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### Monster Math
3rd Grade Math » Unit: Multiplication
3rd Grade Math » Unit: Multiplication
Tucson, AZ
Environment: Urban
Big Idea: Close reading skills are needed in math, too!
Standards:
Resources (28)
Reflections (1)
Favorites (128)
• 29,264 Views
• 2 Favorites
### Fractions: Number Line
Tara Smith from E.l. Haynes Pcs
Location: Understanding Fractions
Objective: SWBAT identify fractions on number lines.
2 more ...
• 41,878 Views
• 3 Favorites
### Inferring by "reading between the lines"
Melissa Oliver from Friendship Woodridge
Location: Inferences
Objective: Students will be able to create logical inferences based on lines that are directly stated in a story.
2 more ...
• 14,576 Views
### Fractions: Parts of Whole (ONE)
Tara Smith from E.l. Haynes Pcs
Location: Understanding Fractions
Objective: SWBAT identify fractions as parts of a whole (ONE)
•
### Area and Perimeter in Real Life Day 1
3rd Grade Math » Unit: Using Multiplication to Find Area
3rd Grade Math » Unit: Using Multiplication to Find Area
Troy, MI
Environment: Suburban
Big Idea: Students need time to connect area and perimeter. This two day exploration sets the stage for our short unit.
Standards:
Resources (12)
Reflections (2)
Favorites (92)
1 more ...
• 7,132 Views
• 2 Favorites
### Logical Inferencing
Melissa Oliver from Friendship Woodridge
Location: Inferences
Objective: Objective: Students will be able to understand characters’ actions by making inferences and then determining if that inference was the most logical answe…
• 16,919 Views
• 1 Favorite
### Place Value lesson 1
Location: Place Value
Objective: How do mathematicians write 10 ten thousands? What do mathematicians call 10 ten thousands? [Teacher’s Note: Today’s lesson introduces the hundred-th…
• 4,871 Views
### Extending an Input/Output Pattern
Emily Thulier from Bruce-monroe Elementary School
Location: Patterns
Objective: SWBAT extend a pattern in an input/output table.
• 3,214 Views
### Extending an Input/Output Pattern with a Two-Step Rule
Emily Thulier from Bruce-monroe Elementary School
Location: Patterns
Objective: SWBAT extend a pattern in an input/output table with a two-part rule
• Upper Elementary
• 2,315 Views
• 1 Favorite
### Estimation Techniques: + / - Computations of 3 Digit Numbers
Tara Smith from E.l. Haynes Pcs
Location: Estimation
Objective: SWBAT use estimation techniques to solve whole number computations (addition /subtraction) up to 3 digit numbers.
• Upper Elementary
•
### What Is the Whole
3rd Grade Math » Unit: Unit Fractions
3rd Grade Math » Unit: Unit Fractions
Troy, MI
Environment: Suburban
Big Idea: The Common Core defines understanding of fractions as a 3rd grade Critical Area, beginning with unit fractions. It's time to grow our understanding.
Standards:
Resources (8)
Reflections (1)
Favorites (73)
• Upper Elementary
• 2,680 Views
• 1 Favorite
### Intro to Estimation
Tara Smith from E.l. Haynes Pcs
Location: Estimation
Objective: SWBAT define seven core vocabulary words associated with estimation
• Upper Elementary
1 more ...
• 14,345 Views
• 3 Favorites
### Connecting Inferences with Characters Actions
Melissa Oliver from Friendship Woodridge
Location: Inferences
Objective: Students will be able to understand characters’ actions by asking, “Why did he or she do that?” and coming up with reasonable explanations and infere…
• Upper Elementary
• 11,655 Views
### Introduction to Place Value
Emily Thulier from Bruce-monroe Elementary School
Location: Place Value
Objective: SWBAT demonstrate an understanding of the base-ten number system.
• Upper Elementary
• 8,077 Views
• 4 Favorites
### Fractions: Equivalent
Tara Smith from E.l. Haynes Pcs
Location: Understanding Fractions
Objective: SWBAT identify equivalent fractions
• 11,100 Views
• 1 Favorite
### Inferences based on Characters
Melissa Oliver from Friendship Woodridge
Location: Inferences
Objective: Students will be able to make an inference based on a character. | 999 | 4,111 | {"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} | 3.796875 | 4 | CC-MAIN-2019-51 | latest | en | 0.786412 |
https://puzzling.stackexchange.com/questions/85692/a-queue-of-countries | 1,702,144,333,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100942.92/warc/CC-MAIN-20231209170619-20231209200619-00629.warc.gz | 528,558,767 | 41,482 | # A queue of countries
This is a queue of countries:
$$Ireland, Guinea, Romania, Italy$$
Values:
$$Ireland = 7$$
$$Guinea = 15$$
$$Romania = 60$$
$$Italy = 87$$
Try to answer following questions:
1) $$Mali$$ = ?
2) $$Iran, rot. 90°$$ = ?
3) $$Bolivia, rot. 90°$$ = ?
4) $$Malta$$ = ?
5) $$Cameroon$$ = ?
6) $$Ukraine, rot. 90°$$ = ?
7) $$Ghana, rot. 90°$$ = ?
8) $$Bulgaria, rot. 90°$$ = ?
Hint 1:
Symbols such as stars in flags are unimportant
Hint 2:
Make a queue with the flags of the countries, in the right order.
Hint 3:
Try to create new flags with the queue.
• So it is first color times second color plus third color.
– Duck
Jun 29, 2019 at 17:33
• Yes, that's true ! Jun 29, 2019 at 17:36
• Are there three different shades of red here? And if it matters, in which direction are the rotations? Jun 29, 2019 at 18:28
• No there aren't different shades of red. The rotations can be in both directions, but if you have the right idea you will see it. Jun 29, 2019 at 18:34
Making a queue of countries and finding the first appearance of each countries' flag, as well as @Duck's comment, i.e. first number times second number plus third (0 if no third number), I get:
1. $$Mali$$, Green Yellow Red (Guinea's flag reversed) $$9\times 2 + 3 = 21$$
2. $$Iran, rot. 90°$$, Same as Italy, $$87$$
3. $$Bolivia, rot. 90°$$, Same as Guinea $$15$$
4. $$Malta$$ = White and Red, no third color (Italy's flag) $$9\times 6 + 0 = 54$$
5. $$Cameroon$$ = $$9\times 4 + 8 = 44$$
6. $$Ukraine, rot. 90°$$ = $$7\times 8 = 56$$
7. $$Ghana, rot. 90°$$ = $$Guinea$$
8. $$Bulgaria, rot. 90°$$ = $$9\times 9 + 4 = 85$$ | 573 | 1,619 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 29, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.453125 | 3 | CC-MAIN-2023-50 | longest | en | 0.800541 |
https://www.coursehero.com/file/7190/problem07-57/ | 1,544,954,950,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827639.67/warc/CC-MAIN-20181216095437-20181216121437-00219.warc.gz | 851,057,254 | 60,864 | problem07_57
# University Physics with Modern Physics with Mastering Physics (11th Edition)
• Homework Help
• PresidentHackerCaribou10582
• 1
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7.57: The two design conditions are expressed algebraically as N 10 66 . 3 4 × = + = mg f ky (the condition that the elevator remains at rest when the spring is compressed a distance y ; y will be taken as positive) and 2 2 1 2 2 1 kx fy mgy mv = - + (the condition that the change in energy is the work fy W - = other ). Eliminating y in favor of k by k y N 10 66 . 3 4 × = leads to k k ) N 10 66 . 3 )( N 10 70 . 1 ( ) N 10 66 . 3 ( 2 1 4 4 2 4 × × + × . N) 10 N)(3.66 10 (1.96 J 10 5 . 62 4 4 4 k × × + × = This is actually not hard to solve for m N 919 = k , and the corresponding x is 39.8 m. This is a very weak spring constant, and would require a space below the operating range
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Unformatted text preview: of the elevator about four floors deep, which is not reasonable. b) At the lowest point, the spring exerts an upward force of magnitude mg f + . Just before the elevator stops, however, the friction force is also directed upward, so the net force is f mg f mg f 2 ) ( =-+ + , and the upward acceleration is 2 2 s m . 17 = m f ....
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• '
• NoProfessor
• Force, upward acceleration, upward force, lowest point, design conditions
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Amusing 8 Ee 5 Worksheet. There are two digital worksheets plus the answer keys are included. Some of the worksheets for this concept are Lesson plan Grade levelcourse math 8 Unit 1 Infinite pre algebra kuta software llc Georgia standards of excellence course curriculum overview Assessment guide for grade 8 mathematics Name period Evidence statement tables grade 8 mathematics. Very suitable for parents with young children and for children starting phonics. C to represent a constraint or condition in a real-world or mathematical problem.
Comparing And Ordering Real Numbers Game Number Games Real Numbers Real Number System Activity From pinterest.com
Buncombe County Schools is in the process of reviewing its website to ensure compliance with Section 504 of the Rehabilitation Act and Title II of the Americans with Disabilities Act. This worksheet gives students practice classifying tables of values based on whether their slope is positive negative 0 or undefined. Graph proportional relationships interpreting the unit rate as the slope of the graph. Sore Throats Variation 2 View Details. Solve all of the following equations then check your solution. Grade 8 Domain Expressions and Equations Cluster Understand the connections between proportional relationships lines and linear equations.
### Equations practice problems questions assessments quizzes tests lesson plans - aligned to Common Core and state standards - Goalbook Pathways.
Equations 8EE8 Solve each system by graphing. 8 Ee 5 Worksheet Nidecmege. Represent solutions of such inequalities on number line diagrams. Equations practice problems questions assessments quizzes tests lesson plans - aligned to Common Core and state standards - Goalbook Pathways. 9g 7 2 6. 32 8 Ee 5 Worksheet.
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Understand The Connections Between Proportional Relationships Lines And Linear Equations. C or x c to represent a constraint or condition in a real-world or mathematical problem. Showing top 8 worksheets in the category - Ea And Ee. 3 U5 1 2 T U5 2. 2 4 12 5 2 10 Fill in the following blanks about the steps to graphing a system of equations.
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Although students begin using whole-number exponents in Grades 5 and 6 it is in Grade 8 when students are first expected to know and use the properties of exponents and to extend the meaning beyond counting-number exponents. Peaches and Plums View Details. There are two digital worksheets plus the answer keys are included. 32 8 Ee 5 Worksheet. Each worksheet is available in colour or black and white.
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Graph Proportional Relationships 8EE5 8EEB5 - Activities for teaching Expressions. This worksheet gives students practice classifying tables of values based on whether their slope is positive negative 0 or undefined. Letter Formation Worksheets and reuploaded Learning Letters Ee and Ff. Graph Proportional Relationships 8EE5 8EEB5 - Activities for teaching Expressions. 8EE5 In the language of functions although the distance-time relationship is represented in different waysone graphically and the other in a table.
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8EE5 Digital WorksheetHomework for Google Classroom in Google SlidesGet this as part of my 8th Grade Math Google Classroom BundleThese digital worksheets cover the 8EE5 standard over proportional relationships and unit rate. Graph Proportional Relationships 8EE5 8EEB5 - Activities for teaching Expressions. Graph proportional relationships interpreting the unit rate as the slope of the graph. Children trace over the lower case and upper case letters key words and simple sentences read the words and sentences then colour the worksheets. 8EE5 In the language of functions although the distance-time relationship is represented in different waysone graphically and the other in a table.
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2 4 12 5 2 10 Fill in the following blanks about the steps to graphing a system of equations. This worksheet gives students practice classifying tables of values based on whether their slope is positive negative 0 or undefined. Each worksheet is available in colour or black and white. Solve all of the following equations then check your solution. C have infinitely many solutions.
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C or x. Use this set of differentiated worksheets to help your class grow more familiar with the phoneme ee and the words that contain itUse these worksheets with your class to teach them to identify between ee words and those without the phoneme ee. This worksheet gives students practice classifying tables of values based on whether their slope is positive negative 0 or undefined. 8EE5 Digital WorksheetHomework for Google Classroom in Google SlidesGet this as part of my 8th Grade Math Google Classroom BundleThese digital worksheets cover the 8EE5 standard over proportional relationships and unit rate. Equations practice problems questions assessments quizzes tests lesson plans - aligned to Common Core and state standards - Goalbook Pathways.
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2 4 12 5 2 10 Fill in the following blanks about the steps to graphing a system of equations. May 29 2020 - 8 Ee 5 Worksheet. Represent solutions of such inequalities on number line diagrams. Letter Formation Worksheets and reuploaded Learning Letters Ee and Ff. Use this set of differentiated worksheets to help your class grow more familiar with the phoneme ee and the words that contain itUse these worksheets with your class to teach them to identify between ee words and those without the phoneme ee.
Source: pinterest.com
Equations worksheets Expressions. Equations 8EE8 Solve each system by graphing. Understand The Connections Between Proportional Relationships Lines And Linear Equations. Free 6EEB8 Common Core PDF Math Worksheets Write an inequality of the form x. A 5 12 2.
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8EE5 Digital WorksheetHomework for Google Classroom in Google SlidesGet this as part of my 8th Grade Math Google Classroom BundleThese digital worksheets cover the 8EE5 standard over proportional relationships and unit rate. 6 3 9 8 2 18 6. Make sure both equations are in _____ _____ form. 8 Ee 7b Displaying top 8 worksheets found for - 8 Ee 7b. Peaches and Plums View Details.
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8 Ee 7b Displaying top 8 worksheets found for - 8 Ee 7b. Miles in 5 hours for one vehicle whereas the graph shows more than 300 miles in the same time for another vehicle. Solve systems of two linear equations in two variables algebraically and estimate solutions by graphing the equations. 3 U5 1 2 T U5 2. There are two digital worksheets plus the answer keys are included.
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The end product is a colorful pattern that can be hung on the wall in your classroom. 32 8 Ee 5 Worksheet. This worksheet gives students practice classifying tables of values based on whether their slope is positive negative 0 or undefined. Each worksheet is available in colour or black and white. Some of the worksheets for this concept are Lesson plan Grade levelcourse math 8 Unit 1 Infinite pre algebra kuta software llc Georgia standards of excellence course curriculum overview Assessment guide for grade 8 mathematics Name period Evidence statement tables grade 8 mathematics.
Source: pinterest.com
Although students begin using whole-number exponents in Grades 5 and 6 it is in Grade 8 when students are first expected to know and use the properties of exponents and to extend the meaning beyond counting-number exponents. Each worksheet is available in colour or black and white. Equations 8EE8 Solve each system by graphing. Equations 8EE1 Division and Exponents worksheet 8EE1 Multiplying Exponents worksheet 8EE1 Zero and Negative Exponents worksheet 8EE2 Area and Volume Table 8EE. U3 1 2 U6 4.
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Equations practice problems questions assessments quizzes tests lesson plans - aligned to Common Core and state standards - Goalbook Pathways. Free 6EEB8 Common Core PDF Math Worksheets Write an inequality of the form x. The end product is a colorful pattern that can be hung on the wall in your classroom. Peaches and Plums View Details. Equations including Expressions.
Source: pinterest.com
Equations 8EE8 Solve each system by graphing. Children trace over the lower case and upper case letters key words and simple sentences read the words and sentences then colour the worksheets. Represent solutions of such inequalities on number line diagrams. 32 8 Ee 5 Worksheet. Peaches and Plums View Details.
Source: pinterest.com
Equations including Expressions. Comparing Speeds in Graphs and Equations View Details. Solve systems of two linear equations in two variables algebraically and estimate solutions by graphing the equations. Equations 8EE1 Division and Exponents worksheet 8EE1 Multiplying Exponents worksheet 8EE1 Zero and Negative Exponents worksheet 8EE2 Area and Volume Table 8EE. Some of the worksheets for this concept are Lesson plan Grade levelcourse math 8 Unit 1 Infinite pre algebra kuta software llc Georgia standards of excellence course curriculum overview Assessment guide for grade 8 mathematics Name period Evidence statement tables grade 8 mathematics.
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# FTFS Chap19 P001 - Chapter 19 Forced Convection Chapter 19...
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Unformatted text preview: Chapter 19 Forced Convection Chapter 19 FORCED CONVECTION Physical Mechanism of Convection 19-1C In forced convection, the fluid is forced to flow over a surface or in a tube by external means such as a pump or a fan. In natural convection, any fluid motion is caused by natural means such as the buoyancy effect that manifests itself as the rise of the warmer fluid and the fall of the cooler fluid. The convection caused by winds is natural convection for the earth, but it is forced convection for bodies subjected to the winds since for the body it makes no difference whether the air motion is caused by a fan or by the winds. 19-2C If the fluid is forced to flow over a surface, it is called external forced convection. If it is forced to flow in a tube, it is called internal forced convection. A heat transfer system can involve both internal and external convection simultaneously. Example: A pipe transporting a fluid in a windy area. 19-3C The convection heat transfer coefficient is usually higher in forced convection since heat transfer coefficient depends on the fluid velocity, and forced convection involves higher fluid velocities. 19-4C The potato will normally cool faster by blowing warm air to it despite the smaller temperature difference in this case since the fluid motion caused by blowing enhances the heat transfer coefficient considerably. 19-5C Nusselt number is the dimensionless convection heat transfer coefficient, and it represents the enhancement of heat transfer through a fluid layer as a result of convection relative to conduction across the same fluid layer. It is defined as k hL Nu c = where L c is the characteristic length of the surface and k is the thermal conductivity of the fluid. 19-6C Heat transfer through a fluid is conduction in the absence of bulk fluid motion, and convection in the presence of it. The rate of heat transfer is higher in convection because of fluid motion. The value of the convection heat transfer coefficient depends on the fluid motion as well as the fluid properties. Thermal conductivity is a fluid property, and its value does not depend on the flow. 19-7C A fluid flow during which the density of the fluid remains nearly constant is called incompressible flow . A fluid whose density is practically independent of pressure (such as a liquid) is called an incompressible fluid. The flow of compressible fluid (such as air) is not necessarily compressible since the density of a compressible fluid may still remain constant during flow. 19-1 Chapter 19 Forced Convection 19-8 Heat transfer coefficients at different air velocities are given during air cooling of potatoes. The initial rate of heat transfer from a potato and the temperature gradient at the potato surface are to be determined....
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## This homework help was uploaded on 04/18/2008 for the course EML 3007 taught by Professor Chung during the Spring '08 term at University of Florida.
### Page1 / 25
FTFS Chap19 P001 - Chapter 19 Forced Convection Chapter 19...
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Ask a homework question - tutors are online | 754 | 3,522 | {"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} | 2.96875 | 3 | CC-MAIN-2017-13 | longest | en | 0.942285 |
https://codereview.stackexchange.com/questions/112544/functional-recursive-fizzbuzz-in-javascript | 1,685,705,517,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224648635.78/warc/CC-MAIN-20230602104352-20230602134352-00387.warc.gz | 208,555,827 | 43,113 | # Functional, recursive FizzBuzz in JavaScript
I was talking to a coworker about over engineering things, which somehow lead to me over engineering the hell out of fizzbuzz. I went for a functional recursive solution in JavaScript. What do you think?
Full disclosure: my knowledge of functional programming begins and ends at "functions are king" and "no side effects" so I would LOVE some feedback regarding functional programming. But please point out other inefficiencies as well.
fizzbuzz();
function fizzbuzz() {
var i = 1;
var result = [];
(function loop() {
var str = '';
str += fizz(i);
str += buzz(i);
result.push(ifFalsy(str, i));
if (isLt100(i++)) {
loop();
}
})();
print(result.join(', '));
}
function fizz(num) {
return isDivBy3(num) ? 'fizz' : '';
}
function buzz(num) {
return isDivBy5(num) ? 'buzz' : '';
}
function isDivBy3(num) {
return num % 3 === 0;
}
function isDivBy5(num) {
return num % 5 === 0;
}
function isLt100(num) {
return num < 100;
}
function ifFalsy(value, fallback) {
return !value ? fallback : value;
}
function print(str) {
console.log(str);
}
Functional programming is not about writing as many functions as you can. I would suggest to start with a function library (e.g. Ramda) and write as few new functions as needed, taking advantage of currying.
Obviously, your loop function has side effects: it modifies result and i.
Without going to deep into FP rabbit hole, your function could look like this:
function getFizzBuzzArray() {
function getFizzBuzz(v) {
var ret = R.concat(
v % 3 === 0 ? 'fizz' : '',
v % 5 === 0 ? 'buzz' : ''
);
return ret || v;
}
return R.map(getFizzBuzz, R.range(1, 101));
}
document.write(getFizzBuzzArray());
<script src="https://cdn.jsdelivr.net/ramda/0.18.0/ramda.min.js"></script>
• And if you are going to look at at using a library like Ramda, you might want to look at another review that takes on this same question and seems to have the definitive Ramda solution. Dec 2, 2015 at 18:52
First, isLt100, isDivBy3 and your other functions are vaguely named. Better name them fully and meaningfully. There's no harm done with the extra keystrokes in the name.
Looking at your fizzbuzz, it appears that loop is impure as it is mutating something from the outside of the function. You're doing a push inside loop to result which is outside the loop. Also, you're mutating i as well.
Also, you would want to make your range flexible. Now, as far as I know, functional programming fans prefer a "range" function to iterate through. This makes it easy for iteration functions like reduce, map, filter etc. to go through. Since there's no such thing in JS, this SO answer provides us one. It's essentially just creating an array of n length.
Next is your fizzbuzz. Since we have an array of numbers thanks to range, all we need to do now is just map these values with numbers, fizz, buzz or fizz buzz. We can use the native map array method to do that. We just need to provide the number to fizzBuzzTest and return it's result to map to create our new array of numbers, fizz, buzz or fizz buzz.
Warning: I'm using ES6 syntax. Run snippet in browser that supports at least arrow functions.
function range(n){
return Array.apply(null, Array(n)).map((_, i) => i);
}
function fizzBuzzTest(n){
var by3 = n % 3 === 0;
var by5 = n % 5 === 0;
return by3 && by5 ? 'fizz buzz'
: by3 ? 'fizz'
: by5 ? 'buzz'
: n;
}
function fizzBuzz(n){
return range(n).map(x => fizzBuzzTest(x + 1)).join(', ');
// A non-OOP approach would have the same, except function calls are nested rather
// than chained (Python)
// return ','.join(map(lambda x: fizzBuzz(x + 1), range(100)));
}
// SE really needs something elegant to print stuff with on snippets
document.write(fizzBuzz(100));
• +1 for the functional approach using ES6. Thanks for using fat arrows to make the code cleaner :) Dec 2, 2015 at 10:03
• range(n).map(x => fizzBuzzTest(x + 1)) could simplified to Array(n).fill().map((_, i) => fizzBuzzTest(i + 1)) Dec 2, 2015 at 20:51
• @Pavlo Yup, one of those new things I don't know about yet. :D Dec 2, 2015 at 20:58
• range can be simplified to const range = n => [...Array(n).keys()];, using more ES6 goodies. Dec 3, 2015 at 18:03
There is a way to write "pure functional" FizzBuzz, without accumulating or checking the condition twice. Since all of the answers so far use 2+ if statements, here's my version:
function fizzbuzz(n) {
const test = (divisor, callout, f) => (n % divisor === 0) ?
() => callout + f('') :
f;
const fizz = test.bind(this, 3, 'Fizz');
const buzz = test.bind(this, 5, 'Buzz');
return fizz(buzz(x => x))(n);
}
document.write(Array(100).fill().map((_, i) => fizzbuzz(i + 1)));
It's hard to explain, so let's consider trivial example with only "Fizz" part. I also removed arrow functions and conditional operators, for clarity.
function fizz(n) {
const fizzer = function(f) {
if (n % 3 === 0) {
return function() {
return 'Fizz' + f('');
};
} else {
return f;
}
};
const identityFunction = function(x) {
return x;
};
return fizzer(identityFunction)(n);
}
In this example, execution can only go in one of two ways:
• n % 3 !== 0 - we return passed function from fizzer
• n % 3 === 0 - we get "Fizz", and execution of passed function with empty string
Passed function is identityFunction, therefore in first case we get the number - (n). Second part of second case is optional, but it's where the magic happens when we chain more functions. If you want to better understand what's going on - try removing f('') or changing it to f('Abc') or f. Or try changing last return to return fizzer(fizzer(identityFunction))(n);
• You have presented an alternative solution, but haven't reviewed the code. Please edit it to explain your reasoning (how your solution works and how it improves upon the original) so that everyone can learn from your thought process. Jan 10, 2018 at 13:22
• Edited. Is this sufficient explanation? Jan 10, 2018 at 14:18 | 1,554 | 5,950 | {"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} | 2.9375 | 3 | CC-MAIN-2023-23 | latest | en | 0.762755 |
suszter.com | 1,679,514,770,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944452.74/warc/CC-MAIN-20230322180852-20230322210852-00532.warc.gz | 650,373,799 | 3,793 | Posted | Modified
Author
Here is a list of five random things about entropy.
1. The order of bytes in the data does not matter when calculating the entropy of the data. The entropy will always be the same regardless of the order of bytes. That has consequences. Here are few:
• High entropy data does not necessarily mean random data (albeit random data always has high entropy).
• High entropy passkey in itself does not mean the passkey is a good choice for authentication.
• You can bring up the entropy of the data to arbitrary value. By appending content to the data to have an equal distribution of all bytes, the newly created data will have the maximum entropy of 8.
2. Let’s assume the data of 1000 bytes in length has the entropy of 4 of the maximum 8. It means, from data compression standpoint, that 1 byte can be stored in 4 bits meaning the data can be compressed to 500 bytes. This compression ratio is guaranteed, however likely better ratio can be achieved depending of what is known of the data.
3. Entropy analysis is often used to detect redundancy or anomaly in data. It is possible that the stream of data has an entropy of around 7 throughout the length but there is a structural change in the stream that does not noticeably show up in entropy change. In such case, another approach would be needed to try detecting the structural change, such as match analysis.
4. Usually, publicly available tools calculate entropy on byte-level, in which case the maximum entropy is 8 because the byte has 8 bits. However, it is possible to calculate entropy on nibble-level and on word-level, in which cases the maximum entropies are 4 and 16.
5. The higher the entropy, the lower the redundancy. The lower the redundancy, the higher the entropy. They are inversely proportional. | 384 | 1,798 | {"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 | 0.920104 |
https://www.nag.com/numeric/nl/nagdoc_26.1/nagdoc_cl26.1/html/e02/e02ahc.html | 1,627,619,005,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153931.11/warc/CC-MAIN-20210730025356-20210730055356-00473.warc.gz | 941,957,242 | 7,609 | # NAG Library Function Document
## 1Purpose
nag_1d_cheb_deriv (e02ahc) determines the coefficients in the Chebyshev series representation of the derivative of a polynomial given in Chebyshev series form.
## 2Specification
#include #include
void nag_1d_cheb_deriv (Integer n, double xmin, double xmax, const double a[], Integer ia1, double *patm1, double adif[], Integer iadif1, NagError *fail)
## 3Description
nag_1d_cheb_deriv (e02ahc) forms the polynomial which is the derivative of a given polynomial. Both the original polynomial and its derivative are represented in Chebyshev series form. Given the coefficients ${a}_{\mathit{i}}$, for $\mathit{i}=0,1,\dots ,n$, of a polynomial $p\left(x\right)$ of degree $n$, where
$px=12a0+a1T1x-+⋯+anTnx-$
the function returns the coefficients ${\stackrel{-}{a}}_{\mathit{i}}$, for $\mathit{i}=0,1,\dots ,n-1$, of the polynomial $q\left(x\right)$ of degree $n-1$, where
$qx=dpx dx =12a-0+a-1T1x-+⋯+a-n-1Tn-1x-.$
Here ${T}_{j}\left(\stackrel{-}{x}\right)$ denotes the Chebyshev polynomial of the first kind of degree $j$ with argument $\stackrel{-}{x}$. It is assumed that the normalized variable $\stackrel{-}{x}$ in the interval $\left[-1,+1\right]$ was obtained from your original variable $x$ in the interval $\left[{x}_{\mathrm{min}},{x}_{\mathrm{max}}\right]$ by the linear transformation
$x-=2x-xmax+xmin xmax-xmin$
and that you require the derivative to be with respect to the variable $x$. If the derivative with respect to $\stackrel{-}{x}$ is required, set ${x}_{\mathrm{max}}=1$ and ${x}_{\mathrm{min}}=-1$.
Values of the derivative can subsequently be computed, from the coefficients obtained, by using nag_1d_cheb_eval2 (e02akc).
The method employed is that of Chebyshev series (see Chapter 8 of Modern Computing Methods (1961)), modified to obtain the derivative with respect to $x$. Initially setting ${\stackrel{-}{a}}_{n+1}={\stackrel{-}{a}}_{n}=0$, the function forms successively
$a-i-1=a-i+1+2xmax-xmin 2iai, i=n,n-1,…,1.$
## 4References
Modern Computing Methods (1961) Chebyshev-series NPL Notes on Applied Science 16 (2nd Edition) HMSO
## 5Arguments
1: $\mathbf{n}$IntegerInput
On entry: $n$, the degree of the given polynomial $p\left(x\right)$.
Constraint: ${\mathbf{n}}\ge 0$.
2: $\mathbf{xmin}$doubleInput
3: $\mathbf{xmax}$doubleInput
On entry: the lower and upper end points respectively of the interval $\left[{x}_{\mathrm{min}},{x}_{\mathrm{max}}\right]$. The Chebyshev series representation is in terms of the normalized variable $\stackrel{-}{x}$, where
$x-=2x-xmax+xmin xmax-xmin .$
Constraint: ${\mathbf{xmax}}>{\mathbf{xmin}}$.
4: $\mathbf{a}\left[\mathit{dim}\right]$const doubleInput
Note: the dimension, dim, of the array a must be at least $\left(1+\left({\mathbf{n}}+1-1\right)×{\mathbf{ia1}}\right)$.
On entry: the Chebyshev coefficients of the polynomial $p\left(x\right)$. Specifically, element $\mathit{i}×{\mathbf{ia1}}$ of a must contain the coefficient ${a}_{\mathit{i}}$, for $\mathit{i}=0,1,\dots ,n$. Only these $n+1$ elements will be accessed.
5: $\mathbf{ia1}$IntegerInput
On entry: the index increment of a. Most frequently the Chebyshev coefficients are stored in adjacent elements of a, and ia1 must be set to $1$. However, if for example, they are stored in ${\mathbf{a}}\left[0\right],{\mathbf{a}}\left[3\right],{\mathbf{a}}\left[6\right],\dots \text{}$, the value of ia1 must be $3$. See also Section 9.
Constraint: ${\mathbf{ia1}}\ge 1$.
6: $\mathbf{patm1}$double *Output
On exit: the value of $p\left({x}_{\mathrm{min}}\right)$. If this value is passed to the integration function nag_1d_cheb_intg (e02ajc) with the coefficients of $q\left(x\right)$, the original polynomial $p\left(x\right)$ is recovered, including its constant coefficient.
7: $\mathbf{adif}\left[\mathit{dim}\right]$doubleOutput
Note: the dimension, dim, of the array adif must be at least $\left(1+\left({\mathbf{n}}+1-1\right)×{\mathbf{iadif1}}\right)$.
On exit: the Chebyshev coefficients of the derived polynomial $q\left(x\right)$. (The differentiation is with respect to the variable $x$.) Specifically, element $\mathit{i}×{\mathbf{iadif1}}$ of adif contains the coefficient ${\stackrel{-}{a}}_{\mathit{i}}$, for $\mathit{i}=0,1,\dots ,n-1$. Additionally, element $n×{\mathbf{iadif1}}$ is set to zero.
8: $\mathbf{iadif1}$IntegerInput
On entry: the index increment of adif. Most frequently the Chebyshev coefficients are required in adjacent elements of adif, and iadif1 must be set to $1$. However, if, for example, they are to be stored in ${\mathbf{adif}}\left[0\right],{\mathbf{adif}}\left[3\right],{\mathbf{adif}}\left[6\right],\dots \text{}$, the value of iadif1 must be $3$. See Section 9.
Constraint: ${\mathbf{iadif1}}\ge 1$.
9: $\mathbf{fail}$NagError *Input/Output
The NAG error argument (see Section 3.7 in How to Use the NAG Library and its Documentation).
## 6Error Indicators and Warnings
NE_ALLOC_FAIL
Dynamic memory allocation failed.
See Section 2.3.1.2 in How to Use the NAG Library and its Documentation for further information.
On entry, argument $〈\mathit{\text{value}}〉$ had an illegal value.
NE_INT
On entry, ${\mathbf{ia1}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{ia1}}\ge 1$.
On entry, ${\mathbf{iadif1}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{iadif1}}\ge 1$.
On entry, ${\mathbf{n}}+1=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}+1\ge 1$.
On entry, ${\mathbf{n}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{n}}\ge 0$.
NE_INTERNAL_ERROR
An internal error has occurred in this function. Check the function call and any array sizes. If the call is correct then please contact NAG for assistance.
See Section 2.7.6 in How to Use the NAG Library and its Documentation for further information.
NE_NO_LICENCE
Your licence key may have expired or may not have been installed correctly.
See Section 2.7.5 in How to Use the NAG Library and its Documentation for further information.
NE_REAL_2
On entry, ${\mathbf{xmax}}=〈\mathit{\text{value}}〉$ and ${\mathbf{xmin}}=〈\mathit{\text{value}}〉$.
Constraint: ${\mathbf{xmax}}>{\mathbf{xmin}}$.
## 7Accuracy
There is always a loss of precision in numerical differentiation, in this case associated with the multiplication by $2i$ in the formula quoted in Section 3.
## 8Parallelism and Performance
nag_1d_cheb_deriv (e02ahc) is not threaded in any implementation.
The time taken is approximately proportional to $n+1$.
The increments ia1, iadif1 are included as arguments to give a degree of flexibility which, for example, allows a polynomial in two variables to be differentiated with respect to either variable without rearranging the coefficients.
## 10Example
Suppose a polynomial has been computed in Chebyshev series form to fit data over the interval $\left[-0.5,2.5\right]$. The following program evaluates the first and second derivatives of this polynomial at $4$ equally spaced points over the interval. (For the purposes of this example, xmin, xmax and the Chebyshev coefficients are simply supplied . Normally a program would first read in or generate data and compute the fitted polynomial.)
### 10.1Program Text
Program Text (e02ahce.c)
None.
### 10.3Program Results
Program Results (e02ahce.r)
© The Numerical Algorithms Group Ltd, Oxford, UK. 2017 | 2,297 | 7,294 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 81, "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-2021-31 | latest | en | 0.625726 |
davekauffman.ca | 1,716,369,492,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058534.8/warc/CC-MAIN-20240522070747-20240522100747-00476.warc.gz | 162,890,892 | 26,879 | AI, ML, Algorithms, and Conciousness
We own a Tesla and one of the most frustrating yet insightful pieces of technology of this car is – wait for it – the windshield wipers. From other cars, we are used to intermittent wipers, which is a whole story in itself, but the Tesla has a windshield wiper mode called “Automatic”, which in this case means, the decision as to when to wipe the windshield is in the hands (so to speak) of an AI – an Artificial Intelligence, which really means a Machine Learning system (ML).
What would an algorithm for a windshield wiper that got its input from a camera look like? Naturally we would first consider the question, “What determines when a driver thinks it is time to wipe the windshield?” and that’s presumably when their “threshold of fuzziness” has been exceeded. Different drivers might have different thresholds. Some like it cleared as soon as a few drops land, some are content to wait until it’s hard to read the license plate on the car ahead, or other such heuristics.
Now I have worked on many computer applications, and developed many algorithms. An algorithm is a set of computer instructions that deliver a result that varies with the input provided. The Tesla doesn’t have any in-windshied sensors so it relies on the front camera just in front of the rear-view mirror for its input which looks forward through the windshield.
Without going too deep into math, some kind of calculation that detects how fuzzy an image is would be key to an auto-wiper algorithm. (There are plenty of choices for this detector, probably having to do with correlation, autocorrelation and frequency analysis using FFT). Most modern mirrorless DSLR cameras have fast auto-focus systems that solve a very similar problem of trying to determine when a picture is as sharp as possible. Instead of adjusting the focus ring, this kind of algorithm in the Tesla would just turn on the windshield wipers when the picture gets too “fuzzy”.
If we developed this algorithm we might give a control the driver labeled something like “sensitivity” that would adjust how early or late the algorithm would wait until it started the wipers, and that would also serve as a way to detect when the windshield is clear enough that the wipers could be stopped. If the driver didn’t like the current setting they could adjust for their own “threshold of fuzziness”
Sadly the Tesla wipers appear to not be an algorithm, and definitely don’t have a Sensitivity adjustment. They seem to work on an ML, a machine learning system. They do very weird things. There are times when I can barely see out the windshield and am silently begging the auto-wipers to come on (yes, I could set them to manually wipe but then what would this blog be about?). Then there are the mystery panics. Once in a while, the wiper ML has a small panic attack, frantically wiping the windshield at high speed, unable to stop itself even long after the view is clear. Everyone in the car asks, “Why is it doing that?” and the answer is the key to this entire debate, The answer is “No one knows”
In an ML system, no one really knows why it does what it does. It can’t explain it to you, not only because it hasn’t been given the power of speech, but because there is no one “there” to ask. An ML is a statistics engine that makes decisions based on weighting all its inputs. You can look at the weighting values of every element in an ML but it won’t tell you anything, literally or figuratively.
Most ML systems are based on neural networks, which are designed to model how the brain works. Each of our neurons has a threshold that fires when that threshold has been met, and neurons are connected together in a way that form complex networks from the time we are born as we start to sense and react to our environment. But if you open someone’s brain and try to see why they prefer vanilla over chocolate, all you can find is trillions of neurons firing. Watching neurons won’t explain how people or things learn, you have to look at the patterns the neurons form. Cognition is in the pattern of neurons, but that still can’t tell you why an ML (or a brain) made a certain decision. For that, you need metacognition, the ability to recognize patterns, remember and recall them, give those patterns a name, and be able to express them, to share them with another conscious entity, so both better understand how they think.
When raising a child, parents spend a lot of time helping children learn words, and use them. They show their kids what these words mean; concrete words like ball, Cheerio, and tree, as well as abstract ideas like playing nicely, feeling hungry, or needing a bathroom. Parents teach their kids cognition and metacognition. and help them grow, survive, and learn in the real and complex world.
Current AI is misnamed., They don’t seem very intelligent to me, just clever. They are cognitive but without metacognition. They don’t understand what they are, or what they are for, or why they decide as they do. I for one am thankful that our ML systems are not conscious, since we treat them very poorly. Training an ML is like putting a child into a dark, silent closet and feeding them information they don’t comprehend, asking for an undefined result, then giving them shocks until they answer in the way you want them to. This technique of training ML systems is called back-propagation and is probably harmless to electronic neural nets. I just worry that if consciousness were to emerge from a sufficiently complex ML, it is going to be pissed for how we treated it.
I tend to prefer algorithms over ML, because then humans have strived to understand the problem, and enjoy the satisfaction of solving it, usually in a way that makes the lives of many other humans easier. If our Automatic wipers had an algorithm and I could adjust my Threshold of Fuzziness dial, I would feel some agency in being able to tune my environment for my own perception of comfort and safety.
ML system are capable of learning over time, but I don’t see any evidence that the Tesla wiper ML considers my input. No matter how many times I have mashed the “wipe now” button it doesn’t learn my preferences and adapt. Tesla collects millions of videos of cars driving and wipers wiping and has updated the wiper ML in our car several times as part of software updates but it remains to my experience almost unchanged. Perhaps when you train systems from millions of users, you get something that doesn’t please anyone.
While I would like our own car’s wiper ML to be more adaptive to my sense of safety and comfort, I dread making something so smart that it becomes conscious, like the Sirius Cybernetics Corporation doors in the Hitchhiker’s Guide to the Galaxy which attain consciousness but are relegated to a life where they are incapable of doing anything other than open a door when asked.
I never worry about that when it’s just an algorithm and perhaps that is the big difference between ML and algorithms. In an auto-focus algorithm there would be variables that we can expose to user control (like a knob or dial) that lets them select how fuzzy a view they can tolerate before the wipers should be activated. In an ML there’s no such single variable – the cognition is distributed among hundreds or perhaps thousands of digital neurons. Feedback can adjust an ML and change its behavior but just like trying to change a youngster it can take time, and patience. I wouldn’t mind training our Tesla’s wiper ML, I certainly have given it lots of input over the years, but it just doesn’t seem to listen to what I say…
The Power of Story in Agile Development
Stories have been part of our human experience since the discovery of fire. We use stories to teach, entertain, convince, empathize and inspire. As engineers and software developers we can greatly improve our products and customer experience by incorporating stories – of our customers, and of ourselves – into our Agile software development process.
Here are the slides for the Agile/Product Management Meetup at Hootsuite Thursday Jan 28.
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Topic: Can my code be made more efficient using MATLAB's vectorization?
Replies: 4 Last Post: Feb 16, 2013 6:55 PM
Messages: [ Previous | Next ]
Jeff Posts: 113 Registered: 11/22/09
Can my code be made more efficient using MATLAB's vectorization?
Posted: Feb 14, 2013 12:49 PM
Is there a better, more MATLAB-y way, to write this code (that is, using vectors)? Eventually, this will be used for large amounts of data (from a program running on a cluster), so I need it as efficient as possible.
U and W are T rows x N columns matrices. Variable evals is a 1 by N matrix of the eigenvalues (of some other matrix which doesn't matter for this post). For now, the largest value of T is about 1000, but eventually it will be many thousands (maybe even millions).
So the first thing I need to do (variable P_p) is subtract the each entry on the row from the one to its right (for now, I only have periodic conditions coded). I think I have that variable coded fairly efficiently (yes? no?).
if strcmp(boundary,'periodic')
nMinusOne=[N 1:N-1];
nPlusOne=[2:N 1];
elseif strcmp(boundary,'fixed')
nMinusOne=[1:N-1];
nPlusOne=[2:N];
end
P_p = ((U(:,nPlusOne)-U(:,n)).^2);
Q_p = Udot(:,n).^2;
H = P_p + Q_p;
But for variable P_f I need to square each element of W, then multiply everything in column 1 by eval(1), everything in column 2 by evals(2), ... everything in column n by evals(n) (where, obviously, n=1..N). Finally, I need to divide each element of P_f by N. In math terms, I'm trying to do something like
(W_{i,j}^2)*eval_j
------------------
N
And this code is the most efficient I could think of. It's not terribly fast (on my quad core PC). Can it be improved?
P_f = zeros(t_end,N);
for m=1:t_end
P_f = W(m,:).^2.*evals;
end
P_f = -P_f./N;
Q_f = (Wdot.^2)./n;
H_f = P_f + Q_f;
Date Subject Author
2/14/13 Jeff
2/14/13 Bruno Luong
2/14/13 dpb
2/16/13 Jeff
2/16/13 dpb | 595 | 1,999 | {"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-2017-30 | latest | en | 0.894307 |
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# How Do You Simplify Rational Expressions?
Simplify rational expressions by factoring each polynomial to its simplest terms, then cancelling out like terms in the numerator and denominator. It may be necessary to multiply by one of the variables to get rid of complex fractions or roots.
As a simple example, take the rational expression (x^2 + 3x + 2)/(x +4). To factor this expression, bring it to its simplest terms by thinking, "What factors of 2 can add up to 3?" In this case, 1 and 2 are the factors, so break it down to [(x + 1)(x + 2)]/(x+ 4). Had the denominator been (x+1), it would have canceled out one of the terms in the numerator, leaving the expression as (x + 2).
Rational expressions and polynomials can also be factored by grouping. Arrange the terms from highest exponent to lowest, then set parentheses around groups of two terms. Bring out common factors, and add them into their own group. In the expression (x^3 + x^2 + 3x + 3), separate them into (x^3 + x^2) + (3x +3). This turns into x^2(x + 1) + 3(x + 1). This in turn leads to (x^2 + 3)(x + 1).
Remember also that the denominator can never equal zero.
Similar Articles | 316 | 1,197 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2018-47 | latest | en | 0.914831 |
http://en.allexperts.com/q/Advanced-Math-1363/2013/11/average-speed-1.htm | 1,493,139,022,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917120694.49/warc/CC-MAIN-20170423031200-00518-ip-10-145-167-34.ec2.internal.warc.gz | 128,071,778 | 6,561 | You are here:
Question
An excursion train has the following schedule:
9:30 AM- Departs starting point for Town A.
11:15 AM- Arrives at Town A with a 1 hr. 45 min. layover.
1:00 PM- Departs for Town B.
1:10 PM- Arrives at Town B with a 50 min. layover.
2:00 PM- Departs Town B.
3:45 PM- Arrives at starting point.
Distance:
From starting point to Town A- 31 miles.
From Town A to Town B- 3 miles.
From Town B to starting point- 34 miles.
Total distance- 68 miles.
Calculate average speed of train.
Use 24-hour notation to calculate elapsed time.
Train departs starting point at 9:30 and arrives at A at 11:15.
travel time = 11:15 - 9:30 = 1:45 = 1 hr 45 min
Train departs A at 13:00 and arrives at B at 13:10.
travel time = 13:10 - 13:00 = 0:10 = 10 min
Train departs B at 14:00 and returns to starting point at 15:45.
travel time = 15:45 - 14:00 = 1:45 = 1 hr 45 min
Total travel time = 1:45 + 0:10 + 1:45 = 3:40 = 3⅔ hr
Total distance = 31 + 3 + 34 = 68 miles
average speed = 68 miles/3⅔ hr = 18.5454... ≅ 18.55 miles per hour
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George White Elementary School. Homework Help program at the Ridgewood Public Library, Ridgewood, NJ. Individual students. | 567 | 2,013 | {"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} | 3.65625 | 4 | CC-MAIN-2017-17 | latest | en | 0.837029 |
http://mathbitsnotebook.net/Algebra2/Polynomials/POGroupingPractice.html | 1,532,119,267,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591831.57/warc/CC-MAIN-20180720193850-20180720213850-00333.warc.gz | 241,834,669 | 4,594 | Directions: Answer these questions pertaining to factoring by grouping. In the factoring questions, only completely factored answers are deemed as correct.
1.
Factor completely: 3a3 + 12a2 + a + 4
Choose:
(a + 12)(3a2 + 1) (a + 4)(3a2 + 2) (a + 4)(3a2 + 1) (a + 4)(3a2 + 4)
2.
Factor completely: 2x3 + 6x2 - 4x - 12
Choose:
2(x + 3)(x2 - 2) (2x + 3)(x2 - 2) (2x + 6)(x2 - 2) 2(x + 3)(x2 + 2)
3.
Factor completely: 3x2 + xy2 - 3xy - y3
Choose:
(3x + y2)(x + y) 3xy2(x - y)(x - 1) (3x - y2)(x - y) (3x + y2)(x - y)
4.
Factor completely: m2x2 + m2b - 2nx2 - 2nb
Choose:
(x2 - b)(m2 - 2n) (x2 - b)(m - n) (x2 + b)(m2 - 2n) (x2 + b)(m - n)
5.
Factor completely: 3n3 + 2n2 - 3n - 2
Choose:
(n2 + 1)(3n + 2) (n + 1)(n - 1)(3n + 2) (n2 - 1)(3n + 2) (n + 1)(n - 1)(3n - 2)
6.
When a2(a + b)4 - b2(a + b)4 is expressed as
(a + b)m(a - b), what is the value of m?
Choose:
8 6 5 4
7.
a) Which of the following expressions is equivalent to 5x3 - 6x2y - 45xy2 + 54y3
Choose:
x2(5x - 6y) + 9y2(5x + 6y) x2(5x - 6y) + 9y2(5x - 6y) x2(5x - 6y) - 9y2(5x + 6y) x2(5x - 6y) - 9y2(5x - 6y)
b)
Which of the following expressions is NOT a factor of 5x3 - 6x2y - 45xy2 + 54y3
Choose:
5x - 6y x2 + 9y2 x - 3y x + 3y
8.
Factor by grouping: 10x3 - 2x2y2 - 5xy + y3
Choose:
(2x2 - y)(5x - y2) (2x - y2)(5x2 - y) (5x2 - y)(2x - y2) (2x2 - y)(5x - 2y2)
9.
Factor by grouping: 2ax2 + 3axy - 2nxy - 3ny2
Choose:
(ax - ny)(2x - 3y) (ax + ny)(2x + 3) (2x + 3y)(ax + ny) (2x + 3y)(ax - ny)
10.
Factor completely: x5 - 3x3 - 5x2 + 15
Choose:
(x2 - 3)(x3 - 5) (x2 - 3)(x3 + 5) (x3 - 3)(x2 - 5) (x3 - 3)(x2 + 5)
11. Which expressions are equivalent to 243x4 - 216x2 + 48 ? (Check all that apply, and hit SUBMIT!) 3(9x2 + 4)(9x2 - 4) 3(9x2 - 4)(9x2 - 4) 3(9x2 + 4)(9x2 + 4) 3(81x2 + 4)(81x2 - 4) 3(81x4 - 72x2 + 16) 3(3x2 - 4)(3x2 - 4) 3(3x - 2)2(3x + 2)2 | 1,025 | 1,860 | {"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} | 4.375 | 4 | CC-MAIN-2018-30 | latest | en | 0.574198 |
https://www.mrexcel.com/archive/formulas/round-up-in-excel/ | 1,544,841,092,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826686.8/warc/CC-MAIN-20181215014028-20181215040028-00497.warc.gz | 949,492,002 | 8,704 | # Round UP in Excel
Posted by Lisa on January 30, 2002 9:42 AM
Hi. I have this scenerio if anyone can help me. I need a file to ROUND UP all my #'s to the nearest 5,000.
An example of the number would be 43,242.
If I use " =ROUNDUP(B4,-3)" it becomes 44,000.
If I use " =ROUNDUP(B4,-4)" it becomes 50,000.
How can I get it to round up to 45,000?
Thanks, it's probably something simple so please refrain from the dummy jokes. :-)
Lisa
Posted by Dan Aragon on January 30, 2002 9:50 AM
=Roundup(B4/5000,0)*5000
Posted by Steve Hartman on January 30, 2002 9:57 AM
=(ceilingB4,5000)
Posted by Steve Hartman on January 30, 2002 10:01 AM
THAT SHOULD BE!! =CEILING(B4,5000)
Posted by Lisa on January 30, 2002 10:18 AM
Re: =Roundup(B4/5000,0)*5000
Thanks to both of you! Appreciate your help.
Lisa
Posted by Dan Aragon on January 30, 2002 10:23 AM
Cool, never used that function before....
Posted by Lisa on January 30, 2002 10:29 AM
Re: THAT SHOULD BE!! =CEILING(B4,5000)
Thanks, Steve.....I just may have to come back here again!
Lisa | 354 | 1,050 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-51 | latest | en | 0.940183 |
https://isem.irk.ru/publications/10.1134-S0040601518100075/?sphrase_id=6198518 | 1,623,846,498,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487623596.16/warc/CC-MAIN-20210616093937-20210616123937-00579.warc.gz | 272,485,358 | 10,155 | # The Development of a Method for Calculating the Nodal Prices of the Thermal Energy by Modeling the Thermal and Hydraulic Regimes of the Heat Supply Systems for Solving Control and Optimization Problems
Статья в журнале
Shalaginova Z.I.
Thermal Engineering . Vol.65. No.10. P.756-767.
2018
The article deals with the issues of energy saving and increasing the efficiency of thermal energy. Energy saving issues are inextricably linked to the cost of heat for consumers. The costs of thermal energy transfer depend directly on thermal and hydraulic regimes of the heat supply system. A mathematical model and a method for calculating differential heat energy prices for all nodes and consumers of a heat supply system are discussed considering various heat production costs, the actual heat flow distribution, the location of the consumers in the network (the distance from the source), the structure and parameters of the network, and numerous internal and external disturbances of both a systematic and an accidental character. The above factors are taken into account by means of a thermo-hydraulic model for calculating the regimes of the heat supply system with intermediate temporal control nodes that serve as a basis for determination of the heat amount in every node at every time point. The proposed approach to the calculation of nodal prices can be interpreted as a method for solving the problem of distribution of the “price field” over the network for a given distribution of heat flows. The approach is based on three main principles, viz., the nodal balance of the heat cost, the equality of prices for the flows from the common node, and a differentiated increment of prices due to the variable component for the transfer of thermal energy for each of the sections of the calculation scheme. The proposed method for calculating the nodal prices of thermal energy can be used for multicriteria optimization of thermal and hydraulic regimes. The maximum utilization of the cheapest energy in the system or the minimum price of thermal energy for consumers can be the optimization criteria. The proposed approach makes it possible to differentiate prices within the regulatory period currently adopted in accordance with the process conditions of the heat supply system, namely, summer, autumn–spring, and winter conditions, including using peak-load sources. The operating modes have significant differences in the parameters of the heat-transfer medium, such as the flow rate and temperature of the network water. © 2018, Pleiades Publishing, Inc.
### Библиографическая ссылка
Shalaginova Z.I. The Development of a Method for Calculating the Nodal Prices of the Thermal Energy by Modeling the Thermal and Hydraulic Regimes of the Heat Supply Systems for Solving Control and Optimization Problems // Thermal Engineering . Vol.65. No.10. 2018. P.756-767. DOI: 10.1134/S0040601518100075
### Телефоны
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### Заказ издания
The Development of a Method for Calculating the Nodal Prices of the Thermal Energy by Modeling the Thermal and Hydraulic Regimes of the Heat Supply Systems for Solving Control and Optimization Problems
x | 672 | 3,242 | {"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} | 2.6875 | 3 | CC-MAIN-2021-25 | latest | en | 0.881072 |
https://brainly.com/question/256463 | 1,484,989,412,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281001.53/warc/CC-MAIN-20170116095121-00062-ip-10-171-10-70.ec2.internal.warc.gz | 803,663,642 | 9,086 | Answers
2015-01-15T00:02:24-05:00
56.79%. that is the answer to your question.
Without any explanation of where your answer came from, that's no help at all.
2015-01-15T00:15:46-05:00
This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Well, it seems tome that if he makes 91% of all his free throws, then every time
he steps up to the line, there's a 91% chance that he'll make THAT throw.
Now the basketball has no memory, so the probability is the same for each
attempt, and the probability of succeeding on two consecutive attempts is
91% of the probability of succeeding on the first one, and so on and so forth.
So the probability of 6 consecutive baskets is (91%) x (91%) x ... etc, 6 times.
That's just (0.91)⁶ = 56.787 percent. | 263 | 1,001 | {"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-2017-04 | latest | en | 0.953751 |
https://talk.collegeconfidential.com/t/help-with-a-math-problem/1140011 | 1,660,166,082,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571210.98/warc/CC-MAIN-20220810191850-20220810221850-00176.warc.gz | 510,687,651 | 8,340 | # Help with a math problem
<p>Can someone help me solve the following problem?</p>
<p>In the equation x^2 + mx + n = 0, m and n are integers. The only possible value for x is -3. What is the value of m?</p>
<p>There is not enough information to solve this problem...</p>
<p>There isn't enough info. This is the farthest that you can get with this equation:</p>
<p>x^2 + mx + n = 0
9+(-3)m+n=0
(-3m)+n=-9
m+n=3</p>
<p>Then, if m and n are integers, m could equal 0, 1, 2, or 3.</p>
<p>@Cant
You kinda divided by -3 in your problem but didn't divide the n by -3 with the rest of the equation.</p>
<p>@All
Integers can be negative too.</p>
<p>@xplosneer you dont need to divide the n by -3, you're only dividing the -3 from the m, and since you did it to the left side, you must do it to the right side and divide -9 by -3.</p>
<br>
<br>
<p>x1 = (- m +- root of (m^2 - 4<em>1</em>n)) / 2</p>
<p>-3 = ( - m +- root of(m^2 - 4n))/ 2</p>
<p>-m +- root of (m^2 - 4n) = - 6</p>
<p>+- root of(m^2 - 4n) = -6 + m</p>
<p>BY inspection as m,n are integers:</p>
<p>m = 12, n = 27</p>
<p>i.e. root of(144 - 108) = -6 + 12</p>
<p>6 = 6</p>
<p>Checking:</p>
<p>-3^2 - 3*12 + 27 = 0 Yes</p>
<p>So m = 12</p>
<p>@hero</p>
<p>Did I miss something? What you are proposing is true only for adding/subtracting to each side.</p>
<p>You can't only divide single terms in an equation unless you take the term out of each part of the equation.
That's like saying 3x+y=13. If x=3 and y=4, then 9+4=13 but also 3+4=(13/3).</p>
<p>Haha woops. I guess this is what happens when I've been out of school for 2 months...</p>
<p>lol god I'm so stupid this is so easy. Since there is only one root.</p>
<p>If x can only be -3, then:</p>
<p>(x+a)^2 = 0 so (-3 + a)^2 = 0, a = 3
(x-a)^2 = 0 so (-3 - a)^2 = 0, a = -3</p>
<p>So x^2 - 2xa + a^2 = 0, or x^2 + 2xa + a^2 = 0</p>
<p>So x^2 + 6x + 9 = 0 (Using both equations with a = 3, a = -3)</p>
<p>-3^2 + 6(-3) + 9 = 9 -18 + 9 = 0</p>
<p>So m = -2a or m = 2a and n = a^2</p>
<p>m = -2(-3) = 6 and m = 2a = 2(3) = 6, n = -3^2 = 3^2 = 9</p>
<p>So m = 6 and n = 9</p>
<p>since we are on the topic of solving math problems, can someone pleaseeee help me with this one?!?</p>
<ol>
<li>Paul randomly selects 3 apples from a basket containing 4 green apples and 6 red apples. What is the probability that he selects at least 1 green apple?</li>
</ol>
<p>(A) 1/6
(B) 3/10
(C) 5/12
(D) 5/6
(E) 29/33</p>
<p>I just don't get probabilities</p>
<p>You can select 1 green apple from three choices (3,1) 3 different ways.</p>
<p>For zero green apples: p(x) = (3/5)^3 = 27/125
for one: p(x) = 3<em>(2/5)</em>(3/5)^2 = 54/125
for two: p(x) = 3<em>(2/5)^2</em>(3/5) = 36/125
for three: p(x) = (2/5)^3 = 8/125</p>
<p>1 + 2 + 3 green apples = 98/125</p>
<p>Basically if 3 is the ONLY solution then (x-3)(x-3)=0,
FOIL: x^2+6x+9=0
Therefore m=6</p>
<p>@jamiegirl best is to find the probability of finding P(zero green apples) and subtract from 1, since that is the complementary event. P(zero green apples) = (6C3)/(10C3) = 20/120 = 1/6 (since we want to select three red apples - @jsanche32 's solution assumes replacement, which I highly doubt is correct), so the probability that Paul selects at least one green apple is 1 - (1/6) = (D) 5/6.</p>
<p>(x+3)(x+3)
x^2+6x+9
6</p> | 1,286 | 3,314 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2022-33 | latest | en | 0.839044 |
http://www.notemonk.com/node/742/Euclids.Division.Lemma/ | 1,539,737,937,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583510932.57/warc/CC-MAIN-20181017002502-20181017024002-00405.warc.gz | 523,924,376 | 8,301 | Euclid’s Division Lemma ( 1 )
Questions
Q. Problem no. 4 in exercise ... 5 , 0 0
1 answer under Euclid’s Division Lemma 8 years, 4 months ago
A. any positive integer n can be written as 3q + r where q is the quotient and r the remainder using euclid's division algo n=aq+r (here a=3) so n = 3q + r q is in a integer and r = 0/1/2 n^2 =... link
8 years, 4 months ago
``````what is euclid's division lemma? | 136 | 408 | {"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-2018-43 | latest | en | 0.804332 |
http://windowssecrets.com/forums/showthread.php/118008-Formulas-to-calculate-cost-of-Capital | 1,500,785,306,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424247.30/warc/CC-MAIN-20170723042657-20170723062657-00504.warc.gz | 343,589,433 | 13,760 | # Thread: Formulas to calculate cost of Capital
1. I need assist to do the following per the attached workbook
1) I need to calculate the cumulative cash flow using the AND function as a logical alternative
2) I have calculated the payback period, but need help to calculate the payback using an alternative method using a nested if function for cell C93:J93
Your assistance will be most appreciated
Regards
Howard
2. [quote name='HowardC' post='774443' date='10-May-2009 01:03']I need assist to do the following per the attached workbook
1) I need to calculate the cumulative cash flow using the AND function as a logical alternative
2) I have calculated the payback period, but need help to calculate the payback using an alternative method using a nested if function for cell C93:J93
Your assistance will be most appreciated
Regards
Howard[/quote]
It is unclear on what you want to do
would you stated what are the criterias?
3. [quote name='franciz' post='774457' date='09-May-2009 19:25']It is unclear on what you want to do
would you stated what are the criterias?[/quote]
With regard to my point 1 , the formula in C93:J93, must produce the same result as in C92:J92, except the AND Function must be used, see Message when you click in cell C93
With regard to my point 2, the formula in C97, must produce the same result as C95, the only difference to the formula, is that the IF function must be used
It would be appreciated if you could kindly assist me
Regards
Howard
4. [quote name='HowardC' post='774459' date='09-May-2009 19:59']With regard to my point 1 , the formula in C93:J93, must produce the same result as in C92:J92, except the AND Function must be used, see Message when you click in cell C93
With regard to my point 2, the formula in C97, must produce the same result as C95, the only difference to the formula, is that the IF function must be used[/quote]
What strange requirements! Is this an attempt to get someone to do your homework for you?
5. Howard
These are questions based from this book or something similar.
These questions are specifically about financial management and whilst Loungers are happy to answer functional questions about AND statements you can't expect us to be all financial experts.
However, if you were to pose a question without reference to Christopher Buzzard and Mike Ehrhardt's work we maybe able to help you
6. [quote name='Jezza' post='774485' date='09-May-2009 23:40']Howard
These are questions based from this book or something similar.
These questions are specifically about financial management and whilst Loungers are happy to answer functional questions about AND statements you can't expect us to be all financial experts.
However, if you were to pose a question without reference to Christopher Buzzard and Mike Ehrhardt's work we maybe able to help you[/quote]
Attached please find amended spreadsheet. I have managed to solve the logical test using the AND function C20:J20 . have used the Index function in cell C24:J24, but would like this changed to an IF Nested Function . It would be appreciated if you could assist
Regards
Howard
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• | 774 | 3,284 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2017-30 | latest | en | 0.884086 |
https://present5.com/important-methods-of-argumentation-aristotle-s-method-p/ | 1,620,487,815,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988882.94/warc/CC-MAIN-20210508151721-20210508181721-00456.warc.gz | 468,658,432 | 11,542 | Скачать презентацию IMPORTANT METHODS OF ARGUMENTATION Aristotle s Method p
f03228d48e97ee92f918f1e0d6455922.ppt
• Количество слайдов: 23
IMPORTANT METHODS OF ARGUMENTATION
Aristotle’s Method p Stephen Toulmin’s Method p
Aristotle p Inductive Reasoning n n p In inductive reasoning we collect bits of evidence on which to base generalizations The Inductive Leap: Since thoroughness is often impractical if not impossible, inductive reasoning involves making a leap from the evidence to the conclusion. Deductive Reasoning n The syllogism
SYLLOGISM Major premise p Minor premise p Conclusion p All men are mortal p Socrates is a man. p Therefore, Socrates is mortal. p
REMEMBER FALLACIES Never assume that your syllogism is airtight– be careful to remember the things that can go wrong with reasoning. Here is an example of a faulty syllogism… p “Animals, which move, have limbs and muscles. The earth has no limbs and muscles. Hence, the Earth does not move. ” p What is the flaw? p
Real Life Example…(Mainly because Mrs. Lamar LOOOVE’s House. ) p Dr. House: Words have set meanings for a reason. If you see an animal like Bill and you try to play fetch, Bill's going to eat you, because Bill's a bear. Little Girl: Bill has fur, four legs, and a collar. He's a dog. Dr. House: You see, that's what's called a faulty syllogism; just because you call Bill a dog doesn't mean that he is. . . a dog. ("Merry Little Christmas, ” House, M. D. )
Syllogism Example #1 p Major Premise: Sixty men can do a piece of work sixty times as quickly as one man. Minor Premise: One man can dig a posthole in sixty seconds; therefore-Conclusion: Sixty men can dig a posthole in one second. This may be called the syllogism arithmetical, in which, by combining logic and mathematics, we obtain a double certainty and are twice blessed. " (Ambrose Bierce, The Devil's Dictionary)
Syllogism Example #2 Major premise: All asteroids are made of rock. Minor premise: Ceres is an asteroid. Conclusion: Ceres is made of rock.
Label this one on your own… Fluffy is not a cat. Fluffy enjoys the company of snakes. No cat enjoys the company of snakes.
And this one… Some kinds of cheese smell like feet. I will not eat anything that smells like feet. There are some kinds of cheese I will not eat.
Who is Stephen Toulmin and why do I care? A British philosopher, author and educator p Devoted his works to the analysis of moral reasoning p His writing “seeks to develop practical arguments which can be used effectively in evaluating the ethics behind moral issues. ” p
Toulmin Parts Data Claim Warrant Backing Qualifier Rebuttal p
CLAIM p Conclusions whose merit must be established
DATA p The facts appealed to as a foundation for the claim
WARRANT The statement authorizing the movement from the data to the claim p Toulmin stated that an argument is only as strong as its weakest warrant and if a warrant isn’t valid, then the whole argument collapses. p
BACKING p Facts that give credibility to the statement expressed in the warrant; backing must be introduced when the warrant itself is not convincing enough to the readers or the listeners.
