk, and the device is spun for a short time.
Blood analysis using conventional centrifuges is impractical in many circumstances, due to high equipment costs or to lack of electric power. The paperfuge is suitable for point-of-care medical diagnostics in under-developed regions as it does not require any electric power. It is hand-powered, and can separate whole blood samples within minutes.
This research is a good example of frugal science. It shows how the complex dynamics of a simple toy can be exploited for valuable global health applications. The elementary design means that mass production is feasible, providing a low-cost solution suitable for medical applications in resource-poor regions.
Peter Lynch is emeritus professor at the school of mathematics & statistics, University College Dublin. He blogs at thatsmaths.com. | 709 | 3,565 | {"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} | 2.765625 | 3 | CC-MAIN-2024-18 | latest | en | 0.94249 |
https://www.jiskha.com/display.cgi?id=1349029175 | 1,516,426,609,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084889325.32/warc/CC-MAIN-20180120043530-20180120063530-00598.warc.gz | 912,199,909 | 3,666 | # cal
posted by .
The height, h, of a cylinder is 3 times its radius, r. Which of the following represents the rate of change of the volume, V, of the cylinder with respect to its height, h?
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https://www.jiskha.com/display.cgi?id=1409700337 | 1,516,731,593,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00483.warc.gz | 910,669,569 | 3,956 | # math
posted by .
a ship sails 145 kms on a bearing of 215 degrees. how far south of its starting point is the ship.
• math -
^%\$*&@ math teachers think they are navigators. The "heading" is 215, NOT the "bearing". You take a bearing on something outside the ship, for example it might be the direction of a lighthouse.
south is 180
so
35 degrees west of south
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2. ### Trig
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https://www.teacherspayteachers.com/Product/Absolute-Value-Graphs-TranslationReflectionDilation-Matching-Activity-176427 | 1,495,783,641,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463608648.25/warc/CC-MAIN-20170526071051-20170526091051-00331.warc.gz | 1,187,152,982 | 22,257 | # Main Categories
Total:
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# Absolute Value Graphs-Translation,Reflection,Dilation ~ Matching Activity
Product Rating
4.0
7 ratings
File Type
PDF (Acrobat) Document File
1 MB|13 pages
Product Description
This sort (matching game) has 3 parts and 4 levels
Level 1: The graphs and equations are only translations: horizontal, vertical, or both. This level includes graphs, equation, and descriptions of the translations. Students can match equations to graphs, graphs to descriptions of the shifts, or a combination of the 3
Level 2: The graphs and equations show translations and reflections across the x-axis. This level includes graphs, equation, and descriptions of the translations.
Level 3: The graphs and equations show translations, reflections across the x-axis and dilations (stretching and shrinking). This level includes only graphs and equations.
Level 4: Uses the same graphs as Set A2 but the equations must be rewritten before graphing.
This matching sort is designed to reduce chances of getting it right by guessing. The cards are very similar. For instance, 1 card shifts to the right 4, another to left 4, another up 4, and another down 4. Students must know which direction to shift based on the equation and not because it was the only one that moved 4 units. Also I have switched some values around to make sure student understand where the shift is up or to the right (right 2 up 3 or right 3 up 2). You will have a good idea if students understand the concept after they have completed the matching exercise.
Differentiation Options
Great for differentiated instruction. As a class you may be on reflections, but some students are still struggling with translations. So when the class is doing the level 2 sort, those students can be completing the level 1 sort. As far as everyone is concerned they are completing the same assignment, they don�t realize that they are working at different levels. Having students work at their own level increases their confidence.
Possible Uses
- Review station for test
- Math Station for student that have completed their work
- Mid-Lesson Practice
- End of Lesson Check for understanding
- Alternative to homework
Keys included
Total Pages
13 pages
N/A
Teaching Duration
45 minutes
• Product Q & A
\$2.00
\$2.00 | 516 | 2,329 | {"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-2017-22 | longest | en | 0.935208 |
http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-451-principles-of-digital-communication-ii-spring-2005/lecture-notes/lecture-11-reed-solomon-codes/mnkTn0Y6GsU.srt | 1,440,703,498,000,000,000 | text/plain | crawl-data/CC-MAIN-2015-35/segments/1440644059455.0/warc/CC-MAIN-20150827025419-00078-ip-10-171-96-226.ec2.internal.warc.gz | 149,095,806 | 30,055 | 1 00:00:00,000 --> 00:00:00,390 2 00:00:00,390 --> 00:00:02,060 PROFESSOR: So OK, what did we do last time? 3 00:00:02,060 --> 00:00:05,334 Thanks for coming back by the way. 4 00:00:05,334 --> 00:00:08,870 So last time, we did MDS codes. 5 00:00:08,870 --> 00:00:15,760 6 00:00:15,760 --> 00:00:17,600 MDS codes, and we looked at this. 7 00:00:17,600 --> 00:00:21,425 And we derived some generic properties of these things if 8 00:00:21,425 --> 00:00:22,380 they existed. 9 00:00:22,380 --> 00:00:28,025 And then eventually, we moved to Reed-Solomon codes. 10 00:00:28,025 --> 00:00:30,730 11 00:00:30,730 --> 00:00:32,759 And so indeed, they exist. 12 00:00:32,759 --> 00:00:36,170 We talked a little bit about existence altogether, the main 13 00:00:36,170 --> 00:00:39,425 conjection, MDS codes, I just wrote it down. 14 00:00:39,425 --> 00:00:41,120 And like I said, you want to be rich and 15 00:00:41,120 --> 00:00:43,890 famous, solve this. 16 00:00:43,890 --> 00:00:50,050 There's people doing geometry in math that would be 17 00:00:50,050 --> 00:00:52,460 deliriously happy if you solved it. 18 00:00:52,460 --> 00:00:54,780 So if for nothing else. 19 00:00:54,780 --> 00:00:59,240 So the last thing I wrote down last time was something about 20 00:00:59,240 --> 00:01:01,800 a matrix which somebody recognized as some Fourier 21 00:01:01,800 --> 00:01:02,890 transform matrix. 22 00:01:02,890 --> 00:01:06,530 And that's where I want to pick up here, at least quickly 23 00:01:06,530 --> 00:01:08,460 make that connection. 24 00:01:08,460 --> 00:01:12,500 So let me rewrite the definition of Reed-Solomon 25 00:01:12,500 --> 00:01:14,330 codes again. 26 00:01:14,330 --> 00:01:19,870 And so a Reed-Solomon code, the way we defined it last 27 00:01:19,870 --> 00:01:22,475 time was all field elements, we would evaluate the 28 00:01:22,475 --> 00:01:24,880 polynomial in all field elements. 29 00:01:24,880 --> 00:01:26,130 Now we do it slightly different. 30 00:01:26,130 --> 00:01:29,980 31 00:01:29,980 --> 00:01:34,590 Let me make it Solomon-Stein in order to denote that 32 00:01:34,590 --> 00:01:35,320 difference. 33 00:01:35,320 --> 00:01:46,150 And let's say this is beta 0, beta n. 34 00:01:46,150 --> 00:01:59,420 35 00:01:59,420 --> 00:02:00,730 So that's the old definition. 36 00:02:00,730 --> 00:02:11,530 Only the, let's say, the beta 0, beta 1, up to beta n, they 37 00:02:11,530 --> 00:02:13,730 are the non-zero elements now. 38 00:02:13,730 --> 00:02:17,390 And just for the heck of it, so we talked about punctured 39 00:02:17,390 --> 00:02:19,360 Reed-Solomon codes last time that it would 40 00:02:19,360 --> 00:02:20,600 still be MDS codes. 41 00:02:20,600 --> 00:02:22,185 So nothing has changed there. 42 00:02:22,185 --> 00:02:25,510 In particular, all the arguments would be the same. 43 00:02:25,510 --> 00:02:32,170 In particular, just to humor me, let's say beta i is omega 44 00:02:32,170 --> 00:02:42,920 to the i, where omega is primitive in Fq. 45 00:02:42,920 --> 00:02:46,790 So what that means, you remember that primitive means 46 00:02:46,790 --> 00:02:48,460 that the powers of omega cycle through all 47 00:02:48,460 --> 00:02:51,040 the non-zero elements. 48 00:02:51,040 --> 00:02:54,760 Remember the non-zero elements are a multiplicative group. 49 00:02:54,760 --> 00:02:55,830 And it's a cyclic group. 50 00:02:55,830 --> 00:02:57,080 And it's the generator of the group. 51 00:02:57,080 --> 00:02:59,970 52 00:02:59,970 --> 00:03:01,220 Good. 53 00:03:01,220 --> 00:03:06,020 54 00:03:06,020 --> 00:03:19,180 So once we write it like this, we can actually write down the 55 00:03:19,180 --> 00:03:23,680 whole Reed-Solomon code in some sort of transfer domain 56 00:03:23,680 --> 00:03:25,855 description in the Fourier transform. 57 00:03:25,855 --> 00:03:28,710 And the way this goes is the following. 58 00:03:28,710 --> 00:03:36,000 So from now on, I will make implicit identification 59 00:03:36,000 --> 00:03:37,950 without making them. 60 00:03:37,950 --> 00:03:39,680 So this is an identification which I 61 00:03:39,680 --> 00:03:41,360 just jump around between. 62 00:03:41,360 --> 00:03:42,790 A vector -- 63 00:03:42,790 --> 00:03:49,280 so this one would be a Fqn. 64 00:03:49,280 --> 00:03:57,604 And Fx would be maybe 0 to MDS 1. 65 00:03:57,604 --> 00:04:10,370 66 00:04:10,370 --> 00:04:11,310 Let's leave it at that. 67 00:04:11,310 --> 00:04:12,080 So I make this -- 68 00:04:12,080 --> 00:04:13,820 AUDIENCE: [INAUDIBLE]. 69 00:04:13,820 --> 00:04:18,430 PROFESSOR: Yeah, let's make this identification free. 70 00:04:18,430 --> 00:04:21,570 It's just a vector of length n and a polynomial of length n. 71 00:04:21,570 --> 00:04:24,240 And so whenever I want this to have a low degree, the 72 00:04:24,240 --> 00:04:29,010 polynomial, then I just write that expression. 73 00:04:29,010 --> 00:04:32,490 So this identification we make freely now. 74 00:04:32,490 --> 00:04:36,620 And then how can we write F? 75 00:04:36,620 --> 00:04:48,820 Let F now be code word in the Reed-Solomon code. 76 00:04:48,820 --> 00:04:50,000 And then I say Fi. 77 00:04:50,000 --> 00:04:51,620 What is Fi? 78 00:04:51,620 --> 00:04:54,390 The i is element. 79 00:04:54,390 --> 00:05:03,400 Well, we know that this is F of omega to the i. 80 00:05:03,400 --> 00:05:07,200 So this is, at the moment, counted from 0. 81 00:05:07,200 --> 00:05:09,100 So this is -- 82 00:05:09,100 --> 00:05:10,730 you write it out -- 83 00:05:10,730 --> 00:05:16,820 84 00:05:16,820 --> 00:05:32,340 Fj omega to the i to the j, which is omega i j. 85 00:05:32,340 --> 00:05:33,530 And now, at this point in time, 86 00:05:33,530 --> 00:05:34,780 everybody recognizes this. 87 00:05:34,780 --> 00:05:38,000 88 00:05:38,000 --> 00:05:41,670 This would be the discrete Fourier transform of a vector. 89 00:05:41,670 --> 00:05:45,940 This is an element of order n. 90 00:05:45,940 --> 00:05:47,840 Usually, there's e to the j and then an 91 00:05:47,840 --> 00:05:48,970 element of order n. 92 00:05:48,970 --> 00:05:50,240 Now it's in the finite field. 93 00:05:50,240 --> 00:05:53,100 But what's the difference? 94 00:05:53,100 --> 00:05:55,180 This would be the Fourier transform. 95 00:05:55,180 --> 00:05:58,450 So a code word would be the Fourier transform of a vector 96 00:05:58,450 --> 00:06:03,350 F. Let's just make sure it also goes the other direction. 97 00:06:03,350 --> 00:06:12,410 So I claim that Fl would be -- so what would be the Fourier 98 00:06:12,410 --> 00:06:16,630 transform if it, indeed, is the Fourier transform? 99 00:06:16,630 --> 00:06:25,680 It is a minus, and then it goes n minus 1 100 00:06:25,680 --> 00:06:31,450 Fj omega minus jl. 101 00:06:31,450 --> 00:06:34,560 So the question is is that true? 102 00:06:34,560 --> 00:06:37,400 Because that would be the inverse. 