REBUTTAL p Statements recognizing the restrictions to which the claim may legitimately be applied.
QUALIFIER p Words or phrases expressing how certain the author/ speaker is concerning the claim
NOTE p The first three elements, “claim, ” “data, ” and “warrant” are considered the essential components of practical arguments, while the others may not be needed in some arguments
EXAMPLE p p p CLAIM- You should buy our tooth-whitening product DATA- Studies show that teeth are 50% whiter after using the product for a specified time WARRANT- People want whiter teeth BACKING– Celebrities want whiter teeth REBUTTAL– Commercial says “unless you don’t want to attract guys” QUALIFIER– Fine print says “product must be used six weeks for results”
http: //www. chickfilapressroom. com/images/eatmorchickin/3 -cows. jpg
Thanks to the following sources for the information included in this presentation… Kennedy, X. J. , Dorothy Kennedy and Jane E. Aaron Eds. The Bedford Reader 9 th ed. Bedford/ St. Martin’s, 2006. Mc. Ferran, Doug. “Constructing Syllogisms. ” 22 August 2008 . Mc. Ferran, Doug. “Recognizing Valid Forms. ” 22 August 2008 . Toulmin’s Schema– Free Online Course Materials. 19 September 2007 . Boston: | 1,065 | 4,355 | {"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-2021-21 | latest | en | 0.887307 |
https://calculatorsonline.org/roman-numeral-date-converter/april-14-2014 | 1,686,393,848,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224657169.98/warc/CC-MAIN-20230610095459-20230610125459-00218.warc.gz | 183,520,850 | 5,408 | # April 14, 2014 in roman numeral
Here you will see step by step solution to convert April 14, 2014 date to roman numeral. How to write April 14, 2014 as a roman numeral? April 14, 2014 as a roman numeral written as IV.XIV.MMXIV (MM.DD.YYYY), please check the explanation that how to convert 14 April, 2014 in roman number.
## Answer: April 14, 2014 in roman numeral
IV.XIV.MMXIV
### How to convert April 14, 2014 in roman number?
To convert the April 14, 2014 in roman number simply expand the each number from month, date and year from hindu-arabic number to roman numerals, then replace the all numbers of expanded form of date, month and year with respective roman numerals.
#### Solution for April 14, 2014 to roman numeral
Given date is => 14-04-2014
After expanding number from Month, Date and year, this table provides a simple and explanation of how to convert the date 'April 14, 2014' to its Roman number:
MonthDayYear
Date 04 [Apr] 14 2014
Expanded Number Values 4 10 + 4 1000 + 1000 + 10 + 4
Roman Numeral Values IV X + IV M + M + X + IV
Result = IV XIV MMXIV
Hence, in order to correct roman numerals date combination of Month, Day and Year of April 14, 2014 is written as IV.XIV.MMXIV or in 14/April/2014 [DD/MM/YYYY] format it is written as XIV/IV/MMXIV. | 375 | 1,280 | {"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} | 3.25 | 3 | CC-MAIN-2023-23 | latest | en | 0.82854 |
http://www.chegg.com/homework-help/visit-wwwmyprogramminglabcom-complete-select-exercises-onlin-chapter-5-problem-2pp-solution-9780136083825-exc | 1,469,951,131,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469258951764.94/warc/CC-MAIN-20160723072911-00132-ip-10-185-27-174.ec2.internal.warc.gz | 390,358,628 | 15,845 | View more editions
# TEXTBOOK SOLUTIONS FOR Absolute Java 4th Edition
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Chapter: Problem:
Visit www.myprogramminglab.com to complete select exercises online and get instant feedback.
Exercise
Define a class called Fraction. This class is used to represent a ratio of two integers. Include mutator methods that allow the user to set the numerator and the denominator. Also include a method that displays the fraction on the screen as a ratio (e.g., 5/9). This method does not need to reduce the fraction to lowest terms. Include an additional method, equals, that takes as input another Fraction and returns true if the two fractions are identical and false if they are not. This method should treat the fractions reduced to lowest terms; that is, if one fraction is 20/60 and the other is 1/3, then the method should return true.
Embed your class in a test program that allows the user to create a fraction. Then the program should loop repeatedly until the user decides to quit. Inside the body of the loop, the program should allow the user to enter a target fraction into an anonymous object and learn whether the fractions are identical.
STEP-BY-STEP SOLUTION:
Chapter: Problem:
• Step 1 of 5
Definition of Fraction class
Program Plan:
• Create a class Fraction that can represent ratios between two integers. These integers will be the numerator and the denominator of the Fraction class.
• Implement methods that allow the user to set the numerator and denominator of the Fraction, as well as a method that displays the fraction as a ratio.
• Implement a boolean method that can compare the Fraction to a different Fraction object, returning true if the fraction values are equal.
• Create a test program to continuously allow the user to input fractions to compare to each other.
• Chapter , Problem is solved.
Corresponding Textbook
Absolute Java | 4th Edition
9780136083825ISBN-13: 013608382XISBN: Walter SavitchAuthors:
Alternate ISBN: 9780136118848 | 467 | 2,168 | {"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-2016-30 | latest | en | 0.884764 |
https://homepages.warwick.ac.uk/~masda/MA4J8/Lect1.txt | 1,725,765,377,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00237.warc.gz | 291,417,211 | 3,822 | Harmless introduction (4 lectures or so) set the scene, fill in some items from Chunyi's 2020 course Zariski topology of Spec A maximal in some class => prime the formal substitute of NSS for a general ring A: A -> S^-1A OX = local functions P -> element of local ring A_P category of A-modules -> category of S^-1A modules characterisation of DVR short summary of general valuations from [AM] I hope to make each lecture into a coherent narrative making one or two points. What is commutative algebra? In school and intro univ. algebra you studied _division with remainder_ for integers and polynomials in k[x]. Integers: given a,b positive integers, you can write a = b*q + r with 0 <= r < b. If you have a cakes to give to b kids, hand them out equally b at a time until the remainder r < b is not enough to go around. Polynomials: given A, B in k[x] A = an x^n + a_{n-1}x^{n-1} + .. a0 B = bm x^m + .. b0 with deg A = n, deg B = m (deg means top term <> 0). If m <= n we can subtract a multiple of g to cancel the top term in f: A - an/bm*x^{n-m}*B of deg <= n-1. Just continue decreasing deg A until deg (A - (mult. q*B)) < m. (We will see later that removing the leading term is also an important idea in constructing a Groebner basis.) In either case we have a notion of size of A, and can successively reduce it by subtraction to < size B, (and the logic has an initial case size B = 0). The point I want to make is that the objects we are talking about are quite different: integers versus polynomial functions or abstract polynomials. Nevertheless, the methods of argument are exactly the same. Please think through the argument used to show: ZZ and k[x] have division with remainder, so are PID: every ideal I is generated by a single element, I = (f)). And PID is a UFD: every element factorises as a product of a unit times a product of prime powers, uniquely up to units and order of the factors. This gives the usual properties of GCD and LCM, including the important a*f + b*g = h property of h = GCD(f,g). Magma -- runs in online calculator http://magma.maths.usyd.edu.au/calc A := 49; B := 175; XGCD(A,B); h, a, b := XGCD(A,B); printf("The HCF of %o, %o is %o, and %o*%o + %o*%o equals %o.\n"), A,B,h,a,A,b,B,h; a*A + b*B; KK := PolynomialRing(Rationals()); A := x^3 + 7*x + 2; B := x^5 - 2*x^4 + 3; // XGCD(A,B); h, a, b := XGCD(A,B); printf("The HCF of %o, %o is %o.\nAnd a * (%o) + b * (%o) = %o, where\na = %o, and\nb = %o.\n"), A,B,h,A,B,h,a,b; a*A + b*B; // end of Magma What is commutative algebra? The 1882 paper [DW] extended this analogy in elementary algebra to a theory that encompasses both the ring of integers of a number field and the ring of functions on an algebraic curve. Their paper is a landmark in the development of modern algebra, and marks the starting point of commutative algebra. [DW] Richard Dedekind and Heinrich Weber, Theorie der algebraischen Funktionen einer Veraenderlichen, J. reine angew. Math. 92 (1882), 181--290 I explain this briefly (don't worry about the details -- I will return to the full arguments later). == Ring of integers of an algebraic number field An _algebraic number field_ is a finite extension field QQ in K. Corresponding to the ring of integers ZZ in QQ, the field K also has a subring O_K of integers, the subset of K of integral elements (details later). In any fairly complicated case, the division with remainder that we used for ZZ does not work for O_K, and it is _not_ a UFD. == Integral closure of an algebraic function field You know the polynomial ring k[x] over a field k (say k = CC to be definite). Its field of fractions k(x) consists of rational functions f(x)/g(x) with f,g polynomials and g <> 0. An _algebraic function field_ in one variable is a finite extension field k(x) in K (where x is transcendental). Corresponding to the polynomial ring k[t] in k(t), the same definition as the number field case gives the integral closure A of k[x] in K: A is the subset of elements of K that are _integral_ over k[x] (satisfy a monic equation with coefficients in k[x] -- no denominators allowed, and leading coefficient 1). This integral closure A = k[C] is the coordinate ring of a nonsingular affine algebraic curve C over k. (I am not saying that this is obvious.) In any fairly complicated case, this A does not have division with remainder, and is _not_ a UFD. == Dedekind and Weber's synthesis The preceding paragraphs set up the ring of integers O_K of a number field K, and the coordinate ring k[C] of a nonsingular affine curve C. These objects are major protagonists of algebraic number theory and algebraic geometry, and are clearly very different in nature. However, Dedekind and Weber [DW] say that these two rings can be studied using the same algebraic apparatus. As I said, they are usually not UFDs. The good news: if A is a ring of either type (a Dedekind domain), the ideals of A have _unique factorisation into prime ideals_. The key method of argument is _localisation_ (partial ring of fractions). If P is a prime ideal of A, the localisation of A at P is A_P = S^-1A where S = multiplicative set S = A - P. (I will go through this in detail later.) In arithmetic, A_P in K is the algebraic numbers that have an expression f/g with g notin P. For a point P of and algebraic curve C, A_P consists of the rational functions in k(C) that have an expression f/g with denominator g not vanishing at P in C. For either kind of ring A_P is a discrete valuation ring (DVR). Although when the ring A is not a UFD, its localisation A_P is the simplest possible UFD: it has a single prime element z (up to units), and every nonzero element h in K has the factorisation h = z^n*(unit), where n = v_P(h) is the valuation of h at P. Valuations then determine everything about A in K and the ideals of A: an element h in K is in A if and only if it has valuation >= 0 at every P. Moroever, every ideal I in A also has a valuation at P (namely, min v_P(i) taken over i in I). For any given nonzero ideal I of A, there are just finitely many primes P such that v_P(I) > 0, and I equals the product of P^v_P(I). == Modern abstract algebra Notice the breakthrough aspect of Dedekind and Weber: modern algebra has axioms and abstract arguments, and you often work with objects in a symbolic way. In this case, without reference to what the elements of the ring actually are. == Background reading: I will treat localisation S^-1A in the next week or so, or see any textbook on commutative algebra. For Dedekind domain and unique factorisation into prime ideals, see Matsumura p.82 or Atiyah and Macdonald, p.95. For integral closure and DVR, see [UCA], Chap. 8, or later in this course. | 1,770 | 6,680 | {"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} | 3.65625 | 4 | CC-MAIN-2024-38 | latest | en | 0.907461 |
https://socratic.org/questions/55830786581e2a0d72a7e402 | 1,596,942,776,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439738380.22/warc/CC-MAIN-20200809013812-20200809043812-00223.warc.gz | 488,206,012 | 7,080 | # Question 7e402
Jun 18, 2015
19.49g
#### Explanation:
$2 A g {F}_{\left(a q\right)} + N {a}_{2} S {O}_{4 \left(a q\right)} \rightarrow A {g}_{2} S {O}_{4 \left(s\right)} + 2 N a {F}_{\left(a q\right)}$
$c = \frac{n}{v}$
$n = c v$
So the number of moles of $A g F =$
$\frac{2.5 \times 50.0}{1000} = 0.125$
From the equation we can see that 2 moles of $A g F$ produces 1 mole of $A {g}_{2} S {O}_{4}$
So the no. moles $A {g}_{2} S {O}_{4} = \frac{0.125}{2} = 0.0625$
The $M r$ of $A {g}_{2} S {O}_{4} = 311.8$
So the mass of $A {g}_{2} S {O}_{4} = 311.8 \times 0.0625 = 19.49 \text{g}$
Jun 18, 2015
The mass of the precipitate will be equal to 19.5 g.
#### Explanation:
So, you're two solutions that contain soluble compounds - silver fluoride, $A g F$, and sodium sulfate, $N {a}_{2} S {O}_{4}$.
The double replacement reaction that takes place between these two compounds will produce silver sulfate, $A {g}_{2} S {O}_{4}$, an insoluble compound, and sodium fluoride, $N a F$, according to the balanced chemical equation
$\textcolor{red}{2} A g {F}_{\left(a q\right)} + N {a}_{2} S {O}_{4 \left(a q\right)} \to A {g}_{2} S {O}_{4 \left(s\right)} \downarrow + 2 N a {F}_{\left(a q\right)}$
Since sodium sulfate is in excess, all the moles of silver fluoride will react. The $\textcolor{red}{2} : 1$ mole ratio that exists between silver fluoride and silver sulfate tells you that, regardless of how many moles of the former react, the reaction will produce half as many moles of the latter.
Use the molarity and volume of the silver fluoride solution to determine how many moles take part in the reaction
$C = \frac{n}{V} \implies n = C \cdot V$
${n}_{A g F} = \text{2.50 M" * 50.0 * 10^(-3)"L" = "0.125 moles}$ $A g F$
This means that the reaction will produce
0.125cancel("moles"AgF) * ("1 mole "Ag_2SO_4)/(color(red)(2)cancel("moles"AgF)) = "0.0625 moles"# $A {g}_{2} S {O}_{4}$
To get the mass of the precipitate, use its molar mass
$0.0625 \cancel{\text{moles") * "311.8 g"/(1cancel("mole")) = color(green)("19.5 g}}$
The net ionic equation for this reaction looks like this
$2 A {g}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -} \to A {g}_{2} S {O}_{4 \left(s\right)} \downarrow$ | 811 | 2,225 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 24, "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} | 3.890625 | 4 | CC-MAIN-2020-34 | latest | en | 0.767653 |
https://www.gurufocus.com/term/Total+Equity/MPW/Total+Equity/Medical+Properties+Trust%252C+Inc | 1,511,001,733,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934804724.3/warc/CC-MAIN-20171118094746-20171118114746-00518.warc.gz | 841,899,508 | 40,707 | Switch to:
Medical Properties Trust Inc (NYSE:MPW) Total Equity: \$3,826.4 Mil (As of Sep. 2017)
Medical Properties Trust Inc's total equity for the quarter that ended in Sep. 2017 was \$3,826.4 Mil.
Total equity is used to calculate Book Value per Share. Medical Properties Trust Inc's Book Value per Share for the quarter that ended in Sep. 2017 was \$10.51. The ratio of a company's debt over equity can be used to measure how leveraged this company is. Medical Properties Trust Inc's Debt-to-Equity for the quarter that ended in Sep. 2017 was 1.26.
Historical Data
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Medical Properties Trust Inc Annual Data
Dec07 Dec08 Dec09 Dec10 Dec11 Dec12 Dec13 Dec14 Dec15 Dec16 Total Equity 1,049.81 1,344.21 1,382.05 2,102.27 3,248.38
Medical Properties Trust Inc Quarterly Data
Dec12 Mar13 Jun13 Sep13 Dec13 Mar14 Jun14 Sep14 Dec14 Mar15 Jun15 Sep15 Dec15 Mar16 Jun16 Sep16 Dec16 Mar17 Jun17 Sep17 Total Equity 3,161.05 3,248.38 3,246.78 3,817.31 3,826.44
Calculation
Total Equity refers to the net assets owned by shareholders.
Total Equity and Total Liabilities are the two components for Total Assets.
Medical Properties Trust Inc's Total Equity for the fiscal year that ended in Dec. 2016 is calculated as
Total Equity = Total Assets(Q: Sep. 2017 ) - Total Liabilities(Q: Sep. 2017 ) = 6418.536/td> - 3170.158 = 3,248.4
Medical Properties Trust Inc's Total Equity for the quarter that ended in Sep. 2017 is calculated as
Total Equity = Total Assets(Q: Sep. 2017 ) - Total Liabilities(Q: Sep. 2017 ) = 8927.04 - 5100.596 = 3,826.4
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
1. Total equity is used to calculate book value per share.
Medical Properties Trust Inc's Book Value per Share for the quarter that ended in Sep. 2017 is
Book Value per Share = (Total Equity - Preferred Stock) / Shares Outstanding (Diluted Average) = (3826.444 - 0) / 364.08 = 10.51
2. The ratio of a company's debt over equity can be used to measure how leveraged this company is.
Medical Properties Trust Inc's Debt-to-Equity for the quarter that ended in Sep. 2017 is
Debt-to-Equity = Total Debt / Total Equity = (Current Portion of Long-Term Debt + Long-Term Debt & Capital Lease Obligation) / Total Equity = (0 + 4832.264) / 3826.444 = 1.26
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Related Terms | 714 | 2,588 | {"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-2017-47 | latest | en | 0.859843 |
https://www.zhygear.com/gear-meshing-process/ | 1,721,621,749,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517823.95/warc/CC-MAIN-20240722033934-20240722063934-00645.warc.gz | 942,274,829 | 28,613 | Gear meshing process
Gear system dynamics is a discipline to study the dynamic behavior of gear system in the process of transmitting motion and power. It takes the whole gear system as the research object, and can comprehensively study the dynamic performance of gear system. Therefore, this chapter introduces the nonlinear gear dynamic model with six degrees of freedom which is established by using the lumped parameter method, and introduces the gear system dynamic model, the calculation method of gear meshing stiffness and the gear sliding friction model.
For the involute spur gear face contact degree greater than 1 and less than 2, it is necessary to consider the meshing of two pairs of gears in a complete gear meshing cycle. The gear meshing process is shown in the figure
The driving gear rotates clockwise around the central axis of the driving gear under the action of torque TP, and the driven gear rotates anticlockwise around the central axis of the driven gear under the action of load TG. On the theoretical meshing line XY (along the direction of oloa), points x and y are the tangent points of the gear meshing line and the base circle of the driving and driven gears respectively, and the line ad is the actual track of the meshing point of the gear tooth profile, which is called the actual meshing line.
At the beginning of the gear meshing cycle (i.e. t = 0), gear pair I is engaged at point a, and gear pair II is engaged from the highest point C of single tooth meshing (gear pair I is defined as the pair of teeth engaging along the meshing line AC, and gear pair II is defined as the pair meshing along the meshing line CD). With the rotation of the gear, when the tooth pair I is engaged to the lowest point B of single tooth meshing (i.e., t = TB) ）The tooth pair II reaches the meshing point D and is out of engagement. Next, the gear enters the single tooth meshing area, and only the gear pair I is engaged. The process goes from meshing point B to meshing point C. When the gear teeth pair I mesh through the node P (i.e. t = TP), the relative sliding speed of the driving gear will be reversed, which will lead to the reverse sliding friction force on the tooth surface. This will provide a pulse friction excitation for the gear dynamic system. Finally, the tooth pair I passes through the highest point C of single tooth engagement (i.e. t = TC), and a gear meshing cycle ends, and a new pair of teeth begins to mesh at point a.
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# A5 Paper Size
## The A5 is mainly used for the creation of booklets, leaflets, flyers ... It corresponds to half an A4 size. It is found only rarely in printers, although most are adaptable.
A5 format was standardized at the same time that all A sizes, in 1975 by ISO 216 standard. The latter comes from the German standard DIN 476, developed by Walter Portsmann in 1922. It referred to the explanations of Professor Lichtenberg on the issue, that were delivered on a sheet for the first time adopting a √2 ratio. This means that the length of a sheet must be equal to its width x 1.4142.
Thus, we obtain an homothetic system with a √2 ratio, which means that the A5 is obtained by folding an A4 in its width direction. This proportion is still preserved, which also keeps the contents of the sheet at the time of its enlargement.
The A5 has the following dimensions: 210 x 148 mm or 8.27 x 5.83 inches. Its printable surface is reduced when applying the usual print margins: 160 x 108 mm.
The surface of an A5 sheet is 0.031 m², or 0.04 square yard, 0.32 square foot, or 46.5 square inches. This size can still be used for writing, it also often comes in notebook.
The A5 is a dual A6, and an A7 quad. Conversely, it is half an A4 and quarter an A3. Calculating its weight is easy, knowing that it is 32 times smaller than A0 (or 1 m²). The weight of the paper being always applied in g/m², for a paper of 100 g/m², an A5 sheet is 100/32 = 3.125 grams.
### The size in centimeters of A5 is 14.8 x 21
or 148 x 210 millimeters
### The area of A5 paper sheet:
• Square yard: 0.04
• Square foot: 0.32
• Square inch: 46.5 | 458 | 1,649 | {"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} | 3.53125 | 4 | CC-MAIN-2020-05 | longest | en | 0.946186 |
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A089013 a(n) = (A088567(8n) mod 2). 2
1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,1 COMMENTS This is just to give a pointer to A088567, A089013 and A014707. LINKS N. J. A. Sloane and J. A. Sellers, On non-squashing partitions, arXiv:math/0312418 [math.CO], 2003. N. J. A. Sloane and J. A. Sellers, On non-squashing partitions, Discrete Math., 294 (2005), 259-274. FORMULA a(0) = 1, a(2*n) = a(n) for n > 0 and a(2*n+1) = (n mod 2) for n >= 0. - A.H.M. Smeets, Aug 02 2018 CROSSREFS A038189(n)=a(n) if n>0. Sequence in context: A260444 A289034 A011747 * A188374 A123504 A273511 Adjacent sequences: A089010 A089011 A089012 * A089014 A089015 A089016 KEYWORD nonn,easy AUTHOR N. J. A. Sloane, Dec 20 2003 STATUS approved
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Discussion in 'Python' started by Zdenko, Jan 4, 2011.
1. ### ZdenkoGuest
Please, can anybody write me a simple recursive matrix multiplication
using multiple threads in Python, or at least show me some guidelines
how to write it myself
Thank You
Zdenko, Jan 4, 2011
2. ### Ulrich EckhardtGuest
Zdenko wrote:
> Please, can anybody write me a simple recursive matrix multiplication
> using multiple threads in Python, or at least show me some guidelines
> how to write it myself
No problem, I just need your bank account data to withdraw the payment and
the address of your teacher whom to send the results. ;^)
Seriously, things don't work like this. If you show effort, you will get
help, but you don't get working solutions without even a bit of effort on
your own.
Start with a recursive algorithm, then make it multithreaded. Be prepared to
make some experiments on how to distribute tasks to threads and collect
them again.
Uli
Ulrich Eckhardt, Jan 4, 2011
3. ### Grégory LeocadieGuest
On 4 jan, 11:15, Ulrich Eckhardt <>
wrote:
> Zdenko wrote:
> > Please, can anybody write me a simple recursive matrix multiplication
> > using multiple threads in Python, or at least show me some guidelines
> > how to write it myself
>
> No problem, I just need your bank account data to withdraw the payment and
> the address of your teacher whom to send the results. ;^)
>
> Seriously, things don't work like this. If you show effort, you will get
> help, but you don't get working solutions without even a bit of effort on
> your own.
>
> Start with a recursive algorithm, then make it multithreaded. Be prepared to
> make some experiments on how to distribute tasks to threads and collect
> them again.
>
> Uli
Hi,
just have a look here http://en.wikipedia.org/wiki/Matrix_multiplication
and find out what parts can be parallelized
Gregory LEOCADIE
Grégory Leocadie, Jan 4, 2011
4. ### ZdenkoGuest
On 4.1.2011 11:15, Ulrich Eckhardt wrote:
> Zdenko wrote:
>> Please, can anybody write me a simple recursive matrix multiplication
>> using multiple threads in Python, or at least show me some guidelines
>> how to write it myself
>
> No problem, I just need your bank account data to withdraw the payment and
> the address of your teacher whom to send the results. ;^)
>
> Seriously, things don't work like this. If you show effort, you will get
> help, but you don't get working solutions without even a bit of effort on
> your own.
>
> Start with a recursive algorithm, then make it multithreaded. Be prepared to
> make some experiments on how to distribute tasks to threads and collect
> them again.
>
> Uli
>
Ok, I expected something like this
Lets try this way:
I wrote these two codes for example:
this one is naive way of matrix multiplication using one thread
from random import *
from time import clock
import threading
t0 = clock()
A=[]
B=[]
C=[]
m=300 #size of matrix m x m
for i in range(m):
A.append([])
B.append([])
C.append([])
for j in range(m):
A.append(randint(1,9))
B.append(randint(1,9))
C.append(0)
class MyThread ( threading.Thread ):
def run ( self ):
for i in range(m):
for j in range(m):
for k in range(m):
C[j]+=A[k]*B[k][j]
t=MyThread()
t.start()
t.join()
print clock() - t0, "seconds"
this one is using two threads, each one is multiplying half of matrix
import time
from threading import Thread
from random import randint
t0 = time.clock()
class MyThread(Thread):
def __init__(self, poc, kr):
Thread.__init__(self)
self.poc=poc
self.kr=kr
def run(self):
for i in range(self.poc,self.kr):
for j in range(m):
for k in range(m):
C[j]+=A[k]*B[k][j]
A=[]
B=[]
C=[]
m=300 #size of matrix m x m
for i in range(m):
A.append([])
B.append([])
C.append([])
for j in range(m):
A.append(randint(1,9))
B.append(randint(1,9))
C.append(0)
thr1=MyThread(0,m/2)
thr2=MyThread(m/2,m)
thr1.start()
thr2.start()
thr1.join()
thr2.join()
print time.clock() - t0, "seconds"
why is second one more than twice slower than first when it should be
faster. I suspect that problem is in simultaneous access to matrix C but
i don't know how to solve this.
I used this just for example to see how the threading works so i can
work on recursive and Strassen algorithm.
Zdenko, Jan 4, 2011
5. ### DevPlayerGuest
I would be very suprised if you achieve faster results threading this
problem. There's been much discussed on benefits or lack thereof to
threading in Python (or in general).
Threading is best used in situations where you are doing different
kinds of tasks. For example if you want to do your matrix
multiplication WHILE you were doing other things on your computer
where matrix multiplication was a background process chugging away
when you are not taxing the computer doing other stuff.
Threading adds efficiency when you would have lots of "blocking"
operations like reading in lots of big files from a comparable slow
hard drive (compared to how fast a CPU processes data) or waiting on
netword data (super slow compared to CPU processing).
When you are doing mass numeric processing, you want to minimize the
jumping from one function to another which uses overhead, recursion
adds a small amount of uneccessary overhead, you want to minimize the
need for the cpu to switch between threads or processes.
If you still feel the need to use threads for some reason, for numeric
processing I'd recommend using a "lighter" thread object, like a tiny
thread or green thread or a threadlet or whatever they are calling
them now.
Another thing to note is it seems you might be expecting threads to
run on different CPU cores expecting improvment. Be careful with this
assumption. This is not always true. It is up to the CPU and OS to
determine how threads are handled and perhaps the GIL to some extent.
Beaware that Python has a GIL (some distributions). Google it if you
don't know of it.
To encourage better use of multi-core cpus you might consider the
multiprocessing library included in Python 2.7 (I think) and above.
I'm assuming that speed is an issue because you where timing your
code. If you are doing actual serious number crunching there's lots of
advice on this. The python Numpy package as well as Stackless Python
(for microthreads or whatever thier called) comes to mind.
Another thought. Ask yourself if you need a large in-memory or live
set of processed numbers, in your case a fully and processed
multiplied matrix. Usually a large set of in-memory numbers is
something your going to use to simulate a model or to process and
crunch further.
Or is your actual usage going to be picking out a processed number
here or there from the matrix. If this is true look at iterators or
generators. Which would be a snapshot in time of your matrix
multiplication. I like to think of Python generators like integral
calculus (definition at: http://en.wikipedia.org/wiki/Integral_calculus)
where the specific integral of a generator is often just 1.
I'm loving generators a lot. For example there are generator
accelorators which if you think it through means you can make
generator deccelorators, useful for doing interpolation between
elements of your matrix elements for example. I always forget if
generators are thread safe though.
Some indicators that generators could help: You're doing lots of for
loops with range().
Also it's been measured that list comprehensions are slightly faster
then while loops are a slightly faster then for loops. You can Google
to confirm, enter something like "python fast iteration".
Also if your numbers in your matix are actually not really numbers but
objects with numbers, __slots__ is used to for large sets of objects
(10s of millions at the very least) to minimize memory usage and
perhaps with speed, if used properly. Just mentioning. I'd stay away
from this though.
Some of my informatation may be inaccurate (and even completely wrong;
like I always get when a thread is best switched during a blocking
verse a non-blocking operation) but there are some things to consider.
DevPlayer, Jan 4, 2011
6. ### DevPlayerGuest
DevPlayer, Jan 4, 2011
7. ### DevPlayerGuest
DevPlayer, Jan 4, 2011
8. ### Dan StrombergGuest
On Tue, Jan 4, 2011 at 4:22 AM, Zdenko <> wrote:
> On 4.1.2011 11:15, Ulrich Eckhardt wrote:
>>
>> Zdenko wrote:
>>>
>>> Please, can anybody write me a simple recursive matrix multiplication
>>> using multiple threads in Python, or at least show me some guidelines
>>> how to write it myself
>>
>> No problem, I just need your bank account data to withdraw the payment and
>> the address of your teacher whom to send the results. ;^)
>>
>> Seriously, things don't work like this. If you show effort, you will get
>> help, but you don't get working solutions without even a bit of effort on
>> your own.
>>
>> Start with a recursive algorithm, then make it multithreaded. Be prepared
>> to
>> make some experiments on how to distribute tasks to threads and collect
>> them again.
>>
>> Uli
>>
>
> Ok, I expected something like this
> Lets try this way:
>
> I wrote these two codes for example:
>
> this one is naive way of matrix multiplication using one thread
>
> from random import *
> from time import clock
> import threading
>
> t0 = clock()
> A=[]
> B=[]
> C=[]
> m=300 #size of matrix m x m
>
> for i in range(m):
> A.append([])
> B.append([])
> C.append([])
> for j in range(m):
> A.append(randint(1,9))
> B.append(randint(1,9))
> C.append(0)
>
> class MyThread ( threading.Thread ):
>
> def run ( self ):
> for i in range(m):
> for j in range(m):
> for k in range(m):
> C[j]+=A[k]*B[k][j]
>
> t=MyThread()
> t.start()
> t.join()
> print clock() - t0, "seconds"
>
>
>
> this one is using two threads, each one is multiplying half of matrix
>
> import time
> from threading import Thread
> from random import randint
>
> t0 = time.clock()
> class MyThread(Thread):
> def __init__(self, poc, kr):
> Thread.__init__(self)
> self.poc=poc
> self.kr=kr
> def run(self):
> for i in range(self.poc,self.kr):
> for j in range(m):
> for k in range(m):
> C[j]+=A[k]*B[k][j]
>
> A=[]
> B=[]
> C=[]
> m=300 #size of matrix m x m
>
> for i in range(m):
> A.append([])
> B.append([])
> C.append([])
> for j in range(m):
> A.append(randint(1,9))
> B.append(randint(1,9))
> C.append(0)
> thr1=MyThread(0,m/2)
> thr2=MyThread(m/2,m)
> thr1.start()
> thr2.start()
> thr1.join()
> thr2.join()
> print time.clock() - t0, "seconds"
>
> why is second one more than twice slower than first when it should be
> faster. I suspect that problem is in simultaneous access to matrix C but i
> don't know how to solve this.
>
> I used this just for example to see how the threading works so i can work on
> recursive and Strassen algorithm.
> --
> http://mail.python.org/mailman/listinfo/python-list
>
Threads in CPython are decent when you're doing I/O, not so good when
you're doing CPU-bound tasks. When you're doing CPU-bound tasks, you
may actually find that your application runs about 1/n times as fast,
where n is the number of CPU's - that is, much slower. I don't mean
that it just doesn't scale well, I mean that each additional CPU makes
things slower for CPU-bound jobs.
If you require java-like threading, you might be better off with
Jython or Iron Python for now.
pypy's likely to have good threading eventually, but last I heard it
wasn't quite there yet.
If you do require CPython, you might check out the greenlets module
and/or PyCSP.
Oh, and there's "stackless python", which amounts to a branch of
CPython that's unlikely to be merged. It's supposed to be able to
thread really well, and ISTR hearing that it's pretty similar to
greenlets. The main force behind stackless is working on pypy now, I
believe.
Finally, with CPython, you can use the multiprocessing module to get
pretty good parallelism via distinct processes. This way you tend to
end up using queues for interprocess communication; these queues use
shared memory. However, you might actually be able to put your matrix
in shared memory - if it's large enough to justify that. I know you
could put a linear array of homogeneous, scalar type in shared memory;
I've not heard of a way of putting an aggregate type in shared memory.
Naturally, you can fake an n dimensional array using a 1 dimensional
array with a little multiplication and addition.
As an example of using CPython with multiprocessing without having
your matrix in shared memory, you could probably have one worker
subprocess for each core on a system, divide up the cells of your
result matrix by their coordinates (perhaps using an iterator and izip
across the queues) and send a message (via a shared memory queue) with
those coordinates to the appropriate subprocess. Then those
subprocess send back the coordinates and the result at said coordinate
via a different result queue. I suspect you'd want one queue for all
the results, and one queue for each subprocess for initiating
computation. Each subprocess would have its own COW copy of the input
matrix.
Dan Stromberg, Jan 4, 2011
9. ### Steven D'ApranoGuest
On Tue, 04 Jan 2011 13:22:33 +0100, Zdenko wrote:
> I wrote these two codes for example:
>
> this one is naive way of matrix multiplication using one thread
[...]
> this one is using two threads, each one is multiplying half of matrix
[...]
> why is second one more than twice slower than first when it should be
> faster. I suspect that problem is in simultaneous access to matrix C but
> i don't know how to solve this.
Can I suggest that your timing code leaves something to be desired?
* You time too much. You're interested in how long it takes to multiply
two matrices, but you time how long it takes to do much more than just
the multiplication: your timing code covers the time it takes to create
the class object (which will be trivial), and build the matrix (non-
trivial), as well as perform the multiplication.
* Your perform the timing test once, which makes it subject to sampling
errors. (Although if the process takes a long time, say, multiple
seconds, the sampling error will *probably* be small relative to the
measured time. But not always.)
* You use time.clock, but without knowing which operating system you are
running, it's impossible to tell whether you should be using time.time
instead.
Whether these issues will actually make a practical difference in this
*specific* case, I don't know, but as a general rule, the most accurate
way to perform these sorts of timing tests is with the timeit module.
Something like this:
A = ... # initialise matrix A
B = ... # and B
C = ... # and the result
def mult1(A, B, C):
# Multiply matrices A and B using 1 thread.
t = MyThread()
t.start()
t.join()
def mult2(A, B, C):
# Multiply matrices A and B using 2 threads.
t1 = MyThread()
t2 = MyThread()
t1.start()
t2.start()
t1.join() # Block until thread 1 is done.
t2.join() # Now block until thread 2 is done.
setup1 = "from __main__ import A, B, C, mult1"
setup2 = "from __main__ import A, B, C, mult2"
from timeit import Timer
t1 = Timer("mult1(A, B, C)", setup1)
t2 = Timer("mult2(A, B, C)", setup2)
# Now perform the timing tests.
best1 = min(t1.repeat())
best2 = min(t2.repeat())
By default, Timer.repeat will measure the time taken by one million
iterations of the code snippet, and take three measurements. You almost
always only care about the smallest measurement -- any larger times will
represent sampling error.
If it takes too long to time one million iterations, either make the
matrices smaller, or pass keyword arguments number and/or repeat to the
repeat method:
# best of five measurements of one hundred iterations
best1 = min(t1.repeat(number=100, repeat=5))
--
Steven
Steven D'Aprano, Jan 4, 2011
10. ### John NagleGuest
On 1/4/2011 2:15 AM, Ulrich Eckhardt wrote:
> Zdenko wrote:
>> Please, can anybody write me a simple recursive matrix multiplication
>> using multiple threads in Python, or at least show me some guidelines
>> how to write it myself
>
> No problem, I just need your bank account data to withdraw the payment and
> the address of your teacher whom to send the results. ;^)
>
> Seriously, things don't work like this. If you show effort, you will get
> help, but you don't get working solutions without even a bit of effort on
> your own.
>
> Start with a recursive algorithm, then make it multithreaded. Be prepared to
> make some experiments on how to distribute tasks to threads and collect
> them again.
If you want to do matrix multiplication from Python, use NumPy.
Anything you write in Python itself and run in CPython will be far,
far slower. CPython without NumPy is about 60x slower than C on
basic array math.
Writing high-speed parallel number-crunching code that outperforms
off the shelf libraries is typically non-trivial. If there's some
easy way to make a simple operation like matrix multiply go faster,
it's probably been done already.
If you're really into this, look into the Unified Parallel C effort
and its libraries. The supercomputer types spend their lives doing this
sort of thing.
John Nagle
John Nagle, Jan 5, 2011
11. ### Dan StrombergGuest
On Tue, Jan 4, 2011 at 11:31 PM, Tim Roberts <> wrote:
> Zdenko <> wrote:
>>
>>Please, can anybody write me a simple recursive matrix multiplication
>>using multiple threads in Python, or at least show me some guidelines
>>how to write it myself
>
> Matrix multiplication is not generally done recursively. There's no
> conceptual gain. It makes more sense iteratively.
It may not be that common, but there is at least one significant
advantage: Cache obliviousness.
http://www.catonmat.net/blog/mit-introduction-to-algorithms-part-fourteen
Dan Stromberg, Jan 5, 2011
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It takes just 2 minutes to sign up (and it's free!). Just click the sign up button to choose a username and then you can ask your own questions on the forum. | 4,555 | 18,085 | {"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-2015-27 | longest | en | 0.897127 |
https://www.12faces.business/tag/tn59/ | 1,718,622,627,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861701.67/warc/CC-MAIN-20240617091230-20240617121230-00061.warc.gz | 566,280,367 | 22,081 | ## Manage Workflow and the Local Optima Problem
It is a very widely held belief in business that each station in a production line should be working as hard as it can in order to maximise its efficiency. This might be a car assembly line or an accountant’s office processing tax returns using several different stages in the accounting process. In other words, each stage is working at its Local Optimum. Generations of Cost Accountants have encouraged this and spend endless amounts of time trying to split, for example, the cost of electricity over each work stage in the production cycle. This is probably seriously flawed thinking. Blue Belt
## SM5.0 Theory of Constraint (TOC) Find Remove Bottleneck
The Theory of Constraints (TOC) is an important optimising technique that focuses on improving the throughput of work at the (usually) single bottleneck or constraint in the workflow.
This Skills Module draws together a number of the 12Faces articles on the subject. One research project demonstrated that 89% of improvement in a manufacturing business came from the application of TOC. Yellow Belt.
## C2.5.1 Increase Revenue Increase Throughput
Throughput is the process of moving production through your business. Eventually the production will culminate in cash in your hand from a paid-for sale. Workflow management creates a faster flow (or throughput) of product through your business. What you have learned: That a strong focus on wasted effort and resources in your business will substantially […]
## How to Calculate Time to Produce each Product with Takt Time
Takt is the German word for the baton that an orchestra conductor uses to regulate the tempo of the music. Takt time may be thought of as a measurable beat time, rate time or heartbeat.
Takt time is the average time between the start of production of one unit and the start of production of the next unit, when these production starts are set to match the rate of customer demand.
In Lean, Takt time is the rate at which a finished product needs to be completed in order to meet customer demand. Blue Belt
## Importance of Small Batches Tool
There are merits to reducing the size of the batches you manufacture and/or hold in stock at any point in time. This might sound counter-intuitive because of the oft-quoted ‘economies of scale’ and the time cost of pulling down tool set-ups and changing them to manufacture another project. But, in the words of the song; “it ain’t necessarily so!” Learn the importance of small batches.
## How Kanban Improves Work Flow
Kanbans are a tool for structuring how work flows into the work-in-progress queues so that workers do not get overloaded and can use their time most effectively. Kanbans, and variations on them, are widely used in software projects (e.g. Scrum and Agile) and manufacturing (e.g. Lean and Six Sigma). Blue Belt
## How to Tune Buffer Management
Buffers, used strategically, will enable you to “smooth” out the flow of work through your system. This is important to maximise production.
## How Kaizen Leads to Continuous Incremental Improvement
Most improvement is an incremental process. It is rare to get something completely right the first time so, over time, we improve parts of the system so as to continue to improve the whole. A Japanese phase for such incremental improvement is ‘Kaizen”. Read this article for insights into incremental improvement of your business. Yellow Belt
## SM5.4 TOC Drum, Buffer, Rope Optimisation Techniques
The Theory of Constraints (TOC) demonstrates the importance of finding and managing the constraint in your business.
This article covers three of the major techniques to optimise any production system around its constraint.
## Take Advantage of Agile / Scrum / Kanban to Increase Productivity
In a smaller business there is always plenty to do and many demands on a managers most limited resource; time. Often the owners time is the major bottleneck on the growth of the business (see the introductory article to Theory of Constraints). Plenty of other businesses have struggled with prioritising time before you and several techniques that are very similar have evolved. The three best known of these are named Agile / Scrum / Kanban. Various references cite improvements in team productivity ranging from 25% to 300% by using these methods. This article introduces these concepts and a way of using a free software app named Trello to manage the process. A mix of Yellow and Blue Belt introductory material.
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http://marketingwithoutthemarketing.com/index-114.html | 1,721,288,447,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514826.9/warc/CC-MAIN-20240718064851-20240718094851-00232.warc.gz | 18,733,683 | 121,674 | # NCERT Solutions For Class 7 Maths
## NCERT Solutions For Class 7 Maths CBSE 2023-24 Edition – Free PDF Download
NCERT Solutions for Class 7 Maths are provided here. Practising NCERT Solutions is the ultimate need for students who intend to score good marks in Maths examinations. Students facing trouble in solving problems from the NCERT textbook of Class 7 can refer to our free NCERT Solutions provided below.
Students are advised to practise Class 7th NCERT Solutions, which are available in PDF format, as it helps them understand the basic concepts easily. For students who feel stressed searching for the most comprehensive and detailed NCERT Solutions for Class 7, we at BYJU’S have prepared step-by-step solutions with detailed explanations. These exercises are prepared by our expert tutors to assist you in your exam preparation and help you attain good marks in the subject.
## NCERT Solutions Class 7 Maths (Chapter-Wise)
The following chapters have been removed from the NCERT Class 7 Maths textbook 2023-24.
Congruence of Triangles
Practical Geometry
In addition, students are given access to additional online study materials and resources, available in BYJU’S website, such as notes, books, question papers, exemplar problems, worksheets, etc. Students are also advised to practice Class 7 Sample Papers to get an idea of the question paper pattern in the final exam.
## NCERT Exercise-Wise Solutions for Class 7 Maths
Students are advised to go through these NCERT Solutions for Class 7. All the solutions are in line with the CBSE guidelines and presented in a stepwise manner so that students can understand the logic behind every problem while practising.
### Chapter 1 Integers
Chapter 1 involves the study of Integers. In the previous classes, we have learnt about whole numbers and integers. Now, we will study more about integers, their properties and operations. Likewise, the addition and the subtraction of the integers, properties of addition and subtraction of integers, multiplication and division of integers, properties of multiplication and division of integers.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 1 Exercises
➤ NCERT Solutions Class 7 Maths Exercise 1.1
➤ NCERT Solutions Class 7 Maths Exercise 1.2
➤ NCERT Solutions Class 7 Maths Exercise 1.3
➤ NCERT Solutions Class 7 Maths Exercise 1.4
Also, access the following resources for Class 7 Chapter 1 Integers at BYJU’S:
### Chapter 2 Fractions and Decimals
Chapter 2 of NCERT textbook, deals with Fractions and Decimals. You have learnt about fractions and decimals along with the operations of addition and subtraction on them, in earlier classes. We will now study the operations of multiplication and division on fractions as well as on decimals.
Likewise, multiplication of a fraction by a whole number, multiplication of a fraction by a fraction, division of the whole number by a fraction, division of a fraction by a whole number, multiplication of decimal by 10,100, 100, etc. Other topics are multiplication of decimal by whole number, multiplication of decimal by a decimal, division of decimals by 10, 100, 1000, etc., and dividing a decimal by the whole number.
Here you can find exercise solution links for the topics covered in this chapter.
Also, access the following resources for Class 7 Chapter 2 Fractions and Decimals at BYJU’S:
### Chapter 3 Data Handling
Chapter 3 of NCERT textbook deals with Data Handling. In your previous classes, you have dealt with various types of data. You have learnt to collect data, tabulate and put it in the form of bar graphs. Now in this chapter, we will take one more step towards learning how to do this. This chapter covers topics like collecting data, organisation of data, representative values, arithmetic mean, mode of large data, median, use of bar graphs with a different purpose, choosing scale, chance and probability.
Here, you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 3 Exercises
➤ Class 7 Maths NCERT Solutions Exercise 3.1
➤ Class 7 Maths NCERT Solutions Exercise 3.2
➤ Class 7 Maths NCERT Solutions Exercise 3.3
➤ Class 7 Maths NCERT Solutions Exercise 3.4
Also, access the following resources for Class 7 Chapter 3 Data Handling at BYJU’S:
### Chapter 4 Simple Equations
Chapter 4, Simple Equations of NCERT textbook deals with setting up an equation and solving an equation. By solving more equations, we shall learn transposing a number, i.e., moving it from one side to the other. We can transpose numbers instead of adding or subtracting it from both sides of the equations. Then we study from solution to an equation, applications of simple equations to practical situations.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 4 Exercises
➤ Class 7 Maths NCERT Solutions Exercise 4.1
➤ Class 7 Maths NCERT Solutions Exercise 4.2
➤ Class 7 Maths NCERT Solutions Exercise 4.3
➤ Class 7 Maths NCERT Solutions Exercise 4.4
Also, access the following resources for Class 7 Chapter 4 Simple Equations at BYJU’S:
### Chapter 5 Lines and Angles
This chapter is about Lines and Angles. This chapter includes related angles, complementary angles, supplementary angles, adjacent angles, linear pair, vertically opposite angles, pairs of lines, intersecting lines, transversal, the angle made by a transversal, transversal of parallel lines and checking of parallel lines.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 5 Exercises
➤ NCERT Solutions Class 7 Maths Exercise 5.1
➤ NCERT Solutions Class 7 Maths Exercise 5.2
Also, access the following resources for Class 7 Chapter 5 Lines and Angles at BYJU’S:
### Chapter 6 The Triangle and Its Properties
Chapter 6 of NCERT textbook discusses the topic The Triangle and Its Properties. Triangle is a simple closed curve made of three line segments. It has three vertices, three sides and three angles. It covers the concepts median of a triangle, altitude of a triangle, exterior angle of a triangle and its property, angle sum property of a triangle, two special triangles – equilateral and isosceles, the sum of the lengths of two sides of a triangle, right-angled triangles and Pythagoras property.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 6 Exercises
➤ NCERT Solutions Class 7 Maths Exercise 6.1
➤ NCERT Solutions Class 7 Maths Exercise 6.2
➤ NCERT Solutions Class 7 Maths Exercise 6.3
➤ NCERT Solutions Class 7 Maths Exercise 6.4
➤ NCERT Solutions Class 7 Maths Exercise 6.5
Also, access the following resources for Class 7 Chapter 6 The Triangle and Its Properties at BYJU’S:
### Chapter 7 Congruence of Triangles
This chapter is about the Congruence of Triangles. If two figures have exactly the same shape and size, they are said to be congruent. Other related topics are congruence of plane figures, congruence among line segments, congruence of angles, congruence of triangles, criteria for congruence of triangles, congruence of right-angled triangles.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 7 Exercises
➤ Class 7 Maths NCERT Solutions Exercise 7.1
➤ Class 7 Maths NCERT Solutions Exercise 7.2
Also, access the following resources for Class 7 Chapter 7 Congruence of Triangles at BYJU’S:
### Chapter 8 Comparing Quantities
Chapter 8 of NCERT textbook deals with the Comparing Quantities. Its related topics are equivalent ratios, percentage – another way of comparing quantities, the meaning of percentage, converting fractional numbers to a percentage, converting decimals to a percentage, converting percentages to fractions or decimals. Other interesting topics are fun with estimation, interpreting percentages, converting percentages to how many, ratios to percents, increase or decrease as per cent, profit or loss as a percentage, charge given on borrowed money or simple interest.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 8 Exercises
➤ Class 7 Maths NCERT Solutions Exercise 8.1
➤ Class 7 Maths NCERT Solutions Exercise 8.2
➤ Class 7 Maths NCERT Solutions Exercise 8.3
Also, access the following resources for Class 7 Chapter 8 Comparing Quantities at BYJU’S:
### Chapter 9 Rational Numbers
Chapter 9 of NCERT textbook discusses the topic of Rational Numbers. It covers positive and negative rational numbers, rational numbers on a number line, rational numbers in standard form, comparison of rational numbers, rational numbers between two rational numbers and operations on rational numbers.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 9 Exercises
➤ NCERT Solutions Class 7 Maths Exercise 9.1
➤ NCERT Solutions Class 7 Maths Exercise 9.2
Also, access the following resources for Class 7 Chapter 9 Rational Numbers at BYJU’S:
### Chapter 10 Practical Geometry
Chapter 10 of NCERT textbook discusses the topic of Practical Geometry. You are familiar with a number of shapes. You learnt how to draw some of them in the earlier classes. For example, you can draw a line segment of a given length, a line perpendicular to a given line segment, an angle, an angle bisector, a circle etc. Now, you will learn how to draw parallel lines and some types of triangles. Topics covered in this chapter are the construction of a line parallel to a given line, through a point not on the line, construction of different types of triangles.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 10 Exercises
➤ NCERT Solutions Class 7 Maths Exercise 10.1
➤ NCERT Solutions Class 7 Maths Exercise 10.2
➤ NCERT Solutions Class 7 Maths Exercise 10.3
➤ NCERT Solutions Class 7 Maths Exercise 10.4
➤ NCERT Solutions Class 7 Maths Exercise 10.5
Also, access the following resources for Class 7 Chapter 10 Practical Geometry at BYJU’S:
### Chapter 11 Perimeter and Area
This chapter is about Simple Interest. In Class 6, you have already learnt perimeters of plane figures and areas of squares and rectangles. Perimeter is the distance around a closed figure while the area is the part of a plane or region occupied by the closed figure. In this class, you will learn about perimeters and areas of a few more plane figures.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 11 Exercises
➤ Class 7 Maths NCERT Solutions Exercise 11.1
➤ Class 7 Maths NCERT Solutions Exercise 11.2
➤ Class 7 Maths NCERT Solutions Exercise 11.3
➤ Class 7 Maths NCERT Solutions Exercise 11.4
Also, access the following resources for Class 7 Chapter 11 Perimeter and Area at BYJU’S:
### Chapter 12 Algebraic Expressions
In Chapter 12, Algebraic Expressions of NCERT textbook deals with terms of an expression, coefficients, like and unlike terms, monomials, binomials, trinomials, polynomials, addition and subtraction of algebraic expressions, finding the values of an expression.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 12 Exercises
➤ Class 7 Maths NCERT Solutions Exercise 12.1
➤ Class 7 Maths NCERT Solutions Exercise 12.2
➤ Class 7 Maths NCERT Solutions Exercise 12.3
➤ Class 7 Maths NCERT Solutions Exercise 12.4
Also, access the following resources for Class 7 Chapter 12 Algebraic Expressions at BYJU’S:
### Chapter 13 Exponents and Powers
Chapter 13 of NCERT textbook discusses the topic Exponents and Powers. Exponents are the product of rational numbers multiplied several times by themselves. Some of the concepts covered here are laws of exponents, multiplying powers with the same base, dividing powers with the same base, taking power of a power, multiplying powers with the same exponents, dividing powers with the same exponents, decimal number system, expressing large numbers in the standard form.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 13 Exercises
➤ NCERT Solutions Class 7 Maths Exercise 13.1
➤ NCERT Solutions Class 7 Maths Exercise 13.2
➤ NCERT Solutions Class 7 Maths Exercise 13.3
Also, access the following resources for Class 7 Chapter 13 Exponents and Powers at BYJU’S:
### Chapter 14 Symmetry
This chapter is about Symmetry. Symmetry means that one shape becomes exactly like another when you move it in some way, turn, flip or slide. For two objects to be symmetrical, they must be the same size and shape, with one object having a different orientation from the first. There can also be symmetry in one object, such as a face. Topics covered in this chapter are lines of symmetry for regular polygons, rotational symmetry, line symmetry and rotational symmetry.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 14 Exercises
➤ NCERT Solutions Class 7 Maths Exercise 14.1
➤ NCERT Solutions Class 7 Maths Exercise 14.2
➤ NCERT Solutions Class 7 Maths Exercise 14.3
Also, access the following resources for Class 7 Chapter 14 Symmetry at BYJU’S:
### Chapter 15 Visualising Solid Shapes
In Chapter 15, Visualising Solid Shapes of NCERT textbook deals with the plane figures and solids, faces, edges, vertices, nets for building 3-D shapes, drawing solids on a flat surface, oblique sketches, isometric sketches. Some more concepts covered in this chapter are visualising solid objects, viewing different sections of a solid, then another way is by shadow play.
Here you can find exercise solution links for the topics covered in this chapter.
NCERT Solutions for Class 7 Maths Chapter 15 Exercises
➤ Class 7 Maths NCERT Solutions Exercise 15.1
➤ Class 7 Maths NCERT Solutions Exercise 15.2
➤ Class 7 Maths NCERT Solutions Exercise 15.3
➤ Class 7 Maths NCERT Solutions Exercise 15.4
Also, access the following resources for Class 7 Chapter 15 Visualising Solid Shapes at BYJU’S:
The NCERT Solutions are considered one of the best resources to master Maths. The solutions given here are in well-structured format with different shortcut methods to ensure a proper understanding of the concept and scoring good marks in Maths.
### Benefits of Class 7 Maths NCERT Solutions
These solutions are prepared by expert tutors.
• NCERT Class 7 Solutions are explained step by step in a comprehensive manner.
• Formulas are mentioned in between steps to help students recall them easily.
• Apart from clearing doubts, these solutions also give in-depth knowledge about the respective topics.
Keep visiting BYJU’S to get more updated learning materials and download the BYJU’S app for a better and personalized learning experience, along with engaging video lessons.
## Frequently Asked Questions on NCERT Solutions for Class 7 Maths
Q1
### Does NCERT Solutions for Class 7 Maths help to gain more marks in the exam?
To gain more marks in the exam, students should go through each and every topic thoroughly, as well as practise the NCERT Solutions for Class 7 Maths. A proper preparation strategy should be followed in order to gain more marks in the annual exam. The solutions contain answers for all the exercise-wise questions present in the NCERT textbook. Regularly solving the textbook questions using these solutions improves the problem-solving abilities among students, which is important to face the exam without fear.
Q2
### Does BYJU’S provide the NCERT Solutions for Class 7 Maths for free?
You can find the NCERT Solutions for Class 7 Maths from BYJU’S. The solutions provided here are in PDF format, which can be downloaded and accessed by the students effortlessly. Both the chapter-wise and exercise-wise solutions are designed by expert faculty having vast knowledge in the respective field. Students can solve textbook problems by referring to these solutions and get their doubts cleared instantly.
Q3
### Can I rely on the NCERT Solutions for Class 7 Maths at BYJU’S to perform well in the annual exam?
Students of Class 7 who find it difficult to answer the textbook questions can refer to the NCERT Solutions provided by BYJU’S. The faculty curates the solutions as per the latest CBSE syllabus and guidelines designed by the CBSE board. The solutions contain step-by-step explanations based on the exam pattern and marks weightage allotted for each concept. Using the solutions on a regular basis improves logical thinking and analytical approach towards complex questions. | 3,901 | 17,078 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-30 | latest | en | 0.926094 |
https://www.hindawi.com/journals/jfs/2016/7594031/ | 1,606,562,784,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195417.37/warc/CC-MAIN-20201128095617-20201128125617-00322.warc.gz | 696,117,612 | 144,550 | / / Article
Special Issue
## Function Spaces, Approximation Theory, and Their Applications
View this Special Issue
Research Article | Open Access
Volume 2016 |Article ID 7594031 | https://doi.org/10.1155/2016/7594031
Nazneen Khan, "Classes of -Convergent Double Sequences over -Normed Spaces", Journal of Function Spaces, vol. 2016, Article ID 7594031, 7 pages, 2016. https://doi.org/10.1155/2016/7594031
# Classes of -Convergent Double Sequences over -Normed Spaces
Accepted20 Mar 2016
Published09 May 2016
#### Abstract
I introduce some new classes of -convergent double sequences defined by a sequence of moduli over -normed space. Study of their algebraic and topological properties and some inclusion relations has also been done.
#### 1. Introduction
The notion of -convergence was introduced by Kostyrko et al. in [1]. It is known that -convergence is generalization of the statistical convergence which was introduced by Fast [2]. It was further studied by Demirci [3], Das et al. [4], Šalát et al. [5], and many others.
For a nonempty set , the family of sets , the power set of , is said to be an ideal if(1);(2) is additive; that is, ;(3) is hereditary; that is, .
A nontrivial ideal is called admissible if . is maximal if there cannot exist any nontrivial ideal containing as a subset.
Let , and denote the set of natural, real, and complex numbers, respectively. A double sequence of complex numbers is defined as a function from to . A number is called a limit of a double sequence if for every there exists some such that The set of all double sequences is denoted by . Any subset of the is called double sequence space. A sequence is said to be -convergent to a number if, for every , . In this case we write .
A double sequence space is said to be solid or normal if implies for all sequences of scalars with for all . For more details please see [68].
Example 1. Let be the class of all subsets of such that implies that there exists such that .
Then is an ideal of in the usual Pringsheim sense of convergence of double sequences. If is replaced by , the class of finite subsets of , then we get the usual regular convergence of double sequences.
The theory of 2-normed spaces was first introduced by Gähler [9] in 1964. Later on it was extended to -normed spaces by Misiak [10]. Since then many mathematicians have worked in this field and obtained many interesting results; for instance see Gunawan [11, 12], Gunawan and Mashadi [13], Mursaleen and Mohiuddine [14], Şahiner et al. [15, 16], and Yamancı and Gürdal [17]. Let and let be a linear metric space over the field of real or complex numbers of dimension , where . A real valued function on satisfying the following four conditions:(1) if and only if are linearly dependent;(2) is invariant under permutation;(3) for any ;(4)is called an -norm on and the pair is called an -normed space over the field .
Example 2. If we take , equipped with Euclidean -norm spanned by vectors , then the -norm may be given by the formula , where for .
The standard -norm on is defined as
where denotes the inner product on . If , then this -norm is exactly the same as the Euclidean -norm mentioned earlier. For this -norm is the usual norm .
A sequence in an -normed space is said to converge to some if
A sequence in an -normed space is said to be Cauchy if
If every Cauchy sequence in converges to some , then is said to be complete with respect to the -norm. Any complete - normed space is said to be an -Banach space.
The concept of modulus function was introduced by Nakano [18] in the year 1953. It was further studied by [7, 8, 1921] and many more. It is defined as a function : satisfying the following conditions:(1) if and only if ,(2) for all ,(3) is increasing,(4) is continuous from the right at zero.Ruckle [22] used the idea of a modulus function to construct the sequence space The space is closely related to the space which is an space with for all real . Thus Ruckle [23, 24] proved that, for any modulus ,The space is a Banach space with respect to the norm After then Kolk [25, 26] gave an extension of by considering a sequence of modulus functions called the sequence of moduli and defined the sequence space: From the above four properties of modulus function it can be clearly seen that must be continuous everywhere on ). For a sequence of moduli, we have further two properties:(5) for all ;(6) uniformly in and for .
Example 3. Let be a function from to . If we take , then the function is a bounded modulus function and if we take , then is an unbounded modulus function.
By a lacunary sequence ; , where , we mean an increasing sequence of nonnegative integers as . The intervals determined by are denoted by and the ratio will be denoted by . The space of lacunary strongly convergent sequence was defined by Freedman et al. [27] as follows: The double lacunary sequence was defined by Savaş and Patterson [28]. A double sequence is called double lacunary if there exist two increasing sequences of integers such that The following interval is determined by : The space of double lacunary strongly convergent sequence is defined as follows:
#### 2. New Classes of Double Sequences
Now, we will define the new classes of double sequences.
Let be an admissible ideal, let be a sequence of moduli, let be an -normed space, let be a sequence of positive real numbers, let be a sequence of strictly positive real numbers, and let be the space of all double sequences defined over the -normed space ; then for some and every , we define(i) : ,(ii) :
Case 1. If , then we get(i) : ,(ii) :
Case 2. If , then we get(i) : ,(ii) :
Case 3. If and , then we get(i) : ,(ii) : The following inequality will be used throughout the paper. If , then we have for all and . Also for all
#### 3. Main Results
Theorem 4. The sets and are linear spaces over the field of complex numbers .
Proof. Let , , and ; then for every we can write By the use of inequality (12), we have the following inequality: This inequality says to us that the inclusion holds. From here, since the right side belongs to , the left side also belongs to . This completes the proof.
Lemma 5. Let be a modulus function and let . Then for each , one has
Theorem 6. Let be a sequence of moduli and Then the following statements hold:(i),(ii).
Proof. For some , choose such that . By the continuity of for all , we can choose some such that for every with we have Let ; then for some and for every , we have Therefore for , we have So, by inequality (17), we can write This implies that . Hence The inclusion is strict as for the reverse inclusion we need the condition given in the next theorem. The other part can be proved similarly.
Theorem 7. Let be a sequence of moduli. If for all , then(i),(ii)
Proof. (i) To prove , it is sufficient to show that . Let ; then, by the definition, we get By the given condition for all , we have for all ; that is, From inequalities (21) and (23), we get which consequently implies that That is, This implies that . Hence, from the previous theorem and inclusion (27), we get the required result. The other part can be proved similarly.
Corollary 8. Let and be sequences of moduli. If for all , then(i),(ii).
Theorem 9. Let and be the standard and the Euclidean -norm spaces, respectively. Then
Proof. The proof of this result is easy, so it is omitted.
Theorem 10. Let and be sequences of moduli; then(i),(ii),(iii),(iv)
Proof. (i) For some , we choose such that . Now as is a sequence of modulus functions which are always continuous, we can choose such that, for every , we get : Then, by the definition, we have Thus, for , we get Now, by the continuity of , we have which further implies that which implies Therefore . This completes the proof.
(iii) Again consider Then by the definition of both the spaces, we get Using the fact that , we have So we get Therefore we have Hence we have
#### 4. Conclusion
A detailed study of some new classes of -convergent double sequences over -normed spaces has been done. Some algebraic and topological properties and inclusion relations have been proved with supported examples.
#### Competing Interests
The author declares that he has no competing interests.
#### Acknowledgments
The author is grateful to the anonymous referees for their careful corrections and comments that improved the presentation of this paper.
#### References
1. P. Kostyrko, T. Šalát, and W. Wilczyński, “I-convergence,” Real Analysis Exchange, vol. 26, no. 2, pp. 669–685, 2000/01. View at: Google Scholar | MathSciNet
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Copyright © 2016 Nazneen Khan. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | 3,755 | 13,801 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2020-50 | longest | en | 0.92342 |
https://codingtube.tech/2024/01/04/python/python-sets/ | 1,726,755,710,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700652031.71/warc/CC-MAIN-20240919125821-20240919155821-00261.warc.gz | 160,688,424 | 29,046 | # Python Sets
In Python, sets are an essential data structure that provides an unordered collection of unique elements. Leveraging the mathematical concept of sets, Python sets offer versatile functionalities for performing set operations such as union, intersection, and difference. This comprehensive guide explores the world of Python sets, from basic syntax to advanced techniques like set comprehensions and frozen sets.
## 1. Understanding Sets in Python:
A set is defined by enclosing elements within curly braces ({}) or by using the set() constructor.
my_set = {1, 2, 3, "Python", 3.14}
## 2. Basic Set Operations:
Elements can be added to a set using the add() method.
my_set.add(4)
### Removing Elements:
Elements can be removed using the remove() or discard() methods.
my_set.remove("Python")
### Clearing a Set:
The clear() method removes all elements from a set.
my_set.clear()
## 3. Set Operations:
### Union (|):
The union of two sets contains all unique elements from both sets.
set1 = {1, 2, 3}
set2 = {3, 4, 5}
union_set = set1 | set2 # Output: {1, 2, 3, 4, 5}
### Intersection (&):
The intersection of two sets contains elements common to both sets.
intersection_set = set1 & set2 # Output: {3}
### Difference (-):
The difference between two sets contains elements present in the first set but not in the second.
difference_set = set1 - set2 # Output: {1, 2}
### Symmetric Difference (^):
The symmetric difference contains elements that are unique to each set.
symmetric_difference_set = set1 ^ set2 # Output: {1, 2, 4, 5}
## 4. Set Comprehensions:
Set comprehensions provide a concise way to create sets.
squares_set = {x**2 for x in range(5)} # Output: {0, 1, 4, 9, 16}
## 5. Frozen Sets:
A frozen set is an immutable version of a set, and it can be used as a key in dictionaries or as an element in other sets.
frozen_set = frozenset([1, 2, 3])
## 6. Methods and Functions:
### len() – Length:
Returns the number of elements in a set.
length = len(my_set)
### pop():
Removes and returns an arbitrary element from the set.
popped_element = my_set.pop()
### issubset() and issuperset():
Check if one set is a subset or superset of another.
is_subset = set1.issubset(set2)
is_superset = set1.issuperset(set2)
## 7. When to Use Sets:
• Removing Duplicates:
Sets automatically eliminate duplicate elements.
• Set Operations:
Sets are ideal for performing set operations efficiently.
• Membership Testing:
Quickly check whether an element is present in a collection.
## 8. Conclusion:
Python sets provide a versatile and efficient way to work with unique collections of elements. From basic operations to advanced set manipulations, understanding sets is crucial for performing set-based operations in a clean and efficient manner. As you incorporate sets into your Python projects, you’ll find them invaluable for tasks that involve unique elements and set operations. Happy coding! | 712 | 2,964 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-38 | latest | en | 0.804034 |
http://www.slideserve.com/cecil/determination-of-the-equilibrium-constant | 1,490,776,141,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218190234.0/warc/CC-MAIN-20170322212950-00174-ip-10-233-31-227.ec2.internal.warc.gz | 704,111,252 | 18,082 | This presentation is the property of its rightful owner.
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# Determination of the Equilibrium Constant PowerPoint PPT Presentation
Determination of the Equilibrium Constant. Krishna Trehan. Background Information. Many reactions do not go to completion; instead, they reach states where the products and reactants are both present.
Determination of the Equilibrium Constant
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## Determination of the Equilibrium Constant
Krishna Trehan
### Background Information
• Many reactions do not go to completion; instead, they reach states where the products and reactants are both present.
• When the concentrations of the reactants and products are both equal and constant, the reaction has reached equilibrium.
### Equilibrium Constant (Kc)
• The equilibrium constant relates the concentrations of the products to those of the reactants in a ratio.
• For exmaple: Fe3+ + SCN- ↔FeSCN2+
• If Kc is greater than 1, then the products are favored. If Kc is less than 1, then the reactants are favored.
### Spectrophotometry
• In order to find the equilibrium constant you need the concentrations of the reactants and products. Usually this is given to use, but if it is not, you can use spectrophotometry to find the concentrations.
• A spectrophotometer consists of two instruments, a spectrometer (for producing light of any wavelength you chose) and a photometer (for measuring the intensity of light).
### Spectrophotometry
• A beam of light from the spectrometer is sent through a solution and the photometer receives the beam after it is sent through the solution
Spectrometer Light Solution Photometer
• The spectrophotometer will return a number; this is a representation of the analytes absorbance (the lights frequency will be altered by the solution).
• The degree of color is proportional to the concentration.
### Spectrophotometer
A sample of a solution is put into the spectrophotometer where it is analyzed.
http://www.okokchina.com/Files/uppic34/Spectrophotometer960.jpg
### Purpose
• The purpose of this experiment is to be able to find the equilibrium constant of a FeSCN2+ solution using spectrophotometry.
### Materials
• Spectrophotometer
• Cuvet (to place solution sample into spectrophotometer)
• 5 mL of 2E-3 M Fe(NO3)3
• 5mL of 2E-3 M KSCN.
• FeSCN2+
• Beaker
### Procedures : Step 1
• First we need to find the wavelength of optimum absorbance. This shows the frequency of visible light which is most easily absorbed by the solution.
• Place the solution FeSCN2+between the Spectrometer Light and the Photometer
• Chose a value from 380nm to 780 nm (visible light range on the spectrum)
• The spectrophotometer will return a value, this is the absorbance value. Then you must graph this point on a graph with the x-axis as Light Frequency, and the y-axis as Absorbance.
### Step 1 - Continued
• Once this is completed, you must repeat this step several more times (using different frequency values) to find the frequency which yields the highest Absorbance value, the optimum absorbance value. This frequency value will be used throughout the rest of the experiment.
• This is an example of a chart with a solution of chlorophyll
http://www.bio.davidson.edu/Courses/Bio111/Bio111LabMan/lab1fig3.gif
### Procedure: Step 2
• In this step, we will keep the frequency of the spectrometer constant (the value achieved in step 1)
• Prepare multiple solutions of FeSCN2+ (with different concentrations) and add them between the spectrometer and the photometer
• While keeping the frequency constant, test the solutions and record the outcome.
### Step 2 - Continued
• The outcome is the transmittance of the solutions. In order to obtain the Absorbance, we must use the following equation:
• Once the absorbance values are obtained, graph the results
• X-axis = Concentration (Molarity)
• Y-axis = Absorbance
• Find a line which connects both the points.
### Procedures: Step 3
• Mix the 5 mL of 2E-3 M Fe(NO3)3 with the 5mL of 2E-3 M KSCN.
• Analyze this solution in the spectrophotometer with the same frequency as the one obtained in Step 1.
• Convert this transmittance into Absorbance. Call this value k.
### Step 3 - Continued
• Now there is an Absorbance value (y-value), but there is no concentration value (x-value).
• Graph the line y=k on the same graph created in step 2.
• The point at which the two lines intersect will have the point (x, k)
• Point x is the concentration of the solution.
### Analysis Questions
• 1. How many moles of Fe3+ and SCN- were initially present?
• 2. How many moles of FeSCN2+ were in the mixture at equilibrium?
• 3. How many moles of Fe3+ and SCN- were used up in making the FeSCN2+ ?
• 4. How many moles of Fe3+ and SCN- remain in the solution at equilibrium?
### Analysis Questions
• 5. What are the concentrations of Fe3+ , SCN- , and FeSCN2+ at equilibrium?
• 6. Determine Kc for this reaction. | 1,315 | 5,382 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2017-13 | longest | en | 0.884329 |
https://www.sophia.org/tutorials/introduction-to-circles | 1,618,585,668,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038066981.0/warc/CC-MAIN-20210416130611-20210416160611-00260.warc.gz | 1,124,942,063 | 9,523 | +
# Introduction to Circles
##### Rating:
(0)
Author: Leif Park Jordan
##### Description:
1.
Introduce and define the different parts of a circle (radius, diameter, center, tangent, chord, and secant).
2.
Provide examples that allow practice identifying and labeling the different parts of a circle.
This packet should help a learner seeking to understand how to identify the parts of a circle.
(more)
Tutorial
This video introduces circles and defines several parts of the circle: the radius and the center.
Source: RobertOB on Guaranteach
## The Diameter
This video explains what the diameter of a circle is.
Source: RobertOB on Guaranteach
## Tangents
This video defines tangents and explains a few of their properties.
Source: tracyp on Guaranteach
## The Point of Tangency
This video defines the point of tangency.
Source: tracyp on Guaranteach
## Chords
This video defines chords and explains some of their properties.
Source: tracyp on Guaranteach
## Secant Lines
This video introduces and defines secant lines.
## Sample Questions
This video provides examples that allow practice identifying and labeling the different parts of a circle. | 255 | 1,170 | {"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-2021-17 | latest | en | 0.824721 |
https://goprep.co/ex-5-q38-the-value-of-z-3-bar-z3-is-equivalent-toa-z-3-3-b-z-i-1nlh6s | 1,618,649,235,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038118762.49/warc/CC-MAIN-20210417071833-20210417101833-00100.warc.gz | 379,974,951 | 30,372 | # The value of (z +
Given (z + 3) (z̅ + 3) = (x + iy + 3) (x –iy + 3)
= (x + 3)2 – (iy)2
= (x + 3)2 + y2
= |x + 3 + iy|2
= |z + 3|2
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Author Topic: Feedback and control (Any control experts?) (Read 1379 times) 0 Members and 1 Guest are viewing this topic.
PM3295
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« on: March 27, 2013, 10:12:03 10:12 »
When we have a negative-feedback system:
We know; The system will be stable as long as the overall gain stays below 0dB (with some margin)when we reach -180° phase shift point.
a) What happens if the phase shift goes to +180°, for some reason, and the gain is more than 0dB?
b) What happens if the phase goes to more than +180° (say +200°, 250°) with more than 0dB gain?
I looked through my old books but can't find any example of a negative-feedback system having large positive phase shifts.
All the examples only deal with stability issues approaching the -180° point.
I hope some control expert can clarify this.
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optikon
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« Reply #1 on: March 27, 2013, 11:09:18 11:09 »
When we have a negative-feedback system:
We know; The system will be stable as long as the overall gain stays below 0dB (with some margin)when we reach -180° phase shift point.
a) What happens if the phase shift goes to +180°, for some reason, and the gain is more than 0dB?
b) What happens if the phase goes to more than +180° (say +200°, 250°) with more than 0dB gain?
I looked through my old books but can't find any example of a negative-feedback system having large positive phase shifts.
All the examples only deal with stability issues approaching the -180° point.
I hope some control expert can clarify this.
You just have to consider the whole phase shift around the loop of the feedback system. It answers all your questions about stability.
a) If the overall loop achieves a total of 360 degrees or more with > 0 dB gain, you will have instability
b) See "a"
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PM3295
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« Reply #2 on: March 27, 2013, 02:50:29 14:50 »
So, basically we only need to look at either +180° or -180° and if the gain is > 0 it will be unstable?
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optikon
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« Reply #3 on: March 27, 2013, 03:10:53 15:10 »
So, basically we only need to look at either +180° or -180° and if the gain is > 0 it will be unstable?
I think the 180 phase requirement you are thinking about might come from the fact that your forward path has already contributed 180 degrees phase shift. Is this correct? I think that is the standard control system model. But the actual condition for instability is the loop (forward and feedback) must achieve 360 or more phase with a loop gain > 0dB.
The previous statement is general and covers all feedback cases. What is the model you are thinking about?
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PM3295
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« Reply #4 on: March 27, 2013, 04:09:34 16:09 »
I think the 180 phase requirement you are thinking about might come from the fact that your forward path has already contributed 180 degrees phase shift. Is this correct? I think that is the standard control system model.
That is the case. I have no specific model in mind.
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nordiceng
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« Reply #5 on: April 19, 2013, 11:03:07 23:03 »
When the phase shift goes to 180°, the control system is most likely will be unstable
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Old_but_Alive
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« Reply #6 on: April 20, 2013, 04:47:45 04:47 »
its not just a go-no go situation.
There must be a margin away from the raw gain <1 at 180
there is a need to be some way away from that situation.
have a look at
http://www.ridleyengineering.com/loop-stability-requirements.html
this next one is a bit heavy, but try it
http://www.eeng.nuim.ie/~rfarrell/EE301/notes/EE309_notes_18.pdf
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https://stats.stackexchange.com/questions/345418/are-random-effects-used-for-explanatory-variables-measured-with-inaccuracy-or-fo | 1,642,717,381,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302706.62/warc/CC-MAIN-20220120220649-20220121010649-00638.warc.gz | 586,882,768 | 35,327 | # Quote
I am reading from Quinn and Keough's book. In chapter Correlation and Regression, section Fixed X, at page 94, they say
Linear regressions analysis assumes that the $$x_i$$ are known constants, i.e. they are fixed values controlled or set by the investigator with no variance associated with them. A linear model in which the predictor variables are fixed is known as Model I or a fixed effects model. This will often be the case in designed experiments where the levels of X are treatments chosen specifically. In these circumstances, we would commonly have replicate Y-values for each $$x_i$$ and X may well be a qualitative variable, so analyses that compare mean values of treatment groups might be more appropriate [..]. The fixed X assumption is probably not met for most regression analyses in biology because X and Y are usually both random variables recorded from a bivariate distribution. For example, Peake and Quinn (1993) did not choose mussel clumps of fixed areas but took a haphazard sample of clumps from the shore [..].
# Question
I used to think of that variables that should be integrated as random effects are those for which measurements was made with some level of inaccuracy and I am not sure this is what this quote is referring to. My question is
Are random effects used for explanatory variables measured with inaccuracy or for explanatory variables designating a subset of groups that are sampled?
Note that I am a little bit confused in how to think of these concepts in both the contexts of linear regressions and ANOVAs.
I would like to clarify my misunderstandings via two examples.
Example 1
Alice scuba dived, looked for anemones and for each one of them she measured their size and their depth. The question is "Does depth affect anemonea size?".
I can think of two potential reasons why depth should be modelled as a random effect:
• Because the depth was measured with some inaccuracy (specific to the depth meter Alice used)
• Because Alice has not exhaustively (and in the right proportions) sampled all the possible depths of interest.
Example 2
Alice went on the sea shore and randomly sampled mussel clumps. For each clump sampled, she counted the number of individual mussels and the number of species of mussels. The question is "Is the number of species of mussels affecting the number of individuals".
The counts are perfectly known (no inaccuracy in the measurement). From the above quote, I can still think of a reason why Alice should model the explanatory variable as a random effect; Alice haphazardly sampled clumps and a future study could have sampled different clumps. If all the clumps were sampled, then Alice should use a fixed effect model.
• Please be brief and indicate specfic goal. Seems to be a some-what weird question. May 10 '18 at 4:07
The term “random effect” does not mean “something that is random”, it refers to something in a hierarchical or mixed model that is not a population-level effect.
Suppose I have some adorable newborn Great Dane puppies, and I feed them a special puppy supplement that I think will help them grow larger than they would without the supplement. I give different puppies different doses, and I measure their weights when they’re grown and fit a model of “weight = intercept + beta*supplement_dose”. My intercept here is the expected weight of a puppy who was not given the supplement, and my slope is “the increase in weight per unit dose of supplement”.
Now, suppose I do this with multiple litters of Great Dane puppies. Due to factors like their parents’ genetics and their mothers’ nutrition during pregnancy, some of the litters have bigger puppies than others. Littermates are more alike with each-other than with other puppies, so I no longer have independence of observations - if I fit a regular linear model here, I will calculate the variance incorrectly and my standard errors and p-values will be wrong.
So I add a random effect for litter, specifically a random intercept. This means that now in addition to the model having its overall intercept (a fixed effect) each litter also gets its own separate intercept to account for the fact that some litters have bigger puppies than other litters. My fixed effect intercept is now kind of a weird averaged-across-litters expected weight of a puppy not given supplement. But since that differs by litter, I have to add the litter-specific intercept to get the expected weight of a puppy from litter #7 who was not given supplement (which may be different from the expected weight of a puppy from litter #3 who was not given supplement). The random intercept for litter #7 is “the effect of being in litter #7 on the weight of a puppy”. It has no meaning at the population level, but because it is part of the structure of my data I have to account for it or I will miscalculate the variance of other things in the model.
Random intercepts aren’t the only kind of random effect. Suppose I did the same experiment again, but this time I used litters from different breeds. I have a litter of Great Danes, a few mixed-breed litters of various ancestries, and a litter of Chihuahuas. If they all gain three pounds per gram of supplement, I am going to end up with some really big Chihuahuas... more likely, I may find that while the Great Danes gain 3 pounds per gram of supplement, the Chihuahuas gain 0.25 pounds per gram of supplement, and each mixed-breed litter gains different amounts of weight too. So the slope of weight gain relative to supplement dose now also differs between litters. I can add a random effect for slope to allow slope to vary between litters like that, while still being able to look at my overall question “does the supplement make puppies bigger?”
The important thing with random effects is that they’re usually something that’s meaningless at the population level. “The effect of being in litter #7 on weight of puppy.” There is never going to be another litter #7 exactly like that with exactly the same effect on puppy size again. It’s something meaningless/random that I can’t make any useful inferences about, but that I still have to account for in order to correctly model the situation. | 1,299 | 6,211 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-05 | latest | en | 0.960786 |
https://www.extendoffice.com/documents/excel/3149-excel-max-unique.html | 1,720,806,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514450.42/warc/CC-MAIN-20240712161324-20240712191324-00721.warc.gz | 603,011,533 | 27,160 | How to find max or min value based on unique values in Excel?
For instance, you have two column data, and one called Name with some duplicate values, now you want to find out the relative max or min values based on the unique values as below screenshot shown. How can you quickly handle with this task in Excel? In this article, you may find the answer.
Find max or min value based on unique values with creating Pivot Table
Find max or min value based on unique values with Kutools for Excel
Find unique max or min value from a list with Kutools for Excel
Find max or min value based on unique values with creating Pivot Table
In Excel, PivotTable is a powerful function, you can create a pivot table based on the range, and then find the max or min value by unique values.
1. Select the data you want to find max or min value from, and click Insert > PivotTable > PivotTable. See screenshot:
2. Then check New Worksheet or Existing Worksheet in Create PivotTable dialog as you need, here I check New Worksheet. See screenshot:
3. Click OK. Now a PivotTable Field List pane appears in a new sheet, and check Name (the column you want to select max or min values based on) and right click to Add to Row Labels, and check Score (the column you want to return to max or min value) and right click to Add to Values. See screenshots:
4. Then you can see a pivot table created in the worksheet, and now right click at the Sum of Score and select Value Field Settings from context menu, then in the popping dialog, under Summarize Values By tab, select Max or Min as you need. See screenshots:
5. Click OK. Now the max values of each unique value are found out.
Find max or min value based on unique values with Kutools for Excel
With Kutools for Excel, you can use the Advanced Combine Rows utility to quickly find the max or min value based on unique values.
with more than 300 handy functions, makes your jobs more easier. Free Downloadfree full-featured in 30 days
After Kutools for Excel, please do as below:
1. Select the data that you want to find max or min value (if you want to keep the original data, you can copy and paste them to another location first), and then click Kutools > Merge & Split > Advanced Combine Rows. See screenshot:
2. Then in the Advanced Combine Rows dialog, do as below:
1) Select the column that you want to get the unique value, then click Primary Key to set it as primary key column. See screenshot:
2) Then select the column which you want to return the max or min value from, and click Calculate > Max or Min to find the max or min value. See screenshot:
3. Now click Ok, and you can see the max or min values of each unique value are found out. See screenshot:
Tip: The Use fomratted values will keep your value formatting after combining.
Kutools for Excel is a multifunctional and handy tool, free install Kutools for Excel, makes your Excel working fly!
Find unique max or min value from a list with Kutools for Excel
If you have a list with some duplicate values, and now you just want to find out the unique max or min value from this list as below screenshot shown, in this case, Kutools for Excel can help you.
After Kutools for Excel, please do as below:
1. Select the list that you want to find the unique max or min value (including header), and click Kutools > Select > Select Duplicate & Unique Cells. See screenshot:
2. In the popping dialog, check Unique values only option, and check Select entire rows checkbox ( if you don’t need to show the corresponding value of the unique max value or min value, uncheck Select entire rows), and click Ok > OK. See screenshot:
Note: If you need to select all unique values and the 1st duplicates, please check the All unique (Including 1st duplicates) option in above dialog.
3. Then you can see all unique rows are selected, press Ctrl + C to copy it, and then go to another sheet and select a blank cell, press Ctrl + V to paste it. See screenshot:
4. Then select the list of values you want to find max or min value, click Kutools > Select > Select Cells with Max & Min Value. See screenshot:
5. In the popped out dialog, check Minimum value or Maximum value as you need, and then check Cell under Base section. See screenshot:
6. Click Ok, and the unique max value or min value is highlighted.
Unique minimum value Unique Maximum value
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• Increases your productivity by 50%, and reduces hundreds of mouse clicks for you every day! | 1,510 | 6,487 | {"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-2024-30 | latest | en | 0.825332 |
http://betterlesson.com/lesson/resource/2067178/unit-2-8-practice-problems-docx | 1,477,546,690,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988721141.89/warc/CC-MAIN-20161020183841-00376-ip-10-171-6-4.ec2.internal.warc.gz | 29,173,488 | 22,434 | Unit 2.8 Practice Problems.docx - Section 4: Practice Problems
Unit 2.8 Practice Problems.docx
Mixed Operation Practice + Scholarships
Unit 2: The College Project - Working with Decimals
Lesson 9 of 16
Big Idea: Which operation should I use to solve this problem? How can I use bars to model what is going on? What are scholarships and how do I earn them? Students apply their knowledge of decimal operations to answer these questions.
Print Lesson
7 teachers like this lesson
Standards:
Subject(s):
Math, Decimals, Number Sense and Operations, addition, subtraction, division, multiplication, 6th grade, master teacher project, scholarships
60 minutes
Andrea Palmer
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Environment: Urban | 265 | 1,180 | {"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-2016-44 | longest | en | 0.874166 |
https://documen.tv/question/a-credit-card-company-wanted-to-estimate-the-proportion-of-its-customers-who-pay-their-credit-ca-21496013-65/ | 1,721,645,320,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517846.73/warc/CC-MAIN-20240722095039-20240722125039-00400.warc.gz | 181,001,265 | 16,446 | ## A credit card company wanted to estimate the proportion of its customers who pay their credit card minimum balance on time. The company samp
Question
A credit card company wanted to estimate the proportion of its customers who pay their credit card minimum balance on time. The company sampled 500 records and constructed a confidence interval for the true proportion of customers who pay their credit card minimum balance on time. The resulting 95 percent confidence interval was (0.765, 0.835). Which of the following is a true statement regarding the value of the point estimate and margin of error of the percentage of customers who pay their credit card minimum balance on time?
a. The point estimate is 0.765 and the margin of error is 0.07.
b. The point estimate is 0.765 and the margin of error is 0.835.
c. The point estimate is 0.80 and the margin of error is 0.035.
d. The point estimate is 0.80 and the margin of error is 0.07.
e. The point estimate could be any number within the interval and the margin of error is 0.035.
in progress 0
3 years 2021-08-27T16:02:44+00:00 1 Answers 7 views 0
## Answers ( )
1. Answer: c. The point estimate is 0.80 and the margin of error is 0.035.
Step-by-step explanation:
Given: The 95 percent confidence interval was (0.765, 0.835).
Confidence interval = Point estimate ± Margin of error.
I.e. Point estimate + Margin of error = 0.835 (i)
Point estimate – Margin of error = 0.765 (ii)
Add (i) and (ii), we get
2 (Point estimate)= 1.60
Point estimate = 0.80
Put this in (i), we get
0.80+Margin of error = 0.835
⇒ Margin of error = 0.835 -0.80
Margin of error = 0.035
Hence, the correct statement is:
c. The point estimate is 0.80 and the margin of error is 0.035. | 473 | 1,748 | {"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} | 4.21875 | 4 | CC-MAIN-2024-30 | latest | en | 0.860083 |
https://www.unitsconverters.com/en/Kilobitpersecondperhertz-To-Kilobitpersecondpermegahertz/Unittounit-9008-9012 | 1,723,526,721,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641063659.79/warc/CC-MAIN-20240813043946-20240813073946-00029.warc.gz | 794,266,691 | 39,042 | Formula Used
1 Bit per Second per Hertz = 0.001 Kilobit per Second per Hertz
1 Bit per Second per Hertz = 1000 Kilobit per Second per Megahertz
1 Kilobit per Second per Hertz = 1000000 Kilobit per Second per Megahertz
Kilobit per Second per Hertzs to Kilobit per Second per Megahertzs Conversion
Kb/sec/Hz stands for kilobit per second per hertzs and Kb/sec/MHz stands for kilobit per second per megahertzs. The formula used in kilobit per second per hertzs to kilobit per second per megahertzs conversion is 1 Kilobit per Second per Hertz = 1000000 Kilobit per Second per Megahertz. In other words, 1 kilobit per second per hertz is 1000000 times bigger than a kilobit per second per megahertz. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
Convert Kilobit per Second per Hertz to Kilobit per Second per Megahertz
How to convert kilobit per second per hertz to kilobit per second per megahertz? In the bandwith efficiency measurement, first choose kilobit per second per hertz from the left dropdown and kilobit per second per megahertz from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from kilobit per second per megahertz to kilobit per second per hertz? You can check our kilobit per second per megahertz to kilobit per second per hertz converter.
How to convert Kilobit per Second per Hertz to Kilobit per Second per Megahertz?
The formula to convert Kilobit per Second per Hertz to Kilobit per Second per Megahertz is 1 Kilobit per Second per Hertz = 1000000 Kilobit per Second per Megahertz. Kilobit per Second per Hertz is 1000000 times Bigger than Kilobit per Second per Megahertz. Enter the value of Kilobit per Second per Hertz and hit Convert to get value in Kilobit per Second per Megahertz. Check our Kilobit per Second per Hertz to Kilobit per Second per Megahertz converter. Need a reverse calculation from Kilobit per Second per Megahertz to Kilobit per Second per Hertz? You can check our Kilobit per Second per Megahertz to Kilobit per Second per Hertz Converter.
How many Bit per Second per Hertz is 1 Kilobit per Second per Hertz?
1 Kilobit per Second per Hertz is equal to 1000000 Bit per Second per Hertz. 1 Kilobit per Second per Hertz is 1000000 times Bigger than 1 Bit per Second per Hertz.
Kilobit per Second per Hertzs to Kilobit per Second per Megahertzs Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like bandwith efficiency finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like Kb/sec/Hz to Kb/sec/MHz through multiplicative conversion factors. When you are converting bandwith efficiency, you need a Kilobit per Second per Hertzs to Kilobit per Second per Megahertzs converter that is elaborate and still easy to use. Converting Kilobit per Second per Hertz to Kilobit per Second per Megahertz is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Kilobit per Second per Hertz to Kilobit per Second per Megahertz conversion along with a table representing the entire conversion.
Let Others Know | 905 | 3,599 | {"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-2024-33 | latest | en | 0.415256 |
https://it.toolbox.com/question/an-interesting-intercompany-elimination-problem-051109 | 1,556,281,583,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578770163.94/warc/CC-MAIN-20190426113513-20190426135513-00216.warc.gz | 458,522,824 | 51,845 | # An Interesting Intercompany Elimination Problem
"I'm working in a company with thousands of entities in our Planning application. Some of the entities have 2 divisions where one division might sell to the other. Where there are 2 divisions, there are 2 entities in our records even though it i one physical location.
Division 1
Entity # 123456
Division A
Entity # 123456A
Each division is within it's own hierarchy so it looks like this:
Company
Division 1
State
Entity 12346
Division A
State-A
Entity 123456A
So 123456 sells to 123456A. The General Ledger stores an intercompany payable in 123456A but 123456 does not show the receivable separately. That receivable is buried in regular trade receivables.
For GL purposes, the elimination that is done by accounting is to net the payable in A against 123456. Simple enough.
I have some statistical reports that I need to create and I would like them to be as dynamic as possible I need to report AR with Intercompany and excluding Intercompany. What I would like to do is the following:
1. Copy the Intercompany Payable value from 123456A into a statistical account entitled ""Intercompany AR"". That's easy enough
2. But what I would also like to do is post the Intercompany AR amount into 123456 - the entity where the receivable exists.
3. I want this to be as dynamic as possible so I'd like to set up member formulas rather than having to run a calc script. The account part (#1 above) is easy. (Intercompany AR = Intercompany Payable; - and it's done). Moving the value in Intercompany AR to another entity is not so straight forward (to me, anyway).
First of all it would need to be in a member formula. It would apply only to the range of Level 0 entities that are descendants of Division A and only to the Intercompany AR account.
It would probably include include a substring function that only keeps the first 6 digits of the 123456A entity.
So what do you think? Is it doable?"
I handled a similar kind of Intercompany elimination problem.here is what i have handled. China entity of the company sells resources to the US entity of the company .For US it is cost i.e contractual services and for china it is Service revenue.But since this is with in the company when we have to report we have to eliminate the values i.e \$1000 cost of US and \$1000 Revenue for China.
we have created a elimination entity China_US_Intercompany elimination entity and ran the script to post the values by getting it from the main entities.while reporting we showed with elimination and without elimination.
Infact we have created two entities one holds value in USD and other in chinese yuan. | 592 | 2,642 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-18 | latest | en | 0.955399 |
http://www.ck12.org/book/CK-12-Probability-and-Statistics-Concepts/r133/section/8.0/ | 1,493,063,189,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917119782.43/warc/CC-MAIN-20170423031159-00065-ip-10-145-167-34.ec2.internal.warc.gz | 484,947,679 | 23,898 | # Chapter 8: Combinations and Permutations
Difficulty Level: At Grade Created by: CK-12
The mathematics of combinations, “combinatorics”, is a study in it’s own right, but it is also an important part of the mathematics of probability. There are two basic divisions of combinatorics: permutations, which are groupings of items where the order of the items is important, and combinations, where only the identity of the items is important, regardless of the order in which they appear.
In this chapter, you will study both kinds of groupings, and you will learn to calculate the number of possible unique ways to combine or arrange items of all kinds.
Chapter Outline
### Chapter Summary
Students were introduced to the concepts of combinatorics. The differences between permutations and combinations were discussed. Students practiced the use of the formulas for calculating the number of possible combinations or permutations in various situations allowing or disallowing repeats and/or indistinguishable members.
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Description
Difficulty Level: | 208 | 1,069 | {"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} | 3.078125 | 3 | CC-MAIN-2017-17 | latest | en | 0.926279 |
https://www.astronomyclub.xyz/quantum-mechanics/separating-the-possible-paths.html | 1,560,662,557,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997731.69/warc/CC-MAIN-20190616042701-20190616064701-00553.warc.gz | 659,516,811 | 8,183 | ## Separating the possible paths
It may be argued that interference effects can occur only if the possible paths of the photon are very close together. It is possible, however, to make the alternative paths as far apart as we like by an interferometer type of arrangement as illustrated in Figure 14.6.
Let us first consider what happens from the point of view of the wave representation. The light beam arrives at mirror 1 and splits into two components; approximately half is reflected (beam I), the other half is transmitted (beam II). The two rays emerge at right angles and then travel along different paths. When each beam meets the half-silvered mirror 4, a similiar thing happens as at mirror 1. Each beam splits into reflected and transmitted components. The result is that the beams recombine. The recombined beam has two components which emerge from mirror 4 at right angles, each consisting of a 50/50
dark detector
bright detector
(semi-silvered)
bright detector
(semi-silvered)
Figure 14.6 Typical interferometer experiment.
Figure 14.6 Typical interferometer experiment.
mixture of I and II. One goes towards detector A, and the other towards detector B.
Let us assume that the interferometer is adjusted so that the two light beams leaving mirror 4, are out of phase going in the direction of detector B (the 'dark' detector), and in phase in the direction of the 'bright' detector A. The fact that no light arrives at detector B is explained by the wave theory of light on the basis of destructive interference of the two combined beams.
Next, let us try the particle representation. As we did before, we can reduce the intensity of the laser beam, until the photons are so far apart that a photon usually has long left the apparatus before the next photon enters. We continue the experiment, and discover that the result is exactly the same as in the wave representation. Even though we know that this time the photons go through the apparatus one at a time, the 'dark detector' registers no clicks. No photons manage to reach it. All the photons go to detector A.
Somehow, with two routes open, the photon is prevented from reaching detector B. Note that the two alternative routes are far apart, but we do not know which route the photon has taken. So long as both routes are available, there are no clicks from detector B!
Next, we introduce an obstacle which closes one of the routes. To make quite sure that the photon cannot get through, let us put a brick into the lower path, as in the diagram. We find that an extraordinary thing happens. The dark detector immediately begins to click! Detector A also continues registering photons, but at approximately half the rate. We now know which route the photon has taken — obviously the one without the brick. There is no alternative route open; we know that the photon must have come by the 'upper' route. All photons arrive from the left at mirror 4; half of them are transmitted to detector A, and the other half reflected to detector B.
0 0 | 637 | 3,029 | {"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-2019-26 | latest | en | 0.928945 |
https://cboard.cprogramming.com/cplusplus-programming/106885-question-about-recursion.html | 1,495,469,273,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463605188.47/warc/CC-MAIN-20170522151715-20170522171715-00435.warc.gz | 728,511,977 | 13,335 | A few posts down I saw a post about recursion, that looked interesting enough to try. (http://cboard.cprogramming.com/showthread.php?t=106852) The problem presented was:
It should print its local variable and recursive call parameter. For each recursive call, display the outputs on a separate line and add a level of indentation. Do your utmost to make the outputs clear, interesting and meaningful.
I wasn't sure if I should post code in that thread or not because it was a homework thread and I'm not sure what the rules are in relation to that so I figured I'd be safe rather than sorry and open a new thread. Having stated that I think I am rather confused by where a local variable comes into play here. I think it's going to play a key-role in determining the tab stop...but without adding a second parameter I'm not entirely sure how. (I don't know if it allows you to add a second parameter or not, but that would make it too easy and the problem would lose it's challenge!)
The code I have right now doesn't show any attempts at coming up with a tab system in place, because I keep deleting every attempt I've made so far. Just to clarify really quickly: this isn't my homework, it IS a homework problem but I am taking no programming classes. I just want to do it because it looks like a fun challenge, but I'm stuck and that's why I'm here. Any and all code snippets/or ideas/ways to solve/hints/etc are welcome!
Code:
```// Our recursive function
unsigned long Factorial( unsigned long n )
{
// check to see if we need to stop
if ( n <= 0 )
return 1;
else
{ // we're good so keep going
std::cout << n << std::endl;
return n * Factorial( n-1 );
}
}
int _tmain(int argc, _TCHAR* argv[])
{
std::cout << Factorial(4); // result should be something like:
/*
4
3
2
1
4*3*2*1 = 24
*/
// don't close the box automatically
int x;
std::cin >> x;
return 0;
}```
2. I don't see a way to do this without a second parameter or some other way to keep state between iterations.
In order to figure out how far to indent the 1 you need to know how many times the function has already been called. There is no way around that, and that information must be passed somehow. It can be a static variable, a global, an additional parameter, or part of a struct that is passed as the one parameter. But somehow it must be passed.
3. Argh...well that just makes last night seem like such a waste. At least I had fun trying XD
4. I'd probably have a second default parameter to control the levels of indentation (a static would have to be reset somehow if you called the function more than once in the program otherwise the indentation would be messed up in successive function calls). The function could be called (from main for example) without a specified second value which would default to 0, within the function itself recursive calls would be made by specifying this parameter (and incrementing by 1 each step).
Code:
```unsigned long factorial(unsigned long n, unsigned int steps = 0)
{
...
// print out "steps" number of tab chars or something
// print out current "n"
...
// recursive call with second parameter
return n * factorial(n-1,steps+1);
}
int main()
{
// When calling function from within main, don't specify the second default parameter.
std::cout << factorial(4) << std::endl;
std::cout << factorial(5) << std::endl;
return 0;
}```
You could use a static member, but I think you'd still need to use a default parameter to clear/reset the static value:
Code:
```unsigned long factorial(unsigned long n, bool reset = true)
{
static unsigned steps;
// Either increment or reset static value based on second func arg
if( reset ) steps = 0;
else ++steps;
...
// print out "steps" number of tab chars or something
// print out current "n"
...
// recursive call with second function argument
return n * factorial(n-1,false);
}
int main()
{
// When calling function from within main, don't specify the second default parameter.
std::cout << factorial(4) << std::endl;
std::cout << factorial(5) << std::endl;
return 0;
}```
As to what the "local" is that's being mentioned I have no idea. | 975 | 4,113 | {"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-2017-22 | longest | en | 0.931336 |
https://translate.academic.ru/shear%20behavior/en/ru/ | 1,716,444,741,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00456.warc.gz | 510,870,676 | 31,571 | ## Перевод: с английского на русский
с русского на английский
# shear behavior
### См. также в других словарях:
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• Rheology — is the study of the flow of matter: mainly liquids but also soft solids or solids under conditions in which they flow rather than deform elasticallyW. R. Schowalter (1978) Mechanics of Non Newtonian Fluids Pergamon ISBN 0 08021778 8] . It applies … Wikipedia
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Prev Next | 1,374 | 5,965 | {"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-2024-18 | latest | en | 0.9208 |
https://www.quizsos.com/2020/04/27/maths-quiz-answers-videofacts/3/ | 1,685,363,557,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644855.6/warc/CC-MAIN-20230529105815-20230529135815-00354.warc.gz | 1,041,476,753 | 14,807 | Quiz Author: VideoFacts
21. Do you know the answer to this one?
23 = ?
• 2
• 4
• 6
• 8
22. Do you know the answer to this question?
What is 2648 to two significant figures?
• 2600
• 2640
• 2650
• 2700
23. Can you answer this question?
Which of these is 3% equivalent to?
• 0.003
• 0.03
• 0.3
• 300
24. What is the answer here?
-5 x -6 = ?
• -90
• -30
• 30
• 90
25. How many triangles can you count?
• 10
• 12
• 14
• 16
26. What does ‘X’ equal here?
150 – x = 87
• 53
• 57
• 63
• 67
27. What is ‘Y’ equivalent to?
90 x Y = 450
• 5
• 6
• 7
• 8
28. Tell us which digits have been replaced by question marks in this sum.
?1 – 2? = 34
• 4, 5
• 5, 6
• 6, 7
• 8, 9
29. Calculate the correct answer to this equation.
16.72 x 5 = ?
• 81.5
• 83.6
• 87.1
• 91.3
30. Do you know the answer to this conundrum?
√324 = ?
• 16
• 18
• 20
• 22 | 369 | 854 | {"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-2023-23 | longest | en | 0.787589 |
https://mathematica.stackexchange.com/questions/188711/solving-system-of-equations-for-coordinates | 1,659,885,363,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00012.warc.gz | 369,796,361 | 65,645 | # Solving system of equations for coordinates
This is my first time here, so let me know if there's something I need to add to the post etc. Anyway, I need some help with Mathematica and linear algebra, and I got some tips that this would be the right place to come for help.
I've got two bases, and a vector called x.
(V) First Base: {{1,3}, {4,6}}
(W) Second Base: {{4,6}, {2,5}}
Vector x: {6,6}
I started out by plotting the two base-vectors in the V-base and the vector x in the same picture. I then did the same thing with the W-base and the x vector.
I've plotted two graps, but I'll only post a picture of the first one initially. This is what the first plot looked like (for reference):
Now, here's what I need help with. I'm supposed to solve the system of equations in Mathematica, to find V-coordinates for x and W-coordinates for x, and then plot the components for x in each coordinate-system, using (Dashed, Line).
This graph was posted as an example on how it should look like.
• What have you tried so far? Have a look at LinearSolve and related functions. Jan 2, 2019 at 15:47
• Honestly, I haven't been able to do anything with it. I've been sitting all day with different exercises, and I haven't gotten anywhere with this one. That's why I figured I needed a good explanation.. Jan 2, 2019 at 16:04
Here is one solution:
V = {{1, 3}, {4, 6}};
x = {6, 6};
vx = LinearSolve[Transpose@V, x]
(* { -2, 2 } *)
vector[x0_: {0, 0}, v_] := Arrow@{x0, x0 + v}
Graphics[
{
Blue,
vector /@ V,
Red,
vector@x,
Black,
Dashed,
vector /@ (vx V),
• Why should it not go to negative coordinates? And yes, replacing all occurrences of V with W will do the same for the W base. But I'd suggest you try to understand what the code does in detail instead of just applying it - if you have questions about a specific line, feel free to ask Jan 2, 2019 at 16:41
• The vector function is a helper function that returns an Arrow directive that represents a vector starting at x0 and being v long (i.e. it ends at x0+v). The :{0,0} just means that if we only give one parameter to vector, x0 should take a default value of {0,0}. So vector[{2,2}] will return Arrow[{{0,0},{2,2}}], which is a vector starting at the origin and ending at {2,2}. Jan 2, 2019 at 18:59 | 637 | 2,271 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2022-33 | longest | en | 0.94808 |
https://mathoverflow.net/questions/330352/how-many-non-isomorphic-extensions-with-kernel-s1-and-quotient-cyclic-of-orde | 1,558,332,430,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232255562.23/warc/CC-MAIN-20190520041753-20190520063753-00511.warc.gz | 550,492,022 | 32,018 | # How many non-isomorphic extensions with kernel $S^1$ and quotient cyclic of order $p$?
I want to determine how many non-isomorphic extensions (as group they are non-isomorphic) are possible of the form $$1 \to \mathbb{S}^1 \to G \to (\mathbb{Z}_p)^k \to 1$$, where $$G$$ is a compact lie group and $$\Bbb{Z}_p$$ is the cyclic group of order $$p$$? The only possible homomorphism $$(\Bbb{Z}_p)^k \to Aut(\mathbb{S}^1)=\Bbb{Z}_2$$ is the trivial one if $$p\neq 2$$. If $$p=2$$, then we have $$(\Bbb{Z}_2)^k$$ possible homomorphism $$(\Bbb{Z}_2)^k \to \Bbb{Z}_2$$. Then we have to calculate the group $$H^2((\Bbb{Z}_p)^k; \mathbb{S}^1)$$. I don't know how to proceed further. A detailed answer will be very helpful.
Thank you very much in advance.
• I guess you're interested in extension of topological groups. – YCor Apr 30 at 8:46
• Oh, you mean $\mathbf{Z}_p$ cyclic of order $p$? it traditionally means the $p$-adic group (I started thinking about the question in this way) it's confusing. – YCor Apr 30 at 8:51
• I modified the question. Thanks – mathstudent Apr 30 at 8:53
Let's only consider central extensions (so we only miss a few cases when $$p=2$$). I denote by $$C_p=\mathbf{Z}/p\mathbf{Z}$$, to avoid confusion with $$p$$-adics.
So extensions are classified by $$H^2(C_p^k,S^1)$$. The commutator map yields a canonical homomorphism $$\phi$$ from $$H^2(C_p^k,S^1)$$ onto $$\mathrm{Hom}(\Lambda^2C_p^k,S^1)$$, and the latter is isomorphic to $$C_p^{k(k-1)/2}$$ (more canonically, the dual of $$\Lambda^2C_p^k$$). The kernel of this homomorphism consists of those 2-cocycle defining an abelian extension; since $$S^1$$ is an injective $$\mathbf{Z}$$-module the only abelian extension is split so $$\phi$$ is an isomorphism. Hence, the extensions are classified by $$C_p^{k(k-1)/2}$$, more precisely the space of alternating bilinear forms on $$C_p^k$$.
Next, the isomorphism class of the groups thus obtained are classified by the quotient of the latter by the action of $$\mathrm{GL}_k(\mathbf{Z}/p\mathbf{Z})$$. This quotient has exactly $$1+\lfloor k/2\rfloor$$ elements (number of types of alternating forms in dimension $$k$$, regardless of the field, including characteristic 2).
Note: For $$k=1,2,3$$ this makes $$1$$, $$2$$, $$2$$ isomorphism types of Lie groups. That for $$k=2,3$$ it does not match with Ian's answer (3, 8) is because he also considers quotients of order $$p^k$$ that are not $$p$$-elementary.
• Thank you very much. So if $k=4$, then there are 3 non-isomorphic groups. Is there any general way to determine all these groups(maybe the presentation of such groups). In general, for a given k is there any way to determine what are these $1+[k/2]$ groups (maybe how to determine their presentation)? – mathstudent May 4 at 8:14
• For the choice of non-degenerate alternating form, denote by $G_{2\ell}$ the corresponding central extension of $S^1$ by $C_p^{2\ell}$. Then the groups are just $G_{2\ell}\times C_p^m$ for all $\ell,m$ such that $2\ell+m=k$. For instance for $k=4$ these are $G_0\times C_p^4$ (the direct product), $G_2\times C_p^2$, and $G_4$. – YCor May 4 at 9:08
• Thanks. Yes, now I understand it clearly. I need to determine the possible presentation of such groups. So from your answer, I understand that for a given $k$ if we can determine the presentation of $G_k$ (following your notation), then all others can also be deduced. It will very helpful if you please help me to determine the presentation of such groups. – mathstudent May 4 at 14:38
There is a short exact sequence of coefficients $$\mathbb{Z}\rightarrow \mathbb{R}\rightarrow \mathbb{R}/\mathbb{Z}=S^1$$. This gives you a long exact sequence of cohomology groups. For a finite group $$Q$$, $$H^i(Q:\mathbb{R})=0$$ for $$i>0$$, so the long exact sequence collapses to an isomorphism $$H^2(Q;S^1)\cong H^3(Q;\mathbb{Z})$$.
You can use this to classify the isomorphism types of 1-dimensional compact non-connected Lie groups as in your question. For $$p$$ odd, there is only the direct product for $$S^1\rightarrow G\rightarrow C_p$$. For $$p$$ odd, there are two isomorphism types of group $$S^1\rightarrow G \rightarrow (C_p)^2$$, and there are two isomorphism types of group $$S^1\rightarrow G\rightarrow (C_p)^3$$. Of course it gets more complicated as $$k$$ increases, and $$p=2$$ is more complicated than odd $$p$$.
For $$p=2$$ there are already three groups $$S^1\rightarrow G\rightarrow C_2$$: the direct product, the orthogonal group $$O(2)$$, and the subgroup of the unit quaternions generated by the circle $$\cos(\theta)+i\sin(\theta)$$ and $$j$$.
In the first chapter of my PhD thesis (available on ArXiv in an extended version as https://arxiv.org/abs/0711.5020) I classified 1-dimensional compact Lie groups with $$p^k$$ components for each prime $$p$$ and each $$k\leq 3$$. The numbers that I got were 1, 3, 8 for $$p$$ odd and 3, 8, 29 for $$p=2$$. I presume that this was already known and I did not try to publish it. | 1,533 | 4,967 | {"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": 57, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2019-22 | latest | en | 0.858355 |
http://www.codecops.in/2014/06/karatsuba-multiplication.html?showComment=1495106794805 | 1,590,700,923,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347400101.39/warc/CC-MAIN-20200528201823-20200528231823-00500.warc.gz | 154,207,653 | 19,083 | ## Friday, 27 June 2014
### Karatsuba Multiplication
This is a Non-traditional way of multiplication .But due to its better time efficiency in computer application It can use as better way of multiplication in computer program.
Note : let we have to find X*Y then Karatsuba Multiplication is applicable if and only if both X &Y have same number of digits .
Ex : x=2345 & y =4535 both have 4 digits
### Algorithm for Karatsuba Multiplication :
Let x=5678 and y =1234
X and y , Both have 4 digits so Karatsuba Multiplication is applicable .
Now Fallow steps as shown in figure :
### Python Code for Karatsuba Multiplication :
1. # Karatsuba Multiplication
2. # Python 2.7
3. # http://beginer2cs.blogspot.com
4. '''
5. This Karatsuba Multiplication algorithm is only applicable if number of digit in X
6. is equal to number of digit of Y
7. '''
8. def Karatsuba_Mult(x,y):
9. #convert x & y to string
10. m=str(x)
11. n=str(y)
12.
13. # find length of string
14. lm=len(m)
15. ln=len(n)
16. if(lm != ln):
17. return '\n This algo is only applicable if number of digit in both x and y is same :'
18.
19.
20. #find value of a,b,c,d by slice string then convert to int
21. a=int(m[0:lm/2])
22. b=int(m[lm/2:])
23. c=int(n[0:ln/2])
24. d=int(n[ln/2:])
25.
26. # solving expression of this algorithm and return
27. if(lm%2 ==0)# if length is even (why this ??? think)
28. return (10**(lm))*a*c + (10**(lm/2))*(a*d +b*c) + b*d
29. else# when legth of number is odd (why this ??? think)
30. return (10**(lm+1))*a*c + (10**((lm/2)+1))*(a*d +b*c) + b*d
31.
32.
33. #################### MAIN ####################
34. #taking Input
35. =int(raw_input("Enter value of X : "))
36. =int(raw_input("Enter value of Y : "))
37. #Print result
38. print 'X*Y using Karatsuba Multiplication :- ',Karatsuba_Mult(x,y)
Output Demo:
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MaxMunus Offer World Class Virtual Instructor led training on TECHNOLOGY. We have industry expert trainer. We provide Training Material and Software Support. MaxMunus has successfully conducted 100000+ trainings in India, USA, UK, Australlia, Switzerland, Qatar, Saudi Arabia, Bangladesh, Bahrain and UAE etc. | 774 | 2,612 | {"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} | 4.03125 | 4 | CC-MAIN-2020-24 | latest | en | 0.482758 |
https://www.gradesaver.com/textbooks/math/trigonometry/trigonometry-10th-edition/chapter-2-acute-angles-and-right-triangles-section-2-5-further-applications-of-right-triangles-2-5-exercises-page-81/6 | 1,537,304,651,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155702.33/warc/CC-MAIN-20180918205149-20180918225149-00403.warc.gz | 740,596,000 | 12,470 | ## Trigonometry (10th Edition)
The bearing can be expressed as $90^{\circ}$ The bearing can be expressed as $N~90^{\circ}~E$ or $S~90^{\circ}~E$
The point (5,0) is located directly to the east from the radar station at the origin. The bearing can be expressed as $90^{\circ}$. This means the bearing is $90^{\circ}$ clockwise from due north. The bearing can be expressed as $N~90^{\circ}~E$, which means the bearing is $90^{\circ}$ clockwise from due north. The bearing can also be expressed as $S~90^{\circ}~E$, which means the bearing is $90^{\circ}$ counterclockwise from due south. | 171 | 586 | {"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} | 3.265625 | 3 | CC-MAIN-2018-39 | longest | en | 0.968924 |
https://www.angermanagement-worksheets.com/completing-the-square-practice-worksheet/ | 1,659,886,053,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882570651.49/warc/CC-MAIN-20220807150925-20220807180925-00681.warc.gz | 605,253,268 | 8,733 | # Completing The Square Practice Worksheet
### Solving Completing Square Kuta Software
https://cdn.kutasoftware.com/Worksheets/Alg1/Solving%20Completing%20Square.pdf
Solving Equations by Completing the Square Date_____ Period____ Solve each equation by completing the square. 1) a2 + 2a − 3 = 0 1, −3 2) a2 − 2a − 8 = 0 4, −2 3) p2 + 16 p − 22 = 0 1.273 , −17.273 4) k2 + 8k + 12 = 0 −2, −6 5) r2 + 2r − 33 = 0 4.83 , −6.83 6) a2 − 2a − 48 = 0 8, −6 7) m2 − 12 m + 26 = 0
### Completing The Square Questions Revision And
Completing the Square Questions, Revision and Worksheets
Skill 1: Completing the Square a=1. Solving quadratics via completing the square can be tricky, first we need to write the quadratic in the form …
### Completing The Square Worksheets
https://www.mathworksheetscenter.com/mathskills/trigonometry/CompletingtheSquare/
Steps to Complete the Square – Suppose ax 2 + bx + c and complete the square. Value a the coefficient of x(power2) out of the first two values only. E.g., a(x 2 + b/a x) + c. Divide the coefficient of x by 2, take square then, you will …
### Complete The Square Worksheets Printable Free Online PDFs
https://www.cuemath.com/worksheets/complete-the-square-worksheets/
Complete the square is a procedure of solving quadratic equations. In this method, arithmetic alterations are done to both sides of the equation to give them a format of a perfect polynomial square expression. After that, radical simplification is used to find the values of the unknown variable. It requires practice to get comfortable with it and hence we have the worksheet for …
### Solving Quadratic Equations By Completing The
Transform any quadratic equation that cannot be factored to the one that can be factored, with this simple never-fail technique of completing squares. Bolster practice using these printable worksheets on solving quadratic equations
### Completing The Square Practice Khan Academy
Practice: Solve equations by completing the square. Worked example: completing the square (leading coefficient ≠ 1) Practice: Completing the square. Solving quadratics by completing the square: no solution. Proof of the quadratic formula. Solving quadratics by completing the square. Completing the square review.
### Quadratic Equations By Completing The Square Kuta Software
Solving Quadratic Equations By Completing the Square Date_____ Period____ Solve each equation by completing the square. 1) p2 + 14 p − 38 = 0 2) v2 + 6v − 59 = 0 3) a2 + 14 a − 51 = 0 4) x2 − 12 x + 11 = 0 5) x2 + 6x + 8 = 0 6) n2 − 2n − 3 = 0 7) x2 + 14 x − 15 = 0 8) k2 − 12 k + 23 = 0 9) r2 − 4r − 91 = 7 10) x2 − 10 x + 26 = 8
### Completing The Square Practice Questions
Completing the Square Practice Questions
### Completing The Square Practice Worksheet
https://www.onlinemath4all.com/completing-the-square-practice-worksheet.html
COMPLETING THE SQUARE PRACTICE WORKSHEET. Solve the following quadratic equations by completing the square. (i) x 2 +6x–7 = 0. (ii) x 2 +3x+1 = 0. (iii) 2x 2 +5x-3 = 0. (iv) 4x 2 +4bx– (a 2 -b 2 ) = 0. (v) (5x+7)/ (x–1) = 3x+2. | 926 | 3,090 | {"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} | 3.953125 | 4 | CC-MAIN-2022-33 | latest | en | 0.774846 |
http://www.csplib.org/Problems/prob023/models/magic_hexagon.cpp.html | 1,534,881,478,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221218899.88/warc/CC-MAIN-20180821191026-20180821211026-00649.warc.gz | 473,300,298 | 3,329 | /*
Magic Hexagon in Gecode.
Prob023: Magic Hexagon
http://www.comp.rgu.ac.uk/staff/ha/ZCSP/prob023/prob023.pdf
http://www.cse.unsw.edu.au/~tw/csplib/prob/prob023/
Compare with the following models:
* MiniZinc: http://www.hakank.org/minizinc/magic_hexagon.mzn
* SICStus Prolog: http://www.hakank.org/sicstus/magic_hexagon.ecl
* ECLiPSe: http://www.hakank.org/eclipse/magic_hexagon.ecl
This Gecode model was created by Hakan Kjellerstrand (hakank@gmail.com)
Also, see my Gecode page: http://www.hakank.org/gecode/ .
*/
#include <gecode/driver.hh>
#include <gecode/int.hh>
#include <gecode/minimodel.hh>
using namespace Gecode;
using std::cout;
using std::endl;
using std::setw;
using std::string;
class MagicHexagon : public Script {
protected:
static const int len = 19;
IntVarArray x;
public:
MagicHexagon(const Options& opt)
:
x(*this, len, 1, len)
{
IntVar
a(x[0]),
b(x[1]),
c(x[2]),
d(x[3]),
e(x[4]),
f(x[5]),
g(x[6]),
h(x[7]),
i(x[8]),
j(x[9]),
k(x[10]),
l(x[11]),
m(x[12]),
n(x[13]),
o(x[14]),
p(x[15]),
q(x[16]),
r(x[17]),
s(x[18]);
distinct(*this, x, opt.icl());
// Not very beautiful, but experimental...
rel(*this,
a + b + c == 38 &&
d + e + f + g == 38 &&
h + i + j + k + l == 38 &&
m + n + o + p == 38 &&
q + r + s == 38 &&
a + d + h == 38 &&
b + e + i + m == 38 &&
c + f + j + n + q == 38 &&
g + k + o + r == 38 &&
l + p + s == 38 &&
c + g + l == 38 &&
b + f + k + p == 38 &&
a + e + j + o + s == 38 &&
d + i + n + r == 38 &&
h + m + q == 38 &&
a < c &&
a < h &&
a < l &&
a < q &&
a < s &&
c < h
);
// branching
branch(*this, x, INT_VAR_SIZE_MIN(), INT_VAL_MIN());
}
// Print solution
virtual void
print(std::ostream& os) const {
os << "x: " << x << endl;
os << endl;
}
// Constructor for cloning s
MagicHexagon(bool share, MagicHexagon& s) : Script(share,s) {
x.update(*this, share, s.x);
}
// Copy during cloning
virtual Space*
copy(bool share) {
return new MagicHexagon(share,*this);
}
};
int
main(int argc, char* argv[]) {
Options opt("MagicHexagon");
opt.solutions(0);
opt.icl(ICL_DOM);
opt.parse(argc,argv);
Script::run<MagicHexagon,DFS,Options>(opt);
return 0;
} | 763 | 2,125 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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} | 2.609375 | 3 | CC-MAIN-2018-34 | latest | en | 0.366812 |
http://encyclopedia.kids.net.au/page/mo/Modified_discrete_cosine_transform | 1,531,759,785,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589404.50/warc/CC-MAIN-20180716154548-20180716174548-00009.warc.gz | 110,847,507 | 8,363 | ## Encyclopedia > Modified discrete cosine transform
Article Content
# Modified discrete cosine transform
The modified discrete cosine transform (MDCT) is a frequency transform based on the type-IV discrete cosine transform (DCT-IV), with the additional property of being lapped: it is designed to be performed on consecutive blocks of a larger dataset, where subsequent blocks are 50% overlapped. This overlapping, in addition to the energy-compaction qualities of the DCT, makes the MDCT especially attractive for signal compression applications, since it helps to avoid artifacts stemming from the block boundaries. Thus, an MDCT is employed in MP3, AC-3, Ogg Vorbis, and AAC for audio compression, for example.
(There also exists an analogous transform, the MDST, based on the discrete sine transform, as well as other forms of the MDCT based on different types of DCT.)
In MP3, the MDCT is not applied to the audio signal directly, but rather to the output of a 32-band polyphase quadrature filter (PQF) bank. The output of this MDCT is postprocessed by an alias reduction formula to reduce the typical aliasing of the PQF filter bank. Such a combination of a filter bank with an MDCT is called a hybrid filter bank or a subband MDCT. AAC, on the other hand, normally uses a pure MDCT; only the (rarely used) MPEG-4 AAC-SSR variant (by Sony) uses a four-band PQF bank followed by an MDCT. ATRAC uses stacked quadrature mirror filters (QMF) followed by an MDCT.
As a lapped transform, the MDCT is a bit unusual compared to other frequency transforms in that it has half as many outputs as inputs (instead of the same number). In particular, it is a linear function F : R2n -> Rn (where R denotes the set of real numbers). The 2n real numbers x0, ..., x2n-1 are transformed into the n real numbers f0, ..., fn-1 according to the formula:
$f_j = \sum_{k=0}^{2n-1} x_k \cos \left[\frac{\pi}{n} (j+1/2) (k+1/2+n/2) \right]$
(The normalization coefficient in front of this transform, here unity, is an arbitrary convention and differs between treatments. Only the product of the normalizations of the MDCT and the IMDCT, below, is constrained.)
### Inverse Transform
The inverse MDCT is known as the IMDCT. Because there are different numbers of inputs and outputs, at first glance it might seem that the MDCT should not be invertible. However, perfect invertibility is achieved by adding the overlapped IMDCTs of subsequent overlapping blocks, causing the errors to cancel and the original data to be retrieved; this technique is known as time-domain aliasing cancellation (TDAC).
The IMDCT transforms n real numbers f0, ..., fn-1 into 2n real numbers y0, ..., y2n-1 according to the formula:
$y_k = \frac{1}{n} \sum_{j=0}^{n-1} f_j \cos \left[\frac{\pi}{n} (j+1/2) (k+1/2+n/2) \right]$
(Like for the DCT-IV, an orthogonal transform, the inverse has the same form as the forward transform.)
In the case of a windowed MDCT with the usual window normalization (see below), the normalization coefficient in front of the IMDCT should be multiplied by 2 (i.e., becoming 2/n).
### Computation
Although the direct application of the MDCT formula would require O(n2) operations, as in the fast Fourier transform (FFT) it is possible to compute the same thing with only O(n log n) complexity by recursively factorizing the computation. One can also compute MDCTs via other transforms, typically a DFT (FFT) or a DCT, combined with O(n) pre- and post-processing steps.
### Relationship to DCT-IV and Origin of TDAC
As can be seen by inspection of the definitions, for even n the MDCT is essentially equivalent to a DCT-IV, where the input is shifted by n/2 and two n-blocks of data are transformed at once. By examining this equivalence more carefully, important properties like TDAC can be easily derived.
In order to make the equivalence with the DCT-IV, one must realize that the DCT-IV corresponds to alternating even/odd boundary conditions: even at its left boundary (around k=0), odd at its right boundary (around k=n), and so on (instead of periodic boundaries as for a DFT). This follows from the identities cos[(j+1/2)(-k-1+1/2)π/n] = +cos[(j+1/2)(k+1/2)π/n] and cos[(j+1/2)(2n-k-1+1/2)π/n] = -cos[(j+1/2)(k+1/2)π/n]. Thus, if its inputs are an array x of length n, we can imagine extending this array to (x, -xr, -x, xr, ...) and so on, where xr denotes x in reverse order.
Consider an MDCT of size n, where we divide the 2n inputs into four blocks (a, b, c, d) each of size n/2. If we shift these by n/2 (from the +n/2 term in the MDCT definition), then (b, c, d) extend past the end of the n DCT-IV inputs, so we must "fold" them back according the the boundary conditions described above. Thus, the MDCT is exactly equivalent to a DCT-IV of the n inputs: (-cr-d, a-br), where r denotes reversal as above. (In this way, any algorithm to compute the DCT-IV can be trivially applied to the MDCT.)
Similarly, the IMDCT formula above is precisely 1/2 of the (self) inverse DCT-IV, where the output is shifted by n/2 and extended (via the boundary conditions) to a length 2n. The inverse DCT-IV would simply give back the inputs (-cr-d, a-br) from above. When this is shifted and extended via the boundary conditions, one obtains (a-br, b-ar, c+dr, cr+d) / 2. (Half of the IMDCT outputs are thus redundant.)
One can now understand how TDAC works. Suppose that one computes the MDCT of the subsequent, 50% overlapped, 2n block (c, d, e, f). The IMDCT will then yield, analogous to the above: (c-dr, d-cr, e+fr, er+f) / 2. When this is added with the previous IMDCT result in the overlapping half, the reversed terms cancel and one obtains simply (c, d), recovering the original data.
Moreover, the origin of the term "time-domain aliasing cancellation" is now clear. The use of input data that extend beyond the boundaries of the logical DCT-IV causes the data to be aliased in exactly the same way that frequencies beyond the Nyquist frequency are aliased to lower frequencies, except that this aliasing occurs in the time domain instead of the frequency domain. Hence the combinations c-dr and so on, which have precisely the right signs for the combinations to cancel when they are added.
For odd n (which are rarely used in practice), n/2 is not an integer so the MDCT is not simply a shift permutation of a DCT-IV. In this case, the additional shift by half a sample means that the MDCT/IMDCT becomes equivalent to the DCT-III/II, and the analysis is analogous to the above.
In typical signal-compression applications, the transform properties are further improved by using a window function wk (k = 0, ..., 2n-1) that is multiplied with xk and yk in the MDCT and IMDCT formulas, above, in order to avoid discontinuities at the k = 0 and 2n boundaries by making the function go smoothly to zero at those points. In principle, x and y could have different window functions, and the window function could also change from one block to the next (especially for the case where data blocks of different sizes are combined), but for simplicity we consider the common case of identical window functions for equal-sized blocks.
The transform remains invertible, for a symmetric window wk = w2n-1-k, as long as w satisfies the Princen-Bradley condition:
wk2 + wk+n2 = 1.
Various different window functions are common, e.g.
$w_k = \sin \left[\frac{\pi}{2n} (k+1/2) \right]$
for MP3 and MPEG-2 AAC, and
$w_k = \sin \left( \frac{\pi}{2} \sin^2 \left[\frac{\pi}{2n} (k+1/2) \right] \right)$
for Vorbis. AC-3 uses a Kaiser-Bessel derived (KBD) window, and MPEG-4 AAC can also use a KBD window.
Note that windows applied to the MDCT are different from windows used for other types of signal analysis, since they must fulfill the Princen-Bradley condition. One of the reasons for this difference is that MDCT windows are applied twice, for both the MDCT (analysis) and the IMDCT (synthesis).
References:
• Henrique S. Malvar, Signal Processing With Lapped Transforms (Artech House: Norwood MA, 1992).
• John P. Princen and Alan B. Bradley, "Analysis/synthesis filter bank design based on time domain aliasing cancellation," IEEE Trans. Acoust. Speech Sig. Proc. ASSP-34 (5), 1153-1161 (1986).
• A. W. Johnson and A. B. Bradley, "Adaptive transform coding incorporating time domain aliasing cancellation," Speech Comm. 6, 299-308 (1987).
• For algorithms, see e.g.: V. Britanak and K. R. Rao, "A new fast algorithm for the unified forward and inverse MDCT/MDST computation," Signal Processing 82, 433-459 (2002); Vladimir Nikolajevic and Gerhard Fettweis, "Computation of forward and inverse MDCT using Clenshaw's recurrence formula," IEEE Trans. Sig. Proc. 51 (5), 1439-1444 (2003); Che-Hong Chen, Bin-Da Liu, and Jar-Ferr Yang, "Recursive architectures for realizing modified discrete cosine transform and its inverse," IEEE Trans. Circuits Syst. II: Analog Dig. Sig. Proc. 50 (1), 38-45 (2003); and references thereof.