103 00:06:37,400 --> 00:06:38,978 Is that true? 104 00:06:38,978 --> 00:06:43,750 Well, let's do what we do with proofs of this type. 105 00:06:43,750 --> 00:06:46,860 106 00:06:46,860 --> 00:06:47,860 I need that space here. 107 00:06:47,860 --> 00:06:48,286 AUDIENCE: [INAUDIBLE]. 108 00:06:48,286 --> 00:06:48,712 PROFESSOR: Sorry? 109 00:06:48,712 --> 00:06:49,990 AUDIENCE: [INAUDIBLE]. 110 00:06:49,990 --> 00:06:50,420 PROFESSOR: Yeah. 111 00:06:50,420 --> 00:06:52,540 So let's do what we do with proofs of this type. 112 00:06:52,540 --> 00:06:56,970 That means we write it out. 113 00:06:56,970 --> 00:06:59,170 That's another beautiful thing about finite fields. 114 00:06:59,170 --> 00:07:02,790 You never have to worry about convergence. 115 00:07:02,790 --> 00:07:07,310 So for the F, we just plug that in here. 116 00:07:07,310 --> 00:07:10,310 Sum, this would be -- 117 00:07:10,310 --> 00:07:11,590 what do we call it -- 118 00:07:11,590 --> 00:07:25,970 119 00:07:25,970 --> 00:07:33,920 Fi omega to the ij. 120 00:07:33,920 --> 00:07:38,860 So this would be Fj. 121 00:07:38,860 --> 00:07:42,600 And then I want omega minus jl. 122 00:07:42,600 --> 00:07:43,535 So is that true? 123 00:07:43,535 --> 00:07:45,930 Is the identity now true? 124 00:07:45,930 --> 00:07:47,420 Well, we see two sums. 125 00:07:47,420 --> 00:07:50,200 What do we do with two sums? 126 00:07:50,200 --> 00:07:52,000 If we're compelled to exchange the order, that's 127 00:07:52,000 --> 00:07:55,570 always what we do. 128 00:07:55,570 --> 00:08:02,150 So Fi goes out. 129 00:08:02,150 --> 00:08:09,900 And then you get omega to the -- 130 00:08:09,900 --> 00:08:12,160 what do we do -- 131 00:08:12,160 --> 00:08:15,290 i minus l times j. 132 00:08:15,290 --> 00:08:23,100 133 00:08:23,100 --> 00:08:25,970 We have this. 134 00:08:25,970 --> 00:08:27,450 And so what is this? 135 00:08:27,450 --> 00:08:30,480 136 00:08:30,480 --> 00:08:33,270 How can we work this out? 137 00:08:33,270 --> 00:08:35,280 Well, you know how to express the sum. 138 00:08:35,280 --> 00:08:42,280 So this part of the sum is really just omega i minus l to 139 00:08:42,280 --> 00:08:49,472 the n minus 1 minus 1. 140 00:08:49,472 --> 00:08:51,690 Well, that's a standard formula for summing 141 00:08:51,690 --> 00:08:52,840 these things up. 142 00:08:52,840 --> 00:08:57,280 But omega is an element of order n. 143 00:08:57,280 --> 00:08:58,900 Omega was an element of order n. 144 00:08:58,900 --> 00:08:59,870 That's how we chose it. 145 00:08:59,870 --> 00:09:00,750 AUDIENCE: [INAUDIBLE]. 146 00:09:00,750 --> 00:09:01,595 PROFESSOR: It's what? 147 00:09:01,595 --> 00:09:04,040 AUDIENCE: [INAUDIBLE]. 148 00:09:04,040 --> 00:09:06,680 PROFESSOR: You multiplied omega -- 149 00:09:06,680 --> 00:09:09,630 it's primitive in the field, so omega -- 150 00:09:09,630 --> 00:09:17,430 we have omega da-da-da-da-da i unequal to 1 for all i less 151 00:09:17,430 --> 00:09:22,300 than n and omega n is equal to 1. 152 00:09:22,300 --> 00:09:23,550 So OK. 153 00:09:23,550 --> 00:09:27,640 154 00:09:27,640 --> 00:09:28,570 That's fine. 155 00:09:28,570 --> 00:09:29,910 So it's an element of order n. 156 00:09:29,910 --> 00:09:31,360 That means the n-th power. 157 00:09:31,360 --> 00:09:32,610 We can interchange this. 158 00:09:32,610 --> 00:09:35,080 159 00:09:35,080 --> 00:09:39,070 So this is 1 to the some power remains 1. 160 00:09:39,070 --> 00:09:40,790 So 1 minus 1 is 0. 161 00:09:40,790 --> 00:09:41,620 The denominator -- 162 00:09:41,620 --> 00:09:44,140 well, actually, we get two results here. 163 00:09:44,140 --> 00:09:51,555 If i is unequal to l, then this is 0. 164 00:09:51,555 --> 00:09:56,720 165 00:09:56,720 --> 00:10:02,590 If i is equal to l, then this was just the sum of n once. 166 00:10:02,590 --> 00:10:05,240 So we get n here. 167 00:10:05,240 --> 00:10:05,970 i equal to 0. 168 00:10:05,970 --> 00:10:13,790 But n was q minus 1, the number of non-zero field 169 00:10:13,790 --> 00:10:17,050 elements because it was primitive in the field. 170 00:10:17,050 --> 00:10:23,990 And this one is just minus 1 in the field. 171 00:10:23,990 --> 00:10:30,190 Because q, we compute q is the power of the prime of p. 172 00:10:30,190 --> 00:10:31,220 So that would be 0. 173 00:10:31,220 --> 00:10:33,220 If we just come out as minus 1, that 174 00:10:33,220 --> 00:10:34,650 explains this minus here. 175 00:10:34,650 --> 00:10:37,000 And we can get rid of the question mark. 176 00:10:37,000 --> 00:10:39,810 177 00:10:39,810 --> 00:10:42,180 Look correct? 178 00:10:42,180 --> 00:10:42,550 All right. 179 00:10:42,550 --> 00:10:43,985 According to popular vote, it is. 180 00:10:43,985 --> 00:10:47,470 181 00:10:47,470 --> 00:10:48,740 So what do we really have here? 182 00:10:48,740 --> 00:10:51,340 We have everything we want from a Fourier transform. 183 00:10:51,340 --> 00:10:53,740 We have a Fourier transform pair. 184 00:10:53,740 --> 00:11:00,130 185 00:11:00,130 --> 00:11:04,780 We have a vector f, which we can think of as being in the 186 00:11:04,780 --> 00:11:06,030 frequency domain. 187 00:11:06,030 --> 00:11:08,800 188 00:11:08,800 --> 00:11:15,700 Then we have a vector F, which is a Fourier 189 00:11:15,700 --> 00:11:17,920 transform of that. 190 00:11:17,920 --> 00:11:31,390 This one would be in the code if the degree of this 191 00:11:31,390 --> 00:11:32,880 polynomial here would be less than k. 192 00:11:32,880 --> 00:11:36,410 193 00:11:36,410 --> 00:11:39,260 So now how can we understand Reed-Solomon codes now from an 194 00:11:39,260 --> 00:11:41,010 engineering point of view? 195 00:11:41,010 --> 00:11:43,150 Classical -- 196 00:11:43,150 --> 00:11:45,110 what about band-limited functions? 197 00:11:45,110 --> 00:11:47,936 What do we know about band-limited functions? 198 00:11:47,936 --> 00:11:54,442 Because in a sense, the degree less than k, this means that 199 00:11:54,442 --> 00:11:58,810 Fi is 0 for all i greater than k. 200 00:11:58,810 --> 00:12:01,320 201 00:12:01,320 --> 00:12:02,820 That's what it means. 202 00:12:02,820 --> 00:12:06,070 So in a sense, it's nothing but a band-limited function if 203 00:12:06,070 --> 00:12:09,060 this is a frequency domain, if you consider this as a 204 00:12:09,060 --> 00:12:10,320 frequency domain. 205 00:12:10,320 --> 00:12:14,600 So what do we know about the band-limited function? 206 00:12:14,600 --> 00:12:18,590 Well, if a function is band limited, 207 00:12:18,590 --> 00:12:20,780 it cannot be impulsive. 208 00:12:20,780 --> 00:12:22,000 We know that. 209 00:12:22,000 --> 00:12:25,940 It's a part of the frequency, and the frequency domain is 210 00:12:25,940 --> 00:12:30,520 inversely relational to the support of the function. 211 00:12:30,520 --> 00:12:32,660 So the whole idea of Reed-Solomon codes, in a 212 00:12:32,660 --> 00:12:35,540 sense, can actually be understood, at least 213 00:12:35,540 --> 00:12:37,960 intuitively, from Fourier transform and the duality 214 00:12:37,960 --> 00:12:40,162 between time and frequency. 215 00:12:40,162 --> 00:12:41,094 Yeah. 216 00:12:41,094 --> 00:12:43,430 AUDIENCE: Was q a prime number? 217 00:12:43,430 --> 00:12:47,722 PROFESSOR: q is not prime. q is a power of a prime. 218 00:12:47,722 --> 00:12:49,000 q to the n. 219 00:12:49,000 --> 00:12:50,250 AUDIENCE: [INAUDIBLE]. 220 00:12:50,250 --> 00:12:52,930 221 00:12:52,930 --> 00:12:54,840 PROFESSOR: There always is. 222 00:12:54,840 --> 00:12:55,400 Oh, sorry. 223 00:12:55,400 --> 00:12:56,350 AUDIENCE: [INAUDIBLE] 224 00:12:56,350 --> 00:12:59,820 or because for any q, there would be always a primitive? 225 00:12:59,820 --> 00:13:02,740 PROFESSOR: Why there's always a primitive element 226 00:13:02,740 --> 00:13:03,990 in the field -- 227 00:13:03,990 --> 00:13:06,350 228 00:13:06,350 --> 00:13:07,360 yeah, sorry. 229 00:13:07,360 --> 00:13:08,820 I thought you had gone through that. 230 00:13:08,820 --> 00:13:11,710 231 00:13:11,710 --> 00:13:15,370 It's because the non-zero elements in a field always 232 00:13:15,370 --> 00:13:18,110 make up a multiplicative group. 233 00:13:18,110 --> 00:13:21,640 And it's always a cyclic multiplicative group. 234 00:13:21,640 --> 00:13:25,690 So it's a cyclic multiplicative group. 235 00:13:25,690 --> 00:13:28,932 And it's just the generator of that group. 236 00:13:28,932 --> 00:13:30,700 AUDIENCE: [INAUDIBLE] it was a -- 237 00:13:30,700 --> 00:13:32,800 PROFESSOR: It's the same statement as saying, a cyclic 238 00:13:32,800 --> 00:13:34,050 group has a generator. 239 00:13:34,050 --> 00:13:38,394 240 00:13:38,394 --> 00:13:43,024 AUDIENCE: If Fq, for example, was a z, zq, 241 00:13:43,024 --> 00:13:46,680 then q has to be prime. 242 00:13:46,680 --> 00:13:47,290 PROFESSOR: Yeah. 243 00:13:47,290 --> 00:13:49,480 Yeah, in order for it to be a field, q has to be prime. 244 00:13:49,480 --> 00:13:49,905 Right. 245 00:13:49,905 --> 00:13:51,180 That's true. 246 00:13:51,180 --> 00:13:52,030 That's true. 247 00:13:52,030 --> 00:13:55,490 But we are not worried about -- 248 00:13:55,490 --> 00:13:58,150 There's also a beautiful theory [UNINTELLIGIBLE PHRASE] 249 00:13:58,150 --> 00:14:01,950 rings if this would not be a prime, very nice theory, 250 00:14:01,950 --> 00:14:03,775 becomes a little bit more technical. 251 00:14:03,775 --> 00:14:05,855 You see, more technical, then more difficult. 252 00:14:05,855 --> 00:14:09,960 253 00:14:09,960 --> 00:14:13,210 I just wanted to give you this correspondence on the Fourier 254 00:14:13,210 --> 00:14:16,170 transform because it's, from an engineering point of view, 255 00:14:16,170 --> 00:14:19,200 a very nice insight. 256 00:14:19,200 --> 00:14:21,510 One of the reasons one can understand that Reed-Solomon 257 00:14:21,510 --> 00:14:24,220 codes have a good minimum distance is because it has a 258 00:14:24,220 --> 00:14:25,790 transform of a band-limited function. 259 00:14:25,790 --> 00:14:29,890 260 00:14:29,890 --> 00:14:31,170 So that's what I wanted to say about the 261 00:14:31,170 --> 00:14:33,240 Fourier transforms here. 262 00:14:33,240 --> 00:14:34,820 I do not want to spend too much time. 263 00:14:34,820 --> 00:14:36,145 I have to get through decoding here. 264 00:14:36,145 --> 00:14:40,280 265 00:14:40,280 --> 00:14:42,820 Since we have now this Fourier transform correspondence, we 266 00:14:42,820 --> 00:14:43,740 can do another thing. 267 00:14:43,740 --> 00:15:00,720 We can, namely, say, well, F is in the code if and only if 268 00:15:00,720 --> 00:15:02,420 the Fourier transform -- 269 00:15:02,420 --> 00:15:05,740 so that was the back transform here. 270 00:15:05,740 --> 00:15:09,340 271 00:15:09,340 --> 00:15:11,480 Also now, the code word, we can interpret the code word as 272 00:15:11,480 --> 00:15:12,180 a polynomial. 273 00:15:12,180 --> 00:15:15,460 And the inverse transform is just evaluating 274 00:15:15,460 --> 00:15:19,980 it at an omega again. 275 00:15:19,980 --> 00:15:26,010 F is in the code if and only the transform of a code word 276 00:15:26,010 --> 00:15:34,385 is 0 for all i greater than k up to n minus 1. 277 00:15:34,385 --> 00:15:34,850 Yeah. 278 00:15:34,850 --> 00:15:36,100 AUDIENCE: [INAUDIBLE]. 279 00:15:36,100 --> 00:15:38,790 280 00:15:38,790 --> 00:15:41,375 PROFESSOR: So it basically just means the support. 281 00:15:41,375 --> 00:15:46,400 282 00:15:46,400 --> 00:15:49,650 If this one is band limited, then this one cannot have 283 00:15:49,650 --> 00:15:51,610 arbitrarily small support. 284 00:15:51,610 --> 00:15:55,110 That's what it means. 285 00:15:55,110 --> 00:15:56,460 That's really all I wanted to say. 286 00:15:56,460 --> 00:16:02,510 287 00:16:02,510 --> 00:16:09,620 i greater than k and less than n. 288 00:16:09,620 --> 00:16:15,470 So we have that F is in the code if and only the 289 00:16:15,470 --> 00:16:20,700 polynomial evaluates to 0 at a bunch of points. 