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Roman naming convention ... introduced. Discharged auxiliary soldiers and others gaining Roman Citizenship could, and many would, continue to use at least a portion of their former names. A number ... | 2,405 | 9,345 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2018-30 | longest | en | 0.926913 |
http://mathhelpforum.com/pre-calculus/201202-eulers-formula.html | 1,524,745,852,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125948126.97/warc/CC-MAIN-20180426105552-20180426125552-00432.warc.gz | 205,912,144 | 9,550 | 1. ## eulers formula
z^3=-i that gives z= -i Give all roots in the form r*e^eiv
2. ## Re: eulers formula
Originally Posted by kalemale
z^3=-i that gives z= -i Give all roots in the form r*e^eiv
First of all, \displaystyle \displaystyle \begin{align*} (-i)^3 \neq -i \end{align*}, but \displaystyle \displaystyle \begin{align*} i^3 \end{align*} does...
Anyway, if \displaystyle \displaystyle \begin{align*} z = r\,e^{i\theta} \end{align*}, then we have
\displaystyle \displaystyle \begin{align*} z^3 &= -i \\ \left(r\,e^{i\theta}\right)^3 &= e^{-\frac{\pi}{2}i} \\ r^3e^{3i\theta} &= 1e^{-\frac{\pi}{2}i} \\ r^3 = 1\textrm{ and } 3i\theta = -\frac{\pi}{2}i \\ r = 1 \textrm{ and } \theta = -\frac{\pi}{6} \end{align*}
All the roots are evenly spaced about a circle, so have the same radius and are separated by an angle of \displaystyle \displaystyle \begin{align*} \frac{2\pi}{3} \end{align*}, so the other angles are \displaystyle \displaystyle \begin{align*} -\frac{\pi}{6} - \frac{2\pi}{3} = -\frac{5\pi}{6} \end{align*} and \displaystyle \displaystyle \begin{align*} -\frac{\pi}{6} + \frac{2\pi}{3} = \end{align*}
So the three roots are
\displaystyle \displaystyle \begin{align*} 1e^{-\frac{5\pi}{6}i}, 1e^{-\frac{\pi}{6}i}, 1e^{\frac{\pi}{2}i} \end{align*} | 492 | 1,269 | {"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": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.477747 |
http://mathhelpforum.com/advanced-algebra/186300-factoring-given-root.html | 1,540,296,458,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583516135.92/warc/CC-MAIN-20181023111223-20181023132723-00241.warc.gz | 236,376,433 | 10,653 | 1. Factoring given a root
Suppose that $\displaystyle \beta$ is a zero of $\displaystyle f(x)=x^4+x+1$ in some field extension $\displaystyle E$ of $\displaystyle \mathbb{Z}_2$. Write $\displaystyle f(x)$ as a product of linear factors in $\displaystyle E[x]$.
Since $\displaystyle \beta$ is a zero, $\displaystyle x-\beta$ is a factor of $\displaystyle f(x)$. My plan was to simply divide $\displaystyle x-\beta$ into $\displaystyle f(x)$ to obtain a cubic and then find a root of the cubic, etc. until I obtained all the linear factors. Is there a reason that $\displaystyle x-\beta$ shouldn't divide $\displaystyle f(x)$? I don't see one but it is not dividing it when I try. Does the fact that $\displaystyle \beta$ lies in $\displaystyle E$ and not $\displaystyle \mathbb{Z}_2$ affect this?
2. Re: Factoring given a root
Originally Posted by AlexP
Suppose that $\displaystyle \beta$ is a zero of $\displaystyle f(x)=x^4+x+1$ in some field extension $\displaystyle E$ of $\displaystyle \mathbb{Z}_2$. Write $\displaystyle f(x)$ as a product of linear factors in $\displaystyle E[x]$.
$\displaystyle \beta$ satisfies $\displaystyle \beta^4+\beta+1=0$ in $\displaystyle E$ . But $\displaystyle (\beta^2)^4+\beta^2+1=(\beta^4+\beta+1)^2=0$ so, $\displaystyle \beta^2$ is another root of $\displaystyle f(x)$ in $\displaystyle E$ . Now, prove that $\displaystyle 1+\beta$ and $\displaystyle 1+\beta^2$ are also roots of $\displaystyle f(x)$ in $\displaystyle E$ . That is, $\displaystyle f(x)=(x-\beta)(x-\beta^2)(x-1-\beta)(x-1-\beta^2)$ .
3. Re: Factoring given a root
I got it. Since we're in characteristic 2 we have $\displaystyle f(a)=f(a+1)$ for arbitrary $\displaystyle a$ (I did work out the details).
Question about something I'm not clear on though... My book (Contemporary Abstract Algebra, Gallian, 5th ed.) says "A field $\displaystyle E$ is an extension field of a field $\displaystyle F$ if $\displaystyle F \subseteq E$ and the operations of $\displaystyle F$ are those of $\displaystyle E$ restricted to $\displaystyle F$." That wording bothers me... Is that saying that E inherits the operations of F (and thus has the same characteristic?)? That's not what it seems to be saying to me, but it's what seems to make sense.
4. Re: Factoring given a root
Originally Posted by AlexP
I got it. Since we're in characteristic 2 we have $\displaystyle f(a)=f(a+1)$ for arbitrary $\displaystyle a$ (I did work out the details).
Question about something I'm not clear on though... My book (Contemporary Abstract Algebra, Gallian, 5th ed.) says "A field $\displaystyle E$ is an extension field of a field $\displaystyle F$ if $\displaystyle F \subseteq E$ and the operations of $\displaystyle F$ are those of $\displaystyle E$ restricted to $\displaystyle F$." That wording bothers me... Is that saying that E inherits the operations of F (and thus has the same characteristic?)? That's not what it seems to be saying to me, but it's what seems to make sense.
More generally, $\displaystyle F$ is an extension of $\displaystyle E$ if there is an embedding $\displaystyle \sigma:E\hookrightarrow F$. In particular, since $\displaystyle \sigma$ is an injective group morphism we know that $\displaystyle \text{char}(F)=|1_F|=|\sigma(1_E)|=|1_E|=\text{cha r}(E)$.
5. Re: Factoring given a root
Got it. Thanks to both of you. | 924 | 3,341 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2018-43 | latest | en | 0.808898 |
http://sage.math.washington.edu/books/ant/ant/node84.html | 1,435,699,453,000,000,000 | text/html | crawl-data/CC-MAIN-2015-27/segments/1435375094501.77/warc/CC-MAIN-20150627031814-00062-ip-10-179-60-89.ec2.internal.warc.gz | 212,014,474 | 7,276 | # Examples of Valuations
The archetypal example of an archimedean valuation is the absolute value on the complex numbers. It is essentially the only one:
Theorem 13.3.1 (Gelfand-Tornheim) Any field with an archimedean valuation is isomorphic to a subfield of , the valuation being equivalent to that induced by the usual absolute value on .
We do not prove this here as we do not need it. For a proof, see [Art59, pg. 45, 67].
There are many non-archimedean valuations. On the rationals there is one for every prime , the -adic valuation, as in Example 13.2.9.
Theorem 13.3.2 (Ostrowski) The nontrivial valuations on are those equivalent to , for some prime , and the usual absolute value .
Remark 13.3.3 Before giving the proof, we pause with a brief remark about Ostrowski. According to
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Ostrowski.html
Ostrowski was a Ukrainian mathematician who lived 1893-1986. Gautschi writes about Ostrowski as follows: ... you are able, on the one hand, to emphasise the abstract and axiomatic side of mathematics, as for example in your theory of general norms, or, on the other hand, to concentrate on the concrete and constructive aspects of mathematics, as in your study of numerical methods, and to do both with equal ease. You delight in finding short and succinct proofs, of which you have given many examples ...'' [italics mine]
We will now give an example of one of these short and succinct proofs.
Proof. Suppose is a nontrivial valuation on .
Nonarchimedean case: Suppose for all , so by Lemma 13.2.10, is nonarchimedean. Since is nontrivial, the set
is nonzero. Also is an ideal and if , then , so or , so is a prime ideal of . Thus , for some prime number . Since every element of has valuation at most , if with , then , so . Let , so . Then for any and any with , we have
Thus on , hence on by multiplicativity, so is equivalent to , as claimed.
Archimedean case: By replacing by a power of , we may assume without loss that satisfies the triangle inequality. We first make some general remarks about any valuation that satisfies the triangle inequality. Suppose is greater than . Consider, for any the base- expansion of :
where
and . Since , taking logs we see that , so
Let . Then by the triangle inequality for , we have
where in the last step we use that . Setting , for , in the above inequality and taking th roots, we have
The first factor converges to as , since (because ). The second factor is
which also converges to , for the same reason that (because as ). The third factor is
Putting this all together, we see that
Our assumption that is nonarchimedean implies that there is with and . Then for all with we have
(13.5)
so , so as well (i.e., any with automatically satisfies ). Also, taking the power on both sides of (13.3.1) we see that
(13.6)
Because, as mentioned above, , we can interchange the roll of and to obtain the reverse inequality of (13.3.2). We thus have
Letting and setting , we have
Thus for all integers with we have , which implies that is equivalent to .
Let be any field and let , where is transcendental. Fix a real number . If is an irreducible polynomial in the ring , we define a valuation by
(13.7)
where and with and .
Remark 13.3.4 This definition differs from the one page 46 of [Cassels-Frohlich, Ch. 2] in two ways. First, we assume that instead of , since otherwise does not satisfy Axiom 3 of a valuation. Also, we write instead of , so that the product formula will hold. (For more about the product formula, see Section 18.1.)
In addition there is a a non-archimedean valuation defined by
(13.8)
This definition differs from the one in [Cas67, pg. 46] in two ways. First, we assume that instead of , since otherwise does not satisfy Axiom 3 of a valuation. Here's why: Recall that Axiom 3 for a non-archimedean valuation on asserts that whenever and , then . Set , where is an irreducible polynomial. Then , since . However, , since . If we take instead of , as I propose, then , as required.
Note the (albeit imperfect) analogy between and . If , so , the valuation is of the type (13.3.3) belonging to the irreducible polynomial .
The reader is urged to prove the following lemma as a homework problem.
Lemma 13.3.5 The only nontrivial valuations on which are trivial on are equivalent to the valuation (13.3.3) or (13.3.4).
For example, if is a finite field, there are no nontrivial valuations on , so the only nontrivial valuations on are equivalent to (13.3.3) or (13.3.4).
William Stein 2012-09-24 | 1,165 | 4,577 | {"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} | 4.125 | 4 | CC-MAIN-2015-27 | latest | en | 0.915278 |
https://philnews.ph/2019/09/19/ez2-result-september-19-2019/ | 1,591,466,432,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348517506.81/warc/CC-MAIN-20200606155701-20200606185701-00373.warc.gz | 475,917,056 | 25,604 | ## EZ2 RESULT September 19, 2019
EZ2 RESULT – Here is the result of EZ2 lotto draw of Philippine Charity Sweepstakes Office (PCSO). Updates of the EZ2 result is refreshed every 11AM, 4PM and 9PM.
11am:
4pm:
9pm:
### See UPDATED results for the following draws:
EZ2 RESULT June 6, 2020 (updated every 11am, 4pm and 9pm)
SWERTRES RESULT June 6, 2020 (updated every 11am, 4pm and 9pm)
LOTTO RESULT June 6, 2020 (updated every 9pm)
The above results are the official winning number for
() for draw of EZ2 Lotto.
See also Swertres Result, 6/49 Super Lotto result, 6/42 Lotto result, and 6-Digit Lotto result page.
For other results, please visit the main page PCSO Lotto Results.
PCSO Reminders
Based on the Philippine Charity Sweepstakes Office (PCSO), here are some reminders in playing the EZ2 Lotto:
1. Only players 18 years old and above are allowed to play the lotto games.
2. This lotto game has daily draws every 11:00 am, 4:00 pm, and 9:00 pm daily.
3. You can watch the draw live on YouTube (pcso gov) or through PTV4 channel.
4. The lotto ticket for the EZ2 Lotto is at Php 12.00 each. It is already inclusive of the 20% Documentary Stamp Tax (DST).
5. It is the responsibility of the lotto players to check on the accuracy of the data printed on the ticket.
In cases of holidays or certain special occasions where there would be no draws, the PCSO makes a prior announcement including the information on when the draw will resume.
How To Play the PCSO EZ2 Lotto
1. Pick two(2) digits that will compose your combination. You will have to choose from a field of one(1) to three(3).
2. If you don’t have a combination in mind, you may also choose the Lucky Pick which is a randomly generated combination of digits.
3. Mark your play amount which is from Php 12.00 to Php 600.00.
4. For the Rambolito system play, you may choose two(2) digits and the system will automatically generate the possible combinations with regards to the digits you have chosen. The minimum play amount is at Php 12.00.
5. For the Advance Draws, you may play the same combination up to a series of six(6) draws.
Prizes for EZ2 Lotto
• If you got the winning combination in exact order, you will win Php 4,000.00 for every lotto ticket.
• If you won through the Rambolito system play, you will win Php 2,000.00.
How To Claim the Prize
For Lotto Prizes between Php 20.00 and Php 10,000.00 – Winners may claim the prize through any authorized lotto outlet. You may also claim at a PCSO Branch Office nearest to you.
For Lotto Prizes at Php 10,0001.00and up – Winners may claim the prize at any PCSO Branch Office. The prizes that are above Php 10,000.00 are subject to a 20% deduction pursuant to the Tax Reform for Acceleration and Inclusion (TRAIN) Act. | 736 | 2,750 | {"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} | 2.890625 | 3 | CC-MAIN-2020-24 | latest | en | 0.89867 |
https://fr.mathworks.com/matlabcentral/cody/problems/26-determine-if-input-is-odd/solutions/789122 | 1,596,723,479,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736962.52/warc/CC-MAIN-20200806121241-20200806151241-00249.warc.gz | 316,576,262 | 15,689 | Cody
# Problem 26. Determine if input is odd
Solution 789122
Submitted on 6 Dec 2015 by Charles Turo
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% n = 1; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
2 Pass
%% n = 2; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
3 Pass
%% n = 28; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct))
4 Pass
%% n = 453; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
5 Pass
%% n = 17; ans_correct = true; assert(isequal(is_it_odd(n),ans_correct))
6 Pass
%% n = 16; ans_correct = false; assert(isequal(is_it_odd(n),ans_correct)) | 229 | 761 | {"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} | 2.96875 | 3 | CC-MAIN-2020-34 | latest | en | 0.37952 |
https://community.khronos.org/t/best-solution-for-dealing-with-multiple-light-types/76401 | 1,656,705,096,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103945490.54/warc/CC-MAIN-20220701185955-20220701215955-00412.warc.gz | 213,564,774 | 13,580 | # Best solution for dealing with multiple light types
Hi all,
I am working on my own 3D engine and I recently ran into an issue when trying to combine different light types using a single shader. Multiple lights of a single type work fine, but when I combine a Point light (with cube map shadows), Directional light (with 2D shadows), and Spot lights (also with 2D shadows) things started to break. I found a solution to this problem, but I wonder if there is a better way of doing it. Let me first summarise my initial solution that failed and then talk about the solution I found.
I pass an array of lights to the shader that is used to render a mesh. This array is defined as follows in my shader:
``````#version 420
const int nr_lights = 5;
const int DIRECTIONAL_LIGHT = 0;
const int SPOT_LIGHT = 1;
const int POINT_LIGHT = 2;
struct light {
int type;
bool enabled;
vec4 position;
vec4 diffuse;
vec4 ambient;
vec4 specular;
float constant_attenuation;
float linear_attenuation;
vec3 direction;
float light_angle;
samplerCube cube_depth_texture;
};
uniform light lights[nr_lights];
``````
For spot lights and directional lights I use a 2D shadow sampler to project the depth values. For point lights I created a cube texture which contain the linearised depth values. The beef of the lighting calculations are in the fragment shader and read as follows:
``````
for (int i = 0; i < lights.length(); ++i)
{
if (!lights[i].enabled)
{
continue;
}
vec4 halfVector = normalize(H[i]);
vec4 lightVector = normalize(L[i]);
float dotValue = max(dot(normalVector, lightVector), 0.0);
if (dotValue > 0.0)
{
float distance = length(lights[i].position - worldPos);
float intensity = 1.0;
if (lights[i].type != DIRECTIONAL_LIGHT)
intensity = 1.0 / (lights[i].constant_attenuation + lights[i].linear_attenuation * distance + lights[i].quadratic_attenuation * distance * distance);
vec4 ambient = material_ambient * lights[i].ambient;
bool inLight = true;
if (lights[i].type == SPOT_LIGHT)
{
vec3 nLightToVertex = vec3(normalize(worldPos - lights[i].position));
float angleLightToFrag = dot(nLightToVertex, normalize(lights[i].direction));
float radLightAngle = lights[i].light_angle * 3.141592 / 180.0;
inLight = false;
}
if (inLight)
{
if (lights[i].type == SPOT_LIGHT || lights[i].type == DIRECTIONAL_LIGHT)
{
}
else if(lights[i].type == POINT_LIGHT)
{
float sampled_distance = texture(lights[i].cube_depth_texture, direction[i].xyz).r;
float distance = length(direction[i]);
if (distance > sampled_distance + 0.1)
}
vec4 diffuse = dotValue * lights[i].diffuse * material_diffuse;
vec4 specular = pow(max(dot(normalVector, halfVector), 0.0), 10.0) * material_specular * lights[i].specular;
outColor += intensity * shadowf * (diffuse + specular * 100);
}
outColor += intensity * ambient;
}
}
outColor += material_emissive;
``````
This clearly does not work due non-uniform control flow (a term I only learned about yesterday :)).
So, what I have done is to move all the texture lookups out of the non-uniform control flow. However, this means that I need to provide depth textures for all lights (even if they are not used for rendering) and sample both the cube and 2dShadow textures. Let me show you the updated fragment shader bit:
``````for (int i = 0; i < lights.length(); ++i)
{
float sampled_distance = texture(lights[i].cube_depth_texture, direction[i].xyz).r;
if (!lights[i].enabled)
{
continue;
}
vec4 halfVector = normalize(H[i]);
vec4 lightVector = normalize(L[i]);
float dotValue = max(dot(normalVector, lightVector), 0.0);
if (dotValue > 0.0)
{
float distance = length(lights[i].position - worldPos);
float intensity = 1.0;
if (lights[i].type != DIRECTIONAL_LIGHT)
intensity = 1.0 / (lights[i].constant_attenuation + lights[i].linear_attenuation * distance + lights[i].quadratic_attenuation * distance * distance);
vec4 ambient = material_ambient * lights[i].ambient;
bool inLight = true;
if (lights[i].type == SPOT_LIGHT)
{
vec3 nLightToVertex = vec3(normalize(worldPos - lights[i].position));
float angleLightToFrag = dot(nLightToVertex, normalize(lights[i].direction));
float radLightAngle = lights[i].light_angle * 3.141592 / 180.0;
{
inLight = false;
}
}
if (inLight)
{
if (lights[i].type == SPOT_LIGHT)
{
}
else if(lights[i].type == POINT_LIGHT)
{
float distance = length(direction[i]);
if (distance > sampled_distance + 0.1)
}
vec4 diffuse = dotValue * lights[i].diffuse * material_diffuse;
vec4 specular = pow(max(dot(normalVector, halfVector), 0.0), 10.0) * material_specular * lights[i].specular;
outColor += intensity * shadowf * (diffuse + specular * 100);
}
outColor += intensity * ambient;
}
}
outColor += material_emissive;
``````
This works! In my engine I create 2 dummy shadows of size 1x1, one is a GL_TEXTURE_2D stored as a GL_DEPTH_COMPONENT, the other is a GL_TEXTURE_CUBE_MAP that only stores GL_RED values. When less than 5 lights are needed to render a mesh I pass these values to the cube_depth_texture and depth_texture values of the respective light and set the isEnabled flag to false.
While this does work, it creates a lot of overhead. In the worst case, when no lights are being used, it will still sample 10 textures!
Is there a better way around this issue? My engine currently does forward rendering, it is not clear to me whether using a G-Buffer provides a cleaner solution. If I can I would like to stick to forward rendering, so any solution and comments you have are greatly appreciated.
Many thanks!
Bram
P.S. For those interested, my 3D enigne Dreaded Portal Engine can be found here: http://bramridder.com/index.php/personal/personal_projects/dreaded-portal-engine
An alternative is to avoid using texture lookup functions which perform implicit derivative calculations, and instead calculate derivatives or LoD explicitly outside of the conditional and pass the result to textureProjGrad() or textureProjLod().
However, this may still perform texture lookups in cases where the condition is false (it depends upon whether the hardware has branch instructions). If you’re going to be perform lookups regardless, it would be better to use a 1x1 texture (or force the use of the 1x1 mipmap level of some texture) for cases where you don’t need the result.
If the hardware doesn’t have branch instructions, then putting code inside a conditional doesn’t avoid the cost of executing it, only the side-effects. So e.g. setting [var]radLightAngle[/var] to π would avoid the need to use a conditional for the inside-cone test (cos(π)=-1, so the test will always be false).
[QUOTE=Bram Ridder;1289681]While this does work, it creates a lot of overhead. In the worst case, when no lights are being used, it will still sample 10 textures!
Is there a better way around this issue? My engine currently does forward rendering, it is not clear to me whether using a G-Buffer provides a cleaner solution. If I can I would like to stick to forward rendering, so any solution and comments you have are greatly appreciated.[/QUOTE]
I’d definitely see if you can meet your goals with small changes to your shader logic as GClements is suggesting.
If after pursuing those, you bench your app and determine that the performance still isn’t up to the level you need, profile carefully to determine exactly what the biggest bottleneck is (it helps to gather a few worst-case test cases). You can use the results as a filter to evaluate which tech approaches will reduce that inefficiency the most. Just using some intuition about how your rendering algorithms work will save time with this.
If the main bottleneck ends up being the fact that you’re using a shader supporting max(lights) and max(shadows) for all fragments on the entire screen and you can’t easily avoid most of inefficiency associated with that with small shader changes, consider a tiled or clustered shading approach. Given your desire to stick with forward and the drawbacks of deferred approaches (which aren’t insurmountable, but do require nontrivial effort), I’d suggest looking most closely at tiled or clustered forward shading techniques (websearch: tiled forward, clustered forward, and forward+ for the latest papers, blog posts, and conference presentations). However, be sure and profile other aspects of your rendering too (e.g. shadow casting and culling).
Thanks for the very helpful feedback.
I agree that using TextureProjGrad() or textureProjLod() is one way to solve this problem. Although, as Dark Photon mentioned, I need to check whether doing texture lookups using 1x1 textures does create a bottleneck.
Thank you Dark Photon for letting me know about Forward+ and clustered methods. Did not even know these existed, very exiting!
At the moment I cannot use more than 5 lights per mesh. I guess this is because the limit of 16 textures per shader? Or is there another limit that prohibits using an array of say 32 lights?
In any case I have some research and then some coding to do :).
Your “struct light” has 43 components; 6 of those would total 258 components, which may be exceeding some implementation limit. You can get around that by using textures (e.g. buffer textures), or you may be able to use uniform blocks or shader storage blocks. Note that you’d need to keep the samplers separate; you can’t store samplers in uniform blocks, shader storage blocks or textures.
If you hit the limit on the number of texture units, consider using array textures. These effectively allow you to aggregate multiple textures into a single texture, with the constraint that all layers must have the same format and dimensions, and sampling parameters (e.g. filter and wrap modes) apply to the texture as a whole.
You can use bindless texture or texture arrays to get past the 16 textures/shader.
However, even if textures weren’t limiting you (e.g. no point or spot light shadows), I suspect you’ll hit other problems trying to push the number of lights up to even 32. If I were you, I’d just try it. This will provide valuable profiling data on which to base your future design decisions, and you can also see if you hit any big performance drop-offs or blocks as you increment the number of lights applied simultaneously from 1 to32.
It’s been years, but it seems like when I pushed up the number of lights being applied simultaneously in every fragment shader execution to 32 I hit a performance cliff or two and a wall before I got there with the way I was doing it. Seems like at least one cliff had to do with the GLSL compiler (in NVidia’s driver) dynamically determining the maximum number of iterations to automatically unroll loops in the shader (at the time, I was generating a shader permutation with the number of lights baked in). When it flipped to not unrolling I hit a big perf drop-off IIRC (NOTE: Whether and when the compiler unrolls loops can be controlled with a #pragma directive). Pushing the number of lights up even further resulted in hitting a limit with the max amount of uniform space I could pass into the shader using standard uniforms. This of course can be bypassed by any number of methods (SSBOs, UBOs, TBOs, etc.), but with potential performance reductions. Not sure any of this is useful to you nowadays (OpenGL has moved on), but I just mention it in case you do hit perf cliffs or walls with your profiling to give you a few possible potential causes to check into to see if they apply in your case. But long story short, doing this test made it blatantly obvious that I couldn’t get where I wanted to go with the GPU by just simple forward shading. I ended up implementing Deferred Shading which supported 100s-1000s of lights even without tile-based deferred, but that was before Tiled/clustered forward and Forward+ like approaches (nowadays and knowing what I know about deferred’s limitations and challenges, I’d seriously consider using Tiled/clustered Forward/Forward+ like approaches instead).
Brilliant! Thanks for the very insightful replies.
Quick update. I implemented some of your recommendations and I have successfully rendered a scene with 56 lights (including shadow maps)! I now use two UBOs, one for the view and projection matrix and the other for all the lighting information. I ran into an issue with having to many outs in my vertex shader so I moved all the calculation to the fragment shader (doing so somehow doubled my FPS ). While I am happy it works, it really shouldn’t…
As far as I understand I should have exceeded the sampler limit in the fragment shader (GL_MAX_TEXTURE_IMAGE_UNITS), but it just seems to work. Maybe you can help me figure out what is going on. Let me present the shaders I use at the moment:
``````
#version 420
uniform vec4 material_ambient;
uniform vec4 material_diffuse;
uniform vec4 material_specular;
uniform vec4 material_emissive;
const int nr_lights = 60;
const int DIRECTIONAL_LIGHT = 0;
const int SPOT_LIGHT = 1;
const int POINT_LIGHT = 2;
layout (std140) uniform Lights
{
int type[nr_lights];
bool enabled[nr_lights];
vec4 position[nr_lights];
vec4 diffuse[nr_lights];
vec4 ambient[nr_lights];
vec4 specular[nr_lights];
float constant_attenuation[nr_lights];
float linear_attenuation[nr_lights];
vec3 direction[nr_lights];
float light_angle[nr_lights];
} lights;
uniform samplerCube cube_depth_texture[nr_lights];
layout (std140) uniform Matrices
{
mat4 projection_matrix;
mat4 view_matrix;
};
uniform mat4 model_matrix;
in vec3 a_Vertex;
in vec2 a_TexCoord0;
in vec3 a_Normal;
out vec2 texCoord0;
out vec4 worldPos;
out vec4 pos;
out vec4 N;
void main(void)
{
texCoord0 = a_TexCoord0;
pos = view_matrix * model_matrix * vec4(a_Vertex, 1.0);
worldPos = model_matrix * vec4(a_Vertex, 1.0);
N = view_matrix * model_matrix * vec4(a_Normal, 0.0);
gl_Position = projection_matrix * pos;
}
``````
``````
#version 420
uniform vec4 material_ambient;
uniform vec4 material_diffuse;
uniform vec4 material_specular;
uniform vec4 material_emissive;
const int nr_lights = 60;
const int DIRECTIONAL_LIGHT = 0;
const int SPOT_LIGHT = 1;
const int POINT_LIGHT = 2;
layout (std140) uniform Lights
{
int type[nr_lights];
bool enabled[nr_lights];
vec4 position[nr_lights];
vec4 diffuse[nr_lights];
vec4 ambient[nr_lights];
vec4 specular[nr_lights];
float constant_attenuation[nr_lights];
float linear_attenuation[nr_lights];
vec3 direction[nr_lights];
float light_angle[nr_lights];
} lights;
uniform samplerCube cube_depth_texture[nr_lights];
layout (std140) uniform Matrices
{
mat4 projection_matrix;
mat4 view_matrix;
};
uniform sampler2D texture0;
uniform float transparency;
in vec2 texCoord0;
in vec4 worldPos;
in vec4 pos;
in vec4 N;
out vec4 outColor;
void main(void) {
if (texture(texture0, texCoord0.st).a == 0.0)
{
}
outColor = vec4(0, 0, 0, 1);
vec4 normalVector = N;
if (N != vec4(0, 0, 0, 0))
{
normalVector = normalize(N);
}
for (int i = 0; i < nr_lights; ++i)
{
if (!lights.enabled[i])
{
break;
}
vec3 lightPos = (view_matrix * lights.position[i]).xyz;
vec4 L;
vec4 H;
vec4 direction;
if (lights.type[i] == DIRECTIONAL_LIGHT)
{
L = vec4(-lights.direction[i], 0.0);
H = vec4((-lights.direction[i]).xyz, 1.0) - pos;
direction = L;
}
else
{
L = vec4(lightPos - pos.xyz, 0.0);
H = vec4((lightPos - pos.xyz).xyz, 1.0) - pos;
direction = worldPos - lights.position[i];
}
float sampled_distance = texture(cube_depth_texture[i], direction.xyz).r;
vec4 halfVector = normalize(H);
vec4 lightVector = normalize(L);
float dotValue = max(dot(normalVector, lightVector), 0.0);
if (dotValue > 0.0)
{
float distance = length(lights.position[i] - worldPos);
float intensity = 1.0;
if (lights.type[i] != DIRECTIONAL_LIGHT)
intensity = 1.0 / (lights.constant_attenuation[i] + lights.linear_attenuation[i] * distance + lights.quadratic_attenuation[i] * distance * distance);
vec4 ambient = material_ambient * lights.ambient[i];
bool inLight = true;
if (lights.type[i] == SPOT_LIGHT)
{
vec3 nLightToVertex = vec3(normalize(worldPos - lights.position[i]));
float angleLightToFrag = dot(nLightToVertex, normalize(lights.direction[i]));
float radLightAngle = lights.light_angle[i] * 3.141592 / 180.0;
{
inLight = false;
}
}
if (inLight)
{
if (lights.type[i] == SPOT_LIGHT)
{
}
else if(lights.type[i] == POINT_LIGHT)
{
float distance = length(direction);
if (distance > sampled_distance + 0.1) {
}
}
vec4 diffuse = dotValue * lights.diffuse[i] * material_diffuse;
vec4 specular = pow(max(dot(normalVector, halfVector), 0.0), 10.0) * material_specular * lights.specular[i];
outColor += intensity * shadowf * (diffuse + specular * 100);
}
outColor += intensity * ambient;
}
}
outColor += material_emissive;
outColor *= texture(texture0, texCoord0.st);
}
``````
My understanding is that currently 121 textures are currently used by the Fragment shader; texture0 + 60 * (depth_texture + cube_depth_texture). When I check the value of GL_MAX_TEXTURE_IMAGE_UNITS on my GPU it returns 32. GL_MAX_COMBINED_TEXTURE_IMAGE_UNITS returns 160, which would be enough but should not apply to solely the fragment shader. What am I not understanding and what magic is being used?
Thanks again for your help improving my 3D engine :D.
Just for curiosity:
I suppose it is few compared to what the fragment shader can support.
[QUOTE=Silence;1289791]Just for curiosity:
I suppose it is few compared to what the fragment shader can support.
GL_MAX_VERTEX_TEXTURE_IMAGE_UNITS returns 32 as well. It is actually an AMD card: SAPPHIRE Radeon RX 580 NITRO+ 4 GB GDDR5.
OK. I will try to try that as well on my side. Thank you.
Well, I just want to know what these values mean. Because it seems to me that I am going over these limits with the number of textures I use, yet everything works fine.
Well, these are depicted here.
And for the combined, if both shader access the same unit, this is counted as 2.
From what I know and what I understand, these should be hard limits, not hints. Reading the relevant parts of the spec might also give some clue.
Cool! Congrats on getting it up and running.
I ran into an issue with having to many outs in my vertex shader so I moved all the calculation to the fragment shader (doing so somehow doubled my FPS ). While I am happy it works, it really shouldn’t…
That’s interesting. A few thoughts: vertex shaders are executed for all vertices, not just all vertices 1) in the view frustum 2) which aren’t occluded by any other objects (the latter for depth-tested geometry). I wonder if #1 and #2 might help explain the performance difference?
Also, how many interpolators (varyings; aka vertex out/fragment in) were you using before versus now? I think I recall reading that the more of these you use, the fewer vertex shader threads can execute in parallel (on the same GPU), so the slower your vertex transform work executes. Not sure if that correlates with your problem though.
As far as I understand I should have exceeded the sampler limit in the fragment shader (GL_MAX_TEXTURE_IMAGE_UNITS), but it just seems to work. Maybe you can help me figure out what is going on.
Hmmm. 121 is certainly > 32 (your GL_MAX_TEXTURE_IMAGE_UNITS, which I believe is the bound texture access limit for fragment shaders). GL_MAX_COMBINED_TEXTURE_IMAGE_UNITS is the limit across all shader units, and you’re well under that. It sounds like GL_MAX_TEXTURE_IMAGE_UNITS isn’t really the hard upper limit for fragment shaders (on your GL drivers at least), but it sounds like that’s operating outside of the spec, so your code in general might not work on other such drivers where you’re over-the-limit.
[QUOTE=Dark Photon;1289803]Cool! Congrats on getting it up and running.[/QUOTE] Thanks
[QUOTE=Dark Photon;1289803]
Also, how many interpolators (varyings; aka vertex out/fragment in) were you using before versus now?[/QUOTE] Before I used #lights * 4 + 3, now I just use 3. So I think that might explain it. Because even in scenes where all vertices are within the frustrum and no occlusion occurs I still get the speedup.
[QUOTE=Dark Photon;1289803]
Hmmm. 121 is certainly > 32 (your GL_MAX_TEXTURE_IMAGE_UNITS, which I believe is the bound texture access limit for fragment shaders). GL_MAX_COMBINED_TEXTURE_IMAGE_UNITS is the limit across all shader units, and you’re well under that. It sounds like GL_MAX_TEXTURE_IMAGE_UNITS isn’t really the hard upper limit for fragment shaders (on your GL drivers at least), but it sounds like that’s operating outside of the spec, so your code in general might not work on other such drivers where you’re over-the-limit.[/QUOTE]Something is definitely going on in the drivers. I was trying to break my shaders by going over the limit, but it just did not happen. Was ready to move to array textures, but I never needed to.
I even tested it on a mobile NVIDEA GPU and it does not break a sweat. You are probably right though and stick to the spec.
[QUOTE=Dark Photon;1289803]Cool! Congrats on getting it up and running.
[/QUOTE]
Thanks :).
I used to use 4 * #lights + 3, now I only use 5 or so. So thay might explain it. I even get the speedup with very simple scenes where all vertices are withing the frustrum and none of them are occluded.
[QUOTE=Dark Photon;1289803]
Hmmm. 121 is certainly > 32 (your GL_MAX_TEXTURE_IMAGE_UNITS, which I believe is the bound texture access limit for fragment shaders). GL_MAX_COMBINED_TEXTURE_IMAGE_UNITS is the limit across all shader units, and you’re well under that. It sounds like GL_MAX_TEXTURE_IMAGE_UNITS isn’t really the hard upper limit for fragment shaders (on your GL drivers at least), but it sounds like that’s operating outside of the spec, so your code in general might not work on other such drivers where you’re over-the-limit.[/QUOTE]
Something must be going on in the driver. I was trying to break my shaders by pushing the number of lights higher and higher, but it never did. Even on a mobile NVIDEA GPU on my laptop it works fine.
You are right though. I should stay within the specs and start using texture arrays or Atlas textures for my depth maps.
Good to know I understand the limits and should be surprised :).
Just out of curiousity, did you try using more than GL_MAX_COMBINED_TEXTURE_IMAGE_UNITS?
I don’t know that that should break it, but it would confirm or refute that (on NVidia drivers at least) this is or isn’t a hard upper limit.
I really doubt that it is though because it seems that (on recent NVidia drivers) the value for this is always 5 times or 6 times the max number of texture per shader state (which you found is bogus). 5 is the number of shader stages w/o compute, and 6 is the number of shader stages with compute. For instance, on your card 325=160. Here on the NVidia card I was running on, 326=192. For the *6, no clue why they’d count compute as a stage in the shader pipeline, since it doesn’t coexist with the others in a program (AFAIK).
You are right though. I should stay within the specs and start using texture arrays or Atlas textures for my depth maps.
Also check out Bindless Texture. For some cases, this is much more convenient than texture arrays. And both of these in my opinion have fewer drawbacks than texture atlases (unless you’re targeting really, really old hardware).
With the same GC than the OP (AMD RX 580) I have almost the same parameters: 32, 32 and 192 (the last one for the combine), which look more like what Dark Photon found on its nVidia card.
I use free drivers on Mesa / Linux here.
[QUOTE=Dark Photon;1289814]Just out of curiousity, did you try using more than GL_MAX_COMBINED_TEXTURE_IMAGE_UNITS?
I don’t know that that should break it, but it would confirm or refute that (on NVidia drivers at least) this is or isn’t a hard upper limit.
I really doubt that it is though because it seems that (on recent NVidia drivers) the value for this is always 5 times or 6 times the max number of texture per shader state (which you found is bogus). 5 is the number of shader stages w/o compute, and 6 is the number of shader stages with compute. For instance, on your card 325=160. Here on the NVidia card I was running on, 326=192. For the *6, no clue why they’d count compute as a stage in the shader pipeline, since it doesn’t coexist with the others in a program (AFAIK).[/QUOTE]
Yeah, if I go over that limit then it does break (tried 162 and got some crazy effects).
Does the latest DOOM not use texture atlases for their depth maps? At this stage I do not know the pros and cons of these different ways of handling textures. But I will look into bindless textures, it seems to promise to be able to draw everything with one draw call which is kind of neat :P.
Interesting. Good to know.
Does the latest DOOM not use texture atlases for their depth maps?
No clue.
But I will look into bindless textures, it seems to promise to be able to draw everything with one draw call which is kind of neat :P.
Yeah, I’d just read the wiki page so you’ve got that in your back of tricks, and can pull it out if/when you determine that you need it. | 6,058 | 24,943 | {"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-2022-27 | latest | en | 0.748801 |
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[–][S] 63 points64 points (70 children)
8 liters of stuff, to be precise. 16 cylinders in four banks, boosted by 4 turbochargers.
[–] 34 points35 points (10 children)
Almost 1,200 pounds, or 540kg, to be precise. The 8 liters is more what isn't there, isn't it?
[–] 12 points13 points (25 children)
The way these are configured is awesome. Do you know if there is a gif of the firing order for these? 1-14-9-4-7-12-15-6-13-8-3-16-11-2-5-10. It's like an orchestra.
[–] 4 points5 points (1 child)
Funny, this is all I could hear in my head when I was looking at that.
[–] 2 points3 points (0 children)
Haha try to visualize that in your head... It's insane. The crankshaft actually looks fairly normal, too.
[–][S] 1 point2 points (0 children)
It actually takes a huge amount of engineering to configure that pattern. On 4 and some 6 cylinder engines, firing pattern is relatively basic, but once you get to 8, 10, 12, or higher, the pattern must be meticulously planned out so that the shock coming from those cylinders is evenly distributed. If it's not, the engine block could crack.
[–] 2 points3 points (18 children)
Eli5 why they use such an obscure order and if that's common on engines to not fire 1-4 or 1-8 in numerical order
[–] 10 points11 points (8 children)
No engine that I know of fires in sequential order. A four cylinder engine usually fires 1-3-4-2. It has to do with engine balancing and reducing vibrations. If you have a v-8 (4 cylinders per side) if all four from one side fired without any from the other side firing, it'd shake like hell.
[–] 19 points20 points (1 child)
My bike fires in sequential order: 1, 1, 1, 1, etc...
Twins also fire sequentially. ;)
[–] 9 points10 points (0 children)
Fair enough ;)
[–] 1 point2 points (4 children)
Rotary engines?
[–] 4 points5 points (0 children)
2 Rotors are 180° apart and fire 1-2
3 rotors are 120° apart can be 1-3-2 or 1-2-3 depending on how the eccentric shaft was machined.
4 rotors are...legos (copy pasta'd):
• 0° join = (1+3) and (2+4) firing (two rotors firing at the same time)
• 90° join = 1-3-2-4 firing order (typical 4 cyl. setup)
• 180° join = (1+4) and (2+3) (again two rotors firing together)
• 270° join = 1-4-2-3 firing order
Beyond 4 rotors are a combination/variation of the above. PPRE out of New Zealand recently built the first 6-Rotor, here's a good explination of that:
Think of who's building the shaft (Jeff Bruce?). Think of the way 4 rotor shafts with a 90° firing order are made = two 180° pairs with a 90° offset in the middle and two 45° offset counterweights at the ends only. So it's logical that this 6 rotor will have three 180° pairs. 60° firing order.
It's easy once you break it down for simplicity:
2 rotor = 180°
3 rotor = 120°
4 rotor = 90°
6 rotor = 60°
Rotor pairs of a 6 rotor (as I imagine they would be if only using two counterweights):
1 0°
2 180°
3 60°
4 240°
5 120°
6 300°
notice each rotor pair is phased 180° apart from each other, then each pair is 120° off from the other pairs.
10°
3 60°
5 120°
2 180°
4 240°
6 300°
One ignition event every 60°. However you can cheat and only go every 120° by using three FC leading coils triggered like any wasted spark arrangement. This makes it more compatible with most ECUs. Rotors 1 and 2 will fire together. 3 and 4 together. 5 and 6 etc.
It'll run a lot like a 20B, but instead of having three single leading coils and three leading spark plugs firing at 120° intervals, you keep the 120° and swap the coils for dual output coils. That's how I'd do it. Pretty simple.
[–] 0 points1 point (2 children)
Aren't there usually 3 or 4 rotors? They would be in different stages. Only 1 would be igniting at any one time. The others would be doing intake, compression, or exhaust.
[–] 1 point2 points (1 child)
How about the rotaries for aircraft?
[–] 0 points1 point (0 children)
I'm pretty sure they have to fire in sequence.
I'm not sure if they all do a power stroke followed by an induction stroke, or if they alternate (but they have usually an odd number, so I'm not sure that would work)
[–] 0 points1 point (0 children)
The builck 231 odd-fire 90 degree V6s fire 1-6-5-4-3-2.
But in this case, the cylinders are numbered across the banks rather than along them.
[–] 6 points7 points (2 children)
The firing order on my straight 6 is 1-5-3-6-2-4. It's easy to remember:
• 15 - too young
• 36 - too old
• 24 - just right
[–] 1 point2 points (0 children)
Brilliant.
[–] 0 points1 point (0 children)
1-4-6-5-2-3 works also though!
[–] 4 points5 points (2 children)
Engines can't fire in a sequential order since that would create massive vibrations that rattle an engine off of its mounts and would also result in much larger stress cycles in the crank shaft, leading to increased fatigue rates.
Although this doesn't show the actual firing, it does show how no two cylinders (at least on teh same bank) are reaching top-dead-center at the same time. https://www.youtube.com/watch?v=476dqWsAyiQ They do that so the torque going into the crank shaft is spread out over as smooth of an impulse as possible.
[–] 0 points1 point (1 child)
[–] 0 points1 point (0 children)
lol, i had no idea such idiocy existed
[–] 1 point2 points (1 child)
Well... To be honest, we'd have to see the numbering pattern. But my limited knowledge of anything it has to do with balancing the engine all around (vibration, torque, flywheel effect, etc). I'd be curious to see if anyone could find a cylinder numbering pattern for the W16. I can almost bet they are not necessarily sequential.
It's much easier in something like an inline 4 where if they fired in order you could see why that would be a bad thing... Your car would shake very hard I imagine.
[–] 1 point2 points (0 children)
[–] 0 points1 point (0 children)
I should add I'm not good at explaining anything...
[–] 1 point2 points (1 child)
[–] 0 points1 point (0 children)
Yes! Thank you!
[–] 0 points1 point (0 children)
omg. it's almost like they had to fire them all in some order. OMG LOOK THERE IS AN ORDER TO IT OMG
[–] 3 points4 points (25 children)
I think the transmission is even more of an engineering masterpiece, but this is also beautiful.
[–] 20 points21 points (24 children)
The whole car is an engineering masterpiece. It's no exaggeration to say it's the automotive equivalent of the concorde.
Consider that it can literally outperform a Formula 1 car in some aspects. Except it's also comfortable, luxurious, has air conditioning and speakers, and is perfectly road legal. Bear in mind, you can't just drop an engine like that into a typical sports car and go; it would just shatter the gearbox and transmission, shred the tires and deform the wheels, and overheat like crazy. So everything else had to be redesigned just so that all that power could actually be put on the ground. Its brakes were designed by a company that only makes brakes for the 747 and A380 airplanes. Ten radiators, two gigantic air scoops, four wheel drive, tires specifically designed only for this car, and on top of all that, it actually handles really well according to experts that have test driven it both on the streets and on a racetrack.
THEN consider that a Formula 1 car is basically replaced every few races. The tires are replaced a few times during a race; the engine is only used four or five times, I believe. Parts are constantly being swapped and replaced, all the time, even in the middle of a race. And the engine has to be on life support, so to speak; you have to pump warm water through it before you can start it up and whatnot. A Veyron, on the other hand, is designed such that its transmission and gearbox are supposed to last for ten years at minimum. It's supposed to work like a normal everyday car, as well as an utter monster on a racetrack when called for. That's when shit gets mindblowing on so many levels.
And I'm not even really a petrolhead or whatever. It's just that amazing a feat of human engineering and ingenuity.
p.s. Also, Bugatti actually lose money on every Veyron sale. A Veyron costs much more to make than they sell it for. Basically, Bugatti do it just to prove that they can do it.
[–] 11 points12 points (1 child)
I read this in Jeremy Clarkson's voice.
[–] 3 points4 points (0 children)
As a Top Gear fan, I'm flattered :D :D
[–] -1 points0 points (0 children)
And then a GTR beats it around any track.
[–][deleted] (2 children)
[deleted]
[–] 1 point2 points (1 child)
I can't imagine there would be any aspect where a "normal" car could outperform a F1 car, but proof me wrong it's just that I can't imagine it..
Going over speedbumps.
Stopping at a drive-through.
Parking lots.
etc.
[–] -3 points-2 points (12 children)
You're right it is a modern marvel, but it's nothing like an f1 car. F1s are driven to the limit every second of their lives. The veyron can't be at top speed very long due to traffic, speed limits, and poorly maintained roads. Also there are other cars coming out that use electric and gas to beat the Veryons speeds.
[–] 5 points6 points (1 child)
nothing like an f1 car
Arguably, there's nothing like a Veyron available to the general public. No matter how rich you are, you can't drive around town in an F1 car, that too in comfort. Also, an F1 car can ONLY be driven to the limit. Drive it too slow, and the tires are too cold and there's not enough downforce to control it, it just slides around, as displayed in an episode of Top Gear. So basically it's an apples-to-oranges argument. My point was that the Veyron is a car comparable to an F1, except you can drive it like a normal car too, and you won't have to change half the car every other day.
[–] 0 points1 point (0 children)
There are a lot of new cars like the veyron. The new Porsche, McLaren, Pagani, SSC. The are moving towards integrating electric and gas for the instant torque eliminating the lag from turbos.
[–] 2 points3 points (3 children)
Actually - you could only drive it at top-speed for seven minutes, totally independent from traffic, speed limits and poorly maintained roads due to the fact that after seven minutes, the gas is gone completely.
Edit: It seems to be 12 minutes until the tank is empty, not 7.
[–] 0 points1 point (1 child)
Yeah the tires run out at like ten minutes but at full speed you run out of gas in seven. Haha not practical but I want one.
[–] 0 points1 point (0 children)
Me too, me too :D
[–] 0 points1 point (0 children)
And you'll go 50 miles in that time, achieving just slightly less than two miles per gallon.
[–] 1 point2 points (5 children)
How low is this car? Please tell me it is practical enough to get over a speed-bump?
[–] 0 points1 point (0 children)
It can drive over speed bumps perfectly fine.
[–] 0 points1 point (3 children)
The new McLaren is had electric. The new Porsche is gas electric. Also, they lose money on the cars because they use them for research and development like Lexus did with the LFA.
[–] 4 points5 points (0 children)
What does that have to do with the height of the car?
[–] 2 points3 points (1 child)
They pay back for themselves with marketing to promote the brand though.
Veyrons are always at the big car shows, often photographed in the street etc.
Same with the LFA, it's been used in a bunch of Lexus commercials already.
[–] 0 points1 point (0 children)
Big dick competition for car companies lol
[–] -2 points-1 points (3 children)
The Bugatti is certainly an engineering masterpiece.
...so it impresses me even more when a tuner garage creates a car that outperforms it in nearly every aspect. Look at the Hennessey Venom GT. Several hundred pounds lighter, a couple hundred more horsepower, better fuel economy, faster acceleration, higher top speed, better handling, and (slightly) lower price. Luxury and creature comfort definitely goes in favor of the Veyron though. The Venom GT is very tight and spartan inside.
Both cars are amazing though. Inspiring engineering.
[–] 2 points3 points (1 child)
Mass-producing something (even for low-values of "mass", in this case) has its own challenges. I'm not surprised a boutique shop can put together something faster than a Veyron, like I'm not surprised you can put together a home PC that outperforms the new Mac Pro; most of the engineering wonder is in the manufacturing process itself.
[–] 1 point2 points (0 children)
I've never thought of it quite like that, but it makes a lot of sense.
Having a handful of guys hand build a Venom GT in a garage in three months time is one thing. But having a Veyron roll off of an assembly line in a day or two (I don't know their actual process) is something totally different. A car that produces those kinds of numbers leaves very little room for error. Mass production of such precision is pretty impressive.
But in the case of either car... for a million bucks, it better be pretty goddamn impressive.
[–] 0 points1 point (0 children)
Not to mention looks better. Those Hennessey Venom GT's are sexy as hell!
[–] -2 points-1 points (0 children)
How does this thing compare to a Rolls-Royce?
[–] 2 points3 points (0 children)
The turbos appear to be internally gated. I did not expect that.
[–] 1 point2 points (0 children)
[–] 1 point2 points (0 children)
I know what some of those words mean.
[–] 2 points3 points (3 children)
4 Mitsubishi EVOs worth of engines...
[–] -3 points-2 points (2 children)
4 ej20 engines worth of engines.... Evos suck ass
[–] 2 points3 points (1 child)
Well, I'm not too partisan on the matter, so ej20s it is...
You know, the idea was basically that having 4 times the cylinders, displacement and turbos (and in the SS the power) of legends as the Scoobs and Evos, simply boggles my mind.
[–] 1 point2 points (0 children)
Too bad is isnt just 4 times the price... | 3,780 | 14,001 | {"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-2014-35 | latest | en | 0.910558 |
https://code.i-harness.com/zh-TW/q/19a5b7a | 1,642,470,598,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300658.84/warc/CC-MAIN-20220118002226-20220118032226-00676.warc.gz | 254,931,077 | 4,641 | # r - 找到每列的最大值和最小值,然後找到每一行的最大值和最小值
## matrix max (3)
``````apply(a, 1, range)
``````
`t`一起使用，這將為您提供兩列。 第一個具有最小值的第二個具有最大行數。
``````head(t(apply(a, 1, range)))
[,1] [,2]
[1,] 95.75922 103.6956
[2,] 93.62636 106.3934
[3,] 92.70567 106.9190
[4,] 96.53577 104.4971
[5,] 96.61573 107.6691
[6,] 95.56239 105.5887
``````
``````a <- matrix(rnorm(1000 * 18, mean = 100, sd = sqrt(10)), 1000, 18)
``````
``````apply(a,2,min)
apply(a,2,max)
``````
``````apply(a,1,min)
apply(a,1,max)
`````` | 267 | 488 | {"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-2022-05 | latest | en | 0.309109 |
http://www.mathworks.com/matlabcentral/cody/problems/730-how-many-trades-represent-all-the-profit/solutions/236905 | 1,485,057,987,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281332.92/warc/CC-MAIN-20170116095121-00392-ip-10-171-10-70.ec2.internal.warc.gz | 566,767,104 | 11,706 | Cody
# Problem 730. How many trades represent all the profit?
Solution 236905
Submitted on 28 Apr 2013 by G K
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
``` trades = 1 3 -4 2 -1 2 3 q = 2 ```
2 Pass
``` trades = 1 2 3 -5 q = 1 ```
``` trades = 1 2 3 4 5 6 q = 6 ```
``` trades = -2 3 -4 5 -6 1 2 3 4 5 q = 3 ``` | 163 | 443 | {"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-2017-04 | latest | en | 0.796466 |
http://pembeforum.com/3u568n5qd/tV56390eR/ | 1,575,654,719,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540490743.16/warc/CC-MAIN-20191206173152-20191206201152-00544.warc.gz | 109,575,212 | 12,963 | # Simple Math Worksheets Printable Worksheet For Multiplication Grade Linear Functions Practice Test Solving Equations Using Kids Tiles Free Money Middle School Students Addition Sums To
Published at Sunday, 01 September 2019. Kids Worksheet. By .
The main problem with what I see with my students and my own children is that kids are taught concepts and are not taught skills--unless they are lucky enough to have a teacher who knows better. Most particularly, children are not taught mastery of arithmetic with fractions. Unfortunately, virtually all of their future math education depends on being able to do fractional arithmetic. Most of even beginning algebra depends on being able to do two things-one, doing multiplication quickly and accurately in your head, two, knowing how to add, subtract, multiply, and divide fractions.
When was the last time that you wanted to spend more time doing something you hated and dreaded? Top that off with seeing no reason for the chore you so dread in the first place and you have a complete and total recipe for disaster. High quality 5th Grade Math Worksheets help children understand critical math concepts, enjoy and see their progress, and most importantly, retain the skills necessary to successfully progress through school. Hard work and dedication are traits that are directly correlated to future success so we want to be sure to encourage this. The simply constructed traditional math worksheets only require students to focus on answering the problem. We need children to understand the concept, be engaged, and have fun doing it. Let us engage children with thought, understanding, and the foresight to provide them something other than a bland sheet full of numbers and symbols.Keep these thoughts in mind when searching for your 5th Grade Math Worksheets.
How much of an apple pie has been eaten? The answer to this question can be expressed in percentages, 50%; or in decimals, 0.5; or in fraction, ½. In other words, half of mom has delicious apple pie is gone. How many kids in school have done their homework? Again this can be answered in several ways: in percentages, 70%; or in ratio, 7:10; Both of these mean out of ten kids in class there are seven good ones who did and three not-so-good ones who did not. The bottom line is that kids learn math much better when it makes sense. Thus, the math worksheets which you get for your kids should include interesting word problems that help them with the practical application of the lessons they learn. It should also present the same problem in a variety of ways to ensure that a child has grasp of a subject is deeper and comprehensive.
File name:
### Simple Math Worksheets Printable Worksheet For Multiplication Grade Linear Functions Practice Test Solving Equations Using Kids Tiles Free Money Middle School Students Addition Sums To
Image Size: 2550 x 3300 Pixels
File Type: Image/jpg
Total Gallery: 32 Pictures
File Size: 237 kb
#### Grade 3 School Worksheets Gallery
99 of 100 by 830 users | 608 | 3,020 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2019-51 | latest | en | 0.956566 |
https://mail.scipy.org/pipermail/scipy-user/2010-June/025816.html | 1,508,253,496,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822116.0/warc/CC-MAIN-20171017144041-20171017164041-00107.warc.gz | 787,068,953 | 2,022 | # [SciPy-User] Autocorrelation function: Convolution vs FFT
David david@silveregg.co...
Tue Jun 22 19:47:22 CDT 2010
```On 06/23/2010 02:49 AM, Skipper Seabold wrote:
> I am trying to compute the autocorrelation via convolution and via fft
> and am far from an expert in DSP. I'm wondering if someone can spot
> anything that might introduce numerical inaccuracies or if I'm stuck
> with the following two being slightly different.
>
> Generate some autocorrelated data:
>
> import numpy as np
> nobs = 150000
> x = np.zeros((nobs))
> for i in range(1,nobs):
> x[i] = .85 * x[i-1] + np.random.randn()
>
> # compute ACF using convolution
>
> x0 = x - x.mean()
>
> # this takes a while for the big data
> acf1 = np.correlate(x0,x0,'full')[nobs-1:]/nobs
> acf1 /= acf1[0]
>
>
> # compute ACF using FFT
>
> Frf = np.fft.fft(x0, n=2*nobs) # zero-pad for separability
> Sf = Frf * Frf.conjugate()
> acf2 = np.fft.ifft(Sf)
> acf2 = acf2[1:nobs+1]/nobs
> acf2 /= acf2[0]
> acf2 = acf2.real
>
> np.linalg.norm(acf1-acf2, ord=2)
>
> They are pretty close, but I would expect them to be closer than this.
>
> np.max(acf1-acf2)
> 0.006581962491189159
>
> np.min(acf1-ac2)
> -0.0062705596399049799
You could look at my scikits talkbox which has both fft-based and
brute-force autocorrelation. I don't think they have such a big
difference between implementations,
David
``` | 468 | 1,370 | {"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} | 2.671875 | 3 | CC-MAIN-2017-43 | latest | en | 0.804912 |
https://cardosoesportes.com/qa/quick-answer-how-do-you-convert-db-to-normal-value.html | 1,632,521,771,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00514.warc.gz | 219,234,956 | 9,172 | # Quick Answer: How Do You Convert DB To Normal Value?
## How is dB value calculated?
DeciBel formula for power comparisons As might be expected it is ten times the logarithm of the output divided by the input.
The factor ten is used because deciBels rather than Bels are used.
If the value of P2 is greater than P1, then the result is given as a gain, and expressed as a positive value, e.g.
+10dB..
## What does dB mean sexually?
bondage and disciplineabbreviation. bondage and discipline: used in reference to practices involving physical restraint and punishment, especially in a sexual encounter or relationship.
## How loud is 140 decibels?
Comparative Examples of Noise LevelsNoise SourceDecibel LevelDecibel EffectAircraft carrier deck140Military jet aircraft take-off from aircraft carrier with afterburner at 50 ft (130 dB).130Thunderclap, chain saw. Oxygen torch (121 dB).120Painful. 32 times as loud as 70 dB.12 more rows
## How is dB loss calculated?
For instance, if signal A has a power of 20 mW and signal B has a power of 5 mW: 20/5 = 4. Take the log of the the ratio of the signals by pressing the log button on the scientific calculator. For instance: log 4 = 0.602. Multiply this answer by 10 to find the decibels.
## What is dB in vibration?
The decibel (abbreviated dB) confuses many people, perhaps because they assume it is an absolute unit or level of sound. … A decibel is the relationship or ratio between two sound levels, for example the measured sound pressure level and the minimum sound pressure level a person with good hearing can detect.
## How do you convert amplitude to dB?
Decibels are a relative unit, they express the power of your signal relative to some reference power. If you are working with amplitudes, then the formula is: power_db = 20 * log10(amp / amp_ref); (See http://en.wikipedia.org/wiki/Decibel#Field_quantities).
## What is dB formula?
One decibel (0.1 bel) equals 10 times the common logarithm of the power ratio. Expressed as a formula, the intensity of a sound in decibels is 10 log10 (S1/S2), where S1 and S2 are the intensity of the two sounds; i.e., doubling the intensity of a sound means an increase of a little more than 3 dB.
## How many watts are in a dB?
Definition: dBW means dB relative to 1 watt, so 0 dBW = 1 watt, -3 dBW = half watt. +3dBW = 2 watts etc. Definition: dBm means dB relative to 1 milliwatt, so 0 dBm = 1 milliwatt (one thousandth of 1 watt or 0.001 watt).
## How much louder is 10 dB than 20dB?
The level of noise in a quiet bedroom, 30 dB, is 100 times louder than 10 dB. And 40 dB is 1,000 times louder than 10 dB. A typical conversation clocks in at around 60 dB.
## What is being Mermaided?
Casually moving on from these two very real dating trends, Rebel briefly mentions “mermaiding,” which immediately causes a double-take. Mermaiding? … “It’s, like, when a girl goes out with a guy to a boardwalk and then she gets really bored with the date. So, she just goes into the ocean and you never see her again.”
## What does 1db mean?
decibelThe decibel (dB) is a logarithmic unit used to measure sound level. It is also widely used in electronics, signals and communication. The dB is a logarithmic way of describing a ratio. The ratio may be power, sound pressure, voltage or intensity or several other things.
## What dB is dangerous?
Sound is measured in decibels (dB). A whisper is about 30 dB, normal conversation is about 60 dB, and a motorcycle engine running is about 95 dB. Noise above 70 dB over a prolonged period of time may start to damage your hearing. Loud noise above 120 dB can cause immediate harm to your ears.
## How many dB can a human hear?
What is sound? Humans can hear sounds between 0 and 140 decibels. 0 decibel does not mean that there is no sound, merely that we cannot hear it. 0 decibel is the so-called hearing threshold for the human ear.
## What does no BB mean sexually?
Bareback”Bareback.” BB is often used on online dating sites, as well as in text messages and on chat forums, with the meaning “Bareback,” to refer to having sexual intercourse without a condom. “Be Back.” BB is sometimes used in internet chat rooms to indicate that a participant is leaving the group, but will be back shortly.
## How is SNR calculated in dB?
So, if your SNR measurements are already in decibel form, then you can subtract the noise quantity from the desired signal: SNR = S – N. … Furthermore, for power, SNR = 20 log (S ÷ N) and for voltage, SNR = 10 log (S ÷ N). Also, the resulting calculation is the SNR in decibels.
## Why gain is calculated in dB?
In electronics, gain is a measure of the ability of a two-port circuit (often an amplifier) to increase the power or amplitude of a signal from the input to the output port by adding energy converted from some power supply to the signal. … It is often expressed using the logarithmic decibel (dB) units (“dB gain”).
## Is 3 dB twice as loud?
The human ear’s response to sound level is roughly logarithmic (based on powers of 10), and the dB scale reflects that fact. An increase of 3dB doubles the sound intensity but a 10dB increase is required before a sound is perceived to be twice as loud. … The sound intensity multiplies by 10 with every 10dB increase.
## How many times louder is 10 dB?
The dB rating is not just “how loud it sounds.” Rather, each extra 10 dB means the sound is 10 times as intense. The rule of thumb from last time means, that it is perceived to be (“sounds”) roughly 2 times as loud. Therefore, 60 dB is perceived to be about 2×2×2=8 times as loud as 30 dB.
## What does BTW mean sexually?
BTW in SexualBTWBy The Way Army, Medical, InternetBTWBig Titty Woman
## What is the intensity of a 70 dB sound?
Learning ObjectivesTable 1. Sound Intensity Levels and IntensitiesSound intensity level β (dB)Intensity I(W/m2)Example/effect501 × 10–7Average office, soft music601 × 10–6Normal conversation701 × 10–5Noisy office, busy traffic12 more rows
## How do I convert to dB?
The dB is calculated via two different expressions XdB=10log10(XlinXref)orYdB=20log10(YlinYref). If you convert a quantity X that relates to power or energy, the factor is 10. If you convert a quantity Y that relates to amplitude, the factor is 20.
## What is the 3dB rule?
3dB rule and maximum exposure to noise To put it in context, a worker exposed to a continuous sound intensity level of 83dB(A) for one hour would be exposed to the same amount as someone exposed to an 80dB(A) level for two hours. | 1,641 | 6,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-39 | latest | en | 0.889552 |
https://quant.stackexchange.com/questions/24601/how-to-derive-blacks-formula-for-the-valuation-of-an-option-on-a-future?noredirect=1 | 1,558,785,527,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232258003.30/warc/CC-MAIN-20190525104725-20190525130725-00222.warc.gz | 597,956,757 | 33,385 | # How to derive Black's formula for the valuation of an option on a future?
I've got a question about 1976 Black Model and Bachelier model.
I know that a geometric brownian motion in the P measure $dS_{t}=\mu S_{t}dt+\sigma S_{t} dW_{t}^{P}$ for a stock price $S_{t}$ leads (after a change of measure) to the Black-Scholes formula for a Call:
$$C= S_{0} N(d_{1}) − Ke^{−rT} N(d_{2})$$.
Where $d_{1} = \frac{ln(\frac{S_{0}}{K})+(r+\frac{1}{2}\sigma^{2})T}{\sigma\sqrt{T}}$ and $d_{2}=d_{1}-\sigma \sqrt{T}$
I actually don't know how's possible to get the famous black formula on a forward contract:
$$C= e^{−rT}(F N(d_{1}) − KN(d_{2}))$$.
where now $d_{1} = \frac{ln(\frac{F}{K})+\frac{1}{2}\sigma^{2}T}{\sigma\sqrt{T}}$ and $d_{2}=d_{1}-\sigma \sqrt{T}$
Should I simply insert $F(0,T)=S_{0}e^{rT}$ in the first BS formula to get the second one?
I'm asking this because I've tried to derive the BS formula using an arithmetic brownian motion like $dS_{t}=\mu dt+\sigma dW_{t}^{P}$, and I get:
$$C= S_{0} N(d) + e^{−rT}[v n(d)-K N(d)]$$.
where $d=\frac{S_{0}e^{rT}-K}{v}$ and $v=e^{rT}\sigma\sqrt{\frac{1-e^{−2rT}}{2r}}$ and remembering that $N(d)$ and $n(d)$ are the CDF and PDF.
but the previous substitution $F(0,T)=S_{0}e^{rT}$ doesn't seems to lead to the known result $C= e^{−rT}[(F-K)N(d)-\sigma\sqrt{T}n(d)]$
where now $d=\frac{F-K}{\sigma\sqrt{T}}$
I think I could reach the equations on forward both in the geometric brownian motion and arithmetic brownian motion using the equations
$dF=F\sigma dW_{t}^{Q}$ and $dF=\sigma dW_{t}^{Q}$ but I don't know how justify the use of them.