290 00:16:20,700 --> 00:16:23,060 Now that leads to something interesting. 291 00:16:23,060 --> 00:16:28,700 Because now we can say, well, so what if we've just 292 00:16:28,700 --> 00:16:34,860 construct our set of vectors F so that they evaluate to 0 293 00:16:34,860 --> 00:16:37,190 somewhere where we want them to zero. 294 00:16:37,190 --> 00:16:40,630 We construct them somewhat differently than here. 295 00:16:40,630 --> 00:16:48,250 And in particular, let's define g as a product of i 296 00:16:48,250 --> 00:16:58,560 equal k up to n minus 1 of x minus omega minus i. 297 00:16:58,560 --> 00:17:00,950 So it's a polynomial. 298 00:17:00,950 --> 00:17:07,369 This is a polynomial which satisfies this. 299 00:17:07,369 --> 00:17:16,210 So in particular, we have g of omega minus i is equal to 0 300 00:17:16,210 --> 00:17:20,481 for all i less than n less than -- 301 00:17:20,481 --> 00:17:21,864 I'm sorry -- 302 00:17:21,864 --> 00:17:23,250 and greater than k. 303 00:17:23,250 --> 00:17:26,540 304 00:17:26,540 --> 00:17:30,700 So this is a code word, right? 305 00:17:30,700 --> 00:17:33,116 We are agreed this is a code word? 306 00:17:33,116 --> 00:17:33,940 Good. 307 00:17:33,940 --> 00:17:36,330 So we found a code word without going through this 308 00:17:36,330 --> 00:17:40,020 here, without going through the evaluation map. 309 00:17:40,020 --> 00:17:42,000 We found it straight away. 310 00:17:42,000 --> 00:17:44,210 But not only did we find this. 311 00:17:44,210 --> 00:17:48,470 This has degree n minus k. 312 00:17:48,470 --> 00:17:51,190 This has degree n minus k. 313 00:17:51,190 --> 00:17:53,990 So maybe there's another interpretation. 314 00:17:53,990 --> 00:17:59,000 Now we can give a whole new definition of the Reed-Solomon 315 00:17:59,000 --> 00:18:08,110 code, namely, as a set of polynomials, which are sort of 316 00:18:08,110 --> 00:18:17,490 implicitly identified with a set of vectors such that times 317 00:18:17,490 --> 00:18:30,830 this g of x and h of x is less then k again. 318 00:18:30,830 --> 00:18:35,330 So what we have here, it's again, you take a vector space 319 00:18:35,330 --> 00:18:38,050 of functions the h. 320 00:18:38,050 --> 00:18:40,800 It's a k dimensional vector space. 321 00:18:40,800 --> 00:18:43,426 And we multiplied with this g. 322 00:18:43,426 --> 00:18:46,280 It's again a linear map, this multiplication with a fixed 323 00:18:46,280 --> 00:18:47,530 polynomial. 324 00:18:47,530 --> 00:18:49,220 325 00:18:49,220 --> 00:18:52,340 A linear map of this vector space gives us, again, a 326 00:18:52,340 --> 00:18:54,810 vector space of dimension k. 327 00:18:54,810 --> 00:19:01,260 And all elements in this vector space evaluate to 0 for 328 00:19:01,260 --> 00:19:04,270 all omega to the minus i. 329 00:19:04,270 --> 00:19:06,140 That's just the same again, just the 330 00:19:06,140 --> 00:19:07,570 Reed-Solomon code again. 331 00:19:07,570 --> 00:19:12,140 So it's a very nice complementary description to 332 00:19:12,140 --> 00:19:17,210 the same Reed-Solomon code we could define up here, we have 333 00:19:17,210 --> 00:19:18,665 found again down here. 334 00:19:18,665 --> 00:19:23,300 335 00:19:23,300 --> 00:19:24,550 What is so nice about this? 336 00:19:24,550 --> 00:19:30,350 337 00:19:30,350 --> 00:19:33,830 Yeah, what do you think is so nice about this? 338 00:19:33,830 --> 00:19:35,080 About this description? 339 00:19:35,080 --> 00:19:39,960 340 00:19:39,960 --> 00:19:42,344 So now you have to think as engineers. 341 00:19:42,344 --> 00:19:43,710 AUDIENCE: You can still evaluate it. 342 00:19:43,710 --> 00:19:47,930 There, we have to evaluate it at end points of function. 343 00:19:47,930 --> 00:19:48,930 PROFESSOR: Yeah, exactly. 344 00:19:48,930 --> 00:19:51,710 This is so easy to implement. 345 00:19:51,710 --> 00:19:55,730 So now we think a little bit about encoder. 346 00:19:55,730 --> 00:20:04,320 An encoder for a Reed-Solomon code. 347 00:20:04,320 --> 00:20:05,570 How would we do that? 348 00:20:05,570 --> 00:20:07,710 349 00:20:07,710 --> 00:20:08,760 You look at this here. 350 00:20:08,760 --> 00:20:09,950 Hey, let's do this. 351 00:20:09,950 --> 00:20:11,550 Let's exactly do this. 352 00:20:11,550 --> 00:20:18,810 Meaning in VLSI, we do something like -- 353 00:20:18,810 --> 00:20:36,720 354 00:20:36,720 --> 00:20:38,450 and here, you would write -- and this is 355 00:20:38,450 --> 00:20:41,700 just a delay element. 356 00:20:41,700 --> 00:20:48,290 This one would be g n minus k. 357 00:20:48,290 --> 00:20:49,540 This one -- 358 00:20:49,540 --> 00:20:54,150 359 00:20:54,150 --> 00:20:56,120 so these would be the coefficients. 360 00:20:56,120 --> 00:20:58,990 This polynomial, we can evaluate once and for all. 361 00:20:58,990 --> 00:21:01,790 We do that once before we start our communication. 362 00:21:01,790 --> 00:21:04,040 We evaluate that polynomial once and for all. 363 00:21:04,040 --> 00:21:06,050 We know these coefficients. 364 00:21:06,050 --> 00:21:10,770 So these are multipliers which multiply with those 365 00:21:10,770 --> 00:21:12,060 corresponding coefficients. 366 00:21:12,060 --> 00:21:19,450 And we feed into here the coefficients of this 367 00:21:19,450 --> 00:21:20,160 polynomial. 368 00:21:20,160 --> 00:21:22,430 This is our information symbols, which we simply feed 369 00:21:22,430 --> 00:21:25,930 into this circuitry. 370 00:21:25,930 --> 00:21:30,410 And into this circuit, out comes F, out comes the 371 00:21:30,410 --> 00:21:32,290 coefficient. 372 00:21:32,290 --> 00:21:34,300 Polynomial multiplication. 373 00:21:34,300 --> 00:21:37,200 If you show that to a VLSI designer, they 374 00:21:37,200 --> 00:21:39,330 are deliriously happy. 375 00:21:39,330 --> 00:21:40,470 They say, you know what? 376 00:21:40,470 --> 00:21:41,690 I implement you this thing. 377 00:21:41,690 --> 00:21:45,940 I implement you in 5,000 gates, which is close to 378 00:21:45,940 --> 00:21:47,190 nothing nowadays. 379 00:21:47,190 --> 00:21:49,760 380 00:21:49,760 --> 00:21:50,240 Maybe not. 381 00:21:50,240 --> 00:21:53,690 Maybe 10,000. 382 00:21:53,690 --> 00:21:56,500 And there you get to the second reason that 383 00:21:56,500 --> 00:21:58,680 Reed-Solomon codes are so extremely 384 00:21:58,680 --> 00:22:00,540 important in practice. 385 00:22:00,540 --> 00:22:04,660 On the one hand, they are MDS codes, meaning they are about 386 00:22:04,660 --> 00:22:07,820 as good as it gets. 387 00:22:07,820 --> 00:22:10,615 And on the other hand, they are algorithms. 388 00:22:10,615 --> 00:22:15,725 They are circuitry for these things whose cost is close to 389 00:22:15,725 --> 00:22:18,970 nothing at least today. 390 00:22:18,970 --> 00:22:21,770 When they were invented, that was quite different. 391 00:22:21,770 --> 00:22:25,250 In the '60s, that would have not been implementable with 392 00:22:25,250 --> 00:22:27,680 transistors on a board or something like that. 393 00:22:27,680 --> 00:22:29,920 But today, the cost is close to nothing. 394 00:22:29,920 --> 00:22:32,380 So the whole algorithmic treatment of Reed-Solomon 395 00:22:32,380 --> 00:22:36,000 codes is very well developed. 396 00:22:36,000 --> 00:22:37,570 The encoder you could do like this. 397 00:22:37,570 --> 00:22:38,886 Yeah. 398 00:22:38,886 --> 00:22:40,983 AUDIENCE: Question about why do you still require that 399 00:22:40,983 --> 00:22:43,546 degree [INAUDIBLE]? 400 00:22:43,546 --> 00:22:45,095 PROFESSOR: You don't need to do that. 401 00:22:45,095 --> 00:22:47,160 But then the mapping is not one-to-one anymore. 402 00:22:47,160 --> 00:22:50,580 403 00:22:50,580 --> 00:22:52,550 Then the mapping is not one-to-one anymore. 404 00:22:52,550 --> 00:22:56,120 405 00:22:56,120 --> 00:22:58,170 Let's put it other ways. 406 00:22:58,170 --> 00:23:00,280 You want code words of length n. 407 00:23:00,280 --> 00:23:03,590 If the degree is larger, you run over. 408 00:23:03,590 --> 00:23:06,250 In order to still get a code word, then you have to take it 409 00:23:06,250 --> 00:23:08,840 modular x to the n minus 1. 410 00:23:08,840 --> 00:23:11,340 And that would fold the coefficients back. 411 00:23:11,340 --> 00:23:13,890 And still, it gives you a valid code word then, but this 412 00:23:13,890 --> 00:23:16,900 is not one-to-one anymore. 413 00:23:16,900 --> 00:23:19,980 Anyway, the short answer is if you allow more, then they 414 00:23:19,980 --> 00:23:22,160 become longer than n. 415 00:23:22,160 --> 00:23:25,410 416 00:23:25,410 --> 00:23:29,860 So what's the time? 417 00:23:29,860 --> 00:23:31,050 OK. 418 00:23:31,050 --> 00:23:32,880 So this is a nice encoder. 419 00:23:32,880 --> 00:23:39,730 There even is a nicer encoder, which I just want to give the 420 00:23:39,730 --> 00:23:41,950 formula for. 421 00:23:41,950 --> 00:23:44,540 The one reason that people still have a problem with this 422 00:23:44,540 --> 00:23:46,860 is it's not systematic. 423 00:23:46,860 --> 00:23:52,185 People like to see the symbols in the code words themselves. 424 00:23:52,185 --> 00:23:54,226 They want the systematic part. 425 00:23:54,226 --> 00:23:56,890 How could we achieve that? 426 00:23:56,890 --> 00:24:00,150 Well, we go from the same description here. 427 00:24:00,150 --> 00:24:06,530 We say, well, let h of x be given. 428 00:24:06,530 --> 00:24:09,970 429 00:24:09,970 --> 00:24:23,060 Then compute x to the n minus k times h of x modular g of x. 430 00:24:23,060 --> 00:24:28,910 So we divide this polynomial by this polynomial g -- 431 00:24:28,910 --> 00:24:29,860 it has a name. 432 00:24:29,860 --> 00:24:32,240 It's called the generator polynomial. 433 00:24:32,240 --> 00:24:34,790 We divide it by g. 434 00:24:34,790 --> 00:24:43,960 And out comes some polynomial r of x of some low degree, 435 00:24:43,960 --> 00:24:45,210 degree less than g. 436 00:24:45,210 --> 00:24:47,960 437 00:24:47,960 --> 00:24:56,900 And degree r of x less than n minus k in particular. 438 00:24:56,900 --> 00:24:59,100 So and then we can form a code word. 439 00:24:59,100 --> 00:25:05,110 I claim F of x, no, F is -- 440 00:25:05,110 --> 00:25:12,040 and here we write r, and here we write h. 441 00:25:12,040 --> 00:25:15,220 The coefficient vector of h and the 442 00:25:15,220 --> 00:25:17,870 coefficient vector of r. 443 00:25:17,870 --> 00:25:20,870 Why is that a code word? 444 00:25:20,870 --> 00:25:21,850 Why is that a code word? 445 00:25:21,850 --> 00:25:23,190 Maybe minus here. 446 00:25:23,190 --> 00:25:25,270 I'm always thinking characteristic 2 anyway. 447 00:25:25,270 --> 00:25:28,180 So why is this a code word? 448 00:25:28,180 --> 00:25:29,430 Anybody, it's clear? 449 00:25:29,430 --> 00:25:32,534 450 00:25:32,534 --> 00:25:42,970 What this is in terms of F of x is r of x plus h of x. 451 00:25:42,970 --> 00:25:45,545 452 00:25:45,545 --> 00:25:48,000 Oh, minus r of x. 453 00:25:48,000 --> 00:25:50,240 That's because we wrote it like this. 454 00:25:50,240 --> 00:25:59,870 If we now take the F of x modular g of x, well, this 455 00:25:59,870 --> 00:26:02,400 part has degree low. 456 00:26:02,400 --> 00:26:05,130 It's not affected by this modular operation. 457 00:26:05,130 --> 00:26:10,820 This would be minus r of x plus this one, modular g of x, 458 00:26:10,820 --> 00:26:13,940 which is r of x. 459 00:26:13,940 --> 00:26:16,860 So the whole thing is 0. 460 00:26:16,860 --> 00:26:22,000 If this is 0, that means g of x is a factor of F. Hence, 461 00:26:22,000 --> 00:26:23,930 it's a code word. 462 00:26:23,930 --> 00:26:25,650 So we have a code word here. 463 00:26:25,650 --> 00:26:32,150 And in the code word, pop up our symbols right away, our 464 00:26:32,150 --> 00:26:33,960 encoded information symbols right away. 465 00:26:33,960 --> 00:26:37,770 So we get some nice systematic encoding going. 