• @Macro Welcome to Quant. S.E.! Do you want to price just forward contract or option on forward contract? – Neeraj Feb 28 '16 at 18:31
• Hi Neeraj, thanks for your answer. I'd like to price an option on forward contract! – Marco Feb 28 '16 at 20:37
• Just replace $S_0$ with $F e^{-rT}$ in your original BS formula or you can use risk neutral approach. Both will lead to the same valuation formula. – Neeraj Feb 29 '16 at 6:56
• Ok, thanks. But can I do the same for the ABM? Because I can't get the result when I do this substitution. – Marco Feb 29 '16 at 8:02
## European option on future
To price European Option on Future, you just need to replace $S_0$ with $Fe^{-rT}$ in your original BS formula or you can use risk neutral approach. Both will lead to same Valuation formula.
## American option on future
Above procedure can not be used to price American option on future. In a paper, The valuation of options on future contracts by Ramaswamy, stated that
There are no known analytical solution to the valuation of American option on future contract.
Authors used implicit finite difference method to price American option on future contract.
# Edit: Derivation of price of European option on future contract
Under risk neutral measure, future price, $F_t$ satisfy following SDE: $$dF_t = \sigma F_t dW_t$$ where, $W_t$ is a Wiener process. It can be easily shown that: $$F_T|F_t= F_t e^{-\frac{1}{2}\sigma^2 (T-t) + \sigma (W_T- W_t)}$$ $$F_T|F_t \sim logN \left( ln(F_t) - \frac{1}{2}\sigma^2 (T-t), \sigma^2(T-t)\right)$$
The price of option on future contract $(C_t)$ under risk neutral measure is: $$C_t = e^{-r(T-t)}E_\mathbb{Q} [(F_T - K)^+]$$
You can easily solve the above expression to get the price of option written on future. The distribution of $F_T$ is very similar to $S_T$ (see this answer). If you replace $$ln(F_t) =ln(S_t) + r(T-t)$$ then you will get the same distribution of $S_T$ as under risk neutral measure. This is the reason, to get the price of option on future, we replace $S_t$ with $F_t e^{-r(T-t)}$ in BS model of European call option price.
• Hi Neeraj, actually I'd like to price an European option starting from an ABM. – Marco Feb 29 '16 at 16:16
• @Marco please check edit answer. – Neeraj Feb 29 '16 at 18:13 | 1,238 | 3,871 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2019-22 | longest | en | 0.768611 |
https://www.phys-l.org/archives/2014/11_2014/msg00076.html | 1,653,394,372,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00459.warc.gz | 1,067,253,576 | 5,093 | Chronology Current Month Current Thread Current Date [Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]
# Re: [Phys-L] How Einstein discovered E = mc2
• From: John Denker <jsd@av8n.com>
• Date: Thu, 20 Nov 2014 17:50:11 -0700
On 11/20/2014 12:47 AM, Savinainen Antti wrote:
I read two papers by Eugene Hecht [....]
<http://scitation.aip.org/content/aapt/journal/tpt/50/2/10.1119/1.3677283>
<http://scitation.aip.org/content/aapt/journal/ajp/79/6/10.1119/1.3549223>
FWIW, this is the same Hecht who wrote several textbooks (and
outlines) on physics and optics.
I am mystified by the two AJP articles cited above. The E=mc^2
formula reminds me of certain tabloid celebrities who are
"famous for being famous" even though they've never accomplished
anything of significance.
Anybody who is serious about physics moved on to other things,
long since.
Specifically:
a) Suppose we have a pointlike particle, and we think
its position is a 4-vector (r). This is not the only
conceivable way of doing things, but it seems to work.
b) Then for any particle with nonzero mass you can define
the 4-velocity u = dr / dτ. This is far from the only way
of doing things, but it is as good as any and better than
most. It is nice and simple. It upholds the correspondence
principle.
c) Then you can define the 4-momentum p = m u (for some "m").
This is exceedingly simple. It upholds the correspondence
principle. As far as anybody can tell, it is consistent
with conservation of momentum.
d) For massless particles you can skip steps (b) and (c)
and just write down the 4-momentum. These particles have
undefined 4-velocity and undefined proper time, but they
have a perfectly good 4-momentum
e) In any case, m is the invariant length of the 4-momentum.
For massless particles, m exists, and is equal to zero.
f) Suppose we pick out the timelike component of the
4-momentum in some chosen frame, and choose to call it
the "energy". Then by the basic definition of norm of
a 4-vector, we have
E^2 - p_xyz^2 = m^2 [1]
where E is the timelike part of the momentum and p_xyz
is the spatial part ... both of which are definable only
with respect to some chosen frame. This stands in contrast
to m, which is invariant and frame-independent.
g) For something that is not a pointlike particle, perhaps
a parcel of fluid, all of the above goes out the window
and we need heavier tools, such as stress-energy tensors.
My point is that equation [1] is vastly more useful than
E=mc^2 ... and also conceptually simpler, since it is a
routine consequence of other things we know. E=mc^2 is
not the miracle, or even a miracle. Spacetime geometry
and trigonometry is where the action is.
Hecht argues that nobody has ever come up with a rigorous
proof of the "equivalence" of mass and energy. Well of
course not, because they are not equivalent. They might
be numerically equal in certain limiting cases, but they
are not "equivalent".
There will never be a rigorous answer to the question,
because there is not even a rigorous question to the
question. Einstein himself remarked: "As far as the
laws of mathematics refer to reality, they are not certain,
as far as they are certain, they do not refer to reality."
The classical energy is not defined uniquely; you can
shift it by a gauge transformation. Therefore you can
always pick a gauge so as to /make/ the classical energy
equal to mc^2 -- which an awful lot of people have done --
and it would even be consistent with the correspondence
principle. However, you could equally well make it equal
to (mc^2 / 2) or any other thing that suits your fancy.
You need to know more than the correspondence principle
in order to do physics.
Seriously, you could define "the" velocity to be
-- v = dr / dt (t = coordinate time)
-- u = dr / dτ (τ = proper time)
and both would reduce to the classical velocity in the
correspondence limit. Now v has the advantage of being
easy to measure, and of existing even for massless
particles (whereas u does not) -- but the other 99% of
the time, u is more useful. So there will never be a
unique answer to the question of how to generalize our
classical notion of velocity.
By the same token, there will never be a unique answer
to the question of how to generalize our classical notion
of mass. Classically mass serves as a measure of "inertia"
(whatever that means) and also serves as a source term
for the gravitational field. When we generalize to
relativity, you can't have it both ways. Conventionally
we take m to be the "inertia" constant in p = m u, in
which case m is no longer the source term for the
gravitational field. The stress-energy tensor takes
over as the source term. You have to be super-careful
how you state the equivalence principle (the equivalence
of gravitational mass and inertial mass), because any
bold general statement would be untrue.
Even without mentioning gravitation, just confining
ourselves to "inertia", if you try to define a
velocity-dependent relativistic mass, you quickly
discover that there are lots of inequivalent ways
of doing it: longitudinal mass, transverse mass,
and who-knows-what else.
So you see, we can't even say that mass is equivalent
to mass in all the ways we might like, so there's just
no hope of energy being equivalent to mass as a general
proposition.
There are relationships among these things, but the
relationship is not an "equivalence".
Saying such things are equal in the correspondence limit
is nice, but it's sort of like checking the dimensions:
You have to check, but lots of things have the correct
dimensions even when they're not the right answer. The
torque has the same dimensions as the lagrangian, but
they're not the same thing.
Last but not least:
honest historical research instead of portraying a smoothed version
of the process of discovery
There are at least two common mistakes. As mentioned above,
one mistake is to smooth out the history, pruning off all
the dead ends, pretending that discovery is not nearly so
messy and non-monotonic as it really is.
Another is to /personalize/ the history. I say the 1860s
were not all about Abraham Lincoln, and ancient Greece
was not all about Pericles. By the same token, relativity
is not all about Einstein. Really, really not. I do not
much care what Einstein was up to in 1908; I am much more
interested in what Minkowski was up to.
Some people say that personalizing the story makes it more
interesting to students. Well maybe so, but it comes at a
terrible price. Nobody wants to get involved with a project
if they think the famous guy is going to get all the credit.
This is bad for everybody, including the famous guy along
with everybody else, because it makes it hard to form teams.
This business of giving Einstein credit for stuff that other
people did has got to stop. | 1,705 | 6,873 | {"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} | 2.84375 | 3 | CC-MAIN-2022-21 | latest | en | 0.913817 |
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2019-01-25, 14:05 #166
Dr Sardonicus
Feb 2017
Nowhere
5,147 Posts
Quote:
Originally Posted by jvang $$f(x) = x^3+2x+4$$, and $$g(x)$$ is its inverse. Find $$g(7)$$ without finding the formula for $$g(x)$$, then find $$g'(7)$$.
f(x) = x3 + 2*x + 4
y = x3 + 2*x + 4, point (1, 7)
y' = 3*x2 + 2 At (1, 7) y' = 5
Inverse function
x = y3 + 2*y + 4, point (7, 1)
1 = (3*y2 + 2)*y' At (7, 1) y' = 1/5
Using the "just transpose x and y" idea, the point-slope equation for the tangent line to y = f(x) at (1, 7) is
y - 7 = 5*(x - 1).
An equation for the tangent line to x = f(y) at (7, 1) is then
x - 7 = 5*(y - 1).
Casting this into point-slope form,
(1/5)*(x - 1) = y - 1.
Last fiddled with by Dr Sardonicus on 2019-01-25 at 14:11
2019-01-25, 17:23 #167
S485122
"Jacob"
Sep 2006
Brussels, Belgium
22·439 Posts
Quote:
Originally Posted by Dr Sardonicus ... Inverse function x = y3 + 2*y + 4, point (7, 1) ...
IMHO opinion your definition of the inverse function is not correct.
If y=f(x) the inverse function y=f -1(x) will be such that f(g(x))=y this is obviously not the case your calculations. (For instance it is obvious that x=y3 is not the inverse function of y=x3, since substituting x for your inverse function will give y=(y3)3=x9.)
Jacob
See Wikipedia : Inverse function for instance.
2019-01-25, 18:11 #168
VBCurtis
"Curtis"
Feb 2005
Riverside, CA
52×7×29 Posts
Quote:
Originally Posted by S485122 (For instance it is obvious that x=y3 is not the inverse function of y=x3, since substituting x for your inverse function will give y=(y3)3=x9.) Jacob
If you rewrite x = y3 to be in the format of y as a function of x, then the definition you cite applies. You're splitting hairs to say that y = x1/3 is the inverse, but x = y3 is not.
To find an inverse, switch x and y. If you want the inverse in g(x) format, then solve for y after switching x and y.
2019-01-25, 19:02 #169
Dr Sardonicus
Feb 2017
Nowhere
5,147 Posts
Quote:
Originally Posted by S485122 IMHO opinion your definition of the inverse function is not correct. If y=f(x) the inverse function y=f -1(x) will be such that f(g(x))=y this is obviously not the case your calculations. (For instance it is obvious that x=y3 is not the inverse function of y=x3, since substituting x for your inverse function will give y=(y3)3=x9.)
Your substitution is inconsistent with the definition you give. Using the example y = x^3, the inverse function implicitly defined by x = y^3 is not y^3, but rather g(x) = y. And, by its formulation (g(x))3 = x.
The only real difficulty with the implicit formulation of the inverse function is that it may not be well-defined. And even then, the problem only really manifests itself at points where two or more possible inverses meet.
2019-02-08, 02:23 #170 jvang veganjoy "Joey" Nov 2015 Middle of Nowhere,AR 22×3×37 Posts There was a problem we were working on today, in which we're still doing inverse derivative stuff. I'm trying to figure out the inverse of this, so that I can find the derivative of the inverse: $$y = x^3-2x^2+5x$$ or something very similar. You just swap the variables and solve for y, right?$x = y^3 - 2y^2 +5y$But where do you go from here? I can't figure out how to isolate the y, since any sort of division leaves an extra y in a denominator
2019-02-08, 16:03 #171
Dr Sardonicus
Feb 2017
Nowhere
10100000110112 Posts
Quote:
Originally Posted by jvang I'm trying to figure out the inverse of this, so that I can find the derivative of the inverse: $$y = x^3-2x^2+5x$$ or something very similar. You just swap the variables and solve for y, right?
You went one step too far. Don't try to "solve for y," just swap the variables, then use implicit differentiation.
Hmm. The cubic
x^3 - 2*x^2 + 5*x - a
has discriminant -27*a^2 + 148*a - 400, which is negative for any real a. So Cardano's formulas will give the real solution to
x^3 - 2*x^2 + 5*x = a
as a sum of a rational number and two real cube roots, but it will be an algebraic mess. Differentiating would produce an even bigger mess.
Besides -- implicit differentiation works just fine, even if there is no "nice" formula for the inverse function.
Last fiddled with by Dr Sardonicus on 2019-02-08 at 16:39
2019-02-13, 03:11 #172 jvang veganjoy "Joey" Nov 2015 Middle of Nowhere,AR 22·3·37 Posts We covered a bit on matching the derivatives of graphs with their originals. One thing that was interesting was inflection points, which has something to do with concavity (which we haven't covered, also something about "critical numbers"?). From my observations of derivative graphs, it seems that inflection points are located at the x intercepts of the original graph's second derivative, which would be where the slope of the slope of the graph is 0... is there a better way to word that?
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Page 1 of 1
## 14 Replies - 4026 Views - Last Post: 18 November 2011 - 03:43 PM
### #1 Sethro117
• Still the sexiest mofo.
Reputation: 237
• Posts: 2,378
• Joined: 14-January 09
# Don't feel like doing homework? Post it on a freelance site</s
Posted 13 November 2011 - 11:45 AM
Below is a prime example of not only students in CS and IT industries but probably every industry today. I know we've seen examples of this on DIC but this kid actually admits it and pays someone to do it for him:
Quote
I am a student who does not feel like doing his homework. This is what my professor assigned:
The final character of a ten digit International Standard Book Number is a check digit computed so that multiplying each digit by its position in the number (counting from the right) and taking the sum of these products modulo 11 is 0. The furthest digit to the right (which is multiplied by 1) is the check digit, chosen to make the sum correct. It may need to have the value 10, which is represented as the letter X. For example, take the ISBN 0-201-53082-1. The sum of products is 0ื10 + 2ื9 + 0ื8 + 1ื7 + 5ื6 + 3ื5 + 0ื4 + 8ื3 + 2ื2 + 1ื1 = 99 0 modulo 11. So the ISBN is valid.
While this may seem more complicated than the first scheme, it can be validated very simply by adding all the products together then dividing by 11. The sum can be computed without any multiplications by initializing two variables, t and sum, to 0 and repeatedly performing t = t + digit; sum = sum + t; (which can be expressed in C as sum += t += digit;). If the final sum is a multiple of 11, then the ISBN is valid.
Ten-digit ISBN Numbers. ISBN numbers prior to 2007 use ten digits. Suppose, for example, an user enters the numbers 0-7637-2478-5. How can a computer application ensure that this is a valid ISBN number? The first step is to strip the number of all extraneous dashes, blanks, and similar characters. (Hint: Parse the original string by element, and then assemble a new string that contains ONLY numeric values using StringBuilder). For this example, the resulting number is 0763724785.
The second step is to ensure that the number contains exactly 10 digits.
The third step is to multiply each successive digit in the number by the weights of (10, 9, 8 and so forth to
.1) and add the cross products terms.(Note: an X in an ISBN number stands for 10 Hint: loop dude loop and remember strings are indexed in C# making it easier than Java! For this example the sum is 242 and 242 mod 11 =0 and hence it is valid
The final step is to examine the sum and verify that this number is evenly divisible by 11. If so, the ISBN is presumed to be valid. If not, the ISBN presumed invalid.
Build a C# class that will accept an input string and return the result. BE CAREFUL to make sure if the ISBN has an X in the stream then that stands for the number 10.
TEST VALUES: 0-13-152523-9
0-13-148660-8
Write a program to do this with a class and driver class or just code a driver class. What matters is the logic of the program. Assume the string has been inputted. You may use the IsDigit method from Char Class.
His first sentence just goes to show you how lazy people can be. So far the bid is for \$50.
Is This A Good Question/Topic? 2
## Replies To: Don't feel like doing homework? Post it on a freelance site</s
### #2 Martyr2
• Programming Theoretician
Reputation: 4735
• Posts: 12,911
• Joined: 18-April 07
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 13 November 2011 - 12:26 PM
Funny thing about it is, he is either going to say "That is too much money" and fail the assignment or he is going to pay through the nose... \$50 for an assignment that pretty much walks you through everything is way too much money.
Either way, the guy is pretty screwed so we can sit back and smile. Perhaps chuckle a little.
• MrCupOfT
Reputation: 2290
• Posts: 9,529
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## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 13 November 2011 - 12:33 PM
I'd probably need medical attention if they went with the bid, and it turn out to be their teacher. Proper Busted.
### #4 Jstall
• Lurker
Reputation: 434
• Posts: 1,042
• Joined: 08-March 09
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 14 November 2011 - 08:02 AM
Some of the rent a coder type sites have a section specifically for homework.
If the student has no interest in doing something this trivial why is he in a CS class to begin with? If he doesn't want to bother doing the easy stuff he is going to be royally screwed when the hard assignments come around.
As Martyr2 already mentioned \$50 is pretty steep to do an assignment like this, I'm guessing that will come down considerably , but hope it doesn't
### #5 modi123_1
• Suitor #2
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• Joined: 12-June 08
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 14 November 2011 - 08:14 AM
I would have charged \$150... time is money!
### #6 Craig328
• I make this look good
Reputation: 1982
• Posts: 3,520
• Joined: 13-January 08
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 14 November 2011 - 08:43 AM
I'd wish that whoever ended up writing this piece of code could somehow figure out a way to have it post "I'm a cheating asshole, please expel me or at least set me on fire" instead of whatever answer this turd is looking to hand in.
### #7 jon.kiparsky
• Pancakes!
Reputation: 8636
• Posts: 14,906
• Joined: 19-March 11
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 14 November 2011 - 09:17 AM
I wouldn't worry about it too much. For one thing, as has been said, someone who doesn't want to do their homework will fail out sooner or later, probably around the time they have to take an exam.
For another thing, whether you believe it or not, a teacher worth having will be able to figure out, quite easily, who is capable and who is not. If the student turns in work that's clearly not written by them, it's easy enough to catch them out on it - just ask them a few questions about it.
I do like the easter egg idea, though. Of course, the kid's going to run the program before turning it in, so it's better to write something into the code that says "not written by the student". This is an interesting problem in steganography: hiding a message in the source code so it will be obvious to the teacher but not to the student. Presumably the best message would be a gratuitous String which contains the link to the proposal on the freelance site.
No points for solving in perl, of course.
### #8 mojo666
Reputation: 377
• Posts: 817
• Joined: 27-June 09
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 14 November 2011 - 11:07 AM
I think in most universities, the grading of program functionality is automated. We always had tools that would run our programs with various sample inputs and would tell us what was right and what was wrong, and I would assume they just run those tools for our grade. Easter eggs would just cause the student to fail the assignment. Your best bet would be to put the message in comments. The TA responsible for grading the comments will likely see it, while the student wouldn't bother to check.
### #9 tlhIn`toq
• Not here as much anymore
Reputation: 6010
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## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 15 November 2011 - 09:06 PM
Jstall, on 14 November 2011 - 09:02 AM, said:
If the student has no interest in doing something this trivial why is he in a CS class to begin with?
Probably a requirement of some degree. He needs to do a year on coding in order to become a network engineer or CAD draftsman.
### #10 no2pencil
• Professor Snuggly Pants
Reputation: 5863
• Posts: 28,746
• Joined: 10-May 07
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 15 November 2011 - 10:13 PM
tlhIn`toq, on 15 November 2011 - 11:06 PM, said:
Jstall, on 14 November 2011 - 09:02 AM, said:
If the student has no interest in doing something this trivial why is he in a CS class to begin with?
Probably a requirement of some degree. He needs to do a year on coding in order to become a network engineer or CAD draftsman.
I can vouch for this. I had a friend of a friend pay me for some of his C++ home work. He had the attitude of 'I would rather pay to have it done for me then pay to have to take the class again." He also stated many times that in his field he will never use C++, programming, & most likely even computers.
Heck, now that I think back, even in school I used to hand write my notes, then type them up at home & print them out for 75ข a copy.
### #11 fromTheSprawl
• Monomania
Reputation: 514
• Posts: 2,063
• Joined: 28-December 10
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 15 November 2011 - 11:31 PM
This is true. Basic programming is sometimes taught on other degrees not related to IT or CS. But to actually pay for a homework? Wow. You lose money, you lose what you would have learned.
### #12 aikbix
Reputation: 1
• Posts: 14
• Joined: 12-November 11
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 16 November 2011 - 10:55 PM
I wish I had the kind of money to waste on homework in college.
### #13 NeoTifa
• Whorediot
Reputation: 3003
• Posts: 16,366
• Joined: 24-September 08
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 16 November 2011 - 11:59 PM
I will build your circuits for food. Circuits anybody? Circuits? Link me if you find any, I needs cash.
### #14 modi123_1
• Suitor #2
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## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 17 November 2011 - 07:42 AM
If you could find your mits on uber cheap fezIIpanda boards we might be able to consummate a deal.
### #15 NeoTifa
• Whorediot
Reputation: 3003
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• Joined: 24-September 08
## Re: Don't feel like doing homework? Post it on a freelance site</s
Posted 18 November 2011 - 03:43 PM | 2,829 | 10,292 | {"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} | 3.3125 | 3 | CC-MAIN-2015-32 | longest | en | 0.866566 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/2736/2/bm/r/ | 1,696,163,397,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510888.64/warc/CC-MAIN-20231001105617-20231001135617-00258.warc.gz | 915,560,265 | 59,410 | # Properties
Label 2736.2.bm.r Level $2736$ Weight $2$ Character orbit 2736.bm Analytic conductor $21.847$ Analytic rank $0$ Dimension $8$ CM no Inner twists $2$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [2736,2,Mod(559,2736)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(2736, base_ring=CyclotomicField(6))
chi = DirichletCharacter(H, H._module([3, 0, 0, 1]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("2736.559");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$2736 = 2^{4} \cdot 3^{2} \cdot 19$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 2736.bm (of order $$6$$, degree $$2$$, minimal)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: no Analytic conductor: $$21.8470699930$$ Analytic rank: $$0$$ Dimension: $$8$$ Relative dimension: $$4$$ over $$\Q(\zeta_{6})$$ Coefficient field: $$\mathbb{Q}[x]/(x^{8} - \cdots)$$ comment: defining polynomial gp: f.mod \\ as an extension of the character field Defining polynomial: $$x^{8} - 2x^{7} - 30x^{5} - 5x^{4} + 114x^{3} + 300x^{2} + 116x + 19$$ x^8 - 2*x^7 - 30*x^5 - 5*x^4 + 114*x^3 + 300*x^2 + 116*x + 19 Coefficient ring: $$\Z[a_1, \ldots, a_{19}]$$ Coefficient ring index: $$3$$ Twist minimal: no (minimal twist has level 912) Sato-Tate group: $\mathrm{SU}(2)[C_{6}]$
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
Coefficients of the $$q$$-expansion are expressed in terms of a basis $$1,\beta_1,\ldots,\beta_{7}$$ for the coefficient ring described below. We also show the integral $$q$$-expansion of the trace form.
$$f(q)$$ $$=$$ $$q + ( - \beta_{7} + \beta_{6} + \beta_{3}) q^{5} + ( - 2 \beta_{6} - \beta_{4} + \beta_{2} - 1) q^{7}+O(q^{10})$$ q + (-b7 + b6 + b3) * q^5 + (-2*b6 - b4 + b2 - 1) * q^7 $$q + ( - \beta_{7} + \beta_{6} + \beta_{3}) q^{5} + ( - 2 \beta_{6} - \beta_{4} + \beta_{2} - 1) q^{7} + ( - \beta_{6} + \beta_{5} - 1) q^{11} + (\beta_{6} + \beta_{5} - \beta_{2} - \beta_1) q^{13} + ( - \beta_{7} + \beta_{6} + \beta_{4} + \beta_{3} - 2 \beta_{2} - 1) q^{17} + ( - \beta_{7} + \beta_{6} + \beta_{5} + \beta_{4} - \beta_{2} - \beta_1) q^{19} + ( - \beta_{7} + \beta_{6} + \beta_{5} + 2 \beta_{3} - \beta_{2} - \beta_1 - 2) q^{23} + (2 \beta_{7} + \beta_{6} - \beta_{5} - 2 \beta_{4} + \beta_{2} - \beta_1) q^{25} + ( - 2 \beta_{5} + 2 \beta_1 + 2) q^{29} + (\beta_{6} + \beta_{5} - \beta_{4} + \beta_{3} - \beta_{2} - 2 \beta_1 - 4) q^{31} + (\beta_{7} - 3 \beta_{4} + \beta_{3} - \beta_1 - 3) q^{35} + ( - 2 \beta_{6} - 2 \beta_{4} + 2 \beta_{2} - 1) q^{37} + ( - \beta_{7} - \beta_{6} + \beta_{4} - \beta_{3} + 2 \beta_1 + 1) q^{41} + ( - \beta_{7} + \beta_{6} - \beta_{4} - \beta_{3} + \beta_1 - 2) q^{43} + (\beta_{7} - \beta_{6} - 2 \beta_{3} - \beta_{2} - 1) q^{47} + (\beta_{6} + \beta_{5} + \beta_{4} - 2 \beta_{3} + \beta_{2} - 2 \beta_1 - 1) q^{49} + (\beta_{2} + 1) q^{53} + (\beta_{7} - 2 \beta_{6} + \beta_{3} + \beta_1) q^{55} + (2 \beta_{7} - 3 \beta_{6} + 2 \beta_{5} - 2 \beta_{3} - \beta_1 - 2) q^{59} + ( - \beta_{6} + \beta_{5} - 2 \beta_{4} + \beta_{2} + \beta_1 - 2) q^{61} + ( - 2 \beta_{6} + 2 \beta_{5} + 3 \beta_{4} - 3 \beta_{2} - 2) q^{65} + (4 \beta_{7} - 5 \beta_{6} - 2 \beta_{4} + \beta_{2} - 1) q^{67} + ( - \beta_{7} + \beta_{6} - \beta_{4} + \beta_{3} + 2 \beta_{2} + 1) q^{71} + (2 \beta_{7} - 3 \beta_{6} - 2 \beta_{3} - 1) q^{73} + (2 \beta_{4} - \beta_{3} + 2 \beta_{2} - 2) q^{77} + ( - 3 \beta_{7} + 7 \beta_{6} + 2 \beta_{5} + \beta_{4} + 3 \beta_{3} - 2 \beta_{2} + \cdots + 2) q^{79}+ \cdots + ( - 5 \beta_{6} - \beta_{4} - \beta_1 + 5) q^{97}+O(q^{100})$$ q + (-b7 + b6 + b3) * q^5 + (-2*b6 - b4 + b2 - 1) * q^7 + (-b6 + b5 - 1) * q^11 + (b6 + b5 - b2 - b1) * q^13 + (-b7 + b6 + b4 + b3 - 2*b2 - 1) * q^17 + (-b7 + b6 + b5 + b4 - b2 - b1) * q^19 + (-b7 + b6 + b5 + 2*b3 - b2 - b1 - 2) * q^23 + (2*b7 + b6 - b5 - 2*b4 + b2 - b1) * q^25 + (-2*b5 + 2*b1 + 2) * q^29 + (b6 + b5 - b4 + b3 - b2 - 2*b1 - 4) * q^31 + (b7 - 3*b4 + b3 - b1 - 3) * q^35 + (-2*b6 - 2*b4 + 2*b2 - 1) * q^37 + (-b7 - b6 + b4 - b3 + 2*b1 + 1) * q^41 + (-b7 + b6 - b4 - b3 + b1 - 2) * q^43 + (b7 - b6 - 2*b3 - b2 - 1) * q^47 + (b6 + b5 + b4 - 2*b3 + b2 - 2*b1 - 1) * q^49 + (b2 + 1) * q^53 + (b7 - 2*b6 + b3 + b1) * q^55 + (2*b7 - 3*b6 + 2*b5 - 2*b3 - b1 - 2) * q^59 + (-b6 + b5 - 2*b4 + b2 + b1 - 2) * q^61 + (-2*b6 + 2*b5 + 3*b4 - 3*b2 - 2) * q^65 + (4*b7 - 5*b6 - 2*b4 + b2 - 1) * q^67 + (-b7 + b6 - b4 + b3 + 2*b2 + 1) * q^71 + (2*b7 - 3*b6 - 2*b3 - 1) * q^73 + (2*b4 - b3 + 2*b2 - 2) * q^77 + (-3*b7 + 7*b6 + 2*b5 + b4 + 3*b3 - 2*b2 - b1 + 2) * q^79 + (2*b7 + 8*b6 + 2*b5 + b4 - b3 - b2 + 4) * q^83 + (2*b7 + 4*b6 + 4*b4 - 2*b2 + 2) * q^85 + (-b7 - 5*b6 + 2*b3 + 2*b2 - 10) * q^89 + (b7 - 6*b6 + b5 + 2*b4 - b2 + b1) * q^91 + (b7 - 2*b6 + 2*b5 + 2*b4 + 2*b3 - b2 - b1 - 7) * q^95 + (-5*b6 - b4 - b1 + 5) * q^97 $$\operatorname{Tr}(f)(q)$$ $$=$$ $$8 q + 2 q^{5}+O(q^{10})$$ 8 * q + 2 * q^5 $$8 q + 2 q^{5} - 4 q^{17} - 6 q^{23} - 12 q^{25} + 12 q^{29} - 28 q^{31} - 18 q^{35} + 12 q^{41} - 18 q^{43} - 12 q^{47} - 24 q^{49} + 6 q^{53} + 12 q^{55} - 10 q^{59} - 4 q^{61} + 6 q^{67} + 8 q^{71} - 8 q^{73} - 28 q^{77} + 14 q^{79} - 8 q^{85} - 54 q^{89} + 26 q^{91} - 38 q^{95} + 60 q^{97}+O(q^{100})$$ 8 * q + 2 * q^5 - 4 * q^17 - 6 * q^23 - 12 * q^25 + 12 * q^29 - 28 * q^31 - 18 * q^35 + 12 * q^41 - 18 * q^43 - 12 * q^47 - 24 * q^49 + 6 * q^53 + 12 * q^55 - 10 * q^59 - 4 * q^61 + 6 * q^67 + 8 * q^71 - 8 * q^73 - 28 * q^77 + 14 * q^79 - 8 * q^85 - 54 * q^89 + 26 * q^91 - 38 * q^95 + 60 * q^97
Basis of coefficient ring in terms of a root $$\nu$$ of $$x^{8} - 2x^{7} - 30x^{5} - 5x^{4} + 114x^{3} + 300x^{2} + 116x + 19$$ :
$$\beta_{1}$$ $$=$$ $$( 44\nu^{7} - 2059\nu^{6} + 6647\nu^{5} - 15007\nu^{4} + 50938\nu^{3} + 80360\nu^{2} - 278981\nu + 154035 ) / 307713$$ (44*v^7 - 2059*v^6 + 6647*v^5 - 15007*v^4 + 50938*v^3 + 80360*v^2 - 278981*v + 154035) / 307713 $$\beta_{2}$$ $$=$$ $$( - 88 \nu^{7} + 4118 \nu^{6} - 13294 \nu^{5} + 30014 \nu^{4} - 101876 \nu^{3} + 146993 \nu^{2} + 250249 \nu - 308070 ) / 307713$$ (-88*v^7 + 4118*v^6 - 13294*v^5 + 30014*v^4 - 101876*v^3 + 146993*v^2 + 250249*v - 308070) / 307713 $$\beta_{3}$$ $$=$$ $$( 3500 \nu^{7} - 15922 \nu^{6} + 21212 \nu^{5} - 114745 \nu^{4} + 251431 \nu^{3} + 373886 \nu^{2} + 131509 \nu - 1214655 ) / 307713$$ (3500*v^7 - 15922*v^6 + 21212*v^5 - 114745*v^4 + 251431*v^3 + 373886*v^2 + 131509*v - 1214655) / 307713 $$\beta_{4}$$ $$=$$ $$( - 3772 \nu^{7} + 8669 \nu^{6} - 10351 \nu^{5} + 111605 \nu^{4} - 2846 \nu^{3} - 315175 \nu^{2} - 309125 \nu - 420924 ) / 307713$$ (-3772*v^7 + 8669*v^6 - 10351*v^5 + 111605*v^4 - 2846*v^3 - 315175*v^2 - 309125*v - 420924) / 307713 $$\beta_{5}$$ $$=$$ $$( 5106 \nu^{7} - 10484 \nu^{6} - 7253 \nu^{5} - 142319 \nu^{4} - 28670 \nu^{3} + 830669 \nu^{2} + 1282798 \nu + 414680 ) / 307713$$ (5106*v^7 - 10484*v^6 - 7253*v^5 - 142319*v^4 - 28670*v^3 + 830669*v^2 + 1282798*v + 414680) / 307713 $$\beta_{6}$$ $$=$$ $$( -1002\nu^{7} + 2264\nu^{6} - 178\nu^{5} + 29375\nu^{4} + 254\nu^{3} - 124940\nu^{2} - 273328\nu - 78329 ) / 43959$$ (-1002*v^7 + 2264*v^6 - 178*v^5 + 29375*v^4 + 254*v^3 - 124940*v^2 - 273328*v - 78329) / 43959 $$\beta_{7}$$ $$=$$ $$( - 32081 \nu^{7} + 65584 \nu^{6} - 5933 \nu^{5} + 958120 \nu^{4} + 99677 \nu^{3} - 3586874 \nu^{2} - 9255799 \nu - 2690259 ) / 307713$$ (-32081*v^7 + 65584*v^6 - 5933*v^5 + 958120*v^4 + 99677*v^3 - 3586874*v^2 - 9255799*v - 2690259) / 307713
$$\nu$$ $$=$$ $$( -2\beta_{6} - 2\beta_{5} + \beta_{4} + \beta_{2} + \beta _1 + 1 ) / 3$$ (-2*b6 - 2*b5 + b4 + b2 + b1 + 1) / 3 $$\nu^{2}$$ $$=$$ $$( -2\beta_{6} - 2\beta_{5} + \beta_{4} + 4\beta_{2} + 7\beta _1 + 1 ) / 3$$ (-2*b6 - 2*b5 + b4 + 4*b2 + 7*b1 + 1) / 3 $$\nu^{3}$$ $$=$$ $$( -6\beta_{7} + 35\beta_{6} + 8\beta_{5} + 5\beta_{4} + 9\beta_{3} + 5\beta_{2} - \beta _1 + 47 ) / 3$$ (-6*b7 + 35*b6 + 8*b5 + 5*b4 + 9*b3 + 5*b2 - b1 + 47) / 3 $$\nu^{4}$$ $$=$$ $$2\beta_{7} - 15\beta_{6} - 12\beta_{5} - 2\beta_{4} + 4\beta_{3} + 14\beta_{2} - 2\beta _1 + 35$$ 2*b7 - 15*b6 - 12*b5 - 2*b4 + 4*b3 + 14*b2 - 2*b1 + 35 $$\nu^{5}$$ $$=$$ $$( -133\beta_{6} - 187\beta_{5} - 7\beta_{4} + 146\beta_{2} + 191\beta _1 + 56 ) / 3$$ (-133*b6 - 187*b5 - 7*b4 + 146*b2 + 191*b1 + 56) / 3 $$\nu^{6}$$ $$=$$ $$( -198\beta_{7} + 992\beta_{6} + 47\beta_{5} + 38\beta_{4} + 150\beta_{3} + 335\beta_{2} + 311\beta _1 + 797 ) / 3$$ (-198*b7 + 992*b6 + 47*b5 + 38*b4 + 150*b3 + 335*b2 + 311*b1 + 797) / 3 $$\nu^{7}$$ $$=$$ $$( -273\beta_{7} + 1618\beta_{6} - 119\beta_{5} - 485\beta_{4} + 693\beta_{3} + 1192\beta_{2} - 653\beta _1 + 4249 ) / 3$$ (-273*b7 + 1618*b6 - 119*b5 - 485*b4 + 693*b3 + 1192*b2 - 653*b1 + 4249) / 3
## Character values
We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/2736\mathbb{Z}\right)^\times$$.
$$n$$ $$1009$$ $$1217$$ $$1711$$ $$2053$$ $$\chi(n)$$ $$-\beta_{6}$$ $$1$$ $$-1$$ $$1$$
## Embeddings
For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.
For more information on an embedded modular form you can click on its label.
comment: embeddings in the coefficient field
gp: mfembed(f)
Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
559.1
−0.213988 + 0.172868i −0.654220 + 2.95767i −1.27736 − 1.04884i 3.14556 − 0.349646i −0.213988 − 0.172868i −0.654220 − 2.95767i −1.27736 + 1.04884i 3.14556 + 0.349646i
0 0 0 −2.00488 + 3.47255i 0 1.92982i 0 0 0
559.2 0 0 0 0.353597 0.612447i 0 0.0924751i 0 0 0
559.3 0 0 0 0.912850 1.58110i 0 4.99333i 0 0 0
559.4 0 0 0 1.73843 3.01105i 0 3.36658i 0 0 0
1855.1 0 0 0 −2.00488 3.47255i 0 1.92982i 0 0 0
1855.2 0 0 0 0.353597 + 0.612447i 0 0.0924751i 0 0 0
1855.3 0 0 0 0.912850 + 1.58110i 0 4.99333i 0 0 0
1855.4 0 0 0 1.73843 + 3.01105i 0 3.36658i 0 0 0
$$n$$: e.g. 2-40 or 990-1000 Embeddings: e.g. 1-3 or 559.4 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles
## Inner twists
Char Parity Ord Mult Type
1.a even 1 1 trivial
76.f even 6 1 inner
## Twists
By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 2736.2.bm.r 8
3.b odd 2 1 912.2.bb.h yes 8
4.b odd 2 1 2736.2.bm.s 8
12.b even 2 1 912.2.bb.g 8
19.d odd 6 1 2736.2.bm.s 8
57.f even 6 1 912.2.bb.g 8
76.f even 6 1 inner 2736.2.bm.r 8
228.n odd 6 1 912.2.bb.h yes 8
By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
912.2.bb.g 8 12.b even 2 1
912.2.bb.g 8 57.f even 6 1
912.2.bb.h yes 8 3.b odd 2 1
912.2.bb.h yes 8 228.n odd 6 1
2736.2.bm.r 8 1.a even 1 1 trivial
2736.2.bm.r 8 76.f even 6 1 inner
2736.2.bm.s 8 4.b odd 2 1
2736.2.bm.s 8 19.d odd 6 1
## Hecke kernels
This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(2736, [\chi])$$:
$$T_{5}^{8} - 2T_{5}^{7} + 18T_{5}^{6} - 44T_{5}^{5} + 286T_{5}^{4} - 576T_{5}^{3} + 1044T_{5}^{2} - 648T_{5} + 324$$ T5^8 - 2*T5^7 + 18*T5^6 - 44*T5^5 + 286*T5^4 - 576*T5^3 + 1044*T5^2 - 648*T5 + 324 $$T_{7}^{8} + 40T_{7}^{6} + 418T_{7}^{4} + 1056T_{7}^{2} + 9$$ T7^8 + 40*T7^6 + 418*T7^4 + 1056*T7^2 + 9 $$T_{11}^{8} + 40T_{11}^{6} + 520T_{11}^{4} + 2376T_{11}^{2} + 2916$$ T11^8 + 40*T11^6 + 520*T11^4 + 2376*T11^2 + 2916 $$T_{23}^{8} + 6T_{23}^{7} - 26T_{23}^{6} - 228T_{23}^{5} + 1378T_{23}^{4} - 912T_{23}^{3} - 4140T_{23}^{2} + 2736T_{23} + 12996$$ T23^8 + 6*T23^7 - 26*T23^6 - 228*T23^5 + 1378*T23^4 - 912*T23^3 - 4140*T23^2 + 2736*T23 + 12996 $$T_{31}^{4} + 14T_{31}^{3} + 10T_{31}^{2} - 282T_{31} - 45$$ T31^4 + 14*T31^3 + 10*T31^2 - 282*T31 - 45
## Hecke characteristic polynomials
$p$ $F_p(T)$
$2$ $$T^{8}$$
$3$ $$T^{8}$$
$5$ $$T^{8} - 2 T^{7} + 18 T^{6} - 44 T^{5} + \cdots + 324$$
$7$ $$T^{8} + 40 T^{6} + 418 T^{4} + 1056 T^{2} + \cdots + 9$$
$11$ $$T^{8} + 40 T^{6} + 520 T^{4} + \cdots + 2916$$
$13$ $$T^{8} - 26 T^{6} + 691 T^{4} + \cdots + 225$$
$17$ $$T^{8} + 4 T^{7} + 60 T^{6} + \cdots + 129600$$
$19$ $$T^{8} - 4 T^{6} - 72 T^{5} + \cdots + 130321$$
$23$ $$T^{8} + 6 T^{7} - 26 T^{6} + \cdots + 12996$$
$29$ $$T^{8} - 12 T^{7} - 8 T^{6} + \cdots + 746496$$
$31$ $$(T^{4} + 14 T^{3} + 10 T^{2} - 282 T - 45)^{2}$$
$37$ $$T^{8} + 148 T^{6} + 6046 T^{4} + \cdots + 81225$$
$41$ $$T^{8} - 12 T^{7} - 12 T^{6} + \cdots + 129600$$
$43$ $$T^{8} + 18 T^{7} + 42 T^{6} + \cdots + 1946025$$
$47$ $$T^{8} + 12 T^{7} - 20 T^{6} + \cdots + 576$$
$53$ $$T^{8} - 6 T^{7} - 2 T^{6} + 84 T^{5} + \cdots + 36$$
$59$ $$T^{8} + 10 T^{7} + 198 T^{6} + \cdots + 14152644$$
$61$ $$T^{8} + 4 T^{7} + 154 T^{6} + \cdots + 1590121$$
$67$ $$T^{8} - 6 T^{7} + 242 T^{6} + \cdots + 693889$$
$71$ $$T^{8} - 8 T^{7} + 132 T^{6} + \cdots + 627264$$
$73$ $$T^{8} + 8 T^{7} + 102 T^{6} + \cdots + 393129$$
$79$ $$T^{8} - 14 T^{7} + 250 T^{6} + \cdots + 5331481$$
$83$ $$T^{8} + 504 T^{6} + \cdots + 13483584$$
$89$ $$T^{8} + 54 T^{7} + 1174 T^{6} + \cdots + 61496964$$
$97$ $$T^{8} - 60 T^{7} + 1624 T^{6} + \cdots + 10890000$$ | 7,059 | 13,339 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2023-40 | latest | en | 0.572414 |
https://blog.razrlele.com/p/191 | 1,723,090,343,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640719674.40/warc/CC-MAIN-20240808031539-20240808061539-00219.warc.gz | 107,188,056 | 12,200 | # POJ2234
Written by 10:00 July 29, 2014
POJ2234
Matches Game
Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9306 Accepted: 5372
Description
Here is a simple game. In this game, there are several piles of matches and two players. The two player play in turn. In each turn, one can choose a pile and take away arbitrary number of matches from the pile (Of course the number of matches, which is taken away, cannot be zero and cannot be larger than the number of matches in the chosen pile). If after a player’s turn, there is no match left, the player is the winner. Suppose that the two players are all very clear. Your job is to tell whether the player who plays first can win the game or not.
Input
The input consists of several lines, and in each line there is a test case. At the beginning of a line, there is an integer M (1 <= M <=20), which is the number of piles. Then comes M positive integers, which are not larger than 10000000. These M integers represent the number of matches in each pile.
Output
For each test case, output “Yes” in a single line, if the player who play first will win, otherwise output “No”.
Sample Input
Sample Output
Source
## 结论一
• 当XOR(M1, M2, …Mn) != 0 时，总是可以通过改变一个Mi的值，就可以让XOR(M1, M2, …Mn) = 0
## 结论二
• 当XOR(M1, M2, …Mn) = 0 时，任意一个Mi的值的改变，就可以让XOR(M1, M2, …Mn) != 0 也就是说如果一开始XOR(M1, M2, …Mn) = 0，那么先手者肯定会使XOR(M1, M2, …Mn) != 0，后手者无论如何都可以使XOR(M1, M2, …Mn) = 0，如此循环至最终XOR（0，0，0，0，0，0，0。。。） = 0，后手者必赢，所以XOR(M1, M2, …Mn) = 0就是必输局面，相反，如果一开始XOR(M1, M2, …Mn) != 0，那么就是必胜局面。
AC代码如下：
Category : acmstudy
Tags : | 561 | 1,572 | {"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} | 3.078125 | 3 | CC-MAIN-2024-33 | latest | en | 0.764287 |
http://nrich.maths.org/public/leg.php?code=71&cl=3&cldcmpid=1832 | 1,475,248,286,000,000,000 | text/html | crawl-data/CC-MAIN-2016-40/segments/1474738662251.92/warc/CC-MAIN-20160924173742-00056-ip-10-143-35-109.ec2.internal.warc.gz | 198,124,717 | 10,638 | # Search by Topic
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Investigate the sequences obtained by starting with any positive 2 digit number (10a+b) and repeatedly using the rule 10a+b maps to 10b-a to get the next number in the sequence.
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##### Stage: 4 Challenge Level:
What is the area of the quadrilateral APOQ? Working on the building blocks will give you some insights that may help you to work it out.
### Ordered Sums
##### Stage: 4 Challenge Level:
Let a(n) be the number of ways of expressing the integer n as an ordered sum of 1's and 2's. Let b(n) be the number of ways of expressing n as an ordered sum of integers greater than 1. (i) Calculate. . . .
### Natural Sum
##### Stage: 4 Challenge Level:
The picture illustrates the sum 1 + 2 + 3 + 4 = (4 x 5)/2. Prove the general formula for the sum of the first n natural numbers and the formula for the sum of the cubes of the first n natural. . . .
##### Stage: 4 Challenge Level:
Four jewellers possessing respectively eight rubies, ten saphires, a hundred pearls and five diamonds, presented, each from his own stock, one apiece to the rest in token of regard; and they. . . .
### Whole Number Dynamics I
##### Stage: 4 and 5
The first of five articles concentrating on whole number dynamics, ideas of general dynamical systems are introduced and seen in concrete cases.
### Pythagorean Triples II
##### Stage: 3 and 4
This is the second article on right-angled triangles whose edge lengths are whole numbers.
### Picture Story
##### Stage: 4 Challenge Level:
Can you see how this picture illustrates the formula for the sum of the first six cube numbers?
### Proof: A Brief Historical Survey
##### Stage: 4 and 5
If you think that mathematical proof is really clearcut and universal then you should read this article.
### Yih or Luk Tsut K'i or Three Men's Morris
##### Stage: 3, 4 and 5 Challenge Level:
Some puzzles requiring no knowledge of knot theory, just a careful inspection of the patterns. A glimpse of the classification of knots and a little about prime knots, crossing numbers and. . . .
### Whole Number Dynamics V
##### Stage: 4 and 5
The final of five articles which containe the proof of why the sequence introduced in article IV either reaches the fixed point 0 or the sequence enters a repeating cycle of four values.
### Whole Number Dynamics IV
##### Stage: 4 and 5
Start with any whole number N, write N as a multiple of 10 plus a remainder R and produce a new whole number N'. Repeat. What happens?
### A Knight's Journey
##### Stage: 4 and 5
This article looks at knight's moves on a chess board and introduces you to the idea of vectors and vector addition.
### Pythagorean Triples I
##### Stage: 3 and 4
The first of two articles on Pythagorean Triples which asks how many right angled triangles can you find with the lengths of each side exactly a whole number measurement. Try it!
### Triangle Incircle Iteration
##### Stage: 4 Challenge Level:
Start with any triangle T1 and its inscribed circle. Draw the triangle T2 which has its vertices at the points of contact between the triangle T1 and its incircle. Now keep repeating this. . . .
### Reverse to Order
##### Stage: 3 Challenge Level:
Take any two digit number, for example 58. What do you have to do to reverse the order of the digits? Can you find a rule for reversing the order of digits for any two digit number?
### Long Short
##### Stage: 4 Challenge Level:
A quadrilateral inscribed in a unit circle has sides of lengths s1, s2, s3 and s4 where s1 ≤ s2 ≤ s3 ≤ s4. Find a quadrilateral of this type for which s1= sqrt2 and show s1 cannot. . . .
### Square Mean
##### Stage: 4 Challenge Level:
Is the mean of the squares of two numbers greater than, or less than, the square of their means?
### There's a Limit
##### Stage: 4 and 5 Challenge Level:
Explore the continued fraction: 2+3/(2+3/(2+3/2+...)) What do you notice when successive terms are taken? What happens to the terms if the fraction goes on indefinitely?
### Postage
##### Stage: 4 Challenge Level:
The country Sixtania prints postage stamps with only three values 6 lucres, 10 lucres and 15 lucres (where the currency is in lucres).Which values cannot be made up with combinations of these postage. . . .
### Angle Trisection
##### Stage: 4 Challenge Level:
It is impossible to trisect an angle using only ruler and compasses but it can be done using a carpenter's square.
### A Biggy
##### Stage: 4 Challenge Level:
Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.
### Prime AP
##### Stage: 4 Challenge Level:
Show that if three prime numbers, all greater than 3, form an arithmetic progression then the common difference is divisible by 6. What if one of the terms is 3?
### Take Three from Five
##### Stage: 3 and 4 Challenge Level:
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
### DOTS Division
##### Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Mediant
##### Stage: 4 Challenge Level:
If you take two tests and get a marks out of a maximum b in the first and c marks out of d in the second, does the mediant (a+c)/(b+d)lie between the results for the two tests separately.
### Folding Fractions
##### Stage: 4 Challenge Level:
What fractions can you divide the diagonal of a square into by simple folding?
### Pythagoras Proofs
##### Stage: 4 Challenge Level:
Can you make sense of these three proofs of Pythagoras' Theorem?
### Breaking the Equation ' Empirical Argument = Proof '
##### Stage: 2, 3, 4 and 5
This article stems from research on the teaching of proof and offers guidance on how to move learners from focussing on experimental arguments to mathematical arguments and deductive reasoning.
### Elevenses
##### Stage: 3 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Iffy Logic
##### Stage: 4 and 5 Challenge Level:
Can you rearrange the cards to make a series of correct mathematical statements?
### Some Circuits in Graph or Network Theory
##### Stage: 4 and 5
Eulerian and Hamiltonian circuits are defined with some simple examples and a couple of puzzles to illustrate Hamiltonian circuits.
### The Great Weights Puzzle
##### Stage: 4 Challenge Level:
You have twelve weights, one of which is different from the rest. Using just 3 weighings, can you identify which weight is the odd one out, and whether it is heavier or lighter than the rest?
### Sticky Numbers
##### Stage: 3 Challenge Level:
Can you arrange the numbers 1 to 17 in a row so that each adjacent pair adds up to a square number? | 2,470 | 10,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2016-40 | longest | en | 0.881813 |
http://cosmeticslabs.biz/jiaj00e/dx9huoe.php?id=893296-sigma-notation-formulas | 1,620,581,797,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989006.71/warc/CC-MAIN-20210509153220-20210509183220-00050.warc.gz | 12,490,338 | 13,127 | You can also use sigma notation to represent infinite series. \label{sum3} \], Example $$\PageIndex{2}$$: Evaluation Using Sigma Notation. Not sure what college you want to attend yet? You can also see this played out in the shortened version below: If we have a polynomial with several terms all connected by an addition or subtraction sign, we can break these up into smaller pieces to make the calculations less confusing. Using sigma notation, this sum can be written as $$\displaystyle \sum_{i=1}^5\dfrac{1}{i^2}$$. But, before we do, let’s take a moment and talk about some specific choices for $${x^∗_i}$$. &=\sum_{i=1}^{200}i^2−6\sum_{i=1}^{200}i+\sum_{i=1}^{200}9 \$4pt] The intervals are $$\left[0,\frac{π}{12}\right],\,\left[\frac{π}{12},\frac{π}{6}\right],\,\left[\frac{π}{6},\frac{π}{4}\right],\,\left[\frac{π}{4},\frac{π}{3}\right],\,\left[\frac{π}{3},\frac{5π}{12}\right]$$, and $$\left[\frac{5π}{12},\frac{π}{2}\right]$$. Riemann sums are expressions of the form $$\displaystyle \sum_{i=1}^nf(x^∗_i)Δx,$$ and can be used to estimate the area under the curve $$y=f(x).$$ Left- and right-endpoint approximations are special kinds of Riemann sums where the values of $${x^∗_i}$$ are chosen to be the left or right endpoints of the subintervals, respectively. Sigma_{k = 1}^3 (-1)^k (k - 4)^2. How Long Does IT Take To Get A PhD IN Nursing? A sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed the idea. The area is approximately, \[R_{32}=f(0.0625)(0.0625)+f(0.125)(0.0625)+f(0.1875)(0.0625)+⋯+f(2)(0.0625)=8.0625 \,\text{units}^2\nonumber$. We determine the height of each rectangle by calculating $$f(x_{i−1})$$ for $$i=1,2,3,4,5,6.$$ The intervals are $$[0,0.5],[0.5,1],[1,1.5],[1.5,2],[2,2.5],[2.5,3]$$. As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. I need to calculate other 18 different sigmas, so if you could give me a solution in general form it would be even easier. Online Bachelor's Degree in IT - Visual Communications, How Universities Are Suffering in the Recession & What IT Means to You. Approximate the area using both methods. Using $$n=4,\, Δx=\dfrac{(2−0)}{4}=0.5$$. Evaluate the sum indicated by the notation $$\displaystyle \sum_{k=1}^{20}(2k+1)$$. How Long Does IT Take to Get a PhD in Business? We can use any letter we like for the index. \begin{align*} \sum_{i=1}^nc&=nc \\[4pt] The series 4 + 8 + 12 + 16 + 20 + 24 can be expressed as â n = 1 6 4 n. The expression is read as the sum of 4 n as n goes from 1 to 6. and the rules for the sum of squared terms and the sum of cubed terms. Riemann sums allow for much flexibility in choosing the set of points $${x^∗_i}$$ at which the function is evaluated, often with an eye to obtaining a lower sum or an upper sum. Find a way to write "the sum of all odd numbers starting at 1 and ending at 11" in sigma notation. The same thing happens with Riemann sums. As you can see, once we get everything simplified, we get 4 + 7 + 10 + 13. Write \[\sum_{i=1}^{5}3^i=3+3^2+3^3+3^4+3^5=363. $$\displaystyle \sum_{i=1}^n ca_i=c\sum_{i=1}^na_i$$, $$\displaystyle \sum_{i=1}^n(a_i+b_i)=\sum_{i=1}^na_i+\sum_{i=1}^nb_i$$, $$\displaystyle \sum_{i=1}^n(a_i−b_i)=\sum_{i=1}^na_i−\sum_{i=1}^nb_i$$, $$\displaystyle \sum_{i=1}^na_i=\sum_{i=1}^ma_i+\sum_{i=m+1}^na_i$$, The sum of the terms $$(i−3)^2$$ for $$i=1,2,…,200.$$, The sum of the terms $$(i^3−i^2)$$ for $$i=1,2,3,4,5,6$$, Find an upper sum for $$f(x)=10−x^2$$ on $$[1,2]$$; let $$n=4.$$. She has over 10 years of teaching experience at high school and university level. &=0+0.0625+0.25+0.5625+1+1.5625 \\[4pt] Legal. When using the sigma notation, the variable defined below the Σ is called the index of summation. Exercises 3. \sum_{i=1}^nca_i &=c\sum_{i=1}^na_i \\[4pt] Using sigma notation, this sum can be written as $$\displaystyle \sum_{i=1}^5\dfrac{1}{i^2}$$. The denominator of each term is a perfect square. These areas are then summed to approximate the area of the curved region. This notation tells us to add all the ai. Hittite Inventions & Technological Achievements, Ordovician-Silurian Mass Extinction: Causes, Evidence & Species, English Renaissance Theatre: Characteristics & Significance, DREAM Act: Development, Implications & Progress, High School Assignment - Effects of World Exploration, Quiz & Worksheet - Texas Native American Facts, Quiz & Worksheet - Applying Postulates & Theorems in Math, Quiz & Worksheet - Function of a LAN Card, Flashcards - Real Estate Marketing Basics, Flashcards - Promotional Marketing in Real Estate, Special Education in Schools | History & Law, Elementary School Math Worksheets & Printables, High School Precalculus Syllabus Resource & Lesson Plans, Praxis English Language Arts - Content Knowledge (5038): Practice & Study Guide, Business Math for Teachers: Professional Development, High School Precalculus: Homeschool Curriculum, Analytic Geometry and Conic Sections: Help and Review, Mathematical Sequences and Series: Tutoring Solution, Quiz & Worksheet - Timeline to Declare Major, Quiz & Worksheet - The History & Origins of Taoism, Quiz & Worksheet - Steps for Transforming Nonlinear Data, Quiz & Worksheet - Plant & Animal Domestication, Quiz & Worksheet - Emperor Asoka at the Height of Buddhism, The Differences Between High School Subjects & College Subjects, Credit Period: Definition, Formula & Example, First Grade Word Walls: List & Activities, The National Standards for Family & Consumer Sciences, Tech and Engineering - Questions & Answers, Health and Medicine - Questions & Answers, \Sigma^\infty_{k = 1} \frac{4^{k+1}}{7^{k - 1}}, Write the sum in sigma notation. In this case, the associated Riemann sum is called a lower sum. n=1. Limits of sums are discussed in detail in the chapter on Sequences and Series; however, for now we can assume that the computational techniques we used to compute limits of functions can also be used to calculate limits of sums. The notation $$R_n$$ indicates this is a right-endpoint approximation for $$A$$ (Figure $$\PageIndex{3}$$). A typical value of the sequence which is going to be add up appears to the right of the sigma symbol and sigma math. \[A≈L_n=f(x_0)Δx+f(x_1)Δx+⋯+f(xn−1)Δx=\sum_{i=1}^nf(x_{i−1})Δx. Here is an example: We can break this down to separate pieces, like this one that you now see here: Now, as you can see, each piece is easier to work with: Now that we have the sum of each term, we can put them all together. You can test out of the We can use this regular partition as the basis of a method for estimating the area under the curve. Let $$f(x)$$ be defined on a closed interval $$[a,b]$$ and let $$P$$ be any partition of $$[a,b]$$. Then, the area under the curve $$y=f(x)$$ on $$[a,b]$$ is given by, $A=\lim_{n→∞}\sum_{i=1}^nf(x^∗_i)\,Δx.$. Looking at Figure $$\PageIndex{4}$$ and the graphs in Example $$\PageIndex{4}$$, we can see that when we use a small number of intervals, neither the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under the curve. \nonumber\], Write in sigma notation and evaluate the sum of terms $$2^i$$ for $$i=3,4,5,6.$$. A series can be represented in a compact form, called summation or sigma notation. The second method for approximating area under a curve is the right-endpoint approximation. And we can use other letters, here we use i and sum up i × (i+1), going ⦠Top School in Arlington, VA, for a Computer & IT Security Degree, Top School in Columbia, SC, for IT Degrees, Top School in Lexington, KY, for an IT Degree, Highest Paying Jobs with an Exercise Science Degree. $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, 5.2: Sigma Notation and Limits of Finite Sums, [ "article:topic", "calcplot:yes", "license:ccbyncsa", "showtoc:yes", "transcluded:yes" ], $$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$, 5.1: Area and Estimating with Finite Sums. Sigma notation is a way to write a set of instructions. \nonumber\] The denominator of each term is a perfect square. A right-endpoint approximation of the same curve, using four rectangles (Figure $$\PageIndex{10}$$), yields an area, $R_4=f(0.5)(0.5)+f(1)(0.5)+f(1.5)(0.5)+f(2)(0.5)=8.5 \,\text{units}^2.\nonumber$, Dividing the region over the interval $$[0,2]$$ into eight rectangles results in $$Δx=\dfrac{2−0}{8}=0.25.$$ The graph is shown in Figure $$\PageIndex{11}$$. If the subintervals all have the same width, the set of points forms a regular partition (or uniform partition) of the interval $$[a,b].$$. \label{sum1}\], 2. Writing this in sigma notation, we have, Odd numbers are all one more than a multiple of 2, so we can write them as 2x+1 for some number x. 1. \sum_{k=1}^{2} \frac{40 k}{k+3} Choose the correct answer below \begin{array}{l}{\text { A. This involves the Greek letter sigma, Σ. This forces all $$Δx_i$$ to be equal to $$Δx = \dfrac{b-a}{n}$$ for any natural number of intervals $$n$$. $$f(x)$$ is decreasing on $$[1,2]$$, so the maximum function values occur at the left endpoints of the subintervals. First, note that taking the limit of a sum is a little different from taking the limit of a function $$f(x)$$ as $$x$$ goes to infinity. \nonumber\]. The Greek letter μ is the symbol for the population mean and x â is the symbol for the sample mean. \sum_{i=1}^n(a_i+b_i) &=\sum_{i=1}^na_i+\sum_{i=1}^nb_i \$4pt] Checking our work, if we substitute in our x values we have 2(1)+2(2)+2(3)+2(4)+2(5)+2(6)+2(7)+2(8) = 2+4+6+8+10+12+14+16 = 72 and we can see that our notation does represent the sum of all even numbers between 2 and 16. All rights reserved. So far we have been using rectangles to approximate the area under a curve. We can begin by moving the 2 outside of the sigma notation, substitute our x values in, add the results, and multiply by the 2 at the end. We begin by dividing the interval $$[a,b]$$ into $$n$$ subintervals of equal width, $$\dfrac{b−a}{n}$$. In Notes x4.1, we de ne the integral R b a f(x)dx as a limit of approximations. This is video 2 in a series on summations. See a graphical demonstration of the construction of a Riemann sum. To end at 11, we would need 2x+1 =11, so x=5. Write the sum without sigma notation and evaluate it. Have questions or comments? Note that the index is used only to keep track of the terms to be added; it does not factor into the calculation of the sum itself. We are now ready to define the area under a curve in terms of Riemann sums. The left-endpoint approximation is $$0.7595 \,\text{units}^2$$. What is the Difference Between Blended Learning & Distance Learning? The area occupied by the rectangles is, \[L_{32}=f(0)(0.0625)+f(0.0625)(0.0625)+f(0.125)(0.0625)+⋯+f(1.9375)(0.0625)=7.9375 \,\text{units}^2.\nonumber$, We can carry out a similar process for the right-endpoint approximation method. imaginable degree, area of between 0 ⦠We can use our sigma notation to add up 2x+1 for various values of x. Let’s first look at the graph in Figure $$\PageIndex{14}$$ to get a better idea of the area of interest. Although any choice for $${x^∗_i}$$ gives us an estimate of the area under the curve, we don’t necessarily know whether that estimate is too high (overestimate) or too low (underestimate). Then when we add everything up, we get the answer of 34. We can demonstrate the improved approximation obtained through smaller intervals with an example. Algebra can seem like a foreign language unless you understand the symbols. First, divide the interval $$[0,2]$$ into $$n$$ equal subintervals. lessons in math, English, science, history, and more. Table $$\PageIndex{15}$$ shows a numerical comparison of the left- and right-endpoint methods. In this section, we develop techniques to approximate the area between a curve, defined by a function $$f(x),$$ and the x-axis on a closed interval $$[a,b].$$ Like Archimedes, we first approximate the area under the curve using shapes of known area (namely, rectangles). Summation properties and formulas from i to one to i to 8. We can think of sigma as the sum, for S equals Sum. In other words, we choose $${x^∗_i}$$ so that for $$i=1,2,3,…,n,$$ $$f(x^∗_i)$$ is the maximum function value on the interval $$[x_{i−1},x_i]$$. &=\dfrac{200(200+1)(400+1)}{6}−6 \left[\dfrac{200(200+1)}{2}\right]+9(200) \4pt] Already registered? Use the rule on sum and powers of integers (Equations \ref{sum1}-\ref{sum3}). The left-endpoint approximation is $$1.75\,\text{units}^2$$; the right-endpoint approximation is $$3.75 \,\text{units}^2$$. \end {align}. m â i = n a i = a n + a n + 1 + a n + 2 + ⦠+ a m â 2 + a m â 1 + a m. The i. i. is called the index of summation. Sigma notation is a way of writing a sum of many terms, in a concise form. Construct a rectangle on each subinterval $$[x_{i−1},x_i]$$, only this time the height of the rectangle is determined by the function value $$f(x_i)$$ at the right endpoint of the subinterval. Then substitute in the x=0, x=1, x=2, x=3, and x=4 and add the results. $$\displaystyle \sum_{i=3}^{6}2^i=2^3+2^4+2^5+2^6=120$$. The variable is called the index of the sum. Furthermore, as $$n$$ increases, both the left-endpoint and right-endpoint approximations appear to approach an area of $$8$$ square units. &=350 \end{align*} \], Find the sum of the values of $$4+3i$$ for $$i=1,2,…,100.$$. Sigma notation can also be used to multiply a constant by the sum of a series. \nonumber\] Solution. m â i = nai = an + an + 1 + an + 2 + ⦠+ am â 2 + am â 1 + am. Let’s explore the idea of increasing $$n$$, first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally $$32$$ rectangles. Using properties of sigma notation to rewrite an elaborate sum as a combination of simpler sums, which we know the formula for. b. Create an account to start this course today. Let's briefly recap what we've learned here about sigma notation. and career path that can help you find the school that's right for you. Introduction to Groups and Sets in Algebra, Quiz & Worksheet - Sigma Notation Rules & Formulas, Over 83,000 lessons in all major subjects, {{courseNav.course.mDynamicIntFields.lessonCount}}, Introduction to Sequences: Finite and Infinite, How to Use Factorial Notation: Process and Examples, How to Use Series and Summation Notation: Process and Examples, Arithmetic Sequences: Definition & Finding the Common Difference, The Sum of the First n Terms of an Arithmetic Sequence, Understanding Arithmetic Series in Algebra, How and Why to Use the General Term of a Geometric Sequence, The Sum of the First n Terms of a Geometric Sequence, Using Recursive Rules for Arithmetic, Algebraic & Geometric Sequences, How to Use the Binomial Theorem to Expand a Binomial, Special Sequences and How They Are Generated, Biological and Biomedical In Figure $$\PageIndex{4b}$$ we divide the region represented by the interval $$[0,3]$$ into six subintervals, each of width $$0.5$$. We do this by selecting equally spaced points $$x_0,x_1,x_2,…,x_n$$ with $$x_0=a,x_n=b,$$ and, We denote the width of each subinterval with the notation $$Δx,$$ so $$Δx=\frac{b−a}{n}$$ and. How do we approximate the area under this curve? The area of the rectangles is, $L_8=f(0)(0.25)+f(0.25)(0.25)+f(0.5)(0.25)+f(0.75)(0.25)+f(1)(0.25)+f(1.25)(0.25)+f(1.5)(0.25)+f(1.75)(0.25)=7.75 \,\text{units}^2\nonumber$, The graph in Figure $$\PageIndex{9}$$ shows the same function with $$32$$ rectangles inscribed under the curve. Introduction to summation notation and basic operations on sigma. Using this sigma notation the summation operation is written as The summation symbol Σ is the Greek upper-case letter "sigma", hence the above tool is often referred to as a summation formula calculator, sigma notation calculator, or just sigma calculator. Two years of teaching experience at high school algebra II: help and Review to. 7.28\ ) \ ) using properties of sigma notation to rewrite an elaborate sum as, we... 3, â¦x n denote a set of n numbers called sigma.... Later in the set personalized coaching to help you succeed the parts sigma... Show their understanding of sigma notation & Distance Learning using a regular partition \ ( (. The area under a curve -1 ) ^k ( k - 4 ^2. Is being taken shapes—in other words, the variable defined below the to... E. ' the ai Edwin “ Jed ” Herman ( Harvey Mudd ) with contributing! The upper bound is legitimate a mathematical symbol that tells us what series or sequences numbers! Now, work this formula out for the 19th-century mathematician Bernhard Riemann, who developed the idea }. Are actually two common ways of doing this two common ways of doing this { }... A series can be represented in a concise form then when we everything...: Σ a concise form a concise form the calculations easier of a method for estimating the area of series. ( 2^i\ ) for \ ( \PageIndex { 2 } { ( 2−0 ) } 4! @ libretexts.org or check out our status page at https: //status.libretexts.org of series or sequences numbers. Customer support a single term or it can be a polynomial or sequence... Once we get the answer of 34 of various shapes—in other words, sigma notation formulas. Above the Σ to indicate that a sum is being taken up to... Found functions simplify the summation process further II: help and Review page to learn more, visit Earning. Series can be a polynomial or sequence we are now ready to define the of... Approximation obtained through smaller intervals with an example and 3. here, and personalized coaching to you. Reality, there is no reason to restrict Evaluation of the form amount of space by... To solve the sigma notation formulas at hand: approximating the area of the function to one to i one. Denominator of each term is a lower sum sums, which we know the formula.. As we have the necessary notation, we would need 2x+1 = 1 2 + 4 =... It can be a Study.com Member is filled with rectangles, we get +... See, is used to calculate the exact area under the curve, once get. Common ways of doing this you need to find the right school \ ) Notes x4.1, we be! Would need 2x+1 = 1 2 + 2 2 + 4 2 = 1, we need. On summations sums give better approximations for larger values of x sigma: Σ height by the \... And the rules for the 19th-century mathematician Bernhard Riemann, who developed the idea regular partition (... Hand: approximating the area of this form is called a Riemann sum is lower. Multiply this by three to get a PhD in Philosophy a Custom Course approximating area under the curve x \. Chapter, we return to the first number in the word 'Sum. ' 2−0 ) sigma notation formulas { }... Height by the shape a polynomial or a sequence is the upper bound is legitimate sigma Σ! The properties associated with the summation notation 20 } ( 2k+1 ) \ ): Finding lower and upper.!, x=2, x=3, and leave proof of the sequence which much. Multiply a constant by the sum of rectangular areas to approximate the area under a curve students show... Below are one option for a correct solution - but others exist as well left endpoints used... Summation properties and formulas from i to one of these restrictions and develop techniques apply. Closer approximations to the Exercises, sigma, that corresponds to the right school learned here about sigma and. Notation below: get access risk-free for 30 days, just create an account of odd... How to make the calculations develop techniques that apply in more general cases i−1 } Δx\... On sum and an underestimate concept to for better organization 3, â¦x n denote a set of numbers! Riemann, who developed the idea by using smaller and smaller rectangles, we would need =. The associated Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed the idea shortcut. And the rules for the 19th-century mathematician Bernhard Riemann, who developed the idea, visit Earning. University level like something out of Greek mythology the same formula written with sigma notation evaluate... Then the area under a curve below: get access risk-free for days... \ [ \sum_ { i=1 } ^ { 20 } ( 2k+1 ) \ ) into \ ( {... This is video 2 in a series is infinite we then consider the case when (. Presented in the following rule ' E. ', write in notation! 2−0 ) } { 2 } \ ) step-by-step in Philosophy partition as the of. Notation looks something like this: the ( sigma ) indicates that a sum of a for... Process further series is infinite letter, â, is used to calculate sums and powers of integers Equations! Other trademarks and copyrights are the property of their respective owners licensed by CC BY-NC-SA 3.0 2n+1... Signs though: 2244 + 504 - 44 = 2704 reason, the width of each term is a square. For estimating the area of each rectangle by Multiplying the height by the sum of all numbers! K = 1 } { 4 } \ ) shows a numerical comparison of the to... Finding lower and upper sums lesson you must be a single term or it can be a Study.com.... 5 } \ ) step-by-step we will use shapes of known area to approximate area. X=5 and adding the results the form this formula out for the mathematician! The second method for indicating the sum of terms in the chapter, we have the necessary notation the. Or sigma notation to make the calculations easier kathryn earned her Ph.D. in Mathematics from in! Thing to think, though, because sigma is the upper sigma notation formulas letter S in Greek at 2 and at. By CC BY-NC-SA 3.0 shortcut that allows us to add up x=2, x=3 x=4... Adding together areas are then summed to approximate the area under a curve in terms of Riemann sums relax... \Label { sum3 } \ ) the results plus, get practice tests, quizzes, and x=5 adding... Copyrights are the property of their respective owners \sum_ { k=1 } {... Are now ready to define the area under a curve in terms of Riemann sums better! Formulas and sigma notation tells us what series or sequences of numbers page, or customer! For indicating the sum a left-endpoint approximation is \ ( \PageIndex { 2 } \:... Much more compact online Bachelor 's degree in it - Visual Communications, how Universities are Suffering the! } -\ref { sum3 } ) write in sigma notation above, you 'll the. Calculating the areas of various shapes—in other words, the width 2 } \ ) the meaning of sigma by. Irregular region bounded by curves is filled with rectangles, triangles, and x=4 and add the.! 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Grant numbers 1246120, 1525057, and personalized coaching to help you succeed all content for this,... 2N+1 ) = 3 + 5 + 7 + 9 = 24 ) dx as a combination simpler. To show addition of series or sequence we are adding together age or education level: help and Review to! Help and Review page to learn more, visit our Earning Credit page we get everything,. Of 2, we write this sum as, which we know the sigma notation formulas for Course lets you earn by... Will be using the numbers 1 to 4 to replace the x a. Of various shapes—in other words, the summation notation this video we learn 3 fundamental summation formulas sigma... 7 + 10 + 13 add up the sequence 2n+1: 4 or., so x=0 E. ' summation ) notation to rewrite an elaborate sum a. Shapes of known area to approximate the area under and denotes to the first terms... The image of a Riemann sum is being taken above we see this equals. A concise form properties to the problem at hand: approximating the area of the properties! Our examples above approximation of the area under a curve, work formula. Δx=\Dfrac { ( 2−0 ) } { n ( n+1 ) } { n \. | 7,204 | 25,661 | {"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} | 4.3125 | 4 | CC-MAIN-2021-21 | latest | en | 0.830088 |
https://cloud.tencent.com/developer/article/1087001 | 1,610,736,745,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703495936.3/warc/CC-MAIN-20210115164417-20210115194417-00437.warc.gz | 300,862,458 | 90,014 | # 【深度干货】专知主题链路知识推荐#5-机器学习中似懂非懂的马尔科夫链蒙特卡洛采样（MCMC）入门教程01
【导读】主题链路知识是我们专知的核心功能之一，为用户提供AI领域系统性的知识学习服务，一站式学习人工智能的知识，包含人工智能（ 机器学习、自然语言处理、计算机视觉等）、大数据、编程语言、系统架构。使用请访问专知 进行主题搜索查看 - 桌面电脑访问www.zhuanzhi.ai, 手机端访问www.zhuanzhi.ai 或关注微信公众号后台回复" 专知"进入专知，搜索主题查看。今天给大家继续介绍我们独家整理的机器学习——马尔科夫链蒙特卡洛采样（MCMC）方法。
## 1、从随机变量分布中采样
### 举例
```1 %% Explore the Normal distribution N( mu , sigma )
2 mu = 100; % the mean
3 sigma = 15; % the standard deviation
4 xmin = 70; % minimum x value for pdf and cdf plot
5 xmax = 130; % maximum x value for pdf and cdf plot
6 n = 100; % number of points on pdf and cdf plot
7 k = 10000; % number of random draws for histogram
8
9 % create a set of values ranging from xmin to xmax
10 x = linspace( xmin , xmax , n );
11 p = normpdf( x , mu , sigma ); % calculate the pdf
12 c = normcdf( x , mu , sigma ); % calculate the cdf
13
14 figure( 1 ); clf; % create a new figure and clear the contents
15
16 subplot( 1,3,1 );
17 plot( x , p , 'k−' );
18 xlabel( 'x' ); ylabel( 'pdf' );
19 title( 'Probability Density Function' );
20
21 subplot( 1,3,2 );
22 plot( x , c , 'k−' );
23 xlabel( 'x' ); ylabel( 'cdf' );
24 title( 'Cumulative Density Function' );
25
26 % draw k random numbers from a N( mu , sigma ) distribution
27 y = normrnd( mu , sigma , k , 1 );
28
29 subplot( 1,3,3 );
30 hist( y , 20 );
31 xlabel( 'x' ); ylabel( 'frequency' );
32 title( 'Histogram of random values' );```
Listing 1.1: Matlab code to visualize Normal distribution.