466 00:26:37,770 --> 00:26:40,500 This division circuit by g is pretty much the 467 00:26:40,500 --> 00:26:41,790 same size as this. 468 00:26:41,790 --> 00:26:45,060 It's not larger at all. 469 00:26:45,060 --> 00:26:46,790 So there, we get beautiful algorithms. 470 00:26:46,790 --> 00:26:48,736 This is actually what's implemented. 471 00:26:48,736 --> 00:26:51,610 If we go to any disc drive [UNINTELLIGIBLE], that is 472 00:26:51,610 --> 00:26:56,610 usually what is implemented in there, exactly this. 473 00:26:56,610 --> 00:26:57,990 Algorithms. 474 00:26:57,990 --> 00:27:00,850 Algorithmic treatment of Reed-Solomon codes. 475 00:27:00,850 --> 00:27:03,590 Do you have any questions about any of this? 476 00:27:03,590 --> 00:27:04,015 Yeah. 477 00:27:04,015 --> 00:27:05,370 AUDIENCE: I have a question about do any these 478 00:27:05,370 --> 00:27:09,555 Reed-Solomon codes map [INAUDIBLE]? 479 00:27:09,555 --> 00:27:11,990 PROFESSOR: They are very costly map. 480 00:27:11,990 --> 00:27:14,940 Usually, it depends a lot on the application. 481 00:27:14,940 --> 00:27:17,180 If you think disc drives, [UNINTELLIGIBLE] as just being 482 00:27:17,180 --> 00:27:21,330 mapped, you take Reed-Solomon codes over a characteristic 2. 483 00:27:21,330 --> 00:27:22,760 Then each field element is 484 00:27:22,760 --> 00:27:25,840 represented as a binary vector. 485 00:27:25,840 --> 00:27:30,460 And that binary vector is mapped into on-off keying. 486 00:27:30,460 --> 00:27:32,230 Straight off the bat. 487 00:27:32,230 --> 00:27:34,560 Nothing more fancy. 488 00:27:34,560 --> 00:27:40,090 That is for disc drives. 489 00:27:40,090 --> 00:27:42,710 And as you do this, the satellite standard, where it's 490 00:27:42,710 --> 00:27:48,510 mapped onto the 256-QAM field elements [UNINTELLIGIBLE]. 491 00:27:48,510 --> 00:27:51,490 So it's many different ways. 492 00:27:51,490 --> 00:27:53,328 Many different ways. 493 00:27:53,328 --> 00:27:55,640 AUDIENCE: But to prove some performance [UNINTELLIGIBLE] 494 00:27:55,640 --> 00:27:58,980 in the [UNINTELLIGIBLE], we need to have mappings, right? 495 00:27:58,980 --> 00:28:01,030 PROFESSOR: In order to prove performance mode, we need to 496 00:28:01,030 --> 00:28:02,110 have mappings. 497 00:28:02,110 --> 00:28:05,700 And the Hamming distance bounds that we get from here 498 00:28:05,700 --> 00:28:09,060 give you bounds on the minimum Euclidean distance. 499 00:28:09,060 --> 00:28:12,400 If these are good bounds or not depends a lot on the 500 00:28:12,400 --> 00:28:14,150 modulation scheme. 501 00:28:14,150 --> 00:28:17,620 And to be perfectly honest, they usually are not. 502 00:28:17,620 --> 00:28:21,850 They usually are not very good, the bounds. 503 00:28:21,850 --> 00:28:26,350 But it's a very difficult problem to design a code or to 504 00:28:26,350 --> 00:28:31,010 find a representation of the fields that maps nicely onto a 505 00:28:31,010 --> 00:28:31,965 modulation scheme. 506 00:28:31,965 --> 00:28:34,725 Very difficult problem. 507 00:28:34,725 --> 00:28:37,210 AUDIENCE: How do we know how to think that 508 00:28:37,210 --> 00:28:38,210 this is a good code? 509 00:28:38,210 --> 00:28:42,030 PROFESSOR: The code itself is excellent in terms of MDS, the 510 00:28:42,030 --> 00:28:43,010 MDS property. 511 00:28:43,010 --> 00:28:45,070 AUDIENCE: But why does MDS mean good codes? 512 00:28:45,070 --> 00:28:46,696 PROFESSOR: In respect to modulation? 513 00:28:46,696 --> 00:28:47,580 AUDIENCE: Yes. 514 00:28:47,580 --> 00:28:49,490 PROFESSOR: It doesn't. 515 00:28:49,490 --> 00:28:52,374 It doesn't. 516 00:28:52,374 --> 00:28:55,390 It's a bit like this. 517 00:28:55,390 --> 00:28:59,080 It's really not easy to define codes in Euclidean space. 518 00:28:59,080 --> 00:29:07,170 So all that we do is we find ways to do that and guarantee 519 00:29:07,170 --> 00:29:12,120 some performance, some sort of performance. 520 00:29:12,120 --> 00:29:15,400 It's not easy to spread out -- 521 00:29:15,400 --> 00:29:15,910 I don't know -- 522 00:29:15,910 --> 00:29:24,080 2 to the 1,000 points in 1,500 dimensions. 523 00:29:24,080 --> 00:29:25,380 These would be typical numbers really. 524 00:29:25,380 --> 00:29:29,090 525 00:29:29,090 --> 00:29:31,240 It's not easy. 526 00:29:31,240 --> 00:29:35,350 And since that problem is practically daunting, it's a 527 00:29:35,350 --> 00:29:39,530 daunting task, you have to develop all sort of crutches 528 00:29:39,530 --> 00:29:40,210 to do that. 529 00:29:40,210 --> 00:29:42,290 And this is really one. 530 00:29:42,290 --> 00:29:46,620 So how to code MDS codes playing a 531 00:29:46,620 --> 00:29:48,120 part in this mapping. 532 00:29:48,120 --> 00:29:52,230 If you want to be more fancy about that, then you put a 533 00:29:52,230 --> 00:29:55,440 convolutional code or some other code also in there and 534 00:29:55,440 --> 00:29:56,690 do a combine scheme. 535 00:29:56,690 --> 00:29:59,670 536 00:29:59,670 --> 00:30:02,910 I think Professor Forney will talk about that more. 537 00:30:02,910 --> 00:30:05,060 So here, it's just the coding theoretic 538 00:30:05,060 --> 00:30:07,550 groundwork of these things. 539 00:30:07,550 --> 00:30:10,610 540 00:30:10,610 --> 00:30:11,090 Anything else? 541 00:30:11,090 --> 00:30:11,565 Yeah. 542 00:30:11,565 --> 00:30:12,900 AUDIENCE: [INAUDIBLE] 543 00:30:12,900 --> 00:30:15,390 of the [INAUDIBLE] n minus 1. 544 00:30:15,390 --> 00:30:17,507 Close, very close [INAUDIBLE]. 545 00:30:17,507 --> 00:30:18,757 PROFESSOR: Yeah, thanks. 546 00:30:18,757 --> 00:30:24,050 547 00:30:24,050 --> 00:30:28,780 So the algorithmic treatment of Reed-Solomon codes is 548 00:30:28,780 --> 00:30:29,980 extremely elegant. 549 00:30:29,980 --> 00:30:34,675 And that's the second main reason they are so much used. 550 00:30:34,675 --> 00:30:37,990 It doesn't cost so much to implement them. 551 00:30:37,990 --> 00:30:40,210 Well, at least we've seen that for the encoder. 552 00:30:40,210 --> 00:30:42,130 That's a fairly small circuit. 553 00:30:42,130 --> 00:30:43,380 So what about decoding? 554 00:30:43,380 --> 00:30:48,950 555 00:30:48,950 --> 00:30:51,650 Decoding. 556 00:30:51,650 --> 00:30:53,350 How do we decode these things? 557 00:30:53,350 --> 00:30:57,990 558 00:30:57,990 --> 00:31:00,400 How could we possibly decode them? 559 00:31:00,400 --> 00:31:02,390 And I give you -- 560 00:31:02,390 --> 00:31:04,840 AUDIENCE: Fourier? 561 00:31:04,840 --> 00:31:05,660 PROFESSOR: Right, that's true. 562 00:31:05,660 --> 00:31:06,930 We could do the Fourier transform. 563 00:31:06,930 --> 00:31:13,026 But it doesn't help us so much because we receive something. 564 00:31:13,026 --> 00:31:16,540 And we receive a vector, say, y. 565 00:31:16,540 --> 00:31:20,962 566 00:31:20,962 --> 00:31:23,323 From now on, let's say x is a code word. 567 00:31:23,323 --> 00:31:27,760 568 00:31:27,760 --> 00:31:28,750 rs, right. 569 00:31:28,750 --> 00:31:32,590 So x, it's usually a code word from now on. 570 00:31:32,590 --> 00:31:37,190 So this is a code word plus an error. 571 00:31:37,190 --> 00:31:39,740 So in particular, if we take the Fourier transform, we take 572 00:31:39,740 --> 00:31:42,110 the Fourier transform of the code word, which is fine. 573 00:31:42,110 --> 00:31:43,590 But then we get the Fourier transform of the error. 574 00:31:43,590 --> 00:31:46,780 So that destroys all the fun. 575 00:31:46,780 --> 00:31:49,370 576 00:31:49,370 --> 00:31:50,110 What else could we do? 577 00:31:50,110 --> 00:31:54,250 Here's typical parameters of a code. 578 00:31:54,250 --> 00:32:01,740 255, 239, 256. 579 00:32:01,740 --> 00:32:04,100 And you immediately see that, OK, any sort of 580 00:32:04,100 --> 00:32:05,350 group force is out. 581 00:32:05,350 --> 00:32:08,820 582 00:32:08,820 --> 00:32:10,490 How many code words do we have here? 583 00:32:10,490 --> 00:32:13,622 We have 8 to the 239 code words. 584 00:32:13,622 --> 00:32:16,042 Now you don't want to search that. 585 00:32:16,042 --> 00:32:18,386 You definitely don't want to search that. 586 00:32:18,386 --> 00:32:22,690 587 00:32:22,690 --> 00:32:25,536 So how could we possibly decode these things? 588 00:32:25,536 --> 00:32:28,790 589 00:32:28,790 --> 00:32:32,070 Turns out, to decode them in some sort of optimal fashion, 590 00:32:32,070 --> 00:32:33,160 maximum likelihood [INAUDIBLE] 591 00:32:33,160 --> 00:32:37,220 actually, it's an NP-hard problem. 592 00:32:37,220 --> 00:32:41,140 Maybe last year, actually, it has been shown that it's an 593 00:32:41,140 --> 00:32:44,190 NP-hard problem to decode Reed-Solomon codes. 594 00:32:44,190 --> 00:32:47,110 It was known that decoding in general was NP-hard. 595 00:32:47,110 --> 00:32:51,930 But this is now the constraint to Reed-Solomon 596 00:32:51,930 --> 00:32:55,200 codes is still hard. 597 00:32:55,200 --> 00:32:56,940 So what do we do? 598 00:32:56,940 --> 00:32:59,640 Yes, OK, what do we do? 599 00:32:59,640 --> 00:33:01,800 Let's say x is a code word. 600 00:33:01,800 --> 00:33:05,170 We know that. 601 00:33:05,170 --> 00:33:09,560 And set rate e, let's just call it t. 602 00:33:09,560 --> 00:33:16,680 603 00:33:16,680 --> 00:33:23,390 So what happens if you don't have an error? 604 00:33:23,390 --> 00:33:24,640 Just thought experiment. 605 00:33:24,640 --> 00:33:28,630 606 00:33:28,630 --> 00:33:34,520 Thought experiment, if you don't have an error, then yi 607 00:33:34,520 --> 00:33:46,640 in all the positions is equal to F of xi for 608 00:33:46,640 --> 00:33:49,500 some F of x of degree. 609 00:33:49,500 --> 00:33:59,000 610 00:33:59,000 --> 00:34:02,520 In particular, since we know the x's, we know 611 00:34:02,520 --> 00:34:03,770 the positions -- 612 00:34:03,770 --> 00:34:09,440 613 00:34:09,440 --> 00:34:14,600 sorry, oh, sorry. 614 00:34:14,600 --> 00:34:17,050 That's not what I wanted to write. 615 00:34:17,050 --> 00:34:18,864 This is definitely not what I wanted to write. 616 00:34:18,864 --> 00:34:21,820 617 00:34:21,820 --> 00:34:38,870 Let's keep that as c as a code word and position i in c is 618 00:34:38,870 --> 00:34:49,679 associated with xi in the fields. 619 00:34:49,679 --> 00:34:57,340 So meaning ci would be F of xi. 620 00:34:57,340 --> 00:34:59,310 That's really what I wanted to write. 621 00:34:59,310 --> 00:35:01,810 It makes more sense. 622 00:35:01,810 --> 00:35:03,710 So thought experiment. 623 00:35:03,710 --> 00:35:04,960 No errors. 624 00:35:04,960 --> 00:35:07,080 625 00:35:07,080 --> 00:35:16,170 Then yi is F of xi for some F. If they ran no errors, then we 626 00:35:16,170 --> 00:35:21,900 could just solve this linear system of equations to find 627 00:35:21,900 --> 00:35:26,360 the coefficient of F. And the coefficient of F, say, were 628 00:35:26,360 --> 00:35:28,286 our information circuits. 629 00:35:28,286 --> 00:35:32,100 Is it clear that this is a linear system of equations? 630 00:35:32,100 --> 00:35:34,380 Yeah? 631 00:35:34,380 --> 00:35:40,680 You could write it out as y0, y1, y2 equal to -- 632 00:35:40,680 --> 00:35:56,490 633 00:35:56,490 --> 00:36:01,260 and here we have f0, f1, up to fk-1. 634 00:36:01,260 --> 00:36:03,110 This is the linear system of equations. 635 00:36:03,110 --> 00:36:04,510 We know the xi's. 636 00:36:04,510 --> 00:36:05,960 This is a linear system of equation we have to solve. 637 00:36:05,960 --> 00:36:11,170 638 00:36:11,170 --> 00:36:13,560 If there are no errors, then life is easy. 639 00:36:13,560 --> 00:36:15,360 That seems to be reasonable. 640 00:36:15,360 --> 00:36:17,820 So what happens if we do have errors? 641 00:36:17,820 --> 00:36:23,810 Somehow, we have to make sure that the errors that we get do 642 00:36:23,810 --> 00:36:27,590 not cause any problem for us. 643 00:36:27,590 --> 00:36:33,950 And then we do something very ingenious. 644 00:36:33,950 --> 00:36:37,825 We define something called an error locator which is a 645 00:36:37,825 --> 00:36:55,470 polynomial x such that x minus xi. 646 00:36:55,470 --> 00:37:01,400 So it's a polynomial which is 0 in all error positions. 647 00:37:01,400 --> 00:37:04,700 648 00:37:04,700 --> 00:37:07,270 Well, you might say, we do not know the error positions. 649 00:37:07,270 --> 00:37:09,750 Well, OK, that's true. 650 00:37:09,750 --> 00:37:12,480 Basically, this is, in the end, what we want to find, 651 00:37:12,480 --> 00:37:13,150 this polynomial. 652 00:37:13,150 --> 00:37:14,835 But nonetheless, this polynomial exists. 653 00:37:14,835 --> 00:37:18,080 654 00:37:18,080 --> 00:37:25,990 We can cast, actually, the coding problem -- 655 00:37:25,990 --> 00:37:29,170 this is form 1 s. 656 00:37:29,170 --> 00:37:32,695 657 00:37:32,695 --> 00:37:35,032 Yes, what? 658 00:37:35,032 --> 00:37:36,282 AUDIENCE: [INAUDIBLE PHRASE]. 659 00:37:36,282 --> 00:37:42,130 660 00:37:42,130 --> 00:37:43,480 PROFESSOR: Yeah, it's an additive error model. 661 00:37:43,480 --> 00:37:46,510 662 00:37:46,510 --> 00:37:48,330 But you can cast pretty much anything in the 663 00:37:48,330 --> 00:37:49,140 [UNINTELLIGIBLE]. 664 00:37:49,140 --> 00:37:52,440 If a position is altered, you can always model that as if 665 00:37:52,440 --> 00:37:53,754 something was added to it. 666 00:37:53,754 --> 00:37:56,480 667 00:37:56,480 --> 00:38:09,670 Decoding problem one is find lambda of x of minimal degree 668 00:38:09,670 --> 00:38:30,800 such that lambda of xi, is 0 for all xi and 669 00:38:30,800 --> 00:38:35,230 degree f less than k. 670 00:38:35,230 --> 00:38:41,750 So I claim if you solve this problem, namely, this is a 671 00:38:41,750 --> 00:38:44,570 problem I give you. 672 00:38:44,570 --> 00:38:47,000 I give you vector y. 673 00:38:47,000 --> 00:38:48,335 I give you vector y. 674 00:38:48,335 --> 00:38:48,890 Here it is. 675 00:38:48,890 --> 00:38:50,380 Here is vector y. 676 00:38:50,380 --> 00:38:53,500 And I give you vector x which corresponds to the field 677 00:38:53,500 --> 00:38:56,310 elements where you evaluated that in order to 678 00:38:56,310 --> 00:38:57,540 get the code word. 679 00:38:57,540 --> 00:39:02,760 And then I said, given y and x, find two polynomials lambda 680 00:39:02,760 --> 00:39:09,180 and f such that f has maximum degree k and lambda has 681 00:39:09,180 --> 00:39:12,730 minimum degree, the smaller degree possible so 682 00:39:12,730 --> 00:39:15,360 that this is true. 683 00:39:15,360 --> 00:39:19,140 And I claim this solves the decoding problem. 684 00:39:19,140 --> 00:39:22,340 This would solve the decoding problem because once we have 685 00:39:22,340 --> 00:39:24,910 found this, then we can take lambda to 686 00:39:24,910 --> 00:39:27,140 be the error locator. 687 00:39:27,140 --> 00:39:32,480 And we can basically read off the information 688 00:39:32,480 --> 00:39:35,180 symbols from the f. 689 00:39:35,180 --> 00:39:38,810 So this decoding formulation now brings it, at least, into 690 00:39:38,810 --> 00:39:42,780 the algebraic ground, brings the whole decoding problem, 691 00:39:42,780 --> 00:39:45,072 makes it something algebraically. 692 00:39:45,072 --> 00:39:49,220 But now the question becomes, is that easy? 693 00:39:49,220 --> 00:39:50,890 Or can we do this? 694 00:39:50,890 --> 00:39:52,140 This problem here. 695 00:39:52,140 --> 00:39:54,440 696 00:39:54,440 --> 00:39:56,670 Do you see any hope for solving this problem? 697 00:39:56,670 --> 00:40:02,270 698 00:40:02,270 --> 00:40:05,290 I guess the only answer that -- 699 00:40:05,290 --> 00:40:06,660 OK, anybody says no? 700 00:40:06,660 --> 00:40:10,072 Anybody does not see any hope? 701 00:40:10,072 --> 00:40:10,950 All right. 702 00:40:10,950 --> 00:40:11,750 This is great. 703 00:40:11,750 --> 00:40:14,940 You all see hope here. 704 00:40:14,940 --> 00:40:18,380 You all see hope here, which seems to make you an 705 00:40:18,380 --> 00:40:20,166 opportunistic bunch. 706 00:40:20,166 --> 00:40:21,360 Not opportunistic. 707 00:40:21,360 --> 00:40:22,350 What's the word? 708 00:40:22,350 --> 00:40:22,930 Optimistic. 709 00:40:22,930 --> 00:40:24,180 Optimistic bunch. 710 00:40:24,180 --> 00:40:26,470 711 00:40:26,470 --> 00:40:27,270 Let's put it like this. 712 00:40:27,270 --> 00:40:29,333 What is the problem in solving this? 713 00:40:29,333 --> 00:40:31,991 714 00:40:31,991 --> 00:40:34,270 It's not linear. 715 00:40:34,270 --> 00:40:35,890 You get the coefficients of lambda. 716 00:40:35,890 --> 00:40:38,390 Multiply the coefficients of f. 717 00:40:38,390 --> 00:40:44,900 That whole thing becomes a nonlinear problem, where we 718 00:40:44,900 --> 00:40:51,490 say in the end, find the solution to a set of 719 00:40:51,490 --> 00:40:57,400 polynomial equations in a field, which is a multivariate 720 00:40:57,400 --> 00:41:00,390 polynomial equations, where the coefficients of lambda and 721 00:41:00,390 --> 00:41:02,680 f are the variables. 722 00:41:02,680 --> 00:41:03,540 You can do that. 723 00:41:03,540 --> 00:41:06,400 You could use techniques like Grobner basis or so, and you 724 00:41:06,400 --> 00:41:07,720 could do that. 725 00:41:07,720 --> 00:41:09,020 But this is very difficult. 726 00:41:09,020 --> 00:41:11,805 This is computationally tedious. 727 00:41:11,805 --> 00:41:15,710 728 00:41:15,710 --> 00:41:18,930 So is that clear why this is nonlinear, and why this is 729 00:41:18,930 --> 00:41:20,780 hard to solve a nonlinear problem here? 730 00:41:20,780 --> 00:41:24,740 731 00:41:24,740 --> 00:41:27,830 If not, then you have to say something now, or forever hold 732 00:41:27,830 --> 00:41:29,080 your peace. 733 00:41:29,080 --> 00:41:36,880 734 00:41:36,880 --> 00:41:38,350 So what do we do with hard problems? 735 00:41:38,350 --> 00:41:41,820 736 00:41:41,820 --> 00:41:44,920 Once you take the optimization classes, there's almost like a 737 00:41:44,920 --> 00:41:47,320 reflex, there is a relax time. 738 00:41:47,320 --> 00:41:49,870 We find the proper relaxation of the problem. 739 00:41:49,870 --> 00:41:53,450 Now the proper relaxation of this problem is the following. 740 00:41:53,450 --> 00:42:00,690 741 00:42:00,690 --> 00:42:03,150 Decoding problem two. 742 00:42:03,150 --> 00:42:17,865 Find lambda of x of minimal degree such that -- 743 00:42:17,865 --> 00:42:21,760 it's almost the same, almost the same -- 744 00:42:21,760 --> 00:42:39,460 lambda fi yi minus h of xi is 0, where the degree of h is 745 00:42:39,460 --> 00:42:48,820 less than k plus degree lambda. 746 00:42:48,820 --> 00:42:52,370 So all that we did from this formulation, which would give 747 00:42:52,370 --> 00:42:55,560 us a clean solution to the whole thing, right now, we 748 00:42:55,560 --> 00:43:00,350 multiply in this lambda which gives here, it keeps the 749 00:43:00,350 --> 00:43:02,060 lambda times y. 750 00:43:02,060 --> 00:43:05,370 And here, we get a new polynomial, lambda times f, 751 00:43:05,370 --> 00:43:11,800 which now has degree at most k plus degree lambda. 752 00:43:11,800 --> 00:43:15,760 And then we say, let's instead solve this problem. 753 00:43:15,760 --> 00:43:18,130 Let's solve this problem. 754 00:43:18,130 --> 00:43:32,800 In particular, the question now becomes well, we do not 755 00:43:32,800 --> 00:43:35,000 require this anymore. 756 00:43:35,000 --> 00:43:39,870 In this relaxed formulation, we do not require this. 757 00:43:39,870 --> 00:43:41,800 And that makes all the difference. 758 00:43:41,800 --> 00:43:45,836 It makes a world of difference because this one -- 759 00:43:45,836 --> 00:43:49,470 look at it -- it's a linear problem. 760 00:43:49,470 --> 00:43:50,760 It's a linear program. 761 00:43:50,760 --> 00:43:51,880 Why is it a linear problem? 762 00:43:51,880 --> 00:43:53,375 Do you see it's a linear problem? 763 00:43:53,375 --> 00:43:56,020 764 00:43:56,020 --> 00:44:00,180 Could you write down the equation, the matrix equation? 765 00:44:00,180 --> 00:44:01,430 It's pretty straight, right? 766 00:44:01,430 --> 00:44:04,690 767 00:44:04,690 --> 00:44:05,730 It's pretty straight. 768 00:44:05,730 --> 00:44:09,340 You could, for example, write it like, here, a 769 00:44:09,340 --> 00:44:14,285 diagonal matrix yn. 770 00:44:14,285 --> 00:44:16,790 771 00:44:16,790 --> 00:44:18,280 Here, you would get something. 772 00:44:18,280 --> 00:44:26,688 773 00:44:26,688 --> 00:44:28,590 I think that seems to be all right. 774 00:44:28,590 --> 00:44:35,710 775 00:44:35,710 --> 00:44:39,070 And here, just the evaluation of the h. 776 00:44:39,070 --> 00:45:00,330 So up to hk plus degree lambda. 777 00:45:00,330 --> 00:45:03,070 That's a linear system of equations. 778 00:45:03,070 --> 00:45:04,330 We can certainly solve this. 779 00:45:04,330 --> 00:45:08,340 Well, we do not know really what the lambda is, what 780 00:45:08,340 --> 00:45:10,180 degree the lambda has. 781 00:45:10,180 --> 00:45:12,720 But we're just hypothesizing on all the possible degrees. 782 00:45:12,720 --> 00:45:15,920 783 00:45:15,920 --> 00:45:21,915 Well, we could say, OK, let's assume the degree lambda is 0. 784 00:45:21,915 --> 00:45:24,740 It's a constant, which would be the same as saying there 785 00:45:24,740 --> 00:45:27,520 are no errors. 786 00:45:27,520 --> 00:45:29,120 Then we can look at the system of equations. 787 00:45:29,120 --> 00:45:30,120 Does it have a solution? 788 00:45:30,120 --> 00:45:31,380 Well, yes, no. 789 00:45:31,380 --> 00:45:34,300 If no, then we say, all right, let's assume it's 1. 790 00:45:34,300 --> 00:45:35,370 Well, does it have a solution? 791 00:45:35,370 --> 00:45:36,250 Yes, no. 792 00:45:36,250 --> 00:45:36,860 And so on. 793 00:45:36,860 --> 00:45:38,110 Then we can move on. 794 00:45:38,110 --> 00:45:41,480 795 00:45:41,480 --> 00:45:43,210 So we can solve this relaxed problem. 796 00:45:43,210 --> 00:45:46,610 797 00:45:46,610 --> 00:45:48,580 Does that help us? 798 00:45:48,580 --> 00:45:50,500 It's nice to solve a relaxed problem. 799 00:45:50,500 --> 00:45:52,540 And in the end, we get two vectors out of it, two 800 00:45:52,540 --> 00:45:57,620 polynomials, lambda and h. 801 00:45:57,620 --> 00:45:58,870 Does it really help us? 802 00:45:58,870 --> 00:46:02,290 803 00:46:02,290 --> 00:46:06,635 Well, yes. 804 00:46:06,635 --> 00:46:08,390 Why? 805 00:46:08,390 --> 00:46:09,240 Because -- 806 00:46:09,240 --> 00:46:10,490 let's put it like this. 807 00:46:10,490 --> 00:46:12,930 808 00:46:12,930 --> 00:46:16,960 We could easily check if this is true. 809 00:46:16,960 --> 00:46:20,860 Once we have our h, we can easily check if this is true. 810 00:46:20,860 --> 00:46:25,760 And if it is true, then we have solved this problem. 811 00:46:25,760 --> 00:46:28,430 If it is true, we have solved this problem, which is the 812 00:46:28,430 --> 00:46:31,350 problem we wanted to solve. 813 00:46:31,350 --> 00:46:32,600 Good. 814 00:46:32,600 --> 00:46:40,170 815 00:46:40,170 --> 00:46:41,870 So is that all we need to know about this? 816 00:46:41,870 --> 00:46:50,390 817 00:46:50,390 --> 00:46:51,950 We want to give guarantees. 818 00:46:51,950 --> 00:46:55,380 We want to give guarantees that we correct up to so many 819 00:46:55,380 --> 00:46:56,991 errors, t errors, right? 820 00:46:56,991 --> 00:46:59,580 821 00:46:59,580 --> 00:47:03,390 So we have to guarantee that if there are not more than t 822 00:47:03,390 --> 00:47:08,600 errors, whatever t will turn out to be, we can guarantee 823 00:47:08,600 --> 00:47:12,460 that A, we will find a solution, and 824 00:47:12,460 --> 00:47:15,570 B, this will hold. 825 00:47:15,570 --> 00:47:17,815 So two things to prove. 826 00:47:17,815 --> 00:47:18,880 Is that clear? 827 00:47:18,880 --> 00:47:21,360 That we have to prove those two things? 828 00:47:21,360 --> 00:47:24,020 So the first one first. 829 00:47:24,020 --> 00:47:27,430 830 00:47:27,430 --> 00:47:34,970 When do we find a solution? 831 00:47:34,970 --> 00:47:36,465 And are we guaranteed to find it? 832 00:47:36,465 --> 00:47:41,200 833 00:47:41,200 --> 00:47:50,280 Can we guarantee the existence of a solution? 