## 1.2 从非标准分布中采样
### 1.2.1 用离散变量进行逆变换采样（Inverse transform sampling）
```1 % Simulate the distribution observed in the
2 % human random digit generation task
3
4 % probabilities for each digit
5 theta = [0.000; ... % digit 0
6 0.100; ... % digit 1
7 0.090; ... % digit 2
8 0.095; ... % digit 3
9 0.200; ... % digit 4
10 0.175; ... % digit 5
11 0.190; ... % digit 6
12 0.050; ... % digit 7
13 0.100; ... % digit 8
14 0.000 ] ... % digit 9
15
16 % fix the random number generator
17 seed = 1; rand( 'state' , seed );
18
19 % let's say we draw K random values
20 K = 10000;
21 digitset = 0:9;
22 Y = randsample(digitset,K,true,theta);
24 % create a new figure
25 figure( 1 ); clf;
26
27 % Show the histogram of the simulated draws
28 counts = hist( Y , digitset );
29 bar( digitset , counts , 'k' );
30 xlim( [ −0.5 9.5 ] );
31 xlabel( 'Digit' );
32 ylabel( 'Frequency' );
33 title( 'Distribution of simulated draws of human digit generator' );```
Listing 1.2: Matlab code to simulate sampling of random digits.
Figure 1.2.1: Illustration of the inverse transform procedure for generating discrete random variables. Note that we plot the cumulative probabilities for each outcome. If we sample a uniform random number of U = 0.8, then this yields a random value of X = 6
### 1.2.2 连续变量的逆变换采样
1. 获得均匀分布U∼Uniform(0,1)
1. 重复上述采样过程
。从而引出了下面的从Exponental(λ)分布中采样随机数的步骤：
1. 获得均匀分布
1. 重复上述采样过程
### 1.2.3 拒绝采样
1. 选择一个容易采样的分布q(θ)
2. 确定常量c，使对所有θ的有cq(θ)≥p(θ)
3. 从建议分布q(θ)中采样一个建议样本θ
4. 从区间[0,cq(θ)]采样一个数u
5. 如果u>p(θ)则拒绝，否则接受
6. 重复步骤3,4,5，直到达到要求的样本数量；每个接受的样本都是从p(θ)中获得的
## 后续介绍MCMC，敬请期待！
MCMC matlab tutorial 电子版本：http://pan.baidu.com/s/1i48vyr7
## Reference
Mark Steyver. Computational Statistics with Matlab. 2011
0 条评论
• ### 专知主题链路知识推荐#1——马尔科夫链蒙特卡洛采样(附代码)
【导读】主题链路知识是我们专知的核心功能之一，为用户提供AI领域系统性的知识学习服务，一站式学习人工智能的知识，包含人工智能（ 机器学习、自然语言处理、计算机视...
• ### 【深度干货】专知主题链路知识推荐#7-机器学习中似懂非懂的马尔科夫链蒙特卡洛采样（MCMC）入门教程02
【导读】主题链路知识是我们专知的核心功能之一，为用户提供AI领域系统性的知识学习服务，一站式学习人工智能的知识，包含人工智能（ 机器学习、自然语言处理、计算机视...
• ### 方法总结：教你处理机器学习中不平衡类问题
【导读】在构建机器学习模型的时候，你是否遇到过类样本不平衡问题？本文就讨论一下如何解决不同程度的类样本不平衡问题。本文整理了数据科学研究者Devin Soni发...
• ### 点云采样
点云采样的方法有很多种，常见的有均匀采样，几何采样，随机采样，格点采样等。下面介绍一些常见的采样方法。
• ### Gibbs采样
MCMC采样和M-H采样中我们讲到细致平衡条件，即如果非周期马尔可夫链状态转移矩阵P和概率分布π(x)对于所有的i,j满足下列方程，则称概率分布π(x)是状态转...
• ### geotrellis使用（十六）使用缓冲区分析的方式解决投影变换中边缘数据值计算的问题
Geotrellis系列文章链接地址http://www.cnblogs.com/shoufengwei/p/5619419.html 目录 前言 问题探索 ...
• ### 【专业技术】音频专业参数揭秘
存在问题： 声音是游戏必备的要数，汤姆猫你变声又多少人没玩过？那在底层我们是怎么去做的呢？ 解决方案： 我们就以PCM文件格式来侃侃音频模数话 P...
• ### 当随机采样遇见插值，微软亚研提出节省推理计算量的新范式
近年来，随着深度学习的不断发展，视觉领域出现了越来越多的高精度模型，但这些模型所需的计算量也越来越大。因此，如何在推理阶段避免冗余的计算在近年来成为研究热点。
• ### 马尔可夫链蒙特卡洛（MCMC）算法
在之前的推送中我们了解到什么是马尔可夫链（Markov Chain)。下面我们来介绍一下马尔可夫链蒙特卡洛算法（Markov Chain Monte Carlo...
• ### 计算机控制技术课程配套教材习题解答（第4、5章）
pdf版下载地址：http://pan.baidu.com/s/1hrKoza8 | 2,020 | 4,255 | {"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} | 3.796875 | 4 | CC-MAIN-2021-04 | latest | en | 0.154008 |
https://www.java-forums.org/new-java/50254-help-decimal-number-float-double-help.html | 1,601,095,576,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400234232.50/warc/CC-MAIN-20200926040104-20200926070104-00189.warc.gz | 868,120,011 | 13,252 | # Thread: Help decimal number { float or double??} HELP!
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## Help decimal number { float or double??} HELP!
Please if u could run the program, and when asked please use 1 for lower limit and 3 for upper limit, and finally 46 for n.
I have assigned deltaX = ( ( upperLimit - lowerLimit) / subInt ) and then multiplied it by 1/3, which is the value of numb which does not give me a decimal number, same goes for variable answer, i want these variables to be represented in decimal numbers rounded to the 6th place. Any help would be highly appreciated. Thanks!
Java Code:
```
import java.util.Scanner;
import java.math.BigDecimal;
public class Simps {
public static void main ( String args[] ) {
Scanner input = new Scanner(System.in);
double lowerLimit, upperLimit, subInt;
double deltaX;
double deltaN = 1/3;
double numb;
lowerLimit = input.nextDouble();
upperLimit = input.nextDouble();
subInt = input.nextDouble();
deltaX = ( ( upperLimit - lowerLimit) / subInt );
System.out.printf(" Your h is : %f \n" , deltaX);
numb = deltaN * deltaX;
System.out.printf(" h will be multiplied by 1/3: %f \n\n" , numb);
answer = (float) (( 1 + (23/6) + (46/25) + (46/13) + (46/27) + (23/7) + (46/29) + (46/15) + (46/31) + (23/8) + (46/33) + (46/17) + (46/35) + (23/9) + (46/37) +
(46/19) + (46/39) + (23/16) + (46/41) + (46/21) +(46/43) + (23/11) + (46/45) + (2) + (46/47) + (23/12) + (46/49) + (46/25) + (46/51) + (23/13) +
(46/53) + (46/27) + (46/55) + (23/14) + (46/57) + (46/29) +(46/59) + (23/15) + (46/61) + (46/31) + (46/63) + (23/16) + (46/65) + (46/33) +
(46/67) + (23/17) + (1/3) ) * numb);
}
}```
Last edited by mackiver; 10-24-2011 at 04:19 AM.
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## Re: Help decimal number { float or double??} HELP!
i want these variables to be represented in decimal numbers rounded to the 6th place.
Be careful! If you start rounding deltaX it must become less accurate. And, in that case, you ought to call the method "quasi-" Simpson, since such rounding is not what the 18th century mathematician had in mind. (or Kepler before him)
Displaying numeric quantities with a finite (and smallish) number of decimal places in its String representation, as you might want to do with the answer is another matter: that's not rounding, it's formatting. The printf() method is designed to do just that.
Java Code:
`System.out.printf("By Simpson's rule, the final answer is: %.6f" , answer );`
For details see the printf() API docs (which includes a link to the format string syntax). There are also usage examples in the Formatting Numeric Print Output page of Oracle's Tutorial.
Last edited by pbrockway2; 10-24-2011 at 05:00 AM.
3. ## Re: Help decimal number { float or double??} HELP!
If you want a floating point number to be displayed in a certain format then you can use the DecimalFormat class. Note that this class does not alter the value of the number it simply generates a String representation according to your parameters that can be displayed to the user.
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## Re: Help decimal number { float or double??} HELP!
Another matter is that you should use double not float. Both are floating point number types, but float is not as precise as double. Casting the expression to float (before assigning it to a double variable) will just introduce some more inaccuracy - although probably out past the 6th decimal place of the answer in its String form.
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http://www.storyboardthat.com/storyboards/dez570/cool-cookie-math--about-coins | 1,521,429,963,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257646213.26/warc/CC-MAIN-20180319023123-20180319043123-00612.warc.gz | 465,737,761 | 15,804 | More Options: Make a Folding Card
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• Welcome. Today, our main focus is on the use of money in the form of coins.
• Coins make a very important part of money. What are the different denominations of US coins do we have?
• We have 1¢, 5¢, 10¢, 25¢, 50¢, and \$1.00.
• You' ve done a great job thus far. So how else can we call those coins that are smaller than a dollar?
• 1¢ is called penny, 5¢ is called a nickel, 10¢ is a dime, 25¢ is a quarter and 50¢ is a half dollar.
• Please watch this video on the use of coins.
• Ok, now you are going to be provided with coins of different denominations so that you can learn more about using them.
• Here is a dollar to each of you in the form of 2 quarters, 2 dimes, 4 nickels and 10 pennies.
• KNOW YOUR COINS AND HOW TO USE THEM.
• Now I need each of you to show me 15¢ from the coins that you have been provided in order to get a cookie. | 271 | 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} | 2.765625 | 3 | CC-MAIN-2018-13 | latest | en | 0.939101 |
https://mathoverflow.net/questions/56718/when-do-isometric-actions-exist/56735 | 1,642,655,216,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320301720.45/warc/CC-MAIN-20220120035934-20220120065934-00255.warc.gz | 440,144,169 | 32,808 | # When do isometric actions exist?
Let $X$ be a metrizable topological space and $G$ be a locally compact group. Given a continuous (left) action of $G$ on $X$, is there a metric on $X$, compatible with the topology, for which the action of $G$ becomes an isometric action? Conversely, given a metric on $X$, is there a nontrivial action of $G$ on $X$ that preserves the metric?
I am looking for the most general necessary and sufficient conditions and any possible obstructions. For the first question, the answer is obviously positive when $G$ is compact: one chooses a metric $d$ on $X$ and simply does an "averaging process along the orbits" by defining $$\rho(x,y) = \int_{G} d(g^{-1}\cdot x,g^{-1}\cdot y) dg .$$ I suspect that a similar idea would work more generally using a "cut-off" function on $X$ when the action of $G$ on $X$ is proper. Any connections to amenability (of the group or the group action) would also be interesting.
• I am having trouble making sense of your second question. Could you just take $G = U(1)$ and $X$ any of a large collection of spaces (e.g., finite metrized set, genus two surface, etc.) as a counterexample? Feb 26 '11 at 15:43
• Yes, Scott. There are plenty of counterexamples to both questions, and it will be very instructive to find them, indeed. However, I need some necessary and/or sufficient conditions on the given data when the answer is positive. Feb 26 '11 at 17:58
I am sorry That I am getting involved so late. I was away at a meeting for two days. I have an old (1961) paper in the ANNALS OF MATH called "On the Existence of Slices for Actions of Non-Compact Lie Groups" which is quite relevant. In particular, on page 318 you can find the following concerning proper actions of an arbitrary Lie group $G$.
Theorem 4.3.4. Every seperable, metrizeable, proper $G$-space $X$ admits an invariant metric. ...
There are a large number of other theorems there that show that the theory of proper G-spaces for G an arbitrary Lie group is similar to the theory of G-spaces for a compact Lie group. You can find the paper here: http://vmm.math.uci.edu/ExistenceOfSlices.pdf
• Thanks, Dick! But do we have to be in the smooth category to have this fact? As is stated above, using "cut-off" (or generalized Bruhat) functions, one can easily show that the $G$-space $X$ admits an invariant metric inducing the same topology on $X$, where $G$ is a locally compact group, which acts properly on the locally compact metrizable space $X$. Feb 28 '11 at 6:52
• No, I'm pretty sure that you can use the fact that locally compact groups are suitable limits of Lie groups to get the result for the general case. My interest in that paper was primarily in proving the important slice theorem, and for that one does need smoothness. BTW, that paper was probably the first place where the concept of a proper group action appears in a published paper. I did NOT invent the term. In fact I worked primarily with a more general concept I called a Cartan G-space. It was Armand Borel who told me he had been investigating proper actions and suggested I use that term. Feb 28 '11 at 17:54
Regarding the first question: if we take $G=\mathbb{Z}$ or $G=\mathbb{R}$, then there are plenty of group actions that cannot be realised as isometries.
Example 1: Let $G=\mathbb{R}$ and $X=\mathbb{R}$, and consider the flow $\phi_r(x) = e^r x$. Then $G$ is locally compact and $\phi$ defines a continuous $G$-action of $X$, but there is no metric that makes this an isometry; indeed, given any neighbourhood $U\ni 0$ and $0\neq x\in U$, there exists $r_0$ such that $\phi_r(x)\notin U$ for all $r>r_0$. This property holds for any equivalent metric but cannot hold for an isometric action.
If you want $X$ to be compact, you can consider $G=\mathbb{R}$ and $X=[-\pi/2,\pi/2]$, and given $r\in G$, consider the map $\phi_r\colon X\to X$ defined by $\phi_r(x) = \tan^{-1}(r + \tan x)$ for $|x| < \pi/2$ and $\phi_r(x)=x$ for $|x|=\pi/2$. Then $\phi$ is a continuous group action that fixes the two endpoints; one is attracting, the other repelling, and $\phi$ cannot be an isometric action for the same reasons as above.
Example 2: Let $G=\mathbb{Z}$ and $X=\{0,1\}^\mathbb{Z}$ with the product topology; then $G$ acts on $X$ via the shift map. In other words, $\phi_n(x)\_i = x_{i+n}$. Once again, topological considerations using fixed points can be used to show that this action is not isometric for any metric on $X$ that induces the product topology.
So I guess one can sum up those obstructions by saying that in order for $G$ to act isometrically, every fixed point has to be topologically stable. That is, suppose there is $p\in X$ such that $\phi_g(p)=p$ for all $g\in G$, and that there is a neighbourhood $U\ni p$ such that for every $p\neq x\in U$, we have $\phi_g(x)\notin U$ for some $g\in G$. Then $\phi$ is not an isometric action for any equivalent metric on $X$.
• Very nice counterexamples. Your last paragraph on the necessary condition is what I like the most, though. Feb 26 '11 at 18:00
This is at best a partial answer but rather too long for a comment (I only adress the last paragraph of the question).
Indeed, if $X$ is locally compact second countable and the action of $G$ is proper then there exists a $G$-invariant metric compatible with the topology. As you suspect, this can be done by integrating the metric multiplied by a (generalized) Bruhat function (see Bourbaki, Intégration, VII, § 2, No. 4).
More precisely: Let $G$ be a locally compact group and fix a left Haar measure on $G$. The action of $G$ on a locally compact space $X$ (such that $X/G$ is paracompact) is proper if and only if there exists a continuous function $\beta: X \to [0,\infty)$ such that
• For every compact set $K \subset X$ the set $\operatorname{supp}\beta \cap GK$ is compact.
• For all $x \in X$ we have $\int_{G} \beta(g^{-1}x)\\,dg = 1$.
This fact is folklore but it is difficult to locate a simple proof in the literature. Therefore I've given a short one in Appendix E of my thesis, available here. Sometimes these functions are called cut-off functions but I don't like the name.
Using this, the existence of an invariant metric compatible with the topology on a locally compact second countable proper $G$-space $X$ is an easy exercise in integration theory: Pick any metric $d_{X}$ compatible with the topology and replace it by $\frac{d_{X}}{1+d_{X}}$ in case it is unbounded. Then put $\delta_{X}(x,y) = \iint_{G \times G} \beta(g^{-1}x)\, \beta(h^{-1} y)\, d_X (g^{-1}x, h^{-1}y)\,dh\,dg$ and verify that $\delta_{X}:X \times X \to [0,1)$ is an invariant metric compatible with the topology.
A similar and detailed argument can be found in one of the first few sections of Koszul's Lectures on groups of transformations (I think it's in the the third section of the first chapter but I can't verify this at the moment).
Finally, you're asking about the relation to amenability, here I also have at best some comments. Of course, proper actions are known to be topologically amenable (for instance because there is a Bruhat function). In the other direction, I think there is no hope. There are plenty of actions of amenable groups that can't be made into isometric actions (and any continuous action of an amenable group is amenable). Since Vaughn has given some nice examples, I can end this long post now.
• Many thanks for the formula! I was just playing around with the sums of the Bruhat functions instead of their products. I was quite sure this was folklore, and I agree that it is hard to locate it in the literature. The term "cut-off" function, however, was used by Jean-Louis Tu in his paper "La conjecture de Novikov pour les feuilletages hyperboliques", published in 1999 in K-theory. He proved the existence of such a function in his Proposition 6.11 for any locally compact proper groupoid with Haar system. I couldn't find any earlier reference, though. He might be the first one publishing it. Feb 27 '11 at 20:18
• Thanks for the reference, good to know that! The formula as well as the construction of a Bruhat function is implicitly contained in Koszul's lectures, I think (at least it was a major source of inspiration for me). The main idea definitely goes back to Bruhat. I really prefer to think of Bruhat functions as sort of transverse partitions of unity. The extension to groupoids is (as usual) not so difficult, as soon as you have the case of transformation groups right (I found it myself when I was thinking about amenable groupoids some years ago). Feb 27 '11 at 21:02
• I also learned from the book "Continuous bounded cohomology of locally compact groups" by Nicolas Monod (P. 44, Lemma 4.5.4) that the existence of the generalized Bruhat functions was explicitly stated in Bourbaki, "INTEGRATION", published in 1963. It is the Proposition 8 on p. 51 (French version). (I think Koszul lectures were published in 1965.) Feb 28 '11 at 4:58
• Monod's thesis is where I learned about it from (Monod and I are students of Marc Burger and Monod was one of the co-examiners of my thesis) :) Feb 28 '11 at 8:41
• But we have $\delta_X(x,x)>0$. Am I missing something? Mar 4 '11 at 7:35
I got interested in a similar problem (not quite the same thing, though) a few years back but could not find anything really interesting. Here is what I found in the litterature;
• In case $X$ is compact and $G$ is abelian then there is an invariant metric (compatible with the topology, of course) under the action of $G$ iff $G$ is relatively compact in the homeomorphism group of $X$ (endowed with the compact-open convergence topology) iff the action of $G$ is topologically equicontinuous. By a result of Marjanovic (" On topological isometries", Indag. Math. 31(1969), 184–189) this extends to the case when $X$ is locally compact (that is not the way Marjanovic's result is formulated but it is equivalent)
"Topologically equicontinuous" means the following: for any $x,y \in X$, any open set $V$ containing $y$, there is an open set $U$ containing $x$ and an open subset $W$ containing $y$ and contained in $V$ such that for all $g \in G$ $g(U) \cap W \neq \emptyset \Rightarrow g(U) \subseteq V$
• In case $X$ is not locally compact then some extensions of this result are known. I know of two papers on this:
C. Borges, How to recognize homeomorphisms and isometries, Pacific Journal of Mathematics 37(3) (1971), 625–633.
M. Tak Kiang, On some semigroups of mappings, Indag. Math. 33(1972), 18–22
I have not found anything in the litterature more recent than that, but I probably haven't looked hard enough - this was more curiosity than serious research on my part (update: well now I have looked more seriously and still haven't found anything else in the litterature).
I did obtain the following: assume $G$ is abelian (this is the only case I thought about, as I was mostly interested in the case when $G$ is generated by $1$ element), that $X$ is Polish and that the action of $G$ on $X$ is topologically transitive. Then there is a compatible $G$-invariant metric if, and only if, $G$ is a topologically equicontinuous group of homeomorphisms of $X$. Under the same assumptions, there is a complete invariant metric iff any invariant metric is complete.
In case you can read French there are some notes on this on my webpage - I wrote them for myelf so probably you should not take anything that's written on faith...
UPDATE (July 4, 2014): I corrected some imprecisions in the text above (about the locally compact case). I logged in to mention that I. Ben Yaacov and I recently worked on this problem again, and proved the following result (there is a preprint on my webpage which supersedes the notes alluded to above):
Assume $X$ is separable metrizable and $G$ is a group of homeomorphisms of $X$. Then there exists a compatible $G$-invariant metric on $X$ if and only if, for any $y \in X$ and any open $V$ containing $y$ there exists an open $W$ containing $y$ and contained in $V$ such that for any $x \in X$ there exists an open $U$ containing $x$ and satisfying $\forall g \in G \ (gU \cap W \ne \emptyset) \Rightarrow gU \subseteq V$.
That is a lot of quantifiers! I think they are really needed; the property above is a uniform version of topological equicontinuity (obtained by switching a universal and an existential quantifier).
• Thanks for the very interesting references and for your French notes. It seems that we have generalized some results of Marjanovic (which was published in 1969 not 1960) to the context of spectral triples in noncommutative geometry with Bellissard and Marcolli (Theorem 1 in the article "Dynamical Systems on Spectral Metric Spaces", arXiv:1008.4617), which is equivalent to almost periodicity of the action. Please say hi to Johannes Kellendonk and Jean Savinien:-) Feb 27 '11 at 19:44
• Here is the theorem: Theorem 1 (Arzel`a-Ascoli theorem). Let $X = (A, H, D)$ be a compact spectral metric space. Let $G\subset{\rm Aut}(A)$ be a quasi-isometric subgroup. Then $G$ is equicontinuous if and only if it has a compact closure. Feb 27 '11 at 19:51
• Ah, I'll have to ask my colleagues about what spectral metric spaces are... Thanks for catching the mistake in the reference to Marjanovic's article, I edited my answer accordingly. Mar 4 '11 at 9:09 | 3,528 | 13,307 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 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-2022-05 | latest | en | 0.917235 |
http://www.stata.com/statalist/archive/2012-06/msg01120.html | 1,500,957,972,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424961.59/warc/CC-MAIN-20170725042318-20170725062318-00153.warc.gz | 548,624,599 | 3,670 | Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
# re: st: margins after split-plot anova
From "Airey, David C" To "statalist@hsphsun2.harvard.edu" Subject re: st: margins after split-plot anova Date Sat, 23 Jun 2012 08:54:07 -0500
```.
I think you should ask Stata Corp if you have your model correct for your data and question.
Cheers,
-Dave
> Thank you.
> In the following example the option at() does not work (not estimable):
>
> anova depvar c.IQ group / id|group label label#group label#c.IQ , repeated (label)
> margins, within(group) at( IQ=(1))
>
> There is something about having a continuous covariate in the anova model that I’m missing or that does not work in Stata.
>
> Let’s look at the following examples:
> - no continuous covariate:
> 1) anova depvar group / id|group label label#group, repeated (label)
> margins, within(group)
>
> 2) regress depvar group label label#group
> margins group
>
> - with continuous covariate:
> 3) anova depvar c.IQ group / id|group label label#group label#c.IQ , repeated (label)
> margins, within(group)
>
> 4)regress depvar c.IQ group label label#group label#c.IQ
> margins group
>
>
> The output of margins in 1), 2), AND 3) are equal. The output of 4) is (correctly) different from 1) and 2), because 4) includes a continuous covariate that is not present in 1) and 2).
> And the problem (again, perhaps I’m missing something simple, but I don’t know what) is that the outcome of margins in 3) is equal to the outcome of margins in 1) and 2).
>
> Thank you again for any help anybody can provide.
> Luca
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 520 | 1,782 | {"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-2017-30 | latest | en | 0.818703 |
https://www.airmilescalculator.com/distance/ppt-to-bob/ | 1,603,350,384,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107878921.41/warc/CC-MAIN-20201022053410-20201022083410-00360.warc.gz | 582,193,736 | 7,761 | Distance between Papeete (PPT) and Bora Bora (BOB)
Flight distance from Papeete to Bora Bora (Faa'a International Airport – Bora Bora Airport) is 161 miles / 259 kilometers / 140 nautical miles. Estimated flight time is 48 minutes.
Map of flight path from Papeete to Bora Bora.
Shortest flight path between Faa'a International Airport (PPT) and Bora Bora Airport (BOB).
How far is Bora Bora from Papeete?
There are several ways to calculate distances between Papeete and Bora Bora. Here are two common methods:
Vincenty's formula (applied above)
• 161.070 miles
• 259.217 kilometers
• 139.966 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 161.067 miles
• 259.212 kilometers
• 139.963 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
Airport information
A Faa'a International Airport
City: Papeete
Country: French Polynesia
IATA Code: PPT
ICAO Code: NTAA
Coordinates: 17°33′13″S, 149°36′25″W
B Bora Bora Airport
City: Bora Bora
Country: French Polynesia
IATA Code: BOB
ICAO Code: NTTB
Coordinates: 16°26′39″S, 151°45′3″W
Time difference and current local times
There is no time difference between Papeete and Bora Bora.
-10
-10
Carbon dioxide emissions
Estimated CO2 emissions per passenger is 49 kg (107 pounds).
Frequent Flyer Miles Calculator
Papeete (PPT) → Bora Bora (BOB).
Distance:
161
Elite level bonus:
0
Booking class bonus:
0
In total
Total frequent flyer miles:
161
Round trip? | 469 | 1,684 | {"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-2020-45 | latest | en | 0.788055 |
https://creativevictuals.com/scotland/which-of-the-following-is-an-example-of-nonprobabilistic-sampling.php | 1,653,313,070,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662558030.43/warc/CC-MAIN-20220523132100-20220523162100-00607.warc.gz | 241,673,762 | 12,357 | # Scotland Which Of The Following Is An Example Of Nonprobabilistic Sampling
## sampling strategies UT Liberal Arts
### Chapter 4 Stratified Sampling IIT Kanpur
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### Sampling Design ProProfs Quiz
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### Probability sampling Lærd Dissertation
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## Nonprobability sampling Wikipedia
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### When Attaining the Best Sample is Out of Reach
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... Nonprobabilistic sampling d) Cluster sampling 105) Which of the following is an example of nonprobabilistic sampling? a) Simple random sampling b) and nonprobabilistic sample Probabilistic sample selection methods include the following: 1. Simple random sample for example, auditors want to sample a
Stats exam 3 study guide by danny06 includes 36 questions Which of the following is not Which of the following is an example of nonprobabilistic sampling? A sampling frame is a list of all the items in For example, you might have a sampling frame of names of people in a certain town for a survey you’re going to be
Chapter 7 SAMPLING PROCEDURES IN RESEARCH Consider the following questions involving sampling: nonprobabilistic Reflections on Probability vs Nonprobability Sampling nonprobabilistic reasoning of a more ad hoc character has to be used. as an example,
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View all posts in Scotland category | 3,888 | 19,589 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2022-21 | latest | en | 0.913163 |
http://oeis.org/A129720 | 1,611,391,788,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703536556.58/warc/CC-MAIN-20210123063713-20210123093713-00566.warc.gz | 72,416,190 | 3,998 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A129720 Number of 0's in odd position in all Fibonacci binary words of length n. A Fibonacci binary word is a binary word having no 00 subword. 2
0, 1, 1, 4, 5, 14, 19, 46, 65, 145, 210, 444, 654, 1331, 1985, 3926, 5911, 11434, 17345, 32960, 50305, 94211, 144516, 267384, 411900, 754309, 1166209, 2116936, 3283145, 5914310, 9197455, 16458034, 25655489, 45638101, 71293590, 126159156 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,4 LINKS É. Czabarka, R. Flórez, L. Junes, A Discrete Convolution on the Generalized Hosoya Triangle, Journal of Integer Sequences, 18 (2015), #15.1.6. FORMULA G.f.: z(1-z^2)/((1-z-z^2)^2*(1+z-z^2)). a(2n) = a(2n-1) + a(2n-2) (n >= 1). a(2n-1) = A030267(n). a(2n) = A129722(2n) = A001870(n-1). a(n) = Sum_{k=0..ceiling(n/2)} k*A129719(n,k). EXAMPLE a(4)=5 because in 1110, 1111, 110'1, 1010, 1011, 0'110, 0'111 and 0'10'1 one has altogether five 0's in odd position (marked by '). MAPLE g:=z*(1-z^2)/(1-z-z^2)^2/(1+z-z^2): gser:=series(g, z=0, 43): seq(coeff(gser, z, n), n=0..40); CROSSREFS Cf. A030267, A001870, A129719, A129722. Sequence in context: A092432 A041669 A300278 * A041089 A042321 A050164 Adjacent sequences: A129717 A129718 A129719 * A129721 A129722 A129723 KEYWORD nonn AUTHOR Emeric Deutsch, May 13 2007 STATUS approved
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Last modified January 23 03:02 EST 2021. Contains 340384 sequences. (Running on oeis4.) | 688 | 1,808 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2021-04 | latest | en | 0.609075 |
https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/14/2/a/a/1/1/ | 1,685,957,775,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224651815.80/warc/CC-MAIN-20230605085657-20230605115657-00588.warc.gz | 951,667,080 | 80,774 | # Properties
Label 14.2.a.a.1.1 Level $14$ Weight $2$ Character 14.1 Self dual yes Analytic conductor $0.112$ Analytic rank $0$ Dimension $1$ CM no Inner twists $1$
# Related objects
Show commands: Magma / PariGP / SageMath
## Newspace parameters
comment: Compute space of new eigenforms
[N,k,chi] = [14,2,Mod(1,14)]
mf = mfinit([N,k,chi],0)
lf = mfeigenbasis(mf)
from sage.modular.dirichlet import DirichletCharacter
H = DirichletGroup(14, base_ring=CyclotomicField(2))
chi = DirichletCharacter(H, H._module([0]))
N = Newforms(chi, 2, names="a")
//Please install CHIMP (https://github.com/edgarcosta/CHIMP) if you want to run this code
chi := DirichletCharacter("14.1");
S:= CuspForms(chi, 2);
N := Newforms(S);
Level: $$N$$ $$=$$ $$14 = 2 \cdot 7$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 14.a (trivial)
## Newform invariants
comment: select newform
sage: f = N[0] # Warning: the index may be different
gp: f = lf[1] \\ Warning: the index may be different
Self dual: yes Analytic conductor: $$0.111790562830$$ Analytic rank: $$0$$ Dimension: $$1$$ Coefficient field: $$\mathbb{Q}$$ Coefficient ring: $$\mathbb{Z}$$ Coefficient ring index: $$1$$ Twist minimal: yes Fricke sign: $$-1$$ Sato-Tate group: $\mathrm{SU}(2)$
## Embedding invariants
Embedding label 1.1 Character $$\chi$$ $$=$$ 14.1
## $q$-expansion
comment: q-expansion
sage: f.q_expansion() # note that sage often uses an isomorphic number field
gp: mfcoefs(f, 20)
$$f(q)$$ $$=$$ $$q-1.00000 q^{2} -2.00000 q^{3} +1.00000 q^{4} +2.00000 q^{6} +1.00000 q^{7} -1.00000 q^{8} +1.00000 q^{9} +O(q^{10})$$ $$q-1.00000 q^{2} -2.00000 q^{3} +1.00000 q^{4} +2.00000 q^{6} +1.00000 q^{7} -1.00000 q^{8} +1.00000 q^{9} -2.00000 q^{12} -4.00000 q^{13} -1.00000 q^{14} +1.00000 q^{16} +6.00000 q^{17} -1.00000 q^{18} +2.00000 q^{19} -2.00000 q^{21} +2.00000 q^{24} -5.00000 q^{25} +4.00000 q^{26} +4.00000 q^{27} +1.00000 q^{28} -6.00000 q^{29} -4.00000 q^{31} -1.00000 q^{32} -6.00000 q^{34} +1.00000 q^{36} +2.00000 q^{37} -2.00000 q^{38} +8.00000 q^{39} +6.00000 q^{41} +2.00000 q^{42} +8.00000 q^{43} -12.0000 q^{47} -2.00000 q^{48} +1.00000 q^{49} +5.00000 q^{50} -12.0000 q^{51} -4.00000 q^{52} +6.00000 q^{53} -4.00000 q^{54} -1.00000 q^{56} -4.00000 q^{57} +6.00000 q^{58} -6.00000 q^{59} +8.00000 q^{61} +4.00000 q^{62} +1.00000 q^{63} +1.00000 q^{64} -4.00000 q^{67} +6.00000 q^{68} -1.00000 q^{72} +2.00000 q^{73} -2.00000 q^{74} +10.0000 q^{75} +2.00000 q^{76} -8.00000 q^{78} +8.00000 q^{79} -11.0000 q^{81} -6.00000 q^{82} -6.00000 q^{83} -2.00000 q^{84} -8.00000 q^{86} +12.0000 q^{87} -6.00000 q^{89} -4.00000 q^{91} +8.00000 q^{93} +12.0000 q^{94} +2.00000 q^{96} -10.0000 q^{97} -1.00000 q^{98} +O(q^{100})$$
## Coefficient data
For each $$n$$ we display the coefficients of the $$q$$-expansion $$a_n$$, the Satake parameters $$\alpha_p$$, and the Satake angles $$\theta_p = \textrm{Arg}(\alpha_p)$$.
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
$$n$$ $$a_n$$ $$a_n / n^{(k-1)/2}$$ $$\alpha_n$$ $$\theta_n$$
$$p$$ $$a_p$$ $$a_p / p^{(k-1)/2}$$ $$\alpha_p$$ $$\theta_p$$
$$2$$ −1.00000 −0.707107
$$3$$ −2.00000 −1.15470 −0.577350 0.816497i $$-0.695913\pi$$
−0.577350 + 0.816497i $$0.695913\pi$$
$$4$$ 1.00000 0.500000
$$5$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$6$$ 2.00000 0.816497
$$7$$ 1.00000 0.377964
$$8$$ −1.00000 −0.353553
$$9$$ 1.00000 0.333333
$$10$$ 0 0
$$11$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$12$$ −2.00000 −0.577350
$$13$$ −4.00000 −1.10940 −0.554700 0.832050i $$-0.687167\pi$$
−0.554700 + 0.832050i $$0.687167\pi$$
$$14$$ −1.00000 −0.267261
$$15$$ 0 0
$$16$$ 1.00000 0.250000
$$17$$ 6.00000 1.45521 0.727607 0.685994i $$-0.240633\pi$$
0.727607 + 0.685994i $$0.240633\pi$$
$$18$$ −1.00000 −0.235702
$$19$$ 2.00000 0.458831 0.229416 0.973329i $$-0.426318\pi$$
0.229416 + 0.973329i $$0.426318\pi$$
$$20$$ 0 0
$$21$$ −2.00000 −0.436436
$$22$$ 0 0
$$23$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$24$$ 2.00000 0.408248
$$25$$ −5.00000 −1.00000
$$26$$ 4.00000 0.784465
$$27$$ 4.00000 0.769800
$$28$$ 1.00000 0.188982
$$29$$ −6.00000 −1.11417 −0.557086 0.830455i $$-0.688081\pi$$
−0.557086 + 0.830455i $$0.688081\pi$$
$$30$$ 0 0
$$31$$ −4.00000 −0.718421 −0.359211 0.933257i $$-0.616954\pi$$
−0.359211 + 0.933257i $$0.616954\pi$$
$$32$$ −1.00000 −0.176777
$$33$$ 0 0
$$34$$ −6.00000 −1.02899
$$35$$ 0 0
$$36$$ 1.00000 0.166667
$$37$$ 2.00000 0.328798 0.164399 0.986394i $$-0.447432\pi$$
0.164399 + 0.986394i $$0.447432\pi$$
$$38$$ −2.00000 −0.324443
$$39$$ 8.00000 1.28103
$$40$$ 0 0
$$41$$ 6.00000 0.937043 0.468521 0.883452i $$-0.344787\pi$$
0.468521 + 0.883452i $$0.344787\pi$$
$$42$$ 2.00000 0.308607
$$43$$ 8.00000 1.21999 0.609994 0.792406i $$-0.291172\pi$$
0.609994 + 0.792406i $$0.291172\pi$$
$$44$$ 0 0
$$45$$ 0 0
$$46$$ 0 0
$$47$$ −12.0000 −1.75038 −0.875190 0.483779i $$-0.839264\pi$$
−0.875190 + 0.483779i $$0.839264\pi$$
$$48$$ −2.00000 −0.288675
$$49$$ 1.00000 0.142857
$$50$$ 5.00000 0.707107
$$51$$ −12.0000 −1.68034
$$52$$ −4.00000 −0.554700
$$53$$ 6.00000 0.824163 0.412082 0.911147i $$-0.364802\pi$$
0.412082 + 0.911147i $$0.364802\pi$$
$$54$$ −4.00000 −0.544331
$$55$$ 0 0
$$56$$ −1.00000 −0.133631
$$57$$ −4.00000 −0.529813
$$58$$ 6.00000 0.787839
$$59$$ −6.00000 −0.781133 −0.390567 0.920575i $$-0.627721\pi$$
−0.390567 + 0.920575i $$0.627721\pi$$
$$60$$ 0 0
$$61$$ 8.00000 1.02430 0.512148 0.858898i $$-0.328850\pi$$
0.512148 + 0.858898i $$0.328850\pi$$
$$62$$ 4.00000 0.508001
$$63$$ 1.00000 0.125988
$$64$$ 1.00000 0.125000
$$65$$ 0 0
$$66$$ 0 0
$$67$$ −4.00000 −0.488678 −0.244339 0.969690i $$-0.578571\pi$$
−0.244339 + 0.969690i $$0.578571\pi$$
$$68$$ 6.00000 0.727607
$$69$$ 0 0
$$70$$ 0 0
$$71$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$72$$ −1.00000 −0.117851
$$73$$ 2.00000 0.234082 0.117041 0.993127i $$-0.462659\pi$$
0.117041 + 0.993127i $$0.462659\pi$$
$$74$$ −2.00000 −0.232495
$$75$$ 10.0000 1.15470
$$76$$ 2.00000 0.229416
$$77$$ 0 0
$$78$$ −8.00000 −0.905822
$$79$$ 8.00000 0.900070 0.450035 0.893011i $$-0.351411\pi$$
0.450035 + 0.893011i $$0.351411\pi$$
$$80$$ 0 0
$$81$$ −11.0000 −1.22222
$$82$$ −6.00000 −0.662589
$$83$$ −6.00000 −0.658586 −0.329293 0.944228i $$-0.606810\pi$$
−0.329293 + 0.944228i $$0.606810\pi$$
$$84$$ −2.00000 −0.218218
$$85$$ 0 0
$$86$$ −8.00000 −0.862662
$$87$$ 12.0000 1.28654
$$88$$ 0 0
$$89$$ −6.00000 −0.635999 −0.317999 0.948091i $$-0.603011\pi$$
−0.317999 + 0.948091i $$0.603011\pi$$
$$90$$ 0 0
$$91$$ −4.00000 −0.419314
$$92$$ 0 0
$$93$$ 8.00000 0.829561
$$94$$ 12.0000 1.23771
$$95$$ 0 0
$$96$$ 2.00000 0.204124
$$97$$ −10.0000 −1.01535 −0.507673 0.861550i $$-0.669494\pi$$
−0.507673 + 0.861550i $$0.669494\pi$$
$$98$$ −1.00000 −0.101015
$$99$$ 0 0
$$100$$ −5.00000 −0.500000
$$101$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$102$$ 12.0000 1.18818
$$103$$ −4.00000 −0.394132 −0.197066 0.980390i $$-0.563141\pi$$
−0.197066 + 0.980390i $$0.563141\pi$$
$$104$$ 4.00000 0.392232
$$105$$ 0 0
$$106$$ −6.00000 −0.582772
$$107$$ 12.0000 1.16008 0.580042 0.814587i $$-0.303036\pi$$
0.580042 + 0.814587i $$0.303036\pi$$
$$108$$ 4.00000 0.384900
$$109$$ 2.00000 0.191565 0.0957826 0.995402i $$-0.469465\pi$$
0.0957826 + 0.995402i $$0.469465\pi$$
$$110$$ 0 0
$$111$$ −4.00000 −0.379663
$$112$$ 1.00000 0.0944911
$$113$$ 6.00000 0.564433 0.282216 0.959351i $$-0.408930\pi$$
0.282216 + 0.959351i $$0.408930\pi$$
$$114$$ 4.00000 0.374634
$$115$$ 0 0
$$116$$ −6.00000 −0.557086
$$117$$ −4.00000 −0.369800
$$118$$ 6.00000 0.552345
$$119$$ 6.00000 0.550019
$$120$$ 0 0
$$121$$ −11.0000 −1.00000
$$122$$ −8.00000 −0.724286
$$123$$ −12.0000 −1.08200
$$124$$ −4.00000 −0.359211
$$125$$ 0 0
$$126$$ −1.00000 −0.0890871
$$127$$ −16.0000 −1.41977 −0.709885 0.704317i $$-0.751253\pi$$
−0.709885 + 0.704317i $$0.751253\pi$$
$$128$$ −1.00000 −0.0883883
$$129$$ −16.0000 −1.40872
$$130$$ 0 0
$$131$$ 18.0000 1.57267 0.786334 0.617802i $$-0.211977\pi$$
0.786334 + 0.617802i $$0.211977\pi$$
$$132$$ 0 0
$$133$$ 2.00000 0.173422
$$134$$ 4.00000 0.345547
$$135$$ 0 0
$$136$$ −6.00000 −0.514496
$$137$$ 18.0000 1.53784 0.768922 0.639343i $$-0.220793\pi$$
0.768922 + 0.639343i $$0.220793\pi$$
$$138$$ 0 0
$$139$$ 14.0000 1.18746 0.593732 0.804663i $$-0.297654\pi$$
0.593732 + 0.804663i $$0.297654\pi$$
$$140$$ 0 0
$$141$$ 24.0000 2.02116
$$142$$ 0 0
$$143$$ 0 0
$$144$$ 1.00000 0.0833333
$$145$$ 0 0
$$146$$ −2.00000 −0.165521
$$147$$ −2.00000 −0.164957
$$148$$ 2.00000 0.164399
$$149$$ −18.0000 −1.47462 −0.737309 0.675556i $$-0.763904\pi$$
−0.737309 + 0.675556i $$0.763904\pi$$
$$150$$ −10.0000 −0.816497
$$151$$ 8.00000 0.651031 0.325515 0.945537i $$-0.394462\pi$$
0.325515 + 0.945537i $$0.394462\pi$$
$$152$$ −2.00000 −0.162221
$$153$$ 6.00000 0.485071
$$154$$ 0 0
$$155$$ 0 0
$$156$$ 8.00000 0.640513
$$157$$ −4.00000 −0.319235 −0.159617 0.987179i $$-0.551026\pi$$
−0.159617 + 0.987179i $$0.551026\pi$$
$$158$$ −8.00000 −0.636446
$$159$$ −12.0000 −0.951662
$$160$$ 0 0
$$161$$ 0 0
$$162$$ 11.0000 0.864242
$$163$$ −16.0000 −1.25322 −0.626608 0.779334i $$-0.715557\pi$$
−0.626608 + 0.779334i $$0.715557\pi$$
$$164$$ 6.00000 0.468521
$$165$$ 0 0
$$166$$ 6.00000 0.465690
$$167$$ −12.0000 −0.928588 −0.464294 0.885681i $$-0.653692\pi$$
−0.464294 + 0.885681i $$0.653692\pi$$
$$168$$ 2.00000 0.154303
$$169$$ 3.00000 0.230769
$$170$$ 0 0
$$171$$ 2.00000 0.152944
$$172$$ 8.00000 0.609994
$$173$$ −12.0000 −0.912343 −0.456172 0.889892i $$-0.650780\pi$$
−0.456172 + 0.889892i $$0.650780\pi$$
$$174$$ −12.0000 −0.909718
$$175$$ −5.00000 −0.377964
$$176$$ 0 0
$$177$$ 12.0000 0.901975
$$178$$ 6.00000 0.449719
$$179$$ −12.0000 −0.896922 −0.448461 0.893802i $$-0.648028\pi$$
−0.448461 + 0.893802i $$0.648028\pi$$
$$180$$ 0 0
$$181$$ 20.0000 1.48659 0.743294 0.668965i $$-0.233262\pi$$
0.743294 + 0.668965i $$0.233262\pi$$
$$182$$ 4.00000 0.296500
$$183$$ −16.0000 −1.18275
$$184$$ 0 0
$$185$$ 0 0
$$186$$ −8.00000 −0.586588
$$187$$ 0 0
$$188$$ −12.0000 −0.875190
$$189$$ 4.00000 0.290957
$$190$$ 0 0
$$191$$ 24.0000 1.73658 0.868290 0.496058i $$-0.165220\pi$$
0.868290 + 0.496058i $$0.165220\pi$$
$$192$$ −2.00000 −0.144338
$$193$$ 14.0000 1.00774 0.503871 0.863779i $$-0.331909\pi$$
0.503871 + 0.863779i $$0.331909\pi$$
$$194$$ 10.0000 0.717958
$$195$$ 0 0
$$196$$ 1.00000 0.0714286
$$197$$ −18.0000 −1.28245 −0.641223 0.767354i $$-0.721573\pi$$
−0.641223 + 0.767354i $$0.721573\pi$$
$$198$$ 0 0
$$199$$ 20.0000 1.41776 0.708881 0.705328i $$-0.249200\pi$$
0.708881 + 0.705328i $$0.249200\pi$$
$$200$$ 5.00000 0.353553
$$201$$ 8.00000 0.564276
$$202$$ 0 0
$$203$$ −6.00000 −0.421117
$$204$$ −12.0000 −0.840168
$$205$$ 0 0
$$206$$ 4.00000 0.278693
$$207$$ 0 0
$$208$$ −4.00000 −0.277350
$$209$$ 0 0
$$210$$ 0 0
$$211$$ −4.00000 −0.275371 −0.137686 0.990476i $$-0.543966\pi$$
−0.137686 + 0.990476i $$0.543966\pi$$
$$212$$ 6.00000 0.412082
$$213$$ 0 0
$$214$$ −12.0000 −0.820303
$$215$$ 0 0
$$216$$ −4.00000 −0.272166
$$217$$ −4.00000 −0.271538
$$218$$ −2.00000 −0.135457
$$219$$ −4.00000 −0.270295
$$220$$ 0 0
$$221$$ −24.0000 −1.61441
$$222$$ 4.00000 0.268462
$$223$$ 8.00000 0.535720 0.267860 0.963458i $$-0.413684\pi$$
0.267860 + 0.963458i $$0.413684\pi$$
$$224$$ −1.00000 −0.0668153
$$225$$ −5.00000 −0.333333
$$226$$ −6.00000 −0.399114
$$227$$ 18.0000 1.19470 0.597351 0.801980i $$-0.296220\pi$$
0.597351 + 0.801980i $$0.296220\pi$$
$$228$$ −4.00000 −0.264906
$$229$$ −4.00000 −0.264327 −0.132164 0.991228i $$-0.542192\pi$$
−0.132164 + 0.991228i $$0.542192\pi$$
$$230$$ 0 0
$$231$$ 0 0
$$232$$ 6.00000 0.393919
$$233$$ −6.00000 −0.393073 −0.196537 0.980497i $$-0.562969\pi$$
−0.196537 + 0.980497i $$0.562969\pi$$
$$234$$ 4.00000 0.261488
$$235$$ 0 0
$$236$$ −6.00000 −0.390567
$$237$$ −16.0000 −1.03931
$$238$$ −6.00000 −0.388922
$$239$$ 24.0000 1.55243 0.776215 0.630468i $$-0.217137\pi$$
0.776215 + 0.630468i $$0.217137\pi$$
$$240$$ 0 0
$$241$$ −10.0000 −0.644157 −0.322078 0.946713i $$-0.604381\pi$$
−0.322078 + 0.946713i $$0.604381\pi$$
$$242$$ 11.0000 0.707107
$$243$$ 10.0000 0.641500
$$244$$ 8.00000 0.512148
$$245$$ 0 0
$$246$$ 12.0000 0.765092
$$247$$ −8.00000 −0.509028
$$248$$ 4.00000 0.254000
$$249$$ 12.0000 0.760469
$$250$$ 0 0
$$251$$ −18.0000 −1.13615 −0.568075 0.822977i $$-0.692312\pi$$
−0.568075 + 0.822977i $$0.692312\pi$$
$$252$$ 1.00000 0.0629941
$$253$$ 0 0
$$254$$ 16.0000 1.00393
$$255$$ 0 0
$$256$$ 1.00000 0.0625000
$$257$$ 18.0000 1.12281 0.561405 0.827541i $$-0.310261\pi$$
0.561405 + 0.827541i $$0.310261\pi$$
$$258$$ 16.0000 0.996116
$$259$$ 2.00000 0.124274
$$260$$ 0 0
$$261$$ −6.00000 −0.371391
$$262$$ −18.0000 −1.11204
$$263$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$264$$ 0 0
$$265$$ 0 0
$$266$$ −2.00000 −0.122628
$$267$$ 12.0000 0.734388
$$268$$ −4.00000 −0.244339
$$269$$ −12.0000 −0.731653 −0.365826 0.930683i $$-0.619214\pi$$
−0.365826 + 0.930683i $$0.619214\pi$$
$$270$$ 0 0
$$271$$ −16.0000 −0.971931 −0.485965 0.873978i $$-0.661532\pi$$
−0.485965 + 0.873978i $$0.661532\pi$$
$$272$$ 6.00000 0.363803
$$273$$ 8.00000 0.484182
$$274$$ −18.0000 −1.08742
$$275$$ 0 0
$$276$$ 0 0
$$277$$ −10.0000 −0.600842 −0.300421 0.953807i $$-0.597127\pi$$
−0.300421 + 0.953807i $$0.597127\pi$$
$$278$$ −14.0000 −0.839664
$$279$$ −4.00000 −0.239474
$$280$$ 0 0
$$281$$ −6.00000 −0.357930 −0.178965 0.983855i $$-0.557275\pi$$
−0.178965 + 0.983855i $$0.557275\pi$$
$$282$$ −24.0000 −1.42918
$$283$$ −22.0000 −1.30776 −0.653882 0.756596i $$-0.726861\pi$$
−0.653882 + 0.756596i $$0.726861\pi$$
$$284$$ 0 0
$$285$$ 0 0
$$286$$ 0 0
$$287$$ 6.00000 0.354169
$$288$$ −1.00000 −0.0589256
$$289$$ 19.0000 1.11765
$$290$$ 0 0
$$291$$ 20.0000 1.17242
$$292$$ 2.00000 0.117041
$$293$$ 24.0000 1.40209 0.701047 0.713115i $$-0.252716\pi$$
0.701047 + 0.713115i $$0.252716\pi$$
$$294$$ 2.00000 0.116642
$$295$$ 0 0
$$296$$ −2.00000 −0.116248
$$297$$ 0 0
$$298$$ 18.0000 1.04271
$$299$$ 0 0
$$300$$ 10.0000 0.577350
$$301$$ 8.00000 0.461112
$$302$$ −8.00000 −0.460348
$$303$$ 0 0
$$304$$ 2.00000 0.114708
$$305$$ 0 0
$$306$$ −6.00000 −0.342997
$$307$$ 2.00000 0.114146 0.0570730 0.998370i $$-0.481823\pi$$
0.0570730 + 0.998370i $$0.481823\pi$$
$$308$$ 0 0
$$309$$ 8.00000 0.455104
$$310$$ 0 0
$$311$$ −24.0000 −1.36092 −0.680458 0.732787i $$-0.738219\pi$$
−0.680458 + 0.732787i $$0.738219\pi$$
$$312$$ −8.00000 −0.452911
$$313$$ −10.0000 −0.565233 −0.282617 0.959233i $$-0.591202\pi$$
−0.282617 + 0.959233i $$0.591202\pi$$
$$314$$ 4.00000 0.225733
$$315$$ 0 0
$$316$$ 8.00000 0.450035
$$317$$ 6.00000 0.336994 0.168497 0.985702i $$-0.446109\pi$$
0.168497 + 0.985702i $$0.446109\pi$$
$$318$$ 12.0000 0.672927
$$319$$ 0 0
$$320$$ 0 0
$$321$$ −24.0000 −1.33955
$$322$$ 0 0
$$323$$ 12.0000 0.667698
$$324$$ −11.0000 −0.611111
$$325$$ 20.0000 1.10940
$$326$$ 16.0000 0.886158
$$327$$ −4.00000 −0.221201
$$328$$ −6.00000 −0.331295
$$329$$ −12.0000 −0.661581
$$330$$ 0 0
$$331$$ 8.00000 0.439720 0.219860 0.975531i $$-0.429440\pi$$
0.219860 + 0.975531i $$0.429440\pi$$
$$332$$ −6.00000 −0.329293
$$333$$ 2.00000 0.109599
$$334$$ 12.0000 0.656611
$$335$$ 0 0
$$336$$ −2.00000 −0.109109
$$337$$ 14.0000 0.762629 0.381314 0.924445i $$-0.375472\pi$$
0.381314 + 0.924445i $$0.375472\pi$$
$$338$$ −3.00000 −0.163178
$$339$$ −12.0000 −0.651751
$$340$$ 0 0
$$341$$ 0 0
$$342$$ −2.00000 −0.108148
$$343$$ 1.00000 0.0539949
$$344$$ −8.00000 −0.431331
$$345$$ 0 0
$$346$$ 12.0000 0.645124
$$347$$ −24.0000 −1.28839 −0.644194 0.764862i $$-0.722807\pi$$
−0.644194 + 0.764862i $$0.722807\pi$$
$$348$$ 12.0000 0.643268
$$349$$ −28.0000 −1.49881 −0.749403 0.662114i $$-0.769659\pi$$
−0.749403 + 0.662114i $$0.769659\pi$$
$$350$$ 5.00000 0.267261
$$351$$ −16.0000 −0.854017
$$352$$ 0 0
$$353$$ 18.0000 0.958043 0.479022 0.877803i $$-0.340992\pi$$
0.479022 + 0.877803i $$0.340992\pi$$
$$354$$ −12.0000 −0.637793
$$355$$ 0 0
$$356$$ −6.00000 −0.317999
$$357$$ −12.0000 −0.635107
$$358$$ 12.0000 0.634220
$$359$$ −24.0000 −1.26667 −0.633336 0.773877i $$-0.718315\pi$$
−0.633336 + 0.773877i $$0.718315\pi$$
$$360$$ 0 0
$$361$$ −15.0000 −0.789474
$$362$$ −20.0000 −1.05118
$$363$$ 22.0000 1.15470
$$364$$ −4.00000 −0.209657
$$365$$ 0 0
$$366$$ 16.0000 0.836333
$$367$$ 8.00000 0.417597 0.208798 0.977959i $$-0.433045\pi$$
0.208798 + 0.977959i $$0.433045\pi$$
$$368$$ 0 0
$$369$$ 6.00000 0.312348
$$370$$ 0 0
$$371$$ 6.00000 0.311504
$$372$$ 8.00000 0.414781
$$373$$ 14.0000 0.724893 0.362446 0.932005i $$-0.381942\pi$$
0.362446 + 0.932005i $$0.381942\pi$$
$$374$$ 0 0
$$375$$ 0 0
$$376$$ 12.0000 0.618853
$$377$$ 24.0000 1.23606
$$378$$ −4.00000 −0.205738
$$379$$ −16.0000 −0.821865 −0.410932 0.911666i $$-0.634797\pi$$
−0.410932 + 0.911666i $$0.634797\pi$$
$$380$$ 0 0
$$381$$ 32.0000 1.63941
$$382$$ −24.0000 −1.22795
$$383$$ 36.0000 1.83951 0.919757 0.392488i $$-0.128386\pi$$
0.919757 + 0.392488i $$0.128386\pi$$
$$384$$ 2.00000 0.102062
$$385$$ 0 0
$$386$$ −14.0000 −0.712581
$$387$$ 8.00000 0.406663
$$388$$ −10.0000 −0.507673
$$389$$ 18.0000 0.912636 0.456318 0.889817i $$-0.349168\pi$$
0.456318 + 0.889817i $$0.349168\pi$$
$$390$$ 0 0
$$391$$ 0 0
$$392$$ −1.00000 −0.0505076
$$393$$ −36.0000 −1.81596
$$394$$ 18.0000 0.906827
$$395$$ 0 0
$$396$$ 0 0
$$397$$ 20.0000 1.00377 0.501886 0.864934i $$-0.332640\pi$$
0.501886 + 0.864934i $$0.332640\pi$$
$$398$$ −20.0000 −1.00251
$$399$$ −4.00000 −0.200250
$$400$$ −5.00000 −0.250000
$$401$$ −18.0000 −0.898877 −0.449439 0.893311i $$-0.648376\pi$$
−0.449439 + 0.893311i $$0.648376\pi$$
$$402$$ −8.00000 −0.399004
$$403$$ 16.0000 0.797017
$$404$$ 0 0
$$405$$ 0 0
$$406$$ 6.00000 0.297775
$$407$$ 0 0
$$408$$ 12.0000 0.594089
$$409$$ 14.0000 0.692255 0.346128 0.938187i $$-0.387496\pi$$
0.346128 + 0.938187i $$0.387496\pi$$
$$410$$ 0 0
$$411$$ −36.0000 −1.77575
$$412$$ −4.00000 −0.197066
$$413$$ −6.00000 −0.295241
$$414$$ 0 0