834 00:47:50,280 --> 00:47:55,400 835 00:47:55,400 --> 00:47:58,050 Well, when can we do that? 836 00:47:58,050 --> 00:48:00,390 Let's look at this. 837 00:48:00,390 --> 00:48:02,240 These are n constraints. 838 00:48:02,240 --> 00:48:05,390 839 00:48:05,390 --> 00:48:09,350 This is a system of linear equations with n constraints. 840 00:48:09,350 --> 00:48:17,420 If the total number of degrees of freedom exceeds n, then we 841 00:48:17,420 --> 00:48:19,790 are left with something non-trivial after we solve the 842 00:48:19,790 --> 00:48:21,060 system of equations. 843 00:48:21,060 --> 00:48:23,880 So what's the total number of degrees of freedom? 844 00:48:23,880 --> 00:48:26,760 845 00:48:26,760 --> 00:48:28,100 The number of degrees of freedom -- 846 00:48:28,100 --> 00:48:33,100 847 00:48:33,100 --> 00:48:39,920 so we get the lambda as a degree of freedom, so which is 848 00:48:39,920 --> 00:48:44,030 degree lambda plus 1. 849 00:48:44,030 --> 00:48:46,720 And the other one is plus -- 850 00:48:46,720 --> 00:48:49,510 851 00:48:49,510 --> 00:48:51,910 what is this one -- 852 00:48:51,910 --> 00:48:58,930 plus k plus degree lambda. 853 00:48:58,930 --> 00:49:04,140 This is the total number of degrees of freedom here. 854 00:49:04,140 --> 00:49:08,550 And the reason is that this one should be minus 1. 855 00:49:08,550 --> 00:49:09,800 Sorry. 856 00:49:09,800 --> 00:49:11,400 857 00:49:11,400 --> 00:49:13,320 So this is total number of degrees of freedom. 858 00:49:13,320 --> 00:49:18,505 859 00:49:18,505 --> 00:49:21,580 If this is greater than n, then we can guarantee the 860 00:49:21,580 --> 00:49:23,445 existence of a solution, a nontrivial solution. 861 00:49:23,445 --> 00:49:26,690 Then this thing will have a solution. 862 00:49:26,690 --> 00:49:28,620 It's a homogeneous system of equations. 863 00:49:28,620 --> 00:49:30,110 So the 0 is always a solution. 864 00:49:30,110 --> 00:49:33,200 But that's not much good to us. 865 00:49:33,200 --> 00:49:35,458 So what do we get here? 866 00:49:35,458 --> 00:49:37,930 Degree lambda. 867 00:49:37,930 --> 00:49:44,580 To a degree lambda greater than n minus k plus minus 868 00:49:44,580 --> 00:49:49,130 1 or n minus k. 869 00:49:49,130 --> 00:49:52,470 870 00:49:52,470 --> 00:49:58,770 So the degree lambda greater than n minus k over 2 -- does 871 00:49:58,770 --> 00:50:00,100 that remind you of anything here? 872 00:50:00,100 --> 00:50:04,770 873 00:50:04,770 --> 00:50:14,950 So this one is d minus 1 over 2. 874 00:50:14,950 --> 00:50:16,700 So very interesting, right? 875 00:50:16,700 --> 00:50:20,460 Once upon a time, you learned that if you make less than d/2 876 00:50:20,460 --> 00:50:23,760 errors, you can correct that. 877 00:50:23,760 --> 00:50:24,040 Very nice. 878 00:50:24,040 --> 00:50:26,780 It pops up here. 879 00:50:26,780 --> 00:50:29,520 It pops up here out of the blue. 880 00:50:29,520 --> 00:50:32,260 The reason that it pops up here is, of course, the same 881 00:50:32,260 --> 00:50:36,730 statement, that as long as we stay within d/2 errors, we are 882 00:50:36,730 --> 00:50:38,580 guaranteed to be able to decode this. 883 00:50:38,580 --> 00:50:41,200 884 00:50:41,200 --> 00:50:42,030 So this is one. 885 00:50:42,030 --> 00:50:45,630 So we know this one was the number of errors. 886 00:50:45,630 --> 00:50:48,540 If t is greater or equal than this, then -- 887 00:50:48,540 --> 00:50:58,010 888 00:50:58,010 --> 00:51:01,220 let's forget about the t. 889 00:51:01,220 --> 00:51:03,810 If you proceed in this algorithm hypothesizing the 890 00:51:03,810 --> 00:51:09,100 degrees of lambda, once we've reached this number in the 891 00:51:09,100 --> 00:51:11,150 degree, there will be a solution. 892 00:51:11,150 --> 00:51:12,650 And we don't have to go further than that. 893 00:51:12,650 --> 00:51:16,380 894 00:51:16,380 --> 00:51:18,190 So that's the first one. 895 00:51:18,190 --> 00:51:21,880 So the second thing we have to prove is that -- 896 00:51:21,880 --> 00:51:23,130 I shouldn't have done this -- 897 00:51:23,130 --> 00:51:29,190 898 00:51:29,190 --> 00:51:30,320 that for some t -- 899 00:51:30,320 --> 00:51:30,784 Yeah? 900 00:51:30,784 --> 00:51:33,104 AUDIENCE: [INAUDIBLE] 901 00:51:33,104 --> 00:51:36,360 standard degree of freedom or less constraints? 902 00:51:36,360 --> 00:51:38,310 PROFESSOR: You want to guarantee a solution. 903 00:51:38,310 --> 00:51:41,750 If you have more constraints than degrees of freedom, then, 904 00:51:41,750 --> 00:51:44,670 since it's homogeneous, you usually would be stuck with a 905 00:51:44,670 --> 00:51:45,490 zero solution. 906 00:51:45,490 --> 00:51:49,520 Everything's 0, which is no good to us. 907 00:51:49,520 --> 00:51:52,510 It solves this problem, but it doesn't give us any 908 00:51:52,510 --> 00:51:55,305 information. 909 00:51:55,305 --> 00:51:56,622 Is that all right? 910 00:51:56,622 --> 00:52:00,066 911 00:52:00,066 --> 00:52:02,772 AUDIENCE: Why does it have to be the condition that degrees 912 00:52:02,772 --> 00:52:06,970 of freedom has to be greater than the constraints? 913 00:52:06,970 --> 00:52:09,400 PROFESSOR: No, no, the degrees of freedom -- 914 00:52:09,400 --> 00:52:13,460 well, OK, it's true. 915 00:52:13,460 --> 00:52:16,580 We could get lucky. 916 00:52:16,580 --> 00:52:20,030 We could have redundancy in this system of equations. 917 00:52:20,030 --> 00:52:24,780 And if we get lucky, that's just for the better. 918 00:52:24,780 --> 00:52:29,260 And actually, you can show that you do get lucky if very 919 00:52:29,260 --> 00:52:31,250 few errors happen. 920 00:52:31,250 --> 00:52:34,260 Then this would have low rank. 921 00:52:34,260 --> 00:52:35,370 You can show that. 922 00:52:35,370 --> 00:52:39,430 But short of knowing much, we want to guarantee -- 923 00:52:39,430 --> 00:52:42,380 I just want, at this point in time, to guarantee that there 924 00:52:42,380 --> 00:52:44,370 is a solution. 925 00:52:44,370 --> 00:52:47,590 There is a non-zero solution to this system of equations, 926 00:52:47,590 --> 00:52:50,710 and the degree of lambda is not more than d 927 00:52:50,710 --> 00:52:53,460 minus 1 over 2. 928 00:52:53,460 --> 00:52:56,690 There is a solution with a degree of lambda being upper 929 00:52:56,690 --> 00:52:58,324 bounded by this. 930 00:52:58,324 --> 00:52:59,800 AUDIENCE: So [INAUDIBLE] 931 00:52:59,800 --> 00:53:01,280 less constraints? 932 00:53:01,280 --> 00:53:02,530 PROFESSOR: Yeah, no, no. 933 00:53:02,530 --> 00:53:06,550 934 00:53:06,550 --> 00:53:08,660 Once this is satisfied, I guarantee 935 00:53:08,660 --> 00:53:10,260 you there is a solution. 936 00:53:10,260 --> 00:53:14,420 That means, the smallest solution is guaranteed not to 937 00:53:14,420 --> 00:53:15,670 be larger than that. 938 00:53:15,670 --> 00:53:19,090 939 00:53:19,090 --> 00:53:20,150 But there are, of course, many, 940 00:53:20,150 --> 00:53:21,570 many more larger solutions. 941 00:53:21,570 --> 00:53:25,723 There are many solutions of larger degree here, which we 942 00:53:25,723 --> 00:53:27,550 are not interested in since we are interested 943 00:53:27,550 --> 00:53:30,370 in the minimal degree. 944 00:53:30,370 --> 00:53:32,750 So this is an upper bound on the minimal degree lambda. 945 00:53:32,750 --> 00:53:37,200 946 00:53:37,200 --> 00:53:38,300 So now the other one. 947 00:53:38,300 --> 00:53:46,630 So now we want to make the statement about this one here. 948 00:53:46,630 --> 00:53:52,990 When can we guarantee that our relaxation didn't matter? 949 00:53:52,990 --> 00:53:56,930 Despite our relaxation, our solution that we find that in 950 00:53:56,930 --> 00:54:00,110 the solution that we find lambda divides h. 951 00:54:00,110 --> 00:54:03,430 How can we guarantee this? 952 00:54:03,430 --> 00:54:04,860 So how can we guarantee this? 953 00:54:04,860 --> 00:54:06,110 How can we guarantee this? 954 00:54:06,110 --> 00:54:10,080 955 00:54:10,080 --> 00:54:20,310 Let's look at lambda xi yi minus h of xi. 956 00:54:20,310 --> 00:54:24,610 And we know this is 0 for all xi. 957 00:54:24,610 --> 00:54:25,860 We know that. 958 00:54:25,860 --> 00:54:29,210 959 00:54:29,210 --> 00:54:38,780 We know that this guy here can actually be written as ci plus 960 00:54:38,780 --> 00:54:43,630 ei minus h of xi. 961 00:54:43,630 --> 00:54:47,690 962 00:54:47,690 --> 00:54:50,040 This is just expanding this guy. 963 00:54:50,040 --> 00:55:00,790 We also know that we can write lambda xi times ci alone minus 964 00:55:00,790 --> 00:55:08,051 f of xi lambda xi. 965 00:55:08,051 --> 00:55:08,520 This is 0. 966 00:55:08,520 --> 00:55:10,932 We know that this is true too. 967 00:55:10,932 --> 00:55:12,780 AUDIENCE: For some f. 968 00:55:12,780 --> 00:55:13,320 PROFESSOR: For some f. 969 00:55:13,320 --> 00:55:17,220 The text is to f, so that this is true too. 970 00:55:17,220 --> 00:55:22,500 Then let's just subtract these two guys. 971 00:55:22,500 --> 00:55:25,480 Let's just subtract these two things here. 972 00:55:25,480 --> 00:55:36,460 And then we know that lambda xi times ei -- 973 00:55:36,460 --> 00:55:38,250 so the first two cancel -- 974 00:55:38,250 --> 00:55:51,646 minus h of xi minus So we know this is true too. 975 00:55:51,646 --> 00:55:53,550 AUDIENCE: [INAUDIBLE]. 976 00:55:53,550 --> 00:55:56,210 PROFESSOR: Yeah, sorry. 977 00:55:56,210 --> 00:55:57,460 Just in time. 978 00:55:57,460 --> 00:56:01,220 979 00:56:01,220 --> 00:56:02,480 So what can we learn from this? 980 00:56:02,480 --> 00:56:05,510 So what is the degree of this? 981 00:56:05,510 --> 00:56:07,780 What is the degree of this? 982 00:56:07,780 --> 00:56:09,570 Well, this guy here -- 983 00:56:09,570 --> 00:56:10,820 we'll write it somewhere else -- 984 00:56:10,820 --> 00:56:37,550 985 00:56:37,550 --> 00:56:47,320 well, this guy had degree h was less than n minus k plus 986 00:56:47,320 --> 00:56:57,310 degree lambda less than k plus degree lambda. 987 00:56:57,310 --> 00:57:00,760 That's because we set up that problem that way. 988 00:57:00,760 --> 00:57:10,740 And we know that this guy degree lambda. 989 00:57:10,740 --> 00:57:13,780 990 00:57:13,780 --> 00:57:15,740 We know that. 991 00:57:15,740 --> 00:57:28,390 So this whole thing, let's call it S. We know it's 0 for 992 00:57:28,390 --> 00:57:35,360 degree S less than degree lambda. 993 00:57:35,360 --> 00:57:39,080 994 00:57:39,080 --> 00:57:41,730 So why does that help us? 995 00:57:41,730 --> 00:57:44,740 Well, let's look at the vector. 996 00:57:44,740 --> 00:57:48,060 Now let's look at all the positions. 997 00:57:48,060 --> 00:57:53,380 Let's look at the vector where we had, in the vector, we have 998 00:57:53,380 --> 00:57:56,770 lambda xi ei. 999 00:57:56,770 --> 00:58:02,620 1000 00:58:02,620 --> 00:58:05,480 And now the fundamental question. 1001 00:58:05,480 --> 00:58:09,450 What is the rate of this vector? 1002 00:58:09,450 --> 00:58:10,700 At most? 1003 00:58:10,700 --> 00:58:13,400 1004 00:58:13,400 --> 00:58:15,147 What's the rate of this guy at most? 1005 00:58:15,147 --> 00:58:18,883 1006 00:58:18,883 --> 00:58:20,133 It's 12 seconds, [UNINTELLIGIBLE]? 1007 00:58:20,133 --> 00:58:25,740 1008 00:58:25,740 --> 00:58:28,860 Well, the rate of this vector -- 1009 00:58:28,860 --> 00:58:37,030 1010 00:58:37,030 --> 00:58:40,410 it's definitely not more than the rate of the error. 1011 00:58:40,410 --> 00:58:44,260 Because every time the error is 0, the rate drops out. 1012 00:58:44,260 --> 00:58:45,510 That's at most t. 1013 00:58:45,510 --> 00:58:48,890 1014 00:58:48,890 --> 00:58:52,660 What is the rate of this vector? 1015 00:58:52,660 --> 00:59:01,170 So here we have a vector on the other side, S of xi, a 1016 00:59:01,170 --> 00:59:04,240 vector on the other side. 1017 00:59:04,240 --> 00:59:05,490 What's the rate of this guy? 1018 00:59:05,490 --> 00:59:08,790 1019 00:59:08,790 --> 00:59:13,810 That's just the polynomial of degree at most k plus degree 1020 00:59:13,810 --> 00:59:15,640 lambda minus 1. 