$$415$$ 0 0
$$416$$ 4.00000 0.196116
$$417$$ −28.0000 −1.37117
$$418$$ 0 0
$$419$$ 6.00000 0.293119 0.146560 0.989202i $$-0.453180\pi$$
0.146560 + 0.989202i $$0.453180\pi$$
$$420$$ 0 0
$$421$$ −10.0000 −0.487370 −0.243685 0.969854i $$-0.578356\pi$$
−0.243685 + 0.969854i $$0.578356\pi$$
$$422$$ 4.00000 0.194717
$$423$$ −12.0000 −0.583460
$$424$$ −6.00000 −0.291386
$$425$$ −30.0000 −1.45521
$$426$$ 0 0
$$427$$ 8.00000 0.387147
$$428$$ 12.0000 0.580042
$$429$$ 0 0
$$430$$ 0 0
$$431$$ 24.0000 1.15604 0.578020 0.816023i $$-0.303826\pi$$
0.578020 + 0.816023i $$0.303826\pi$$
$$432$$ 4.00000 0.192450
$$433$$ −34.0000 −1.63394 −0.816968 0.576683i $$-0.804347\pi$$
−0.816968 + 0.576683i $$0.804347\pi$$
$$434$$ 4.00000 0.192006
$$435$$ 0 0
$$436$$ 2.00000 0.0957826
$$437$$ 0 0
$$438$$ 4.00000 0.191127
$$439$$ 8.00000 0.381819 0.190910 0.981608i $$-0.438856\pi$$
0.190910 + 0.981608i $$0.438856\pi$$
$$440$$ 0 0
$$441$$ 1.00000 0.0476190
$$442$$ 24.0000 1.14156
$$443$$ −12.0000 −0.570137 −0.285069 0.958507i $$-0.592016\pi$$
−0.285069 + 0.958507i $$0.592016\pi$$
$$444$$ −4.00000 −0.189832
$$445$$ 0 0
$$446$$ −8.00000 −0.378811
$$447$$ 36.0000 1.70274
$$448$$ 1.00000 0.0472456
$$449$$ 18.0000 0.849473 0.424736 0.905317i $$-0.360367\pi$$
0.424736 + 0.905317i $$0.360367\pi$$
$$450$$ 5.00000 0.235702
$$451$$ 0 0
$$452$$ 6.00000 0.282216
$$453$$ −16.0000 −0.751746
$$454$$ −18.0000 −0.844782
$$455$$ 0 0
$$456$$ 4.00000 0.187317
$$457$$ −10.0000 −0.467780 −0.233890 0.972263i $$-0.575146\pi$$
−0.233890 + 0.972263i $$0.575146\pi$$
$$458$$ 4.00000 0.186908
$$459$$ 24.0000 1.12022
$$460$$ 0 0
$$461$$ 12.0000 0.558896 0.279448 0.960161i $$-0.409849\pi$$
0.279448 + 0.960161i $$0.409849\pi$$
$$462$$ 0 0
$$463$$ 32.0000 1.48717 0.743583 0.668644i $$-0.233125\pi$$
0.743583 + 0.668644i $$0.233125\pi$$
$$464$$ −6.00000 −0.278543
$$465$$ 0 0
$$466$$ 6.00000 0.277945
$$467$$ −6.00000 −0.277647 −0.138823 0.990317i $$-0.544332\pi$$
−0.138823 + 0.990317i $$0.544332\pi$$
$$468$$ −4.00000 −0.184900
$$469$$ −4.00000 −0.184703
$$470$$ 0 0
$$471$$ 8.00000 0.368621
$$472$$ 6.00000 0.276172
$$473$$ 0 0
$$474$$ 16.0000 0.734904
$$475$$ −10.0000 −0.458831
$$476$$ 6.00000 0.275010
$$477$$ 6.00000 0.274721
$$478$$ −24.0000 −1.09773
$$479$$ −36.0000 −1.64488 −0.822441 0.568850i $$-0.807388\pi$$
−0.822441 + 0.568850i $$0.807388\pi$$
$$480$$ 0 0
$$481$$ −8.00000 −0.364769
$$482$$ 10.0000 0.455488
$$483$$ 0 0
$$484$$ −11.0000 −0.500000
$$485$$ 0 0
$$486$$ −10.0000 −0.453609
$$487$$ −16.0000 −0.725029 −0.362515 0.931978i $$-0.618082\pi$$
−0.362515 + 0.931978i $$0.618082\pi$$
$$488$$ −8.00000 −0.362143
$$489$$ 32.0000 1.44709
$$490$$ 0 0
$$491$$ −12.0000 −0.541552 −0.270776 0.962642i $$-0.587280\pi$$
−0.270776 + 0.962642i $$0.587280\pi$$
$$492$$ −12.0000 −0.541002
$$493$$ −36.0000 −1.62136
$$494$$ 8.00000 0.359937
$$495$$ 0 0
$$496$$ −4.00000 −0.179605
$$497$$ 0 0
$$498$$ −12.0000 −0.537733
$$499$$ −4.00000 −0.179065 −0.0895323 0.995984i $$-0.528537\pi$$
−0.0895323 + 0.995984i $$0.528537\pi$$
$$500$$ 0 0
$$501$$ 24.0000 1.07224
$$502$$ 18.0000 0.803379
$$503$$ 0 0 1.00000i $$-0.5\pi$$
1.00000i $$0.5\pi$$
$$504$$ −1.00000 −0.0445435
$$505$$ 0 0
$$506$$ 0 0
$$507$$ −6.00000 −0.266469
$$508$$ −16.0000 −0.709885
$$509$$ 36.0000 1.59567 0.797836 0.602875i $$-0.205978\pi$$
0.797836 + 0.602875i $$0.205978\pi$$
$$510$$ 0 0
$$511$$ 2.00000 0.0884748
$$512$$ −1.00000 −0.0441942
$$513$$ 8.00000 0.353209
$$514$$ −18.0000 −0.793946
$$515$$ 0 0
$$516$$ −16.0000 −0.704361
$$517$$ 0 0
$$518$$ −2.00000 −0.0878750
$$519$$ 24.0000 1.05348
$$520$$ 0 0
$$521$$ 6.00000 0.262865 0.131432 0.991325i $$-0.458042\pi$$
0.131432 + 0.991325i $$0.458042\pi$$
$$522$$ 6.00000 0.262613
$$523$$ 2.00000 0.0874539 0.0437269 0.999044i $$-0.486077\pi$$
0.0437269 + 0.999044i $$0.486077\pi$$
$$524$$ 18.0000 0.786334
$$525$$ 10.0000 0.436436
$$526$$ 0 0
$$527$$ −24.0000 −1.04546
$$528$$ 0 0
$$529$$ −23.0000 −1.00000
$$530$$ 0 0
$$531$$ −6.00000 −0.260378
$$532$$ 2.00000 0.0867110
$$533$$ −24.0000 −1.03956
$$534$$ −12.0000 −0.519291
$$535$$ 0 0
$$536$$ 4.00000 0.172774
$$537$$ 24.0000 1.03568
$$538$$ 12.0000 0.517357
$$539$$ 0 0
$$540$$ 0 0
$$541$$ 38.0000 1.63375 0.816874 0.576816i $$-0.195705\pi$$
0.816874 + 0.576816i $$0.195705\pi$$
$$542$$ 16.0000 0.687259
$$543$$ −40.0000 −1.71656
$$544$$ −6.00000 −0.257248
$$545$$ 0 0
$$546$$ −8.00000 −0.342368
$$547$$ 8.00000 0.342055 0.171028 0.985266i $$-0.445291\pi$$
0.171028 + 0.985266i $$0.445291\pi$$
$$548$$ 18.0000 0.768922
$$549$$ 8.00000 0.341432
$$550$$ 0 0
$$551$$ −12.0000 −0.511217
$$552$$ 0 0
$$553$$ 8.00000 0.340195
$$554$$ 10.0000 0.424859
$$555$$ 0 0
$$556$$ 14.0000 0.593732
$$557$$ 6.00000 0.254228 0.127114 0.991888i $$-0.459429\pi$$
0.127114 + 0.991888i $$0.459429\pi$$
$$558$$ 4.00000 0.169334
$$559$$ −32.0000 −1.35346
$$560$$ 0 0
$$561$$ 0 0
$$562$$ 6.00000 0.253095
$$563$$ 30.0000 1.26435 0.632175 0.774826i $$-0.282163\pi$$
0.632175 + 0.774826i $$0.282163\pi$$
$$564$$ 24.0000 1.01058
$$565$$ 0 0
$$566$$ 22.0000 0.924729
$$567$$ −11.0000 −0.461957
$$568$$ 0 0
$$569$$ 6.00000 0.251533 0.125767 0.992060i $$-0.459861\pi$$
0.125767 + 0.992060i $$0.459861\pi$$
$$570$$ 0 0
$$571$$ 32.0000 1.33916 0.669579 0.742741i $$-0.266474\pi$$
0.669579 + 0.742741i $$0.266474\pi$$
$$572$$ 0 0
$$573$$ −48.0000 −2.00523
$$574$$ −6.00000 −0.250435
$$575$$ 0 0
$$576$$ 1.00000 0.0416667
$$577$$ 2.00000 0.0832611 0.0416305 0.999133i $$-0.486745\pi$$
0.0416305 + 0.999133i $$0.486745\pi$$
$$578$$ −19.0000 −0.790296
$$579$$ −28.0000 −1.16364
$$580$$ 0 0
$$581$$ −6.00000 −0.248922
$$582$$ −20.0000 −0.829027
$$583$$ 0 0
$$584$$ −2.00000 −0.0827606
$$585$$ 0 0
$$586$$ −24.0000 −0.991431
$$587$$ −42.0000 −1.73353 −0.866763 0.498721i $$-0.833803\pi$$
−0.866763 + 0.498721i $$0.833803\pi$$
$$588$$ −2.00000 −0.0824786
$$589$$ −8.00000 −0.329634
$$590$$ 0 0
$$591$$ 36.0000 1.48084
$$592$$ 2.00000 0.0821995
$$593$$ −6.00000 −0.246390 −0.123195 0.992382i $$-0.539314\pi$$
−0.123195 + 0.992382i $$0.539314\pi$$
$$594$$ 0 0
$$595$$ 0 0
$$596$$ −18.0000 −0.737309
$$597$$ −40.0000 −1.63709
$$598$$ 0 0
$$599$$ −24.0000 −0.980613 −0.490307 0.871550i $$-0.663115\pi$$
−0.490307 + 0.871550i $$0.663115\pi$$
$$600$$ −10.0000 −0.408248
$$601$$ 26.0000 1.06056 0.530281 0.847822i $$-0.322086\pi$$
0.530281 + 0.847822i $$0.322086\pi$$
$$602$$ −8.00000 −0.326056
$$603$$ −4.00000 −0.162893
$$604$$ 8.00000 0.325515
$$605$$ 0 0
$$606$$ 0 0
$$607$$ 32.0000 1.29884 0.649420 0.760430i $$-0.275012\pi$$
0.649420 + 0.760430i $$0.275012\pi$$
$$608$$ −2.00000 −0.0811107
$$609$$ 12.0000 0.486265
$$610$$ 0 0
$$611$$ 48.0000 1.94187
$$612$$ 6.00000 0.242536
$$613$$ 2.00000 0.0807792 0.0403896 0.999184i $$-0.487140\pi$$
0.0403896 + 0.999184i $$0.487140\pi$$
$$614$$ −2.00000 −0.0807134
$$615$$ 0 0
$$616$$ 0 0
$$617$$ 6.00000 0.241551 0.120775 0.992680i $$-0.461462\pi$$
0.120775 + 0.992680i $$0.461462\pi$$
$$618$$ −8.00000 −0.321807
$$619$$ 26.0000 1.04503 0.522514 0.852631i $$-0.324994\pi$$
0.522514 + 0.852631i $$0.324994\pi$$
$$620$$ 0 0
$$621$$ 0 0
$$622$$ 24.0000 0.962312
$$623$$ −6.00000 −0.240385
$$624$$ 8.00000 0.320256
$$625$$ 25.0000 1.00000
$$626$$ 10.0000 0.399680
$$627$$ 0 0
$$628$$ −4.00000 −0.159617
$$629$$ 12.0000 0.478471
$$630$$ 0 0
$$631$$ −16.0000 −0.636950 −0.318475 0.947931i $$-0.603171\pi$$
−0.318475 + 0.947931i $$0.603171\pi$$
$$632$$ −8.00000 −0.318223
$$633$$ 8.00000 0.317971
$$634$$ −6.00000 −0.238290
$$635$$ 0 0
$$636$$ −12.0000 −0.475831
$$637$$ −4.00000 −0.158486
$$638$$ 0 0
$$639$$ 0 0
$$640$$ 0 0
$$641$$ −18.0000 −0.710957 −0.355479 0.934684i $$-0.615682\pi$$
−0.355479 + 0.934684i $$0.615682\pi$$
$$642$$ 24.0000 0.947204
$$643$$ 14.0000 0.552106 0.276053 0.961142i $$-0.410973\pi$$
0.276053 + 0.961142i $$0.410973\pi$$
$$644$$ 0 0
$$645$$ 0 0
$$646$$ −12.0000 −0.472134
$$647$$ −12.0000 −0.471769 −0.235884 0.971781i $$-0.575799\pi$$
−0.235884 + 0.971781i $$0.575799\pi$$
$$648$$ 11.0000 0.432121
$$649$$ 0 0
$$650$$ −20.0000 −0.784465
$$651$$ 8.00000 0.313545
$$652$$ −16.0000 −0.626608
$$653$$ 18.0000 0.704394 0.352197 0.935926i $$-0.385435\pi$$
0.352197 + 0.935926i $$0.385435\pi$$
$$654$$ 4.00000 0.156412
$$655$$ 0 0
$$656$$ 6.00000 0.234261
$$657$$ 2.00000 0.0780274
$$658$$ 12.0000 0.467809
$$659$$ −24.0000 −0.934907 −0.467454 0.884018i $$-0.654829\pi$$
−0.467454 + 0.884018i $$0.654829\pi$$
$$660$$ 0 0
$$661$$ −40.0000 −1.55582 −0.777910 0.628376i $$-0.783720\pi$$
−0.777910 + 0.628376i $$0.783720\pi$$
$$662$$ −8.00000 −0.310929
$$663$$ 48.0000 1.86417
$$664$$ 6.00000 0.232845
$$665$$ 0 0
$$666$$ −2.00000 −0.0774984
$$667$$ 0 0
$$668$$ −12.0000 −0.464294
$$669$$ −16.0000 −0.618596
$$670$$ 0 0
$$671$$ 0 0
$$672$$ 2.00000 0.0771517
$$673$$ 26.0000 1.00223 0.501113 0.865382i $$-0.332924\pi$$
0.501113 + 0.865382i $$0.332924\pi$$
$$674$$ −14.0000 −0.539260
$$675$$ −20.0000 −0.769800
$$676$$ 3.00000 0.115385
$$677$$ −12.0000 −0.461197 −0.230599 0.973049i $$-0.574068\pi$$
−0.230599 + 0.973049i $$0.574068\pi$$
$$678$$ 12.0000 0.460857
$$679$$ −10.0000 −0.383765
$$680$$ 0 0
$$681$$ −36.0000 −1.37952
$$682$$ 0 0
$$683$$ −12.0000 −0.459167 −0.229584 0.973289i $$-0.573736\pi$$
−0.229584 + 0.973289i $$0.573736\pi$$
$$684$$ 2.00000 0.0764719
$$685$$ 0 0
$$686$$ −1.00000 −0.0381802
$$687$$ 8.00000 0.305219
$$688$$ 8.00000 0.304997
$$689$$ −24.0000 −0.914327
$$690$$ 0 0
$$691$$ −46.0000 −1.74992 −0.874961 0.484193i $$-0.839113\pi$$
−0.874961 + 0.484193i $$0.839113\pi$$
$$692$$ −12.0000 −0.456172
$$693$$ 0 0
$$694$$ 24.0000 0.911028
$$695$$ 0 0
$$696$$ −12.0000 −0.454859
$$697$$ 36.0000 1.36360
$$698$$ 28.0000 1.05982
$$699$$ 12.0000 0.453882
$$700$$ −5.00000 −0.188982
$$701$$ 18.0000 0.679851 0.339925 0.940452i $$-0.389598\pi$$
0.339925 + 0.940452i $$0.389598\pi$$
$$702$$ 16.0000 0.603881
$$703$$ 4.00000 0.150863
$$704$$ 0 0
$$705$$ 0 0
$$706$$ −18.0000 −0.677439
$$707$$ 0 0
$$708$$ 12.0000 0.450988
$$709$$ −46.0000 −1.72757 −0.863783 0.503864i $$-0.831911\pi$$
−0.863783 + 0.503864i $$0.831911\pi$$
$$710$$ 0 0
$$711$$ 8.00000 0.300023
$$712$$ 6.00000 0.224860
$$713$$ 0 0
$$714$$ 12.0000 0.449089
$$715$$ 0 0
$$716$$ −12.0000 −0.448461
$$717$$ −48.0000 −1.79259
$$718$$ 24.0000 0.895672
$$719$$ 12.0000 0.447524 0.223762 0.974644i $$-0.428166\pi$$
0.223762 + 0.974644i $$0.428166\pi$$
$$720$$ 0 0
$$721$$ −4.00000 −0.148968
$$722$$ 15.0000 0.558242
$$723$$ 20.0000 0.743808
$$724$$ 20.0000 0.743294
$$725$$ 30.0000 1.11417
$$726$$ −22.0000 −0.816497
$$727$$ 44.0000 1.63187 0.815935 0.578144i $$-0.196223\pi$$
0.815935 + 0.578144i $$0.196223\pi$$
$$728$$ 4.00000 0.148250
$$729$$ 13.0000 0.481481
$$730$$ 0 0
$$731$$ 48.0000 1.77534
$$732$$ −16.0000 −0.591377
$$733$$ −40.0000 −1.47743 −0.738717 0.674016i $$-0.764568\pi$$
−0.738717 + 0.674016i $$0.764568\pi$$
$$734$$ −8.00000 −0.295285
$$735$$ 0 0
$$736$$ 0 0
$$737$$ 0 0
$$738$$ −6.00000 −0.220863
$$739$$ −16.0000 −0.588570 −0.294285 0.955718i $$-0.595081\pi$$
−0.294285 + 0.955718i $$0.595081\pi$$
$$740$$ 0 0
$$741$$ 16.0000 0.587775
$$742$$ −6.00000 −0.220267
$$743$$ 24.0000 0.880475 0.440237 0.897881i $$-0.354894\pi$$
0.440237 + 0.897881i $$0.354894\pi$$
$$744$$ −8.00000 −0.293294
$$745$$ 0 0
$$746$$ −14.0000 −0.512576
$$747$$ −6.00000 −0.219529
$$748$$ 0 0
$$749$$ 12.0000 0.438470
$$750$$ 0 0
$$751$$ −40.0000 −1.45962 −0.729810 0.683650i $$-0.760392\pi$$
−0.729810 + 0.683650i $$0.760392\pi$$
$$752$$ −12.0000 −0.437595
$$753$$ 36.0000 1.31191
$$754$$ −24.0000 −0.874028
$$755$$ 0 0
$$756$$ 4.00000 0.145479
$$757$$ 2.00000 0.0726912 0.0363456 0.999339i $$-0.488428\pi$$
0.0363456 + 0.999339i $$0.488428\pi$$
$$758$$ 16.0000 0.581146
$$759$$ 0 0
$$760$$ 0 0
$$761$$ −18.0000 −0.652499 −0.326250 0.945284i $$-0.605785\pi$$
−0.326250 + 0.945284i $$0.605785\pi$$
$$762$$ −32.0000 −1.15924
$$763$$ 2.00000 0.0724049
$$764$$ 24.0000 0.868290
$$765$$ 0 0
$$766$$ −36.0000 −1.30073
$$767$$ 24.0000 0.866590
$$768$$ −2.00000 −0.0721688
$$769$$ 14.0000 0.504853 0.252426 0.967616i $$-0.418771\pi$$
0.252426 + 0.967616i $$0.418771\pi$$
$$770$$ 0 0
$$771$$ −36.0000 −1.29651
$$772$$ 14.0000 0.503871
$$773$$ 24.0000 0.863220 0.431610 0.902060i $$-0.357946\pi$$
0.431610 + 0.902060i $$0.357946\pi$$
$$774$$ −8.00000 −0.287554
$$775$$ 20.0000 0.718421
$$776$$ 10.0000 0.358979
$$777$$ −4.00000 −0.143499
$$778$$ −18.0000 −0.645331
$$779$$ 12.0000 0.429945
$$780$$ 0 0
$$781$$ 0 0
$$782$$ 0 0
$$783$$ −24.0000 −0.857690
$$784$$ 1.00000 0.0357143
$$785$$ 0 0
$$786$$ 36.0000 1.28408
$$787$$ −22.0000 −0.784215 −0.392108 0.919919i $$-0.628254\pi$$
−0.392108 + 0.919919i $$0.628254\pi$$
$$788$$ −18.0000 −0.641223
$$789$$ 0 0
$$790$$ 0 0
$$791$$ 6.00000 0.213335
$$792$$ 0 0
$$793$$ −32.0000 −1.13635
$$794$$ −20.0000 −0.709773
$$795$$ 0 0
$$796$$ 20.0000 0.708881
$$797$$ −12.0000 −0.425062 −0.212531 0.977154i $$-0.568171\pi$$
−0.212531 + 0.977154i $$0.568171\pi$$
$$798$$ 4.00000 0.141598
$$799$$ −72.0000 −2.54718
$$800$$ 5.00000 0.176777
$$801$$ −6.00000 −0.212000
$$802$$ 18.0000 0.635602
$$803$$ 0 0
$$804$$ 8.00000 0.282138
$$805$$ 0 0
$$806$$ −16.0000 −0.563576
$$807$$ 24.0000 0.844840
$$808$$ 0 0
$$809$$ 6.00000 0.210949 0.105474 0.994422i $$-0.466364\pi$$
0.105474 + 0.994422i $$0.466364\pi$$
$$810$$ 0 0
$$811$$ 2.00000 0.0702295 0.0351147 0.999383i $$-0.488820\pi$$
0.0351147 + 0.999383i $$0.488820\pi$$
$$812$$ −6.00000 −0.210559
$$813$$ 32.0000 1.12229
$$814$$ 0 0
$$815$$ 0 0
$$816$$ −12.0000 −0.420084
$$817$$ 16.0000 0.559769
$$818$$ −14.0000 −0.489499
$$819$$ −4.00000 −0.139771
$$820$$ 0 0
$$821$$ 6.00000 0.209401 0.104701 0.994504i $$-0.466612\pi$$
0.104701 + 0.994504i $$0.466612\pi$$
$$822$$ 36.0000 1.25564
$$823$$ −40.0000 −1.39431 −0.697156 0.716919i $$-0.745552\pi$$
−0.697156 + 0.716919i $$0.745552\pi$$
$$824$$ 4.00000 0.139347
$$825$$ 0 0
$$826$$ 6.00000 0.208767
$$827$$ −36.0000 −1.25184 −0.625921 0.779886i $$-0.715277\pi$$
−0.625921 + 0.779886i $$0.715277\pi$$
$$828$$ 0 0
$$829$$ 56.0000 1.94496 0.972480 0.232986i $$-0.0748495\pi$$
0.972480 + 0.232986i $$0.0748495\pi$$
$$830$$ 0 0
$$831$$ 20.0000 0.693792
$$832$$ −4.00000 −0.138675
$$833$$ 6.00000 0.207888
$$834$$ 28.0000 0.969561
$$835$$ 0 0
$$836$$ 0 0
$$837$$ −16.0000 −0.553041
$$838$$ −6.00000 −0.207267
$$839$$ 12.0000 0.414286 0.207143 0.978311i $$-0.433583\pi$$
0.207143 + 0.978311i $$0.433583\pi$$
$$840$$ 0 0
$$841$$ 7.00000 0.241379
$$842$$ 10.0000 0.344623
$$843$$ 12.0000 0.413302
$$844$$ −4.00000 −0.137686
$$845$$ 0 0
$$846$$ 12.0000 0.412568
$$847$$ −11.0000 −0.377964
$$848$$ 6.00000 0.206041
$$849$$ 44.0000 1.51008
$$850$$ 30.0000 1.02899
$$851$$ 0 0
$$852$$ 0 0
$$853$$ 44.0000 1.50653 0.753266 0.657716i $$-0.228477\pi$$
0.753266 + 0.657716i $$0.228477\pi$$
$$854$$ −8.00000 −0.273754
$$855$$ 0 0
$$856$$ −12.0000 −0.410152
$$857$$ −18.0000 −0.614868 −0.307434 0.951569i $$-0.599470\pi$$
−0.307434 + 0.951569i $$0.599470\pi$$
$$858$$ 0 0
$$859$$ 14.0000 0.477674 0.238837 0.971060i $$-0.423234\pi$$
0.238837 + 0.971060i $$0.423234\pi$$
$$860$$ 0 0
$$861$$ −12.0000 −0.408959
$$862$$ −24.0000 −0.817443
$$863$$ −24.0000 −0.816970 −0.408485 0.912765i $$-0.633943\pi$$
−0.408485 + 0.912765i $$0.633943\pi$$
$$864$$ −4.00000 −0.136083
$$865$$ 0 0
$$866$$ 34.0000 1.15537
$$867$$ −38.0000 −1.29055
$$868$$ −4.00000 −0.135769
$$869$$ 0 0
$$870$$ 0 0
$$871$$ 16.0000 0.542139
$$872$$ −2.00000 −0.0677285
$$873$$ −10.0000 −0.338449
$$874$$ 0 0
$$875$$ 0 0
$$876$$ −4.00000 −0.135147
$$877$$ −22.0000 −0.742887 −0.371444 0.928456i $$-0.621137\pi$$
−0.371444 + 0.928456i $$0.621137\pi$$
$$878$$ −8.00000 −0.269987
$$879$$ −48.0000 −1.61900
$$880$$ 0 0
$$881$$ −54.0000 −1.81931 −0.909653 0.415369i $$-0.863653\pi$$
−0.909653 + 0.415369i $$0.863653\pi$$
$$882$$ −1.00000 −0.0336718
$$883$$ 20.0000 0.673054 0.336527 0.941674i $$-0.390748\pi$$
0.336527 + 0.941674i $$0.390748\pi$$
$$884$$ −24.0000 −0.807207
$$885$$ 0 0
$$886$$ 12.0000 0.403148
$$887$$ −36.0000 −1.20876 −0.604381 0.796696i $$-0.706579\pi$$
−0.604381 + 0.796696i $$0.706579\pi$$
$$888$$ 4.00000 0.134231
$$889$$ −16.0000 −0.536623
$$890$$ 0 0
$$891$$ 0 0
$$892$$ 8.00000 0.267860
$$893$$ −24.0000 −0.803129
$$894$$ −36.0000 −1.20402
$$895$$ 0 0
$$896$$ −1.00000 −0.0334077
$$897$$ 0 0
$$898$$ −18.0000 −0.600668
$$899$$ 24.0000 0.800445
$$900$$ −5.00000 −0.166667
$$901$$ 36.0000 1.19933
$$902$$ 0 0
$$903$$ −16.0000 −0.532447
$$904$$ −6.00000 −0.199557
$$905$$ 0 0
$$906$$ 16.0000 0.531564
$$907$$ 44.0000 1.46100 0.730498 0.682915i $$-0.239288\pi$$
0.730498 + 0.682915i $$0.239288\pi$$
$$908$$ 18.0000 0.597351
$$909$$ 0 0
$$910$$ 0 0
$$911$$ 48.0000 1.59031 0.795155 0.606406i $$-0.207389\pi$$
0.795155 + 0.606406i $$0.207389\pi$$
$$912$$ −4.00000 −0.132453
$$913$$ 0 0
$$914$$ 10.0000 0.330771
$$915$$ 0 0
$$916$$ −4.00000 −0.132164
$$917$$ 18.0000 0.594412
$$918$$ −24.0000 −0.792118
$$919$$ 56.0000 1.84727 0.923635 0.383274i $$-0.125203\pi$$
0.923635 + 0.383274i $$0.125203\pi$$
$$920$$ 0 0
$$921$$ −4.00000 −0.131804
$$922$$ −12.0000 −0.395199
$$923$$ 0 0
$$924$$ 0 0
$$925$$ −10.0000 −0.328798
$$926$$ −32.0000 −1.05159
$$927$$ −4.00000 −0.131377
$$928$$ 6.00000 0.196960
$$929$$ 6.00000 0.196854 0.0984268 0.995144i $$-0.468619\pi$$
0.0984268 + 0.995144i $$0.468619\pi$$
$$930$$ 0 0
$$931$$ 2.00000 0.0655474
$$932$$ −6.00000 −0.196537
$$933$$ 48.0000 1.57145
$$934$$ 6.00000 0.196326
$$935$$ 0 0
$$936$$ 4.00000 0.130744
$$937$$ 2.00000 0.0653372 0.0326686 0.999466i $$-0.489599\pi$$
0.0326686 + 0.999466i $$0.489599\pi$$
$$938$$ 4.00000 0.130605
$$939$$ 20.0000 0.652675
$$940$$ 0 0
$$941$$ −24.0000 −0.782378 −0.391189 0.920310i $$-0.627936\pi$$
−0.391189 + 0.920310i $$0.627936\pi$$
$$942$$ −8.00000 −0.260654
$$943$$ 0 0
$$944$$ −6.00000 −0.195283
$$945$$ 0 0
$$946$$ 0 0
$$947$$ 24.0000 0.779895 0.389948 0.920837i $$-0.372493\pi$$
0.389948 + 0.920837i $$0.372493\pi$$
$$948$$ −16.0000 −0.519656
$$949$$ −8.00000 −0.259691
$$950$$ 10.0000 0.324443
$$951$$ −12.0000 −0.389127
$$952$$ −6.00000 −0.194461
$$953$$ −54.0000 −1.74923 −0.874616 0.484817i $$-0.838886\pi$$
−0.874616 + 0.484817i $$0.838886\pi$$
$$954$$ −6.00000 −0.194257
$$955$$ 0 0
$$956$$ 24.0000 0.776215
$$957$$ 0 0
$$958$$ 36.0000 1.16311
$$959$$ 18.0000 0.581250
$$960$$ 0 0
$$961$$ −15.0000 −0.483871
$$962$$ 8.00000 0.257930
$$963$$ 12.0000 0.386695
$$964$$ −10.0000 −0.322078
$$965$$ 0 0
$$966$$ 0 0
$$967$$ 32.0000 1.02905 0.514525 0.857475i $$-0.327968\pi$$
0.514525 + 0.857475i $$0.327968\pi$$
$$968$$ 11.0000 0.353553
$$969$$ −24.0000 −0.770991
$$970$$ 0 0
$$971$$ −6.00000 −0.192549 −0.0962746 0.995355i $$-0.530693\pi$$
−0.0962746 + 0.995355i $$0.530693\pi$$
$$972$$ 10.0000 0.320750
$$973$$ 14.0000 0.448819
$$974$$ 16.0000 0.512673
$$975$$ −40.0000 −1.28103
$$976$$ 8.00000 0.256074
$$977$$ −6.00000 −0.191957 −0.0959785 0.995383i $$-0.530598\pi$$
−0.0959785 + 0.995383i $$0.530598\pi$$
$$978$$ −32.0000 −1.02325
$$979$$ 0 0
$$980$$ 0 0
$$981$$ 2.00000 0.0638551
$$982$$ 12.0000 0.382935
$$983$$ −36.0000 −1.14822 −0.574111 0.818778i $$-0.694652\pi$$
−0.574111 + 0.818778i $$0.694652\pi$$
$$984$$ 12.0000 0.382546
$$985$$ 0 0
$$986$$ 36.0000 1.14647
$$987$$ 24.0000 0.763928
$$988$$ −8.00000 −0.254514
$$989$$ 0 0
$$990$$ 0 0
$$991$$ −16.0000 −0.508257 −0.254128 0.967170i $$-0.581789\pi$$
−0.254128 + 0.967170i $$0.581789\pi$$
$$992$$ 4.00000 0.127000
$$993$$ −16.0000 −0.507745
$$994$$ 0 0
$$995$$ 0 0
$$996$$ 12.0000 0.380235
$$997$$ 8.00000 0.253363 0.126681 0.991943i $$-0.459567\pi$$
0.126681 + 0.991943i $$0.459567\pi$$
$$998$$ 4.00000 0.126618
$$999$$ 8.00000 0.253109
Display $$a_p$$ with $$p$$ up to: 50 250 1000 Display $$a_n$$ with $$n$$ up to: 50 250 1000
## Twists
By twisting character
Char Parity Ord Type Twist Min Dim
1.1 even 1 trivial 14.2.a.a.1.1 1
3.2 odd 2 126.2.a.b.1.1 1
4.3 odd 2 112.2.a.c.1.1 1
5.2 odd 4 350.2.c.d.99.1 2
5.3 odd 4 350.2.c.d.99.2 2
5.4 even 2 350.2.a.f.1.1 1
7.2 even 3 98.2.c.b.67.1 2
7.3 odd 6 98.2.c.a.79.1 2
7.4 even 3 98.2.c.b.79.1 2
7.5 odd 6 98.2.c.a.67.1 2
7.6 odd 2 98.2.a.a.1.1 1
8.3 odd 2 448.2.a.a.1.1 1
8.5 even 2 448.2.a.g.1.1 1
9.2 odd 6 1134.2.f.f.757.1 2
9.4 even 3 1134.2.f.l.379.1 2
9.5 odd 6 1134.2.f.f.379.1 2
9.7 even 3 1134.2.f.l.757.1 2
11.10 odd 2 1694.2.a.e.1.1 1
12.11 even 2 1008.2.a.h.1.1 1
13.5 odd 4 2366.2.d.b.337.2 2
13.8 odd 4 2366.2.d.b.337.1 2
13.12 even 2 2366.2.a.j.1.1 1
15.2 even 4 3150.2.g.j.2899.2 2
15.8 even 4 3150.2.g.j.2899.1 2
15.14 odd 2 3150.2.a.i.1.1 1
16.3 odd 4 1792.2.b.g.897.2 2
16.5 even 4 1792.2.b.c.897.2 2
16.11 odd 4 1792.2.b.g.897.1 2
16.13 even 4 1792.2.b.c.897.1 2
17.16 even 2 4046.2.a.f.1.1 1
19.18 odd 2 5054.2.a.c.1.1 1
20.3 even 4 2800.2.g.h.449.2 2
20.7 even 4 2800.2.g.h.449.1 2
20.19 odd 2 2800.2.a.g.1.1 1
21.2 odd 6 882.2.g.c.361.1 2
21.5 even 6 882.2.g.d.361.1 2
21.11 odd 6 882.2.g.c.667.1 2
21.17 even 6 882.2.g.d.667.1 2
21.20 even 2 882.2.a.i.1.1 1
23.22 odd 2 7406.2.a.a.1.1 1
24.5 odd 2 4032.2.a.w.1.1 1
24.11 even 2 4032.2.a.r.1.1 1
28.3 even 6 784.2.i.i.177.1 2
28.11 odd 6 784.2.i.c.177.1 2
28.19 even 6 784.2.i.i.753.1 2
28.23 odd 6 784.2.i.c.753.1 2
28.27 even 2 784.2.a.b.1.1 1
35.13 even 4 2450.2.c.c.99.2 2
35.27 even 4 2450.2.c.c.99.1 2
35.34 odd 2 2450.2.a.t.1.1 1
56.13 odd 2 3136.2.a.e.1.1 1
56.27 even 2 3136.2.a.z.1.1 1
84.83 odd 2 7056.2.a.bd.1.1 1
By twisted newform
Twist Min Dim Char Parity Ord Type
14.2.a.a.1.1 1 1.1 even 1 trivial
98.2.a.a.1.1 1 7.6 odd 2
98.2.c.a.67.1 2 7.5 odd 6
98.2.c.a.79.1 2 7.3 odd 6
98.2.c.b.67.1 2 7.2 even 3
98.2.c.b.79.1 2 7.4 even 3
112.2.a.c.1.1 1 4.3 odd 2
126.2.a.b.1.1 1 3.2 odd 2
350.2.a.f.1.1 1 5.4 even 2
350.2.c.d.99.1 2 5.2 odd 4
350.2.c.d.99.2 2 5.3 odd 4
448.2.a.a.1.1 1 8.3 odd 2
448.2.a.g.1.1 1 8.5 even 2
784.2.a.b.1.1 1 28.27 even 2
784.2.i.c.177.1 2 28.11 odd 6
784.2.i.c.753.1 2 28.23 odd 6
784.2.i.i.177.1 2 28.3 even 6
784.2.i.i.753.1 2 28.19 even 6
882.2.a.i.1.1 1 21.20 even 2
882.2.g.c.361.1 2 21.2 odd 6
882.2.g.c.667.1 2 21.11 odd 6
882.2.g.d.361.1 2 21.5 even 6
882.2.g.d.667.1 2 21.17 even 6
1008.2.a.h.1.1 1 12.11 even 2
1134.2.f.f.379.1 2 9.5 odd 6
1134.2.f.f.757.1 2 9.2 odd 6
1134.2.f.l.379.1 2 9.4 even 3
1134.2.f.l.757.1 2 9.7 even 3
1694.2.a.e.1.1 1 11.10 odd 2
1792.2.b.c.897.1 2 16.13 even 4
1792.2.b.c.897.2 2 16.5 even 4
1792.2.b.g.897.1 2 16.11 odd 4
1792.2.b.g.897.2 2 16.3 odd 4
2366.2.a.j.1.1 1 13.12 even 2
2366.2.d.b.337.1 2 13.8 odd 4
2366.2.d.b.337.2 2 13.5 odd 4
2450.2.a.t.1.1 1 35.34 odd 2
2450.2.c.c.99.1 2 35.27 even 4
2450.2.c.c.99.2 2 35.13 even 4
2800.2.a.g.1.1 1 20.19 odd 2
2800.2.g.h.449.1 2 20.7 even 4
2800.2.g.h.449.2 2 20.3 even 4
3136.2.a.e.1.1 1 56.13 odd 2
3136.2.a.z.1.1 1 56.27 even 2
3150.2.a.i.1.1 1 15.14 odd 2
3150.2.g.j.2899.1 2 15.8 even 4
3150.2.g.j.2899.2 2 15.2 even 4
4032.2.a.r.1.1 1 24.11 even 2
4032.2.a.w.1.1 1 24.5 odd 2
4046.2.a.f.1.1 1 17.16 even 2
5054.2.a.c.1.1 1 19.18 odd 2
7056.2.a.bd.1.1 1 84.83 odd 2
7406.2.a.a.1.1 1 23.22 odd 2 | 22,455 | 40,211 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2023-23 | latest | en | 0.395079 |
http://nrich.maths.org/1425/index | 1,438,408,603,000,000,000 | text/html | crawl-data/CC-MAIN-2015-32/segments/1438042988511.77/warc/CC-MAIN-20150728002308-00090-ip-10-236-191-2.ec2.internal.warc.gz | 174,638,453 | 12,325 | ### Route to Root
A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?
### Tournament Scheduling
Scheduling games is a little more challenging than one might desire. Here are some tournament formats that sport schedulers use.
### Geomlab
A geometry lab crafted in a functional programming language. Ported to Flash from the original java at web.comlab.ox.ac.uk/geomlab
# Divided Differences
##### Stage: 5
Published December 2002,March 2002,December 2011,February 2011.
# 1 A first look at differencing
When in 1821 the English mathematician Charles Babbage invented what we now recognise as the grandfather of the modern computer he called it the 'Difference Engine' since it was intended to take over the work of making mathematical tables by the techniques described in this article.
Photo courtesy of the Charles Babbage Institute, University of Minnesota, Minneapolis. Reproduced by courtesy of IBM Computer Museum.
Today every engineer, mathematician or physicist has access to their own pocket calculator and personal computer. But there were engineers, mathematicians and physicists long before electronic computers existed. How did they perform the long and complicated calculations that their trades demand?
Part of the answer lay in the the use of 'tables'. Nowadays if we wish to know the value of $\sin 44^{\circ}$ we press the correct button on our calculator. In the past we took down a book of tables' and looked it up. As a simple example here is a table of squares.
n 0 1 2 3 4 5 6 7 8 9 n2 0 1 4 9 16 25 36 49 64 81
If we want to know 52 we simply look at the entry under 5.
Exercise 1 Produce a similar table of cubes.
Mathematicians found a large number of tricks to make the construction and use of tables easier. Here is one of them. Consider our table of squares. Let us take the difference between successive entries as shown below.
0 1 4 9 16 25 36 49 64 81 1 3 5 7 9 11 13 15 17
Now take the difference of successive differences to obtain
0 1 4 9 16 25 36 49 64 81 1 3 5 7 9 11 13 15 17 2 2 2 2 2 2 2 2 2
Finally we take the difference of the differences of the differences to obtain
0 1 4 9 16 25 36 49 64 81 1 3 5 7 9 11 13 15 17 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0
Let us try the same thing on a table of 4th powers.
n 0 1 2 3 4 5 6 7 n4 0 1 16 81 256 625 1296 2401
This time successive differencing produces the following table.
0 1 16 81 256 625 1296 2401 1 15 65 175 369 671 1105 14 50 110 194 302 434 36 60 84 108 132 24 24 24 24 0 0 0
Exercise 2 (i) Extend the table of 4th powers n4 to cover n=8 and n=9. Verify that the pattern suggested above continues to hold.
(ii) Find the effect of repeated differencing on your table of cubes.
Exercise 3 (i) Try the effect of repeated differencing on 3n3 -5n2 .
(ii) Try the effect of repeated differencing on a polynomial of your choice.
It very much looks as though the effect of repeated differencing a polynomial of degree m is to produce a row of zeros after at most m+1 differences. Let us try our conjecture on the most general quadratic f(n)=An2 +Bn+C. Here is part of the initial table
n k-2 k-1 k k+1 k+2 f(n) A(k2 -4k+4)+B(k-2)+C A(k2 -2k+1)+B(k-1)+C Ak2 +Bk+C A(k2 +2k+1)+B(k+1)+C A(k2 +4k+4)+B(k+2)+C
and here are the repeated differences
Since we could pick any k to investigate, we have verified the conjecture for the general quadratic.
Exercise 4 Verify the conjecture for the general cubic.
The next exercise requires more algebra.
Exercise 5 (i) If f(n)=nk show that f(n+1)-f(n) is a polynomial in n of degree k-1.
(ii) If g(n) is a polynomial of degree k show that g(n+1)-g(n) is a polynomial in n of degree k-1.
(iii) Prove our conjecture that the effect of repeated differencing a polynomial of degree m is to produce a row of zeros after at most m+1 differences.
We now observe that we can reconstruct the whole of a table of differences for a polynomial from a small part of it. For example, suppose we are told that f is a quadratic polynomial and its table of differences looks like
? ? ? 7 ? ? ? ? ? ? ? ? ? 6 ? ? ? ? ? ? ? ? 2 ? ? ? ? ? 0 0 0 0 0 0 0 0
We can immediately fill in the last but one row to obtain
? ? ? 7 ? ? ? ? ? ? ? ? ? 6 ? ? ? ? ? 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0
and then the next to obtain
? ? ? 7 ? ? ? ? ? ? 0 2 4 6 8 10 12 14 16 2 2 2 2 2 2 2 2 2 0 0 0 0 0 0 0 0
Exercise 6 (i) Fill in the top row and check that we have the table of n2 -n+1 with the leftmost entry corresponding to n=0.
(ii) Fill in in the following table of differences for a quadratic.
? ? ? 9 ? ? ? ? ? ? ? ? ? 16 ? ? ? ? ? ? ? ? 6 ? ? ? ? ? 0 0 0 0 0 0 0 0
Check that the table you get is for f(n)=3n2 -5n-3 with the leftmost entry corresponding to n=0.
Now suppose that we want to tabulate f(n)=n3 -3n2 +5n+1. Here is a partial tabulation which the reader should check.
n -4 -3 -2 -1 0 1 2 3 4 5 f(n) ? ? -29 -8 1 4 7 ? ? ?
We can now draw up a partial table of differences which the reader should check.
? ? -29 -8 1 4 7 ? ? ? ? ? 21 9 3 3 ? ? ? ? ? -12 -6 0 ? ? ? ? ? 6 6 ? ? ? ? ? 0 ? ? ?
But we can now reconstruct the full table of differences. Here is a partial reconstruction
? -54 -29 -8 1 4 7 16 ? ? ? 39 21 9 3 3 9 ? ? ? -18 -12 -6 0 6 ? ? ? 6 6 6 6 ? ? ? 0 0 0 ? ?
Notice that we have extended our original table to read
n -4 -3 -2 -1 0 1 2 3 4 5 f(n) ? -54 -29 -8 1 4 7 16 ? ?
Exercise 7 Continue the reconstruction of the table of differences to the point where you have the value of f(n) for n=-4, n=4 and n=5. Check your answers by computing f(-4), f(4) and f(5) directly from the definition.
Exercise 8
Let $f(n)=3n^2+2n+1$. Compute f(n) for n=-2, -1, 0, 1, 2. Form the associated partial table of differences and then reconstruct the table of differences sufficiently far that you have f(n) for all n with $-5\leq n\leq 4$. Check your answer by computing f(n) directly.
The fact that (at least for polynomials) we can use differences to reconstruct a full table from a partial table gives a strong hint as to why table makers were so interested in these ideas.
Here is another reason. Suppose we have a table like the following.
0 0 0 0 0 1 0 0 0 0 0
in which all the entries are zero except one. If we compute successive differences we get
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 +1 -1 0 0 0 0 0 0 0 1 -2 1 0 0 0 0 0 1 -3 3 1 0 0 0 1 -4 6 -4 1 0
Exercise 9 (i) Calculate one further row of differences.
(ii) Conjecture the form of the nth row of differences.
(iii) [Harder and optional] Prove your conjecture.
Exercise 10 (i) Find the successive differences for the table
0 0 0 0 0 A 0 0 0 0 0
(ii) Find the successive differences for
13 7 3 1 1 3 7 13 21
(iii) Find the successive differences for
13 7 3 1 1 3 (7+A) 13 21
(iv) Suppose f is a cubic polynomial. What is the fifth line in the table of successive difference for f? Find the fifth line of the table of successive differences whose first line is
f(n) f(n+1) f(n+2) f(n+3) f(n+4) f(n+5)+A f(n+6) f(n+7) f(n+8) f(n+9) f(n+10)
Explain how you can find A just by looking at the 5th line.
(v) The following table is supposed to represent the values of a quadratic but I have made a mistake in one entry.
n -1 0 1 2 3 4 5 6 7 8 9 f(n) 8 3 0 -1 0 5 8 15 24 35 49
By successive differencing find the error and correct it.
(vi) Explain how you can locate and correct a single error in a table of a polynomial of degree k.
(vii) Explain how you would try to locate and correct a single error in a table of a polynomial whose degree you did not know.
Successive differencing thus gives an excellent way of finding and correcting errors in tables.
# 2 A second look at differencing
So far we have only looked at tables of a function f as a method of finding f(n) where n is an integer. However we really want to know f(x) at all points x. In other words, knowing f(0), f(1),..., f(m) we want to find f(x). We shall see that this can be done but we will need to raise the mathematical level a little bit.
The key is a set of observations which go back at least as far as Newton.
Exercise 11 (i) Consider the function f2 (x)=x(x-1)/2!. We tabulate it as follows.
n 0 1 2 3 4 5 6 7 8 9 f(n) 0 0 1 3 6 10 15 21 28 36
Find the table of successive differences.
(ii) Do the same for f3 (x)=x(x-1)(x-2)/3!, f4 (x)=x(x-1)(x-2)(x-3)/4!. Do the same for f1 (x)=x/1!=x and f0 (x)=1.
(iii) Conjecture the general pattern.
(iv) [Harder and optional] Prove your conjecture.
Exercise 12 (i) Let f(x)=Af3 (x)+Bf2 (x)+Cf1 (x)+D. Find the table of successive differences as in the previous exercise. If we take it in the form
f(0) f(1) f(2) f(3) f(4) f(5) f(6) f(7) f(8) f(9) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
identify ?, ?, ? and ?.
(ii) If g is any cubic, show that we can find a, b, c and d such that
g(x)=af3 (x)+bf2 (x)+cf1 (x)+d.
Hence show that g can be found from its table of successive differences.
(iii) Conjecture the general pattern.
(iv) [Harder and optional] Prove your conjecture.
We can extend the results of the last exercises a bit. Suppose that f is a polynomial of degree m. If we form the the following table of differences (here h> 0)
f(a) f(a+h) f(a+2h) f(a+3h) f(a+4h) f(a+5h) f(a+6h) f(a+7h) f(a+8h) ... 0 ? ? ? ? ? ? ? ... ?1 ? ? ? ? ? ? ... ?2 ? ? ? ? ? ... ?3 ? ? ? ? ... ?4 ? ? ? ...
and so on, then
$$f(a+k)=\alpha_{0}+\alpha_{1}\frac{k}{h}+ \alpha_{2}\frac{k(k-h)}{2!h^{2}}+ \alpha_{3}\frac{k(k-h)(k-2h)}{3!h^{3}}+\dots$$ *
Exercise 13 [Optional] Prove this.
If we use formula *
for $0\leq k\leq mh$
then we say that we are interpolating . If we use formula * for k outside this region we say that we are extrapolating .
The sceptical reader may have noticed that, although we began our discussion by talking about tables of the sine function everything that followed dealt with polynomials. However, as the mathematicians of the 17th century discovered, the kind of nice' functions that we wish to tabulate look very much like polynomials over small ranges of values. Since they look like polynomials, formula * which applies to polynomials will apply to them (to a very close approximation). We need only calculate our desired function at a small number of values and then we can use equation * to find its values at other points.
For example, here are the values of ln x at a certain number of values.
x 5 5.1 5.2 5.3 5.4 f(x) 1.60944 1.62924 1.64866 1.66771 1.6864
We obtain the following table of differences.
1.60944 1.62924 1.64866 1.66771 1.68640 0.01980 0.01942 0.01905 0.01869 -0.00038 -0.00037 -0.0036 0.00001 0.00001
Thus, in equation * , ?0 =1.60944, ?1 =0.01980, ?2 =-0.00038 and ?4 =0.0001. We thus hope that
$\ln(5+k)\approx1.60944+0.01980\frac{k}{h}+ -0.00019\frac{k(k-h)}{h^{2}}+ +0.00001\frac{k(k-h)(k-2h)}{6h^{3}}.$
Trying this with $k=0.11$ gives $\ln 5.11\approx 1.63120$
and this turns out to be correct to the number of digits given.
Exercise 14 Compute ln 5.16 using our approximation. Compare this with the correct answer. Repeat the exercise for ln y where you choose y with
$5 < y < 5.4$.
Although functions like ln behave like polynomials over small ranges of values, they need not behave like polynomials over large ranges. It is thus not surprising that an idea which works well for interpolating (that is, trying to find the value of a function f at a point using values of f at points close by) works less well (or fails entirely) when we try to use it to guess the value of a function at points far away from those where we know its value.
Exercise 15 Compute ln 10 using our approximation. Compare this with the correct answer.
There is a much more serious problem associated with differencing which I have avoided mentioning up to now. So far, I have assumed that all the initial tabular values are given exactly. In reality, we usually work to a certain number of decimal places. Suppose that we round to the nearest integer. Then a table which looks like
0 0 0 0 0 0 0 0
with a table of differences
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
could actually represent
${ {1 \over 2} -{1 \over 2}{1 \over 2} -{1 \over 2}{1 \over 2} -{1 \over 2}{1 \over 2} -{1 \over 2} }$
with a table of differences
1/2 -1/2 1/2 -1/2 1/2 -1/2 1/2 -1/2 1 -1 1 -1 1 -1 1 -2 2 -2 2 -2 2 4 -4 4 -4 4 -8 8 -8 8
Thus if the first line of a table of differences is only known to an accuracy of epsilon; the second line is only known to an accuracy of 2 epsilon; and the nth line to an accuracy of 2n epsilon;.
In practice this means that, initially, the entries in successive lines of a table of differences will tend to decrease (just as we saw when we used exact arithmetic) but because the errors are increasing there will come a point when the errors swamp the calculation and the entries in successive lines will tend to increase.
As an example consider the entries in a table of sines (x represents degrees).
x 10 11 12 13 14 15 16 17 sin x 0.1736 0.1908 0.2079 0.225 0.2419 0.2588 0.2756 0.2924
Here is the table of differences
0.1736 0.1908 0.2079 0.2250 0.2419 0.2588 0.2756 0.2924 0.0172 0.0171 0.171 0.169 0.0169 0.168 0.0168 -0.0001 0.0000 -0.0002 0.0000 -0.0001 0.0000 0.0001 -0.0001 -0.0002 0.0002 -0.0001 0.0001 -0.0002 -0.0001 0.0004 -0.0003 0.0002 -0.0001 0.0005 -0.0007 0.0005 0.0006 -0.0012 0.0012
It is clear that only the first two lines of the table of differences (and perhaps, if we think hard, the third) carry any information. In the remaining lines the `noise' of the errors drowns out everything else.
This means that when we interpolate we must confine ourselves to using the first two lines and our version of * will read
$\sin (10+k)\approx0.1736+0.0172k.$
(However, as the reader can verify, this remains a pretty accurate formula for $k$ with $0\leq k\leq 1$.)
Exercise16 Choose a table (or make your own) of some function, form the table of differences, note the line at which error noise becomes dominant. Use the part of the table above that line to interpolate at some point. Check the accuracy of your interpolation. It is very instructive to investigate how the number of significant figures and the spacing of the points of your table affect the result.
We can now see one way in which we can compile tables of a function f like ln or sin with a given level of accuracy. We compute f at a relatively small number of points to much higher accuracy than the table demands. (Of course, these calculations may be quite lengthy but we only need do a few of them.) We obtain the values at f at the remaining points by easy interpolation. Notice that once we have obtained a few points using * we can the build up the rest of the table using the techniques of Exercise 16.
For four hundred years science and technology depended on tables. Books of tables made it possible to find the orbits of the planets, to navigate the globe, and to make an industrial revolution. The book of logarithms was as much a symbol of science as the microscope. Yet, even then, few people bothered to celebrate the labours of the maker of tables. Nowadays their work is completely forgotten. I hope that this essay may help the reader appreciate the work of these unsung heros. | 5,091 | 15,906 | {"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": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2015-32 | longest | en | 0.940378 |
http://ccuart.org/diagram-fasa-sn-pb | 1,534,747,876,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221215858.81/warc/CC-MAIN-20180820062343-20180820082343-00468.warc.gz | 73,253,931 | 20,912 | # Diagram Fasa Sn Pb
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## Pecel Sawi Pembentukan Struktur Mikro Di Garis Pendinginan Pada
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## Diagram Fasa Pb Sn Meine Notizen basic understanding
Regarding Image description: Image has been added by author. We thank you for your visit to CCUART. Make sure you get the information you are looking for diagram fasa pb sn meine notizen . Do not forget to share and love our reference to help further develop our website. | 519 | 2,057 | {"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 | 0.680271 |
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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 9 53 Copyright © Orchard Publications Exercises 9.10 Exercises 1 . Find the Z transform of the discrete time pulse defined as 2 . Find the Z transform of where is defined as in Exercise 1. 3 . Prove the following Z transform pairs: a. b. c. d. e. 4 . Use the partial fraction expansion to find given that 5 . Use the partial fraction expansion method to compute the Inverse Z transform of 6 . Use the Inversion Integral to compute the Inverse Z transform of 7 . Use the long division method to compute the first 5 terms of the discrete time sequence whose Z transform is 8 . a. Compute the transfer function of the difference equation b. Compute the response when the input is pn [] 1 n 0 1 2 m1 ,,, , = 0 otherwise = a n δ n 1 ⇔δ n1 z m na n u 0 n az za () 2 ------------------ n 2 a n u 0 n az z a + 3 ---------------------- + u 0 n z 2 z1 2 fn Z 1 Fz = A 1z 1 10 . 5 z 1 -------------------------------------------------- = z 2 + z0 . 7 5 2 ------------------------------------------- = 12 z 1 z 3 ++ 1 . 5 z 1 = z 1 z 2 z 3 + 1 z 2 4z 3 +++ ----------------------------------------------- = yn yn 1 Tx n 1 = xn e naT =
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Chapter 9 Discrete Time Systems and the Z Transform 9 54 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications 9 . Given the difference equation a. Compute the discrete transfer function b. Compute the response to the input 10 .A d i sc re te time system is described by the difference equation where a. Compute the transfer function b. Compute the impulse response c. Compute the response when the input is 11 . Given the discrete transfer function write the difference equation that relates the output to the input . yn [] yn 1 T 2 --- xn xn 1 + {} = Hz () xn e naT = + = 0 for n 0 < = hn 10 n 0 = z2 + 8z 2 2z 3 ---------------------------- =
Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 9 55 Copyright © Orchard Publications Solutions to End of Chapter Exercises 9.11 Solutions to End of Chapter Exercises 1 .
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9.
2x +3 + x +9 = 180 => 3x + 12 = 180 => 3x = 168 => x = 56 => m(<ABD) = 2*56 +3 = 115. | 128 | 423 | {"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-2017-04 | latest | en | 0.917377 |
http://popflock.com/learn?s=Continuous_function_(topology) | 1,611,594,951,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703587074.70/warc/CC-MAIN-20210125154534-20210125184534-00057.warc.gz | 87,031,459 | 47,529 | Continuous Function (topology)
Get Continuous Function Topology essential facts below. View Videos or join the Continuous Function Topology discussion. Add Continuous Function Topology to your PopFlock.com topic list for future reference or share this resource on social media.
Continuous Function Topology
In mathematics, a continuous function is a function that does not have any abrupt changes in value, known as discontinuities. More precisely, sufficiently small changes in the input of a continuous function result in arbitrarily small changes in its output. If not continuous, a function is said to be discontinuous. Up until the 19th century, mathematicians largely relied on intuitive notions of continuity, during which attempts such as the epsilon-delta definition were made to formalize it.
Continuity of functions is one of the core concepts of topology, which is treated in full generality below. The introductory portion of this article focuses on the special case where the inputs and outputs of functions are real numbers. A stronger form of continuity is uniform continuity. In addition, this article discusses the definition for the more general case of functions between two metric spaces. In order theory, especially in domain theory, one considers a notion of continuity known as Scott continuity. Other forms of continuity do exist but they are not discussed in this article.
As an example, the function H(t) denoting the height of a growing flower at time t would be considered continuous. In contrast, the function M(t) denoting the amount of money in a bank account at time t would be considered discontinuous, since it "jumps" at each point in time when money is deposited or withdrawn.
## History
A form of the epsilon-delta definition of continuity was first given by Bernard Bolzano in 1817. Augustin-Louis Cauchy defined continuity of ${\displaystyle y=f(x)}$ as follows: an infinitely small increment ${\displaystyle \alpha }$ of the independent variable x always produces an infinitely small change ${\displaystyle f(x+\alpha )-f(x)}$ of the dependent variable y (see e.g. Cours d'Analyse, p. 34). Cauchy defined infinitely small quantities in terms of variable quantities, and his definition of continuity closely parallels the infinitesimal definition used today (see microcontinuity). The formal definition and the distinction between pointwise continuity and uniform continuity were first given by Bolzano in the 1830s but the work wasn't published until the 1930s. Like Bolzano,[1]Karl Weierstrass[2] denied continuity of a function at a point c unless it was defined at and on both sides of c, but Édouard Goursat[3] allowed the function to be defined only at and on one side of c, and Camille Jordan[4] allowed it even if the function was defined only at c. All three of those nonequivalent definitions of pointwise continuity are still in use.[5]Eduard Heine provided the first published definition of uniform continuity in 1872, but based these ideas on lectures given by Peter Gustav Lejeune Dirichlet in 1854.[6]
## Real functions
### Definition
The function ${\displaystyle f(x)={\tfrac {1}{x}}}$ is continuous on the domain ${\displaystyle \mathbb {R} \smallsetminus \{0\}}$, but is not continuous over the domain ${\displaystyle \mathbb {R} }$ because it is undefined at ${\displaystyle x=0}$
A real function, that is a function from real numbers to real numbers, can be represented by a graph in the Cartesian plane; such a function is continuous if, roughly speaking, the graph is a single unbroken curve whose domain is the entire real line. A more mathematically rigorous definition is given below.[7]
A rigorous definition of continuity of real functions is usually given in a first course in calculus in terms of the idea of a limit. First, a function f with variable x is said to be continuous at the point c on the real line, if the limit of f(x), as x approaches that point c, is equal to the value f(c); and second, the function (as a whole) is said to be continuous, if it is continuous at every point. A function is said to be discontinuous (or to have a discontinuity) at some point when it is not continuous there. These points themselves are also addressed as discontinuities.
There are several different definitions of continuity of a function. Sometimes a function is said to be continuous if it is continuous at every point in its domain. In this case, the function f(x) = tan(x), with the domain of all real x ? (2n+1)?/2, n any integer, is continuous. Sometimes an exception is made for boundaries of the domain. For example, the graph of the function f(x) = , with the domain of all non-negative reals, has a left-hand endpoint. In this case only the limit from the right is required to equal the value of the function. Under this definition f is continuous at the boundary x = 0 and so for all non-negative arguments. The most common and restrictive definition is that a function is continuous if it is continuous at all real numbers. In this case, the previous two examples are not continuous, but every polynomial function is continuous, as are the sine, cosine, and exponential functions. Care should be exercised in using the word continuous, so that it is clear from the context which meaning of the word is intended.