1021 00:59:15,640 --> 00:59:29,940 So we have that the rate of this vector is n minus k plus 1022 00:59:29,940 --> 00:59:35,617 degree lambda minus 1 plus 1 because of the minus. 1023 00:59:35,617 --> 00:59:39,230 AUDIENCE: Greater than or equal to? 1024 00:59:39,230 --> 00:59:41,660 PROFESSOR: It's greater than or equal, sorry. 1025 00:59:41,660 --> 00:59:43,580 Otherwise, I would have been in trouble in a second here. 1026 00:59:43,580 --> 00:59:46,350 1027 00:59:46,350 --> 00:59:47,800 Yeah, so what does that mean? 1028 00:59:47,800 --> 00:59:51,710 So actually, this one is nice to rewrite is equal to d n 1029 00:59:51,710 --> 00:59:55,865 minus t plus 1 d -- 1030 00:59:55,865 --> 01:00:00,320 I really should have done it like this in order not to -- 1031 01:00:00,320 --> 01:00:05,010 it's d minus degree lambda. 1032 01:00:05,010 --> 01:00:08,230 1033 01:00:08,230 --> 01:00:09,550 So what can we learn from that? 1034 01:00:09,550 --> 01:00:12,960 1035 01:00:12,960 --> 01:00:19,300 So we know this is true for all the positions. 1036 01:00:19,300 --> 01:00:24,060 So if this is true for all positions, then -- 1037 01:00:24,060 --> 01:00:30,070 so if d minus degree lambda -- 1038 01:00:30,070 --> 01:00:34,480 actually, this is t because the rate was defined. 1039 01:00:34,480 --> 01:00:37,250 t was the number of errors. 1040 01:00:37,250 --> 01:00:39,830 And the error locator was just defined to have degree t. 1041 01:00:39,830 --> 01:00:43,650 1042 01:00:43,650 --> 01:00:45,390 Right? 1043 01:00:45,390 --> 01:00:49,480 The error locator was just x minus xi over all positions 1044 01:00:49,480 --> 01:00:50,890 where we had an error, so there 1045 01:00:50,890 --> 01:00:51,950 are t of those positions. 1046 01:00:51,950 --> 01:00:55,550 The degree of lambda is equal to t. 1047 01:00:55,550 --> 01:00:56,490 So now we take this. 1048 01:00:56,490 --> 01:01:07,410 If d minus t is greater than t, so if the rate of this 1049 01:01:07,410 --> 01:01:13,720 vector is greater than the rate of this vector, then we 1050 01:01:13,720 --> 01:01:18,030 have a problem because that cannot be. 1051 01:01:18,030 --> 01:01:19,840 Of course, by this identity, the rate of these 1052 01:01:19,840 --> 01:01:22,520 two vectors is equal. 1053 01:01:22,520 --> 01:01:23,770 So what is the way out? 1054 01:01:23,770 --> 01:01:26,660 1055 01:01:26,660 --> 01:01:27,550 What's the way out? 1056 01:01:27,550 --> 01:01:28,800 AUDIENCE: [INAUDIBLE] 1057 01:01:28,800 --> 01:01:33,286 1058 01:01:33,286 --> 01:01:33,770 PROFESSOR: Yeah. 1059 01:01:33,770 --> 01:01:53,090 So if this is true, then S of xi I claim has to be 0 and 1060 01:01:53,090 --> 01:02:01,100 lambda of xi times ei has to be 0 too for all i. 1061 01:02:01,100 --> 01:02:03,860 1062 01:02:03,860 --> 01:02:06,460 That's the only way out of this. 1063 01:02:06,460 --> 01:02:08,760 Because, obviously, this was allowed. 1064 01:02:08,760 --> 01:02:11,760 If we really did find what we wanted to find, namely, an 1065 01:02:11,760 --> 01:02:15,530 error locator here, then this vector -- 1066 01:02:15,530 --> 01:02:17,410 we said it has to be less than t. 1067 01:02:17,410 --> 01:02:18,570 It can be less. 1068 01:02:18,570 --> 01:02:22,460 In particular, it can be 0 if this is, 1069 01:02:22,460 --> 01:02:24,710 indeed, an error locator. 1070 01:02:24,710 --> 01:02:28,820 And if our relaxation didn't matter -- 1071 01:02:28,820 --> 01:02:34,800 that means the h that we find factors as that -- 1072 01:02:34,800 --> 01:02:38,480 if the relaxation didn't matter, then S would be 0 too. 1073 01:02:38,480 --> 01:02:39,830 So that is a fair solution. 1074 01:02:39,830 --> 01:02:41,970 That's a possibility. 1075 01:02:41,970 --> 01:02:48,450 And if d minus t is greater than t, then this proves this 1076 01:02:48,450 --> 01:02:51,300 is the only solution that is feasible. 1077 01:02:51,300 --> 01:02:54,180 That's the only solution you have. 1078 01:02:54,180 --> 01:03:03,480 So what that means is if t less than t/2, then the 1079 01:03:03,480 --> 01:03:20,040 relaxation doesn't matter and h of xi is equal to lambda of 1080 01:03:20,040 --> 01:03:21,568 xi times f of xi. 1081 01:03:21,568 --> 01:03:32,600 1082 01:03:32,600 --> 01:03:35,040 So is that clear? 1083 01:03:35,040 --> 01:03:44,380 That's two very simple, yet nice ways to prove things. 1084 01:03:44,380 --> 01:03:45,750 Here, we guarantee the solution. 1085 01:03:45,750 --> 01:03:49,450 1086 01:03:49,450 --> 01:03:51,610 Here, we guarantee the solution. 1087 01:03:51,610 --> 01:03:55,330 Here, we guarantee that solution is correct, namely, 1088 01:03:55,330 --> 01:03:59,760 if not too many errors happen, if less than t/2 errors 1089 01:03:59,760 --> 01:04:06,370 happens, then relaxing the original problem here to this 1090 01:04:06,370 --> 01:04:08,360 other form, which we could solve, does 1091 01:04:08,360 --> 01:04:11,690 not change the solution. 1092 01:04:11,690 --> 01:04:12,940 That's what's proved there. 1093 01:04:12,940 --> 01:04:16,960 1094 01:04:16,960 --> 01:04:21,270 But about one-third looks puzzled. 1095 01:04:21,270 --> 01:04:23,130 About half looks puzzled, I would think. 1096 01:04:23,130 --> 01:04:29,490 1097 01:04:29,490 --> 01:04:30,900 Half looks puzzled. 1098 01:04:30,900 --> 01:04:33,870 Is there any way I can explain that better? 1099 01:04:33,870 --> 01:04:35,120 Think. 1100 01:04:35,120 --> 01:04:42,790 1101 01:04:42,790 --> 01:04:46,980 Let's just go through the steps quickly. 1102 01:04:46,980 --> 01:04:50,640 We had problem number one. 1103 01:04:50,640 --> 01:04:56,350 Problem number one, which is a nonlinear problem, but you see 1104 01:04:56,350 --> 01:04:59,170 that if you could solve this, you could solve 1105 01:04:59,170 --> 01:05:01,800 the decoding problem. 1106 01:05:01,800 --> 01:05:03,760 So you see that. 1107 01:05:03,760 --> 01:05:06,760 Anybody who doesn't see that? 1108 01:05:06,760 --> 01:05:07,090 All right. 1109 01:05:07,090 --> 01:05:08,340 You're smart guys. 1110 01:05:08,340 --> 01:05:11,130 1111 01:05:11,130 --> 01:05:13,740 If we can solve problem number one, we are home free. 1112 01:05:13,740 --> 01:05:15,300 We have solved the decoding problem. 1113 01:05:15,300 --> 01:05:16,270 Fine, we cannot solve it. 1114 01:05:16,270 --> 01:05:17,640 It's a nonlinear problem. 1115 01:05:17,640 --> 01:05:21,110 Well, we can solve it in exponential time, but that's 1116 01:05:21,110 --> 01:05:23,410 little fun. 1117 01:05:23,410 --> 01:05:24,320 So we relax it. 1118 01:05:24,320 --> 01:05:28,730 We relax it into problem number two. 1119 01:05:28,730 --> 01:05:31,444 All that we do, we multiply things out. 1120 01:05:31,444 --> 01:05:37,510 And we do not require, once we solve the 1121 01:05:37,510 --> 01:05:38,760 problem, this anymore. 1122 01:05:38,760 --> 01:05:46,530 1123 01:05:46,530 --> 01:05:48,490 Now we have solved this problem. 1124 01:05:48,490 --> 01:05:51,520 We have solved this linear system of equations. 1125 01:05:51,520 --> 01:05:59,640 And after doing so, we have found lambda and h. 1126 01:05:59,640 --> 01:06:02,470 They are lying on the table and looking at us. 1127 01:06:02,470 --> 01:06:03,610 Now what? 1128 01:06:03,610 --> 01:06:04,820 Are they any good? 1129 01:06:04,820 --> 01:06:08,950 In particular, since a priori we cannot guarantee that this 1130 01:06:08,950 --> 01:06:13,170 relaxation didn't completely destroy everything. 1131 01:06:13,170 --> 01:06:14,410 We have solved the system. 1132 01:06:14,410 --> 01:06:17,040 Now those two things are lying on the table, looking at us, 1133 01:06:17,040 --> 01:06:19,650 and asking, what am I? 1134 01:06:19,650 --> 01:06:23,380 And in general, if there's an arbitrary amount of errors 1135 01:06:23,380 --> 01:06:27,430 happening near the channel, actually, 1136 01:06:27,430 --> 01:06:29,340 they do not mean much. 1137 01:06:29,340 --> 01:06:32,900 They do not mean much. 1138 01:06:32,900 --> 01:06:38,286 But now I claim that, well, if not too many errors happened, 1139 01:06:38,286 --> 01:06:43,490 but if the number of errors is bounded to be this, less than 1140 01:06:43,490 --> 01:06:47,610 half the minimum distance, then, actually, I claim it did 1141 01:06:47,610 --> 01:06:49,870 not matter if we solved the relaxation or 1142 01:06:49,870 --> 01:06:52,260 the original problem. 1143 01:06:52,260 --> 01:06:55,880 And the argument is roughly this. 1144 01:06:55,880 --> 01:06:58,135 It goes like this. 1145 01:06:58,135 --> 01:06:59,890 Here, we start. 1146 01:06:59,890 --> 01:07:01,460 We have solved this. 1147 01:07:01,460 --> 01:07:06,370 We have two polynomials lambda and h, so that this is true 1148 01:07:06,370 --> 01:07:07,620 for all xi. 1149 01:07:07,620 --> 01:07:10,380 1150 01:07:10,380 --> 01:07:12,440 We know that we can write it like this. 1151 01:07:12,440 --> 01:07:15,290 That's just by definition of yi, just 1152 01:07:15,290 --> 01:07:18,980 expanding the yi into this. 1153 01:07:18,980 --> 01:07:22,400 We know that this is true by the definition of the code. 1154 01:07:22,400 --> 01:07:27,500 The ci is the evaluation of f, of some polynomial f. 1155 01:07:27,500 --> 01:07:31,770 So we can write this just by definition of the code. 1156 01:07:31,770 --> 01:07:36,430 So once we have these two guys here, we can subtract them 1157 01:07:36,430 --> 01:07:37,680 [UNINTELLIGIBLE] here. 1158 01:07:37,680 --> 01:07:39,930 1159 01:07:39,930 --> 01:07:44,500 So we know that we have solved the following problem. 1160 01:07:44,500 --> 01:07:49,000 We have found lambda and h. 1161 01:07:49,000 --> 01:07:57,580 So that there is guaranteed to exist the polynomial f, so 1162 01:07:57,580 --> 01:08:02,580 that this whole thing, namely, S of xi, that S is a 1163 01:08:02,580 --> 01:08:04,470 polynomial of degree at most -- 1164 01:08:04,470 --> 01:08:06,130 what did we have -- 1165 01:08:06,130 --> 01:08:09,790 k plus degree lambda minus 1. 1166 01:08:09,790 --> 01:08:12,880 1167 01:08:12,880 --> 01:08:18,550 So that's a semi-hairy step here. 1168 01:08:18,550 --> 01:08:24,700 Is that clear that we can guarantee the existence of a 1169 01:08:24,700 --> 01:08:34,340 polynomial S of degree at most k plus degree lambda minus 1? 1170 01:08:34,340 --> 01:08:38,320 So that this equation holds. 1171 01:08:38,320 --> 01:08:43,229 1172 01:08:43,229 --> 01:08:45,609 Once we have solved this, we can guarantee the 1173 01:08:45,609 --> 01:08:46,890 existence of this. 1174 01:08:46,890 --> 01:08:48,939 That's what we're saying. 1175 01:08:48,939 --> 01:08:51,029 We can guarantee the existence of this. 1176 01:08:51,029 --> 01:08:53,819 So now look at this. 1177 01:08:53,819 --> 01:08:58,319 If you write this out for all i's and put it in a vector, 1178 01:08:58,319 --> 01:09:00,620 what's the rate of this vector? 1179 01:09:00,620 --> 01:09:05,630 Well, it's at most t because all the other e's are 0. 1180 01:09:05,630 --> 01:09:08,890 If lambda is an error locator, it is 0. 1181 01:09:08,890 --> 01:09:10,740 But we do not know that yet. 1182 01:09:10,740 --> 01:09:12,330 All we know, it's at most t. 1183 01:09:12,330 --> 01:09:15,149 1184 01:09:15,149 --> 01:09:18,939 This is on this side, so that vector has a rate at most t. 1185 01:09:18,939 --> 01:09:25,660 On this side, the rate of this vector is at least this. 1186 01:09:25,660 --> 01:09:28,939 Just by evaluating a polynomial of this degree, 1187 01:09:28,939 --> 01:09:31,193 this is the maximum number of 0's we can get. 1188 01:09:31,193 --> 01:09:34,350 1189 01:09:34,350 --> 01:09:41,740 So now we have two equalities, this one and this one. 1190 01:09:41,740 --> 01:09:45,330 But obviously, the two vectors must be the same. 1191 01:09:45,330 --> 01:09:49,880 So how can these two vectors be the same and still satisfy 1192 01:09:49,880 --> 01:09:53,250 these two inequalities? 