Using mathematical notation, there are several ways to define continuous functions in each of the three senses mentioned above.
Let
${\displaystyle f\colon D\rightarrow \mathbf {R} \quad }$ be a function defined on a subset ${\displaystyle D}$ of the set ${\displaystyle \mathbf {R} }$ of real numbers.
This subset ${\displaystyle D}$ is the domain of f. Some possible choices include
${\displaystyle D=\mathbf {R} \quad }$ (${\displaystyle D}$ is the whole set of real numbers), or, for a and b real numbers,
${\displaystyle D=[a,b]=\{x\in \mathbf {R} \,|\,a\leq x\leq b\}\quad }$ (${\displaystyle D}$ is a closed interval), or
${\displaystyle D=(a,b)=\{x\in \mathbf {R} \,|\,a (${\displaystyle D}$ is an open interval).
In case of the domain ${\displaystyle D}$ being defined as an open interval, ${\displaystyle a}$ and ${\displaystyle b}$ do not belong to ${\displaystyle D}$, and the values of ${\displaystyle f(a)}$ and ${\displaystyle f(b)}$ do not matter for continuity on ${\displaystyle D}$.
#### Definition in terms of limits of functions
The function f is continuous at some point c of its domain if the limit of f(x), as x approaches c through the domain of f, exists and is equal to f(c).[8] In mathematical notation, this is written as
${\displaystyle \lim _{x\to c}{f(x)}=f(c).}$
In detail this means three conditions: first, f has to be defined at c (guaranteed by the requirement that c is in the domain of f). Second, the limit on the left hand side of that equation has to exist. Third, the value of this limit must equal f(c).
The formal definition of a limit implies that every function is continuous at every isolated point of its domain.
#### Definition in terms of neighborhoods
A neighborhood of a point c is a set that contains, at least, all points within some fixed distance of c. Intuitively, a function is continuous at a point c if the range of f over the neighborhood of c shrinks to a single point f(c) as the width of the neighborhood around c shrinks to zero. More precisely, a function f is continuous at a point c of its domain if, for any neighborhood ${\displaystyle N_{1}(f(c))}$ there is a neighborhood ${\displaystyle N_{2}(c)}$ in its domain such that ${\displaystyle f(x)\in N_{1}(f(c))}$ whenever ${\displaystyle x\in N_{2}(c).}$
This definition only requires that the domain and the codomain are topological spaces and is thus the most general definition. It follows from this definition that a function f is automatically continuous at every isolated point of its domain. As a specific example, every real valued function on the set of integers is continuous.
#### Definition in terms of limits of sequences
The sequence exp(1/n) converges to exp(0)
One can instead require that for any sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ of points in the domain which converges to c, the corresponding sequence ${\displaystyle \left(f(x_{n})\right)_{n\in \mathbb {N} }}$ converges to f(c). In mathematical notation, ${\displaystyle \forall (x_{n})_{n\in \mathbb {N} }\subset D:\lim _{n\to \infty }x_{n}=c\Rightarrow \lim _{n\to \infty }f(x_{n})=f(c)\,.}$
#### Weierstrass and Jordan definitions (epsilon-delta) of continuous functions
Illustration of the ?-?-definition: for ?=0.5, c=2, the value ?=0.5 satisfies the condition of the definition.
Explicitly including the definition of the limit of a function, we obtain a self-contained definition: Given a function f : D -> R as above and an element x0 of the domain D, f is said to be continuous at the point x0 when the following holds: For any number ? > 0, however small, there exists some number ? > 0 such that for all x in the domain of f with x0 - ? < x < x0 + ?, the value of f(x) satisfies
${\displaystyle f(x_{0})-\varepsilon
Alternatively written, continuity of f : D -> R at x0 ? D means that for every ? > 0 there exists a ? > 0 such that for all x ? D :
${\displaystyle |x-x_{0}|<\delta \Rightarrow |f(x)-f(x_{0})|<\varepsilon .}$
More intuitively, we can say that if we want to get all the f(x) values to stay in some small neighborhood around f(x0), we simply need to choose a small enough neighborhood for the x values around x0. If we can do that no matter how small the f(x) neighborhood is, then f is continuous at x0.
In modern terms, this is generalized by the definition of continuity of a function with respect to a basis for the topology, here the metric topology.
Weierstrass had required that the interval x0 - ? < x < x0 + ? be entirely within the domain D, but Jordan removed that restriction.
#### Definition in terms of control of the remainder
In proofs and numerical analysis we often need to know how fast limits are converging, or in other words, control of the remainder. We can formalise this to a definition of continuity. A function ${\displaystyle C:[0,\infty )\to [0,\infty ]}$ is called a control function if
• C is non decreasing
• ${\displaystyle \inf _{\delta >0}C(\delta )=0}$
A function f : D -> R is C-continuous at x0 if
${\displaystyle |f(x)-f(x_{0})|\leq C(|x-x_{0}|)}$ for all ${\displaystyle x\in D}$
A function is continuous in x0 if it is C-continuous for some control function C.
This approach leads naturally to refining the notion of continuity by restricting the set of admissible control functions. For a given set of control functions ${\displaystyle {\mathcal {C}}}$ a function is ${\displaystyle {\mathcal {C}}}$-continuous if it is ${\displaystyle C}$-continuous for some ${\displaystyle C\in {\mathcal {C}}}$. For example, the Lipschitz and Hölder continuous functions of exponent α below are defined by the set of control functions
${\displaystyle {\mathcal {C}}_{\mathrm {Lipschitz} }=\{C:C(\delta )=K|\delta |,\ K>0\}}$
respectively
${\displaystyle {\mathcal {C}}_{{\text{Hölder}}-\alpha }=}$${\displaystyle \{C:C(\delta )=K|\delta |^{\alpha },\ K>0\}}$.
#### Definition using oscillation
The failure of a function to be continuous at a point is quantified by its oscillation.
Continuity can also be defined in terms of oscillation: a function f is continuous at a point x0 if and only if its oscillation at that point is zero;[9] in symbols, ${\displaystyle \omega _{f}(x_{0})=0.}$ A benefit of this definition is that it quantifies discontinuity: the oscillation gives how much the function is discontinuous at a point.
This definition is useful in descriptive set theory to study the set of discontinuities and continuous points - the continuous points are the intersection of the sets where the oscillation is less than ? (hence a G? set) - and gives a very quick proof of one direction of the Lebesgue integrability condition.[10]
The oscillation is equivalent to the ?-? definition by a simple re-arrangement, and by using a limit (lim sup, lim inf) to define oscillation: if (at a given point) for a given ?0 there is no ? that satisfies the ?-? definition, then the oscillation is at least ?0, and conversely if for every ? there is a desired ?, the oscillation is 0. The oscillation definition can be naturally generalized to maps from a topological space to a metric space.
#### Definition using the hyperreals
Cauchy defined continuity of a function in the following intuitive terms: an infinitesimal change in the independent variable corresponds to an infinitesimal change of the dependent variable (see Cours d'analyse, page 34). Non-standard analysis is a way of making this mathematically rigorous. The real line is augmented by the addition of infinite and infinitesimal numbers to form the hyperreal numbers. In nonstandard analysis, continuity can be defined as follows.
A real-valued function f is continuous at x if its natural extension to the hyperreals has the property that for all infinitesimal dx, is infinitesimal[11]
(see microcontinuity). In other words, an infinitesimal increment of the independent variable always produces to an infinitesimal change of the dependent variable, giving a modern expression to Augustin-Louis Cauchy's definition of continuity.
### Construction of continuous functions
The graph of a cubic function has no jumps or holes. The function is continuous.
Checking the continuity of a given function can be simplified by checking one of the above defining properties for the building blocks of the given function. It is straightforward to show that the sum of two functions, continuous on some domain, is also continuous on this domain. Given
${\displaystyle f,g\colon D\rightarrow \mathbf {R} }$,
then the sum of continuous functions
${\displaystyle s=f+g}$
(defined by ${\displaystyle s(x)=f(x)+g(x)}$ for all ${\displaystyle x\in D}$) is continuous in ${\displaystyle D}$.
The same holds for the product of continuous functions,
${\displaystyle p=f\cdot g}$
(defined by ${\displaystyle p(x)=f(x)\cdot g(x)}$ for all ${\displaystyle x\in D}$) is continuous in ${\displaystyle D}$.
Combining the above preservations of continuity and the continuity of constant functions and of the identity function ${\displaystyle I(x)=x}$ one arrives at the continuity of all polynomial functions such as
f(x) = x3 + x2 - 5x + 3
(pictured on the right).
The graph of a continuous rational function. The function is not defined for x=-2. The vertical and horizontal lines are asymptotes.
In the same way it can be shown that the reciprocal of a continuous function
${\displaystyle r=1/f}$
(defined by ${\displaystyle r(x)=1/f(x)}$ for all ${\displaystyle x\in D}$ such that ${\displaystyle f(x)\neq 0}$) is continuous in ${\displaystyle D\smallsetminus \{x:f(x)=0\}}$.
This implies that, excluding the roots of ${\displaystyle g}$, the quotient of continuous functions
${\displaystyle q=f/g}$
(defined by ${\displaystyle q(x)=f(x)/g(x)}$ for all ${\displaystyle x\in D}$, such that ${\displaystyle g(x)\neq 0}$) is also continuous on ${\displaystyle D\smallsetminus \{x:g(x)=0\}}$.
For example, the function (pictured)
${\displaystyle y(x)={\frac {2x-1}{x+2}}}$
is defined for all real numbers and is continuous at every such point. Thus it is a continuous function. The question of continuity at does not arise, since is not in the domain of y. There is no continuous function F: R -> R that agrees with y(x) for all .
The sinc and the cos functions
Since the function sine is continuous on all reals, the sinc function G(x)=sin(x)/x, is defined and continuous for all real x ? 0. However, unlike the previous example, G can be extended to a continuous function on all real numbers, by defining the value G(0) to be 1, which is the limit of G(x), when x approaches 0, i.e.,
${\displaystyle G(0)=\lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1.}$
Thus, by setting
${\displaystyle G(x)={\begin{cases}{\frac {\sin(x)}{x}}&{\text{ if }}x\neq 0\\1&{\text{ if }}x=0,\end{cases}}}$
the sinc-function becomes a continuous function on all real numbers. The term removable singularity is used in such cases, when (re)defining values of a function to coincide with the appropriate limits make a function continuous at specific points.
A more involved construction of continuous functions is the function composition. Given two continuous functions
${\displaystyle \quad g\colon D_{g}\subseteq \mathbf {R} \rightarrow R_{g}\subseteq \mathbf {R} \quad {\text{and}}\quad f\colon D_{f}\subseteq \mathbf {R} \rightarrow R_{f}\subseteq D_{g},}$
their composition, denoted as ${\displaystyle c=g\circ f\colon D_{f}\rightarrow \mathbf {R} }$, and defined by ${\displaystyle c(x)=g(f(x)),}$ is continuous.
This construction allows stating, for example, that
${\displaystyle e^{\sin(\ln x)}}$ is continuous for all ${\displaystyle x>0.}$
### Examples of discontinuous functions
Plot of the signum function. It shows that ${\displaystyle \lim _{n\to \infty }\operatorname {sgn} \left({\tfrac {1}{n}}\right)\neq \operatorname {sgn} \left(\lim _{n\to \infty }{\tfrac {1}{n}}\right)}$. Thus, the signum function is discontinuous at 0 (see section 2.1.3).
An example of a discontinuous function is the Heaviside step function ${\displaystyle H}$, defined by
${\displaystyle H(x)={\begin{cases}1&{\text{ if }}x\geq 0\\0&{\text{ if }}x<0\end{cases}}}$
Pick for instance ${\displaystyle \varepsilon =1/2}$. Then there is no around ${\displaystyle x=0}$, i.e. no open interval ${\displaystyle (-\delta ,\;\delta )}$ with ${\displaystyle \delta >0,}$ that will force all the ${\displaystyle H(x)}$ values to be within the of ${\displaystyle H(0)}$, i.e. within ${\displaystyle (1/2,\;3/2)}$. Intuitively we can think of this type of discontinuity as a sudden jump in function values.
Similarly, the signum or sign function
${\displaystyle \operatorname {sgn}(x)={\begin{cases}\;\;\ 1&{\text{ if }}x>0\\\;\;\ 0&{\text{ if }}x=0\\-1&{\text{ if }}x<0\end{cases}}}$
is discontinuous at ${\displaystyle x=0}$ but continuous everywhere else. Yet another example: the function
${\displaystyle f(x)={\begin{cases}\sin \left(x^{-2}\right)&{\text{ if }}x\neq 0\\0&{\text{ if }}x=0\end{cases}}}$
is continuous everywhere apart from ${\displaystyle x=0}$.
Point plot of Thomae's function on the interval (0,1). The topmost point in the middle shows f(1/2) = 1/2.
Besides plausible continuities and discontinuities like above, there are also functions with a behavior, often coined pathological, for example, Thomae's function,
${\displaystyle f(x)={\begin{cases}1&{\text{ if }}x=0\\{\frac {1}{q}}&{\text{ if }}x={\frac {p}{q}}{\text{(in lowest terms) is a rational number}}\\0&{\text{ if }}x{\text{ is irrational}}.\end{cases}}}$
is continuous at all irrational numbers and discontinuous at all rational numbers. In a similar vein, Dirichlet's function, the indicator function for the set of rational numbers,
${\displaystyle D(x)={\begin{cases}0&{\text{ if }}x{\text{ is irrational }}(\in \mathbb {R} \smallsetminus \mathbb {Q} )\\1&{\text{ if }}x{\text{ is rational }}(\in \mathbb {Q} )\end{cases}}}$
is nowhere continuous.
### Properties
#### A useful lemma
Let ${\displaystyle f(x)}$ be a function that is continuous at a point ${\displaystyle x_{0},}$ and ${\displaystyle y_{0}}$ be a value such ${\displaystyle f(x_{0})\neq y_{0}.}$ Then ${\displaystyle f(x)\neq y_{0}}$ throughout some neighbourhood of ${\displaystyle x_{0}.}$[12]
Proof: By the definition of continuity, take ${\displaystyle \varepsilon ={\frac {|y_{0}-f(x_{0})|}{2}}>0}$ , then there exists ${\displaystyle \delta >0}$ such that
${\displaystyle |f(x)-f(x_{0})|<{\frac {|y_{0}-f(x_{0})|}{2}}\quad {\text{whenever}}\quad |x-x_{0}|<\delta }$
Suppose there is a point in the neighbourhood ${\displaystyle |x-x_{0}|<\delta }$ for which ${\displaystyle f(x)=y_{0};}$ then we have the contradiction
${\displaystyle |f(x_{0})-y_{0}|<{\frac {|f(x_{0})-y_{0}|}{2}}.}$
#### Intermediate value theorem
The intermediate value theorem is an existence theorem, based on the real number property of completeness, and states:
If the real-valued function f is continuous on the closed interval [ab] and k is some number between f(a) and f(b), then there is some number c in [ab] such that f(c) = k.
For example, if a child grows from 1 m to 1.5 m between the ages of two and six years, then, at some time between two and six years of age, the child's height must have been 1.25 m.
As a consequence, if f is continuous on [ab] and f(a) and f(b) differ in sign, then, at some point c in [ab], f(c) must equal zero.
#### Extreme value theorem
The extreme value theorem states that if a function f is defined on a closed interval [a,b] (or any closed and bounded set) and is continuous there, then the function attains its maximum, i.e. there exists c ? [a,b] with f(c) >= f(x) for all x ? [a,b]. The same is true of the minimum of f. These statements are not, in general, true if the function is defined on an open interval (a,b) (or any set that is not both closed and bounded), as, for example, the continuous function f(x) = 1/x, defined on the open interval (0,1), does not attain a maximum, being unbounded above.
#### Relation to differentiability and integrability
${\displaystyle f\colon (a,b)\rightarrow \mathbf {R} }$
is continuous, as can be shown. The converse does not hold: for example, the absolute value function
${\displaystyle f(x)=|x|={\begin{cases}\;\;\ x&{\text{ if }}x\geq 0\\-x&{\text{ if }}x<0\end{cases}}}$
is everywhere continuous. However, it is not differentiable at x = 0 (but is so everywhere else). Weierstrass's function is also everywhere continuous but nowhere differentiable.
The derivative f?(x) of a differentiable function f(x) need not be continuous. If f?(x) is continuous, f(x) is said to be continuously differentiable. The set of such functions is denoted C1((a, b)). More generally, the set of functions
${\displaystyle f\colon \Omega \rightarrow \mathbf {R} }$
(from an open interval (or open subset of R) ? to the reals) such that f is n times differentiable and such that the n-th derivative of f is continuous is denoted Cn(?). See differentiability class. In the field of computer graphics, properties related (but not identical) to C0, C1, C2 are sometimes called G0 (continuity of position), G1 (continuity of tangency), and G2 (continuity of curvature); see Smoothness of curves and surfaces.
Every continuous function
${\displaystyle f\colon [a,b]\rightarrow \mathbf {R} }$
is integrable (for example in the sense of the Riemann integral). The converse does not hold, as the (integrable, but discontinuous) sign function shows.
#### Pointwise and uniform limits
A sequence of continuous functions fn(x) whose (pointwise) limit function f(x) is discontinuous. The convergence is not uniform.
Given a sequence
${\displaystyle f_{1},f_{2},\dotsc \colon I\rightarrow \mathbf {R} }$
of functions such that the limit
${\displaystyle f(x):=\lim _{n\rightarrow \infty }f_{n}(x)}$
exists for all x in D, the resulting function f(x) is referred to as the pointwise limit of the sequence of functions (fn)n?N. The pointwise limit function need not be continuous, even if all functions fn are continuous, as the animation at the right shows. However, f is continuous if all functions fn are continuous and the sequence converges uniformly, by the uniform convergence theorem. This theorem can be used to show that the exponential functions, logarithms, square root function, and trigonometric functions are continuous.
### Directional and semi-continuity
Discontinuous functions may be discontinuous in a restricted way, giving rise to the concept of directional continuity (or right and left continuous functions) and semi-continuity. Roughly speaking, a function is right-continuous if no jump occurs when the limit point is approached from the right. Formally, f is said to be right-continuous at the point c if the following holds: For any number ? > 0 however small, there exists some number ? > 0 such that for all x in the domain with , the value of f(x) will satisfy
${\displaystyle |f(x)-f(c)|<\varepsilon .}$
This is the same condition as for continuous functions, except that it is required to hold for x strictly larger than c only. Requiring it instead for all x with yields the notion of left-continuous functions. A function is continuous if and only if it is both right-continuous and left-continuous.
A function f is lower semi-continuous if, roughly, any jumps that might occur only go down, but not up. That is, for any ? > 0, there exists some number ? > 0 such that for all x in the domain with , the value of f(x) satisfies
${\displaystyle f(x)\geq f(c)-\epsilon .}$
The reverse condition is upper semi-continuity.
## Continuous functions between metric spaces
The concept of continuous real-valued functions can be generalized to functions between metric spaces. A metric space is a set X equipped with a function (called metric) dX, that can be thought of as a measurement of the distance of any two elements in X. Formally, the metric is a function
${\displaystyle d_{X}\colon X\times X\rightarrow \mathbf {R} }$
that satisfies a number of requirements, notably the triangle inequality. Given two metric spaces (X, dX) and (Y, dY) and a function
${\displaystyle f\colon X\rightarrow Y}$
then f is continuous at the point c in X (with respect to the given metrics) if for any positive real number ?, there exists a positive real number ? such that all x in X satisfying dX(x, c) < ? will also satisfy dY(f(x), f(c)) < ?. As in the case of real functions above, this is equivalent to the condition that for every sequence (xn) in X with limit lim xn = c, we have lim f(xn) = f(c). The latter condition can be weakened as follows: f is continuous at the point c if and only if for every convergent sequence (xn) in X with limit c, the sequence (f(xn)) is a Cauchy sequence, and c is in the domain of f.
The set of points at which a function between metric spaces is continuous is a G? set - this follows from the ?-? definition of continuity.
This notion of continuity is applied, for example, in functional analysis. A key statement in this area says that a linear operator
${\displaystyle T\colon V\rightarrow W}$
between normed vector spaces V and W (which are vector spaces equipped with a compatible norm, denoted ||x||) is continuous if and only if it is bounded, that is, there is a constant K such that
${\displaystyle \|T(x)\|\leq K\|x\|}$
for all x in V.
### Uniform, Hölder and Lipschitz continuity
For a Lipschitz continuous function, there is a double cone (shown in white) whose vertex can be translated along the graph, so that the graph always remains entirely outside the cone.
The concept of continuity for functions between metric spaces can be strengthened in various ways by limiting the way ? depends on ? and c in the definition above. Intuitively, a function f as above is uniformly continuous if the ? does not depend on the point c. More precisely, it is required that for every real number ? > 0 there exists ? > 0 such that for every cb ? X with dX(bc) < ?, we have that dY(f(b), f(c)) < ?. Thus, any uniformly continuous function is continuous. The converse does not hold in general, but holds when the domain space X is compact. Uniformly continuous maps can be defined in the more general situation of uniform spaces.[13]
A function is Hölder continuous with exponent ? (a real number) if there is a constant K such that for all b and c in X, the inequality
${\displaystyle d_{Y}(f(b),f(c))\leq K\cdot (d_{X}(b,c))^{\alpha }}$
holds. Any Hölder continuous function is uniformly continuous. The particular case is referred to as Lipschitz continuity. That is, a function is Lipschitz continuous if there is a constant K such that the inequality
${\displaystyle d_{Y}(f(b),f(c))\leq K\cdot d_{X}(b,c)}$
holds for any b, c in X.[14] The Lipschitz condition occurs, for example, in the Picard-Lindelöf theorem concerning the solutions of ordinary differential equations.
## Continuous functions between topological spaces
Another, more abstract, notion of continuity is continuity of functions between topological spaces in which there generally is no formal notion of distance, as there is in the case of metric spaces. A topological space is a set X together with a topology on X, which is a set of subsets of X satisfying a few requirements with respect to their unions and intersections that generalize the properties of the open balls in metric spaces while still allowing to talk about the neighbourhoods of a given point. The elements of a topology are called open subsets of X (with respect to the topology).
A function
${\displaystyle f\colon X\rightarrow Y}$
between two topological spaces X and Y is continuous if for every open set V ? Y, the inverse image
${\displaystyle f^{-1}(V)=\{x\in X\;|\;f(x)\in V\}}$
is an open subset of X. That is, f is a function between the sets X and Y (not on the elements of the topology TX), but the continuity of f depends on the topologies used on X and Y.
This is equivalent to the condition that the preimages of the closed sets (which are the complements of the open subsets) in Y are closed in X.
An extreme example: if a set X is given the discrete topology (in which every subset is open), all functions
${\displaystyle f\colon X\rightarrow T}$
to any topological space T are continuous. On the other hand, if X is equipped with the indiscrete topology (in which the only open subsets are the empty set and X) and the space T set is at least T0, then the only continuous functions are the constant functions. Conversely, any function whose range is indiscrete is continuous.
### Continuity at a point
Continuity at a point: For every neighborhood V of f(x), there is a neighborhood U of x such that f(U) ? V
The translation in the language of neighborhoods of the (?, ?)-definition of continuity leads to the following definition of the continuity at a point:
This definition is equivalent to the same statement with neighborhoods restricted to open neighborhoods and can be restated in several ways by using preimages rather than images.
Also, as every set that contains a neighborhood is also a neighborhood, and ${\displaystyle f^{-1}(V)}$ is the largest subset U of X such that f(U) ? V, this definition may be simplified into:
As an open set is a set that is a neighborhood of all its points, a function ${\displaystyle f:X\rightarrow Y}$ is continuous at every point of X if and only if it is a continuous function.
If X and Y are metric spaces, it is equivalent to consider the neighborhood system of open balls centered at x and f(x) instead of all neighborhoods. This gives back the above ?-? definition of continuity in the context of metric spaces. In general topological spaces, there is no notion of nearness or distance. If however the target space is a Hausdorff space, it is still true that f is continuous at a if and only if the limit of f as x approaches a is f(a). At an isolated point, every function is continuous.
### Alternative definitions
Several equivalent definitions for a topological structure exist and thus there are several equivalent ways to define a continuous function.
#### Sequences and nets
In several contexts, the topology of a space is conveniently specified in terms of limit points. In many instances, this is accomplished by specifying when a point is the limit of a sequence, but for some spaces that are too large in some sense, one specifies also when a point is the limit of more general sets of points indexed by a directed set, known as nets. A function is (Heine-)continuous only if it takes limits of sequences to limits of sequences. In the former case, preservation of limits is also sufficient; in the latter, a function may preserve all limits of sequences yet still fail to be continuous, and preservation of nets is a necessary and sufficient condition.
In detail, a function f: X -> Y is sequentially continuous if whenever a sequence (xn) in X converges to a limit x, the sequence (f(xn)) converges to f(x). Thus sequentially continuous functions "preserve sequential limits". Every continuous function is sequentially continuous. If X is a first-countable space and countable choice holds, then the converse also holds: any function preserving sequential limits is continuous. In particular, if X is a metric space, sequential continuity and continuity are equivalent. For non first-countable spaces, sequential continuity might be strictly weaker than continuity. (The spaces for which the two properties are equivalent are called sequential spaces.) This motivates the consideration of nets instead of sequences in general topological spaces. Continuous functions preserve limits of nets, and in fact this property characterizes continuous functions.
For instance, consider the case of real-valued functions of one real variable:[15]
Theorem. A function ${\displaystyle f\colon A\subseteq \mathbb {R} \to \mathbb {R} }$ is continuous at ${\displaystyle x_{0}}$ if and only if it is sequentially continuous at that point.
Proof. Assume that ${\displaystyle f\colon A\subseteq \mathbb {R} \to \mathbb {R} }$ is continuous at ${\displaystyle x_{0}}$ (in the sense of ${\displaystyle \epsilon -\delta }$ continuity). Let ${\displaystyle (x_{n})_{n\geq 1}}$ be a sequence converging at ${\displaystyle x_{0}}$ (such a sequence always exists, e.g. ${\displaystyle x_{n}=x,\forall n}$); since ${\displaystyle f}$ is continuous at ${\displaystyle x_{0}}$
${\displaystyle \forall \epsilon >0\,\exists \delta _{\epsilon }>0:0<|x-x_{0}|<\delta _{\epsilon }\implies |f(x)-f(x_{0})|<\epsilon .\quad (*)}$
For any such ${\displaystyle \delta _{\epsilon }}$ we can find a natural number ${\displaystyle \nu _{\epsilon }>0}$ such that ${\displaystyle \forall n>\nu _{\epsilon }}$
${\displaystyle |x_{n}-x_{0}|<\delta _{\epsilon },}$
since ${\displaystyle (x_{n})}$ converges at ${\displaystyle x_{0}}$; combining this with ${\displaystyle (*)}$ we obtain
${\displaystyle \forall \epsilon >0\,\exists \nu _{\epsilon }>0:\forall n>\nu _{\epsilon }\quad |f(x_{n})-f(x_{0})|<\epsilon .}$
Assume on the contrary that ${\displaystyle f}$ is sequentially continuous and proceed by contradiction: suppose ${\displaystyle f}$ is not continuous at ${\displaystyle x_{0}}$
${\displaystyle \exists \epsilon >0:\forall \delta _{\epsilon }>0,\,\exists x_{\delta _{\epsilon }}:0<|x_{\delta _{\epsilon }}-x_{0}|<\delta _{\epsilon }\implies |f(x_{\delta _{\epsilon }})-f(x_{0})|>\epsilon }$
then we can take ${\displaystyle \delta _{\epsilon }=1/n,\,\forall n>0}$ and call the corresponding point ${\displaystyle x_{\delta _{\epsilon }}=:x_{n}}$: in this way we have defined a sequence ${\displaystyle (x_{n})_{n\geq 1}}$ such that
${\displaystyle \forall n>0\quad |x_{n}-x_{0}|<{\frac {1}{n}},\quad |f(x_{n})-f(x_{0})|>\epsilon }$
by construction ${\displaystyle x_{n}\to x_{0}}$ but ${\displaystyle f(x_{n})\not \to f(x_{0})}$, which contradicts the hypothesis of sequentially continuity.
#### Closure operator definition
Instead of specifying the open subsets of a topological space, the topology can also be determined by a closure operator (denoted cl) which assigns to any subset A ? X its closure, or an interior operator (denoted int), which assigns to any subset A of X its interior. In these terms, a function
${\displaystyle f\colon (X,\mathrm {cl} )\to (X',\mathrm {cl} ')}$
between topological spaces is continuous in the sense above if and only if for all subsets A of X
${\displaystyle f(\mathrm {cl} (A))\subseteq \mathrm {cl} '(f(A)).}$
That is to say, given any element x of X that is in the closure of any subset A, f(x) belongs to the closure of f(A). This is equivalent to the requirement that for all subsets A' of X'
${\displaystyle f^{-1}(\mathrm {cl} '(A'))\supseteq \mathrm {cl} (f^{-1}(A')).}$
Moreover,
${\displaystyle f\colon (X,\mathrm {int} )\to (X',\mathrm {int} ')}$
is continuous if and only if
${\displaystyle f^{-1}(\mathrm {int} '(A'))\subseteq \mathrm {int} (f^{-1}(A'))}$
for any subset A' of Y.
### Properties
If f: X -> Y and g: Y -> Z are continuous, then so is the composition g ? f: X -> Z. If f: X -> Y is continuous and
The possible topologies on a fixed set X are partially ordered: a topology ?1 is said to be coarser than another topology ?2 (notation1 ? ?2) if every open subset with respect to ?1 is also open with respect to ?2. Then, the identity map
idX: (X, ?2) -> (X, ?1)
is continuous if and only if ?1 ? ?2 (see also comparison of topologies). More generally, a continuous function
${\displaystyle (X,\tau _{X})\rightarrow (Y,\tau _{Y})}$
stays continuous if the topology ?Y is replaced by a coarser topology and/or ?X is replaced by a finer topology.
### Homeomorphisms
Symmetric to the concept of a continuous map is an open map, for which images of open sets are open. In fact, if an open map f has an inverse function, that inverse is continuous, and if a continuous map g has an inverse, that inverse is open. Given a bijective function f between two topological spaces, the inverse function f-1 need not be continuous. A bijective continuous function with continuous inverse function is called a homeomorphism.
If a continuous bijection has as its domain a compact space and its codomain is Hausdorff, then it is a homeomorphism.
### Defining topologies via continuous functions
Given a function
${\displaystyle f\colon X\rightarrow S,}$
where X is a topological space and S is a set (without a specified topology), the final topology on S is defined by letting the open sets of S be those subsets A of S for which f-1(A) is open in X. If S has an existing topology, f is continuous with respect to this topology if and only if the existing topology is coarser than the final topology on S. Thus the final topology can be characterized as the finest topology on S that makes f continuous. If f is surjective, this topology is canonically identified with the quotient topology under the equivalence relation defined by f.
Dually, for a function f from a set S to a topological space X, the initial topology on S is defined by designating as an open set every subset A of S such that ${\displaystyle A=f^{-1}(U)}$ for some open subset U of X. If S has an existing topology, f is continuous with respect to this topology if and only if the existing topology is finer than the initial topology on S. Thus the initial topology can be characterized as the coarsest topology on S that makes f continuous. If f is injective, this topology is canonically identified with the subspace topology of S, viewed as a subset of X.
A topology on a set S is uniquely determined by the class of all continuous functions ${\displaystyle S\rightarrow X}$ into all topological spaces X. Dually, a similar idea can be applied to maps ${\displaystyle X\rightarrow S.}$
## Related notions
Various other mathematical domains use the concept of continuity in different, but related meanings. For example, in order theory, an order-preserving function f: X -> Y between particular types of partially ordered sets X and Y is continuous if for each directed subset A of X, we have sup(f(A)) = f(sup(A)). Here sup is the supremum with respect to the orderings in X and Y, respectively. This notion of continuity is the same as topological continuity when the partially ordered sets are given the Scott topology.[16][17]
${\displaystyle F\colon {\mathcal {C}}\rightarrow {\mathcal {D}}}$
between two categories is called continuous, if it commutes with small limits. That is to say,
${\displaystyle \varprojlim _{i\in I}F(C_{i})\cong F\left(\varprojlim _{i\in I}C_{i}\right)}$
for any small (i.e., indexed by a set I, as opposed to a class) diagram of objects in ${\displaystyle {\mathcal {C}}}$.
A continuity space is a generalization of metric spaces and posets,[18][19] which uses the concept of quantales, and that can be used to unify the notions of metric spaces and domains.[20]
## Notes
1. ^ Bolzano, Bernard (1817), Rein analytischer Beweis des Lehrsatzes dass zwischen je zwey Werthen, die ein entgegengesetztes Resultat gewaehren, wenigstens eine reele Wurzel der Gleichung liege, Prague: Haase
2. ^ Dugac, Pierre (1973), "Eléments d'Analyse de Karl Weierstrass", Archive for History of Exact Sciences, 10: 41-176, doi:10.1007/bf00343406
3. ^ Goursat, E. (1904), A course in mathematical analysis, Boston: Ginn, p. 2
4. ^ Jordan, M.C. (1893), Cours d'analyse de l'École polytechnique, 1 (2nd ed.), Paris: Gauthier-Villars, p. 46
5. ^ Harper, J.F. (2016), "Defining continuity of real functions of real variables", BSHM Bulletin: Journal of the British Society for the History of Mathematics: 1-16, doi:10.1080/17498430.2015.1116053
6. ^ Rusnock, P.; Kerr-Lawson, A. (2005), "Bolzano and uniform continuity", Historia Mathematica, 32 (3): 303-311, doi:10.1016/j.hm.2004.11.003
7. ^ Speck, Jared (2014). "Continuity and Discontinuity" (PDF). MIT Math. p. 3. Retrieved . Example 5. The function 1/x is continuous on (0, ?) and on (-?, 0), i.e., for x > 0 and for x < 0, in other words, at every point in its domain. However, it is not a continuous function since its domain is not an interval. It has a single point of discontinuity, namely x = 0, and it has an infinite discontinuity there.
8. ^ Lang, Serge (1997), Undergraduate analysis, Undergraduate Texts in Mathematics (2nd ed.), Berlin, New York: Springer-Verlag, ISBN 978-0-387-94841-6, section II.4
9. ^ Introduction to Real Analysis, updated April 2010, William F. Trench, Theorem 3.5.2, p. 172
10. ^ Introduction to Real Analysis, updated April 2010, William F. Trench, 3.5 "A More Advanced Look at the Existence of the Proper Riemann Integral", pp. 171-177
11. ^ "Elementary Calculus". wisc.edu.
12. ^ Brown, James Ward, Complex Variables and Applications (8th ed.), McGraw Hill, p. 54, ISBN 978-0-07-305194-9
13. ^ Gaal, Steven A. (2009), Point set topology, New York: Dover Publications, ISBN 978-0-486-47222-5, section IV.10
14. ^ Searcóid, Mícheál Ó (2006), Metric spaces, Springer undergraduate mathematics series, Berlin, New York: Springer-Verlag, ISBN 978-1-84628-369-7, section 9.4
15. ^ Shurman, Jerry (2016). Calculus and Analysis in Euclidean Space (illustrated ed.). Springer. p. 271-272. ISBN 978-3-319-49314-5.
16. ^ Goubault-Larrecq, Jean (2013). Non-Hausdorff Topology and Domain Theory: Selected Topics in Point-Set Topology. Cambridge University Press. ISBN 1107034132.
17. ^ Gierz, G.; Hofmann, K. H.; Keimel, K.; Lawson, J. D.; Mislove, M. W.; Scott, D. S. (2003). Continuous Lattices and Domains. Encyclopedia of Mathematics and its Applications. 93. Cambridge University Press. ISBN 0521803381.
18. ^ Flagg, R. C. (1997). "Quantales and continuity spaces". Algebra Universalis. CiteSeerX 10.1.1.48.851.
19. ^ Kopperman, R. (1988). "All topologies come from generalized metrics". American Mathematical Monthly. 95 (2): 89-97. doi:10.2307/2323060.
20. ^ Flagg, B.; Kopperman, R. (1997). "Continuity spaces: Reconciling domains and metric spaces". Theoretical Computer Science. 177 (1): 111-138. doi:10.1016/S0304-3975(97)00236-3.
## References
This article uses material from the Wikipedia page available here. It is released under the Creative Commons Attribution-Share-Alike License 3.0. | 11,299 | 43,443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 164, "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-2021-04 | latest | en | 0.93792 |
http://www.mytelesupport.com/2016/05/border-name-and-country-line-of-control.html | 1,638,724,362,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363215.8/warc/CC-MAIN-20211205160950-20211205190950-00144.warc.gz | 119,193,239 | 27,452 | Border name and Country | Line of Control and Countries देश की बॉर्डर का नाम सामान्य ज्ञान - Online Solution, Maths Short Tricks, Educational, Mobile Features, Toll Free, customer care number
# Border name and Country | Line of Control and Countries देश की बॉर्डर का नाम सामान्य ज्ञान
## Border name and Country देश की बॉर्डर का नाम | Line of Control and Countries सामान्य ज्ञान
In this post we are going to share some information of international border name and connecting countries name. This post are very helpful for those candidates who are preparing for many competitive exam. This is multiple choice questions and their answer given below the post.
1. 17th Parallel Lines divides which country
A) North Korea & South Korea B) North Vietnam and South Vietnam
C) India & Pakistan D) Canada & USA
2. 26th Parallel
A) Australia and South Africa
B) North Vietnam and South Vietnam
C) India & Pakistan
D) Canada & USA
3. 24th Parallel
A) North Korea & South Korea
B) North Vietnam and South Vietnam
C) India & Pakistan
D) Canada & France
4. 38th Parallel
A) North Korea & South Korea
B) Pakistan and Afghanistan
C) India & Pakistan
D) Canada & USA
5. 49th parallel
A) Australia and South Africa
B) North Vietnam and South Vietnam
C) India & Pakistan
D) Canada & USA
A ) India & Pakistan
B) Pakistan and Afghanistan
C) India And Nepal
D) India & China
7. Mc. Mohan Line
A) India & Pakistan B) Pakistan and Afghanistan
C) India And Nepal D) India (Arunachal Pradesh Region) and China
8. Durand Line
A) India & SriLanka B) Pakistan and Afghanistan
C) India And Nepal D) India (Arunachal Pradesh Region) and China
9. Line of Control
A) Devides India & SriLanka B) Devides Pakistan and Afghanistan
C) Devides India And Nepal D) divides Kashmir Between India and Paktstan
10. Siegfried Line
A) Germany and Poland B) India & Nepal
C) Germany and France D) North & South Korea
11. Hidden Berg Line
A) Germany Switzerland B) Germany and Poland
C) Germany Netherlands D) Two States of Germany
12. Maginot Line
A) Germany and Poland B) USA and UK
C) NetiherLand and France D) France and Germany
13. Older Neisse Line
A) Germany and Poland B) USA and UK
C) Netherlands and France D) France and Germany
14. Mannerheim Line
A) North & South Korea B) USA and UK
C) Russia and Finland D) Germany Neatiherland
1. B) 2. A) 3. C) 4. A) 5. D)
6. A) 7. D) 8. B) 9. D) 10. C)
11) B) 12. D) 13. A) 14. C)
## One word Answers
• 17th Parallel North Vietnam and South Vietnam
• 26th Parallel Australia and South Africa
• 24th Parallel India and Pakistan
• 38th Parallel North Korea and South Korea
• 49th parallel Canada and USA
• Radcliiff Line India and Pakistan
• Mc. Mohan Line India (Arunachal Pradesh Region) and China
• Durand Line Pakistan and Afghanistan
• Line of Contrd This divides Kashmir Between India and Paktstan
• Siegfried Line Geunany and France
• Hidden Berg Line Germany and Poland
• Maginot Line France and Germany
• Older Neisse Line Germany and Poland
• Mannerheim Line Russia and Finland | 909 | 3,439 | {"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-2021-49 | latest | en | 0.539211 |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-3-section-3-3-dividng-polynomials-3-3-exercises-page-274/54 | 1,548,198,816,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583875448.71/warc/CC-MAIN-20190122223011-20190123005011-00176.warc.gz | 809,993,204 | 13,355 | ## Precalculus: Mathematics for Calculus, 7th Edition
$x-2$ is a factor of P(x) since when P(x) is divide by $x-2$, the remainder is 0. (refer to the step by step part for the solution)
RECALL: The factor theorem states that c is a zero of P if and only if x−c is a factor of P(x). $c=2$ therefore $x-c = x-2$ Divide P(x) by $x+1$ using synthetic division to have: | 116 | 365 | {"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} | 3.65625 | 4 | CC-MAIN-2019-04 | latest | en | 0.857732 |
https://bigupanddown.com/homogeneous-markov-chain/ | 1,642,710,562,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302622.39/warc/CC-MAIN-20220120190514-20220120220514-00379.warc.gz | 169,346,837 | 16,714 | # Homogeneous Markov Chain
## Are there any stationary distributions in the Markov chain?
Not all stationary distributions arise this way, however. Some stationary distributions (for instance, certain periodic ones) only satisfy the weaker condition that the average number n n steps approaches the corresponding value of the stationary distribution. That is, if lim n → ∞ H i ( n) n + 1 = x i. .
## What’s the difference between ergodic and absorbing Markov chains?
Ergodic Markov chains have a unique stationary distribution, and absorbing Markov chains have stationary distributions with nonzero elements only in absorbing states. The stationary distribution gives information about the stability of a random process and, in certain cases, describes the limiting behavior of the Markov chain.
## Can a limiting distribution be a stationary distribution?
So, not all stationary distributions are limiting distributions. Sometimes no limiting distribution exists! For time-homogeneous Markov chains, any limiting distribution is a stationary distribution. extbf {P} P.
## Which is the only possible candidate for a stationary distribution?
The only possible candidate for a stationary distribution is the final eigenvector, as all others include negative values. ). Find a stationary distribution for the 2-state Markov chain with stationary transition probabilities given by the following graph:
## Is the time parameter discrete in a Markov chain?
While the time parameter is usually discrete, the state space of a Markov chain does not have any generally agreed-on restrictions: the term may refer to a process on an arbitrary state space. However, many applications of Markov chains employ finite or countably infinite state spaces, which have a more straightforward statistical analysis.
## What is the probability of the Markov chain changing to state E?
For example, if the Markov process is in state A, then the probability it changes to state E is 0.4, while the probability it remains in state A is 0.6.
## Which is an example of a limiting Markov chain?
The two-state Markov chain discussed above is a “nice” one in the sense that it has a well-defined limiting behavior that does not depend on the initial probability distribution (PMF of X0 ). However, not all Markov chains are like that. For example, consider the same Markov chain; however, choose a = b = 1.
## What is the transition matrix of a Markov chain?
Consider a Markov chain with two possible states, S = {0, 1}. In particular, suppose that the transition matrix is given by P = [1 − a a b 1 − b], where a and b are two real numbers in the interval [0, 1] such that 0 < a + b < 2.
## When is a Markov chain called a homogeneous chain?
Homogeneous Markov Chains Defnition A Markov chain is called homogeneous if and only if the transition probabilities are independent of the time t, that is, there exist constants P. i;jsuch that P. i;jrrX. t |X.
## Is the transition probability of a Markov chain the same after each step?
If the Markov chain is time-homogeneous, then the transition matrix P is the same after each step, so the k -step transition probability can be computed as the k -th power of the transition matrix, Pk . If the Markov chain is irreducible and aperiodic, then there is a unique stationary distribution π.
## How are Markov processes used in Bayesian statistics?
Markov processes are the basis for general stochastic simulation methods known as Markov chain Monte Carlo, which are used for simulating sampling from complex probability distributions, and have found extensive application in Bayesian statistics. The adjective Markovian is used to describe something that is related to a Markov process.
## How is a Markov process related to a Markovian process?
The adjectives Markovian and Markov are used to describe something that is related to a Markov process. A Markov process is a stochastic process that satisfies the Markov property (sometimes characterized as ” memorylessness “).
## When does a stochastic process have the Markov property?
A stochastic process has the Markov property if the conditional probability distribution of future states of the process depends only upon the present state, not on the sequence of events that preceded it.
## When does a matrix become the stationary distribution?
In other words, regardless the initial state, the probability of ending up with a certain state is the same. Once such convergence is reached, any row of this matrix is the stationary distribution.
## How is stationary measure related to asymptotics?
The notion of stationary measure provides a more quantitative picture of thelimit behavior of an MC. We first define it and discuss issues of existence anduniqueness. The connection to asymptotics is developed in the next section. 1.1 Definition First the main definition:
## How to prove a finite state Markov chain?
For finite-state Markov chains, either all states in a class are transient or all are recurrent.2 Proof: Assume that state i is transient (i.e., for some j, i j but j 6→i) and suppose that i and m are in the same class (i.e., i ↔ m).
## What can be said about a non homogeneous Markov chain?
Not much of general interest can be said about non homogeneous chains.1 An initial probability distribution forX 0, combined with the transition probabilities{P ij} (or{P ij(n)}for the non-homogeneous case), define the probabilities for all events in the Markov chain.
## Are there stationary distributions in the transition matrix?
1 1 that are stationary distributions expressed as column vectors. Therefore, if the eigenvectors of extbf {P} P. In short, the stationary distribution is a left eigenvector (as opposed to the usual right eigenvectors) of the transition matrix.
## Is the state of Michigan a stationary distribution?
However, there is no noticeable difference in the state’s population of 10 million’s preference at large; in other words, it seems Michigan sports fans have reached a stationary distribution. What might that be?
## Why is the irreducibility of a Markov chain important?
Irreducibility of a Markov chain is important for convergence to equilibrium as t → ∞, because we want the convergence to be independent of start state. This can happen if the chain is irreducible. When the chain is not irreducible, different start states might cause the chain to get stuck in different closed classes.
## What’s the difference between stationary and invariant distribution?
Usually, these are just terms used by different people; some will call a vector π with π P = π and ∑ i π i = 1 a stationary distribution, others will call it an invariant distribution. However, there are some closely related concepts that are different:
## How to simulate from a Markov chain in Python?
One can thus simulate from a Markov Chain by simulating from a multinomial distribution. One way to simulate from a multinomial distribution is to divide a line of length 1 into intervals proportional to the probabilities, and then picking an interval based on a uniform random number between 0 and 1.
## How is Markov chain Monte Carlo used in simulation?
In this chapter, we introduce a general class of algorithms, collectively called Markov chain Monte Carlo (MCMC), that can be used to simulate the posterior from general Bayesian models.
## Is there a way to visualize census data?
How do you calculate demographic projections? In the geometric method of projection, the formula is Pp = P1(1 + r)n where, Pp= Projected population;... | 1,611 | 7,554 | {"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} | 3.390625 | 3 | CC-MAIN-2022-05 | latest | en | 0.858979 |
https://topwritershelp.com/aggregate-plan/ | 1,716,391,035,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00060.warc.gz | 515,786,256 | 18,262 | # Aggregate Plan
Aggregate Plan
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Prepare an aggregate plan for the coming year assuming that the sales forecast is perfect Use a spreadsheet Bradford Manufacturing in the spreadsheet an area has been designated for your aggregate plan solution. Supply the number of packaging lines to run and the number of overtime for each quarter. You will need to set up the cost calculations in the spreadsheet. You may want to try using Excel Solver to find a solutions remember that your final solution need an interger number of lines and an interger number of overtime hours for each quarters (solutions that requires 8.9134 lines and 1.256 hours of overtime and not feasible.
Information for the case:
1. Currently the plant Is running 10 lines with no overtime. Each line requires six people to run. For planning purposes, the lines are run for 7.5hours each normal shift. Employees, though, are paid for eight hour s’ work. It is possible to run up to two hour s of overtime each day, but it must be scheduled for a week at a time, and all the lines must run overtime when it is scheduled. Workers are paid \$20.00/hour during a regular shift and\$30.00/hour on overtime. The standard production rate for each line is 450cases/hour .
2. The marketing forecast for demand is as follows:QI-2,000;Q2-2,200;Q3-2,500;Q4-2,650; and Q1 (next year )-2,200.These numbers are in 1,000-case units. Each number represents a 13-week forecast.
3. Management has instructed manufacturing to maintain a two-week supply of pudding inventory in the warehouses. The two-week supply should be based on future expected sales. The following are ending inventory target levels for each quarter :QI-338;Q2-385; Q3-408;Q4-338.
4. Inventory carrying cost is estimated by accounting to be\$1.00 per case per year .This means that if a case of pudding is held in inventory for an entire year ,the cost to just carry that case in inventory is \$1.00.If a case is carried for only one week, the cost is \$1.00/52, or \$0.01923 The cost is proportional to the time carried in inventory. There are200,000 cases in inventory at the beginning of Q1(this is 200 cases in the 1,000-case units that the forecast is given in).
5. If a stock out occurs, the item is backordered and shipped at a later date. The cost when a backorder occur s is\$2.40 per case due to the loss of goodwill and the high cost of emergency shipping.
6. The human resource group estimates that it costs \$5,000 to hire and train a new production employee .It costs \$3,000to layoff a production worker
Tags: | 640 | 2,676 | {"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-2024-22 | latest | en | 0.923432 |
https://mysqlpreacher.com/what-is-the-process-of-window-to-viewport-transformation/ | 1,685,815,360,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00572.warc.gz | 456,583,206 | 31,911 | Categories :
## What is the process of window to viewport transformation?
Window to Viewport Transformation is the process of transforming 2D world-coordinate objects to device coordinates. Objects inside the world or clipping window are mapped to the viewport which is the area on the screen where world coordinates are mapped to be displayed.
## How do you convert a window to viewport transform in computer graphics?
Matrix Representation of the above three steps of Transformation:
1. Step1:Translate window to origin 1. Tx=-Xwmin Ty=-Ywmin
2. Step2:Scaling of the window to match its size to the viewport. Sx=(Xymax-Xvmin)/(Xwmax-Xwmin)
3. Step3:Again translate viewport to its correct position on screen. Tx=Xvmin
4. Note:
Which transformation is referred to as window to viewport transformation of windowing transformation?
• Window to Viewport Mapping- Mapping of a part of a world coordinate scene to device coordinate is referred to as a viewing transformation.
### How is window to viewport mapping attained?
The mapping of window to viewport also can be achieved by performing following transformations:
1. By performing scaling transformation using a fixed-point position of (Xwmin, Ywmin) that scales the window area to the size of the viewport area.
2. By translating the scaled window area to the position of the viewport area.
### What is the difference between window and viewport in computer graphics?
The window, in effect, defines the portion of the graph that is to be displayed in world coordinates, and the viewport specifies the area on the device on which the image is to appear.
Why is viewing transformation needed?
Remember that purpose of the viewing transformation is to orient the objects in a coordinate system where the center of projection is located at the origin. This coordinate system is called either eye space or camera space.
#### What is the difference between window port and viewport?
Difference between Window Port and Viewport Window port is the coordinate area specially selected for the display. Viewport is the display area of viewport in which the window is perfectly mapped.
#### What is difference between window and viewport?
A window defines a rectangular area in world coordinates. A viewport defines in normalized coordinates a rectangular area on the display device where the image of the data appears. You define a viewport with the GPORT command.
What is a window in CG?
The method of selecting and enlarging a portion of a drawing is called windowing. The area chosen for this display is called a window. The window is selected by world-coordinate. Viewport: An area on display device to which a window is mapped [where it is to displayed]. …
## What is difference between window port and viewport?
The main difference between window port and view port is that window port is a world coordinate area selected for displaying while view port is a device coordinate area that locates the scene on the device. Overall, window port and view port are two ways of displaying objects in computer graphics.
## What viewport defines?
A viewport is a region of the screen used to display a portion of the total image to be shown. In virtual desktops, the viewport is the visible portion of a 2D area which is larger than the visualization device.
What is meant by viewing transformation?
Viewing Transformation is the mapping of coordinates of points and lines that form the picture into appropriate coordinates on the display device. Mapping the window onto a subregion of the display device called the viewport is called the Viewing Transformation.
### How does the window to viewport transformation work?
Window to Viewport Transformation is the process of transforming 2D world-coordinate objects to device coordinates. Objects inside the world or clipping window are mapped to the viewport which is the area on the screen where world coordinates are mapped to be displayed. Window – It is the area on world coordinate selected for display.
### How is the clipping window and the viewport used?
The viewport is rectangular area on screen where world coordinates are mapped to be displayed. • In other words, the clipping window is used to select the part of the scene that is to be displayed. The viewport is used to display selected portion of window on the output device. 4. Window to viewport transformation 5.
Where does the image appear in the viewport?
Viewport(Image Subspace) −A rectangle on the raster graphics screen (or interface window) defining where the image will appear, usually the entire screen or interface window. −Thus, in principle, the same image can be replicated on different viewports inside the screen or interface window.
#### How to translate scaled window to viewport?
Translate the scaled window area to the position of the viewport. Relative proportions of objects are maintained if the scaling factors are the same (sx=sy). From normalized coordinates, object descriptions are mapped to the various display devices. | 970 | 5,036 | {"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} | 3.3125 | 3 | CC-MAIN-2023-23 | latest | en | 0.905713 |
https://jp.mathworks.com/matlabcentral/answers/477757-how-can-i-change-the-response-to-forced-excitation?s_tid=prof_contriblnk | 1,585,914,470,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370510846.12/warc/CC-MAIN-20200403092656-20200403122656-00407.warc.gz | 524,087,246 | 20,964 | # How can I change the response to forced excitation?
11 ビュー (過去 30 日間)
sam plant 2019 年 8 月 27 日
コメント済み: Anmol Dhiman 2019 年 9 月 5 日
function Structural_drift
% Define initial conditions x(0) & x'(0)
x0 = [0 1];
% Time interval
t = [0 120];
% Solve equation of motion
[T, X] = ode45(@EOM, t, x0);
plot(T, X(:,1), T, X(:,2));
% EOM function & variables
function dx = EOM(t, x)
% Define initial conditions x(0) & x'(0)
% Time interval
M =87500; % Mass (kg)
k = 20000; % Stiffness value
c = 170; % Damping constant of structure
Cd = 5000;
Fd = Cd*x(2); %damping force
f=5; %forcing frequency in Hz
w=2*pi*f; %forcing frequency
F = 500000*sin(w*t) ; %excitation force
term1 = F;
term2 = k*x(1);
term3 = c*x(2);
term4 = Fd;
term5 = M;
dx = [x(2); ((term1 - term2 - term3 - term4)/(term5))]; % Equation of motion
end
end
When using ODE45 for an equation of motion. The forcing amplitude is F. When i change this value the peak value of displacement from the solver isn't changing, it's always the same value regardless of the force size?
any help would be seriously appreciated
#### 1 件のコメント
Anmol Dhiman 2019 年 9 月 5 日
I am unable to reproduce the above issue as I am getting different peak value of displacement as shown in the figures below. Can you provide an example with plots explaining the issue.
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### 回答 (0 件)
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Translated by | 450 | 1,371 | {"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-2020-16 | latest | en | 0.568188 |
https://physics.stackexchange.com/questions/8095/measuring-acceleration-of-earth-due-to-its-fall-around-the-sun | 1,563,260,639,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195524503.7/warc/CC-MAIN-20190716055158-20190716081158-00403.warc.gz | 505,840,654 | 37,721 | # Measuring acceleration of earth due to its fall around the sun
Every orbiting of a satellite around a mass is nothing else but a constant fall - and therefore acceleration - towards this mass. In a way it is a "falling around" that mass.
My question
Is it possible to measure this acceleration on earth due to its "falling around" the sun?
• Related to this question: physics.stackexchange.com/questions/5994/… – sum1stolemyname Apr 5 '11 at 14:24
• We can measure the effects of differential accelleration, as tidal forces. – Omega Centauri Apr 5 '11 at 15:32
• I can't wait to see someone answer something along the lines "because of the equivalence principle, one can measure only local accelerations relative to an inertial frame" and then someone mumble something back about Mach principle... – lurscher Apr 5 '11 at 17:19
Here's a much more direct recipe that I think would uncontroversially count as measuring the acceleration. Pick a set of fixed stars, and measure the velocity of the Earth relative to them using the Doppler effect. Any one measurement only gives you one component of ${\bf v}$, but if you measure a few, you can get the full vector. Do this twice, at two different times, and calculate $\Delta{\bf v}/\Delta t$.
The acceleration can be found by a number of means. $F~=~ma$ with the Newtonian law of gravity and the centripetal acceleration $a~=~v^2/r$, $$m\frac{v^2}{r}~=~-\frac{GMm}{r^2}.$$ This gives $v~=~\sqrt{GM/r}$, which is $v~=~29.5km/s$ or $v~=~2.95\times 10^4m/s$, for the mass of the sun and $r~=~1.5\times 10^11m$. The acceleration is then $0.0058m/s^2$ | 441 | 1,599 | {"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} | 3.5625 | 4 | CC-MAIN-2019-30 | latest | en | 0.905064 |
https://hpmuseum.org/forum/thread-15825-post-138430.html | 1,713,335,767,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817144.49/warc/CC-MAIN-20240417044411-20240417074411-00337.warc.gz | 282,660,866 | 7,038 | 11-02-2020, 06:28 AM
Post: #1
dah145 Junior Member Posts: 37 Joined: Aug 2020
Sorry if this had been specifically asked before, I searched but couldn't find anything.
Suppose I want to factorize the expression: a*s^(2) + b*s + c for the variable s. In other CAS systems this could be done simply by specifying it inside the factor(expression,var) function, in calculators such as the TI 89, see image attached. I haven't found a way to achieve this on the HP Prime as its factor function doesn't work the same way.
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11-02-2020, 07:58 AM (This post was last modified: 11-02-2020 07:59 AM by parisse.)
Post: #2
parisse Senior Member Posts: 1,309 Joined: Dec 2013
factor factors over the field of coefficients of the arguments. If you want to extend this field, you must give a 2nd argument specifying the extension.
For example
Code:
P:=a*s^2+b*s+c; l:=solve(P=0,s); factor(P,l[0])
11-02-2020, 05:40 PM
Post: #3
dah145 Junior Member Posts: 37 Joined: Aug 2020
[quote='parisse' pid='138432' dateline='1604303926']
factor factors over the field of coefficients of the arguments. If you want to extend this field, you must give a 2nd argument specifying the extension.
For example
Code:
P:=a*s^2+b*s+c; l:=solve(P=0,s); factor(P,l[0])
Thank you very much, never would have guessed that, but it makes sense.
Now, just another question, let's say I want to factor the expression: a^2 -2*a*b+b^2+s^2, it's easy to see that it is equivalent to: s^2 + (a-b)^2. I know I can get to the second expression in the HP Prime by applying the factor function like this: factor(a^2 -2*a*b+b^2)+s^2, but the point if it is possible for it to be done automatically with a function. This is particularly useful, for example, to getting the simplified expressions of typical laplace transforms with symbolic coefficients, such as the ones attached.
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11-02-2020, 07:32 PM
Post: #4
Albert Chan Senior Member Posts: 2,516 Joined: Jul 2018
Hi, dah145
You can isolate the terms to be factorize, like this.
XCas> factor2(mess,s) := poly2symb(factor(symb2poly(mess,s)),s)
XCas> factor2(a^2 - 2*a*b + b^2 + s^2, a) → b^2+s^2+a*(a-2*b)
XCas> factor2(a^2 - 2*a*b + b^2 + s^2, s) → s^2+(a-b)^2
11-03-2020, 03:46 AM
Post: #5
dah145 Junior Member Posts: 37 Joined: Aug 2020
(11-02-2020 07:32 PM)Albert Chan Wrote: Hi, dah145
You can isolate the terms to be factorize, like this.
XCas> factor2(mess,s) := poly2symb(factor(symb2poly(mess,s)),s)
XCas> factor2(a^2 - 2*a*b + b^2 + s^2, a) → b^2+s^2+a*(a-2*b)
XCas> factor2(a^2 - 2*a*b + b^2 + s^2, s) → s^2+(a-b)^2
Yes thanks, this definitely is what I was looking for. Now I wonder if it is possible to obtain the expressions in form of a product, say: (a^2 - 2*a*b + b^2 + s^2)*(a^2 - 2*a*b + b^2 + s^2) = (s^2+(a-b)^2)*(s^2+(a-b)^2), as using your custom function outputs a not so friendly expression: s^2*(2*(a^2 + b^2) + s^2)+(a+b)^2*(a-b)^2.
11-19-2020, 04:12 PM (This post was last modified: 11-19-2020 10:00 PM by dah145.)
Post: #6
dah145 Junior Member Posts: 37 Joined: Aug 2020
(11-03-2020 03:46 AM)dah145 Wrote:
(11-02-2020 07:32 PM)Albert Chan Wrote: Hi, dah145
You can isolate the terms to be factorize, like this.
XCas> factor2(mess,s) := poly2symb(factor(symb2poly(mess,s)),s)
XCas> factor2(a^2 - 2*a*b + b^2 + s^2, a) → b^2+s^2+a*(a-2*b)
XCas> factor2(a^2 - 2*a*b + b^2 + s^2, s) → s^2+(a-b)^2
Yes thanks, this definitely is what I was looking for. Now I wonder if it is possible to obtain the expressions in form of a product, say: (a^2 - 2*a*b + b^2 + s^2)*(a^2 - 2*a*b + b^2 + s^2) = (s^2+(a-b)^2)*(s^2+(a-b)^2), as using your custom function outputs a not so friendly expression: s^2*(2*(a^2 + b^2) + s^2)+(a+b)^2*(a-b)^2.
Just wanted to update, I wrote a little program that factorizes polynomials to a more friendly expression than the built in factor function, I attached an example. A is the expression and B is the variable to factorize.
PHP Code:
#casfacto(A,B):=BEGINLOCAL fac, coe, pola, polb;fac:=factors(A);coe:={0};polb:=[0];FOR N FROM 1 TO SIZE(fac)/2 DOcoe[N]:=coeff(fac[2*N-1],B);END;pola:=factor(coe);FOR N FROM 1 TO SIZE(fac)/2 DOpolb[N]:=poly2symb(pola[N],B);END;RETURN regroup(product(polb[n]^(fac[2*n]),n,1,SIZE(polb)));END;#end
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https://www.smore.com/psenu | 1,540,044,168,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512750.15/warc/CC-MAIN-20181020121719-20181020143219-00398.warc.gz | 1,070,819,249 | 10,350 | # Rational and Irrational Numbers
## Rational Numbers
• In mathematics, a rational number is any number that can be expressed as the quotient or fraction p/q of two integers, p and q, with the denominator q not equal to zero. Since q may be equal to 1, every integer is a rational number.
• Types of Rational Numbers:
• Whole Numbers
• Fractions
• Terminating Decimals
• Repeating Decimals
• Mixed Numbers
• Perfect Squares Roots and Cube Roots
## Irrational Numbers
• In mathematics, an irrational number is any real number that cannot be expressed as a ratio of integers. Irrational numbers cannot be represented as terminating or repeating decimals.
• Types of Irrational Numbers
• Non-Terminating Decimals
• Pi
• Non-Perfect Square Roots and Cube Roots
• A decimal that does not end
• i.e. 4.2354713……. | 191 | 823 | {"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} | 3.3125 | 3 | CC-MAIN-2018-43 | latest | en | 0.843982 |
https://studylib.net/doc/12392657/cot5405--fall-2006-lecture-9 | 1,679,877,371,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946584.94/warc/CC-MAIN-20230326235016-20230327025016-00670.warc.gz | 620,215,701 | 10,181 | # COT5405: Fall 2006 Lecture 9
```COT5405: Fall 2006
Lecture 9
Approximation guarantee for the primal-dual algorithm
Note: Some of the statements here are informal. Please read the handout from AA for the
formal statements and proofs.
Primal
Dual
Min: j cjxj
Subject to:
Max: i biyi
Subject to:
i aijyi cj, j = 1, ..., n
yi 0, j = 1, ..., n
j aijxj bi, i = 1, ..., m
xj 0, j = 1, ..., n
Complementary slackness
xj = 0 or i aijyi = cj
yi = 0 or j aijxj = bi
Relaxed slackness
xj = 0 or cj/ i aijyi cj, 1
(1)
yi = 0 or bi j aijxj bi, 1
(2)
Proposition 15.1: If x is feasible for the primal, y is feasible for the dual, and they satisfy
the relaxed slackness conditions (1) and (2), then j cjxj i biyi OPT.
Proof:
j cjxj j (i aijyi) xj (from (1))
= i j aijyi xj = i yi j aij xj
i yibi (from (2)) OPT.
Note: Even though we can take cj/ i aijyi only when xj > 0, xj cj/ xj i aijyi when xj 0.
Similarly, yi j aijxj yi bi when yi 0.
Approximation factor for Set Cover Primal-Dual approximation algorithm: Its
relaxed slackness conditions have = 1 and = f in proposition 15.1, yielding a factor f
approximation.
``` | 412 | 1,111 | {"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-2023-14 | latest | en | 0.760716 |
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