1193 01:09:53,250 --> 01:09:53,950 Good question. 1194 01:09:53,950 --> 01:09:56,320 How can they be the same and still satisfy this inequality? 1195 01:09:56,320 --> 01:10:10,720 1196 01:10:10,720 --> 01:10:15,070 If t is less than d/2, then the rate of this vector would 1197 01:10:15,070 --> 01:10:18,000 have to be larger than the rate of this. 1198 01:10:18,000 --> 01:10:19,250 If they are non-zero. 1199 01:10:19,250 --> 01:10:22,530 1200 01:10:22,530 --> 01:10:25,030 If they are non-zero, this would have to be larger than 1201 01:10:25,030 --> 01:10:30,250 this because then d minus t would be greater than t. 1202 01:10:30,250 --> 01:10:32,340 If this is true, this implies this. 1203 01:10:32,340 --> 01:10:36,890 1204 01:10:36,890 --> 01:10:40,980 If this is true, this implies this which implies that the 1205 01:10:40,980 --> 01:10:44,270 rate of this vector would be larger than the rate of this 1206 01:10:44,270 --> 01:10:45,670 vector if they are non-zero. 1207 01:10:45,670 --> 01:10:48,840 1208 01:10:48,840 --> 01:10:50,890 But how can that be? 1209 01:10:50,890 --> 01:10:53,210 Answer is cannot. 1210 01:10:53,210 --> 01:10:56,080 Cannot be, hence, they must be 0. 1211 01:10:56,080 --> 01:11:00,150 Both vectors must be 0 completely, which means this 1212 01:11:00,150 --> 01:11:02,250 is an error locator. 1213 01:11:02,250 --> 01:11:08,180 Because otherwise, this wouldn't be 0, and this one -- 1214 01:11:08,180 --> 01:11:09,430 where did I write it? 1215 01:11:09,430 --> 01:11:11,910 1216 01:11:11,910 --> 01:11:13,160 Somewhere I wrote it. 1217 01:11:13,160 --> 01:11:18,640 1218 01:11:18,640 --> 01:11:21,292 Where did I write it? 1219 01:11:21,292 --> 01:11:22,750 Oh, here. 1220 01:11:22,750 --> 01:11:27,300 So this S has to be 0 too. 1221 01:11:27,300 --> 01:11:33,640 If this S is 0, then h of xi is exactly this, means factors 1222 01:11:33,640 --> 01:11:36,870 in the way you want it to. 1223 01:11:36,870 --> 01:11:40,270 So that's all I can say. 1224 01:11:40,270 --> 01:11:43,810 I could say it again, but it's exactly the same words. 1225 01:11:43,810 --> 01:11:47,324 And there's a limited benefit to even the repetition code. 1226 01:11:47,324 --> 01:11:50,290 1227 01:11:50,290 --> 01:11:51,880 So beautiful, right? 1228 01:11:51,880 --> 01:11:54,750 We have brought down the entire decoding problem for 1229 01:11:54,750 --> 01:12:01,300 Reed-Solomon codes to solving a linear system of equations, 1230 01:12:01,300 --> 01:12:06,940 namely, this one here or this one in short form, which, 1231 01:12:06,940 --> 01:12:11,130 well, it's no problem at all in the grand scheme of things. 1232 01:12:11,130 --> 01:12:13,250 And that's the other thing that makes Reed-Solomon codes 1233 01:12:13,250 --> 01:12:14,080 so beautiful. 1234 01:12:14,080 --> 01:12:17,674 They're access encoding, they're access decoding, 1235 01:12:17,674 --> 01:12:19,370 they're everything that we want. 1236 01:12:19,370 --> 01:12:35,380 1237 01:12:35,380 --> 01:12:38,660 Let me do one more thing about the decoding just to bring it 1238 01:12:38,660 --> 01:12:42,850 down a little bit, to talk a little bit about complexity. 1239 01:12:42,850 --> 01:12:50,560 Solving this linear system of equations has been subject to 1240 01:12:50,560 --> 01:12:56,850 research for 30 years, 40 years. 1241 01:12:56,850 --> 01:13:01,520 So it started out with an algorithm which essentially 1242 01:13:01,520 --> 01:13:04,976 solved that linear system directly. 1243 01:13:04,976 --> 01:13:08,030 It was formulated a bit different, but 1244 01:13:08,030 --> 01:13:08,680 that's what it is. 1245 01:13:08,680 --> 01:13:08,980 It's called the 1246 01:13:08,980 --> 01:13:10,750 Peterson-Gorenstein-Zierler algorithm. 1247 01:13:10,750 --> 01:13:14,560 1248 01:13:14,560 --> 01:13:20,850 They realized it was a polynomial time decoding 1249 01:13:20,850 --> 01:13:23,500 algorithm, a nice decoding algorithm. 1250 01:13:23,500 --> 01:13:24,060 Then Berlekamp -- 1251 01:13:24,060 --> 01:13:25,320 I should write down the name. 1252 01:13:25,320 --> 01:13:30,190 1253 01:13:30,190 --> 01:13:31,760 Berlekamp came up with a fast 1254 01:13:31,760 --> 01:13:33,420 algorithm and square algorithm. 1255 01:13:33,420 --> 01:13:40,030 1256 01:13:40,030 --> 01:13:44,700 Massey had his own formulation of that algorithm, which was a 1257 01:13:44,700 --> 01:13:46,095 bit more streamlined I think. 1258 01:13:46,095 --> 01:13:48,810 1259 01:13:48,810 --> 01:13:53,020 Then there was a later version, Berlekamp-Welch. 1260 01:13:53,020 --> 01:13:58,230 1261 01:13:58,230 --> 01:14:01,070 The complexity of these algorithms is all roughly the 1262 01:14:01,070 --> 01:14:03,920 same, is all n squared roughly, 1263 01:14:03,920 --> 01:14:06,430 which solved this here. 1264 01:14:06,430 --> 01:14:08,910 And then there's roughly nothing 1265 01:14:08,910 --> 01:14:10,280 happening for 30 years. 1266 01:14:10,280 --> 01:14:13,170 1267 01:14:13,170 --> 01:14:21,630 And after 30 years, then Madhu Sudan made the next serious 1268 01:14:21,630 --> 01:14:25,820 dent in the decoding history of Reed-Solomon codes. 1269 01:14:25,820 --> 01:14:32,885 So he found a way to solve this problem even beyond half 1270 01:14:32,885 --> 01:14:35,700 the minimum distance. 1271 01:14:35,700 --> 01:14:41,230 And in hindsight, it's a very nice and very simple trick 1272 01:14:41,230 --> 01:14:42,480 that he used. 1273 01:14:42,480 --> 01:14:46,160 1274 01:14:46,160 --> 01:14:54,945 So we started with the following. 1275 01:14:54,945 --> 01:15:03,490 We started that, well, if we have no errors, then the pairs 1276 01:15:03,490 --> 01:15:13,110 of points xi, yi lie on this curve. 1277 01:15:13,110 --> 01:15:16,940 That's another way to formulate what we said that yi 1278 01:15:16,940 --> 01:15:20,760 minus f of xi has to be 0. 1279 01:15:20,760 --> 01:15:23,480 We said, well, if there are no errors, all the points that we 1280 01:15:23,480 --> 01:15:26,260 receive lie on this curve. 1281 01:15:26,260 --> 01:15:31,300 Because there are errors, we have to multiply this with an 1282 01:15:31,300 --> 01:15:39,570 error locator and say this is 0 for all xi, yi. 1283 01:15:39,570 --> 01:15:41,430 So this was problem number one. 1284 01:15:41,430 --> 01:15:57,280 The relaxed form was lambda xy minus h of x 0 xi, yi. 1285 01:15:57,280 --> 01:15:59,270 That's the way we do it. 1286 01:15:59,270 --> 01:16:00,690 So what did Madhu do? 1287 01:16:00,690 --> 01:16:01,330 Very clever. 1288 01:16:01,330 --> 01:16:06,490 He said, all that we have to do is annihilate the effect of 1289 01:16:06,490 --> 01:16:09,380 the error right in this. 1290 01:16:09,380 --> 01:16:12,070 The whole reason for that lambda was to annihilate -- 1291 01:16:12,070 --> 01:16:16,270 so to take out the error influence out of this 1292 01:16:16,270 --> 01:16:20,800 interpolation formula, to allow for a few errors to 1293 01:16:20,800 --> 01:16:23,180 happen, which we can put in lambda. 1294 01:16:23,180 --> 01:16:27,520 Well, Madhu said, well, what about we take a polynomial in 1295 01:16:27,520 --> 01:16:28,770 two variables? 1296 01:16:28,770 --> 01:16:30,570 1297 01:16:30,570 --> 01:16:36,790 A polynomial in two variables, then same thing. 1298 01:16:36,790 --> 01:16:43,030 If no errors happen, then all the points xi, yi will lie on 1299 01:16:43,030 --> 01:16:44,580 that curve. 1300 01:16:44,580 --> 01:16:50,690 If a few errors happen, then let's find a lambda xy, two 1301 01:16:50,690 --> 01:16:55,790 variables of some minimal degree, some very small degree 1302 01:16:55,790 --> 01:16:57,715 so that this is still satisfied. 1303 01:16:57,715 --> 01:17:02,082 1304 01:17:02,082 --> 01:17:04,420 And the problem is entirely the same. 1305 01:17:04,420 --> 01:17:07,820 1306 01:17:07,820 --> 01:17:12,420 And this one, you cannot solve either. 1307 01:17:12,420 --> 01:17:14,160 You cannot solve either. 1308 01:17:14,160 --> 01:17:19,950 But again, you can solve the relaxation minus -- 1309 01:17:19,950 --> 01:17:27,550 1310 01:17:27,550 --> 01:17:29,280 again, you can solve the relaxation. 1311 01:17:29,280 --> 01:17:31,890 This is, again, a linear system of equations. 1312 01:17:31,890 --> 01:17:33,830 And the coefficients of lambda and psi now. 1313 01:17:33,830 --> 01:17:36,960 1314 01:17:36,960 --> 01:17:40,020 Once you solve this linear system of equations -- 1315 01:17:40,020 --> 01:17:43,565 actually, this one you can more handily write simply -- 1316 01:17:43,565 --> 01:17:46,270 now there is no way to distinguish the y anymore 1317 01:17:46,270 --> 01:17:50,090 since both sides anyway depend on y. 1318 01:17:50,090 --> 01:17:56,590 Find a polynomial in two variables such that this is 0 1319 01:17:56,590 --> 01:17:59,930 for all xi, yi. 1320 01:17:59,930 --> 01:18:02,030 This is his central problem. 1321 01:18:02,030 --> 01:18:03,160 This is his relaxation. 1322 01:18:03,160 --> 01:18:05,350 Find a polynomial in two variables, which he 1323 01:18:05,350 --> 01:18:06,600 evaluates to 0. 1324 01:18:06,600 --> 01:18:09,140 1325 01:18:09,140 --> 01:18:15,440 And then he has, essentially, the same proofs we have here, 1326 01:18:15,440 --> 01:18:20,360 a bit more technical, but not much. 1327 01:18:20,360 --> 01:18:22,710 I'm sure you can come up with this if you sit down at home. 1328 01:18:22,710 --> 01:18:26,930 1329 01:18:26,930 --> 01:18:27,610 Let's put it like this. 1330 01:18:27,610 --> 01:18:29,180 There's no heavy machinery in it. 1331 01:18:29,180 --> 01:18:30,870 There's no heavy math in it. 1332 01:18:30,870 --> 01:18:33,710 There's a lot of being clever in it, but there's no heavy 1333 01:18:33,710 --> 01:18:34,960 math in it. 1334 01:18:34,960 --> 01:18:40,690 1335 01:18:40,690 --> 01:18:52,280 He can now guarantee that q of xy, if t is less than n minus 1336 01:18:52,280 --> 01:18:55,700 square root of 2k -- 1337 01:18:55,700 --> 01:18:57,570 is that right -- 1338 01:18:57,570 --> 01:19:06,280 2kn, then you can guarantee that y minus fx is 1339 01:19:06,280 --> 01:19:09,420 a factor q of xy. 1340 01:19:09,420 --> 01:19:12,736 The same thing we wanted to guarantee. 1341 01:19:12,736 --> 01:19:19,170 And like I said, the proof is very clever, but no heavy 1342 01:19:19,170 --> 01:19:19,830 machinery in it. 1343 01:19:19,830 --> 01:19:22,176 It's no heavy algebraic geometry or any of 1344 01:19:22,176 --> 01:19:23,426 this stuff in that. 1345 01:19:23,426 --> 01:19:25,940 1346 01:19:25,940 --> 01:19:29,110 It's high school algebra. 1347 01:19:29,110 --> 01:19:34,200 Well, freshman college algebra. 1348 01:19:34,200 --> 01:19:35,460 So that's what happens here. 1349 01:19:35,460 --> 01:19:39,056 1350 01:19:39,056 --> 01:19:41,310 It took 30 years and a computer scientist to solve 1351 01:19:41,310 --> 01:19:42,560 that problem. 1352 01:19:42,560 --> 01:19:50,380 1353 01:19:50,380 --> 01:19:55,110 That was not all I wanted to say, but that's 1354 01:19:55,110 --> 01:19:57,425 all I have time for. 1355 01:19:57,425 --> 01:19:59,720 There's one minute left, so there's no point in starting 1356 01:19:59,720 --> 01:20:01,060 something now. 1357 01:20:01,060 --> 01:20:02,603 Do you have any questions about any of this? 1358 01:20:02,603 --> 01:20:08,882 1359 01:20:08,882 --> 01:20:14,640 I should say that this 2 one can get rid of, but that takes 1360 01:20:14,640 --> 01:20:15,890 a little bit more machinery. 1361 01:20:15,890 --> 01:20:18,330 1362 01:20:18,330 --> 01:20:20,785 The 2 one can get rid of, but that's a bit more heavy. 1363 01:20:20,785 --> 01:20:24,544 1364 01:20:24,544 --> 01:20:25,040 All right. 1365 01:20:25,040 --> 01:20:27,530 So thanks so much. 1366 01:20:27,530 --> 01:20:28,910 That's it. 1367 01:20:28,910 --> 01:20:30,160 That's it. 1368 01:20:30,160 --> 01:20:36,948 | 36,814 | 90,148 | {"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} | 2.75 | 3 | CC-MAIN-2015-35 | longest | en | 0.914009 |
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