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https://www.genealogy.math.ndsu.nodak.edu/id.php?id=116059
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## Stanislav Aleksandrovich Vinogradov
Ph.D. St. Petersburg State University 1968
Dissertation: Interpolation Problems for Analytic Functions Continuous in the Closed Disc and for Functions with Coefficients in $\ell^p$
Mathematics Subject Classification: 30—Functions of a complex variable
Advisor: Victor Petrovich Havin
Students:
Click here to see the students listed in chronological order.
NameSchoolYearDescendants
Kotochigov, AleksandrSt. Petersburg State University1981
Shamoyan, RomiMoscow State Institute of Electronics and Mathematics2001
Shimorin, SergueiSt. Petersburg State University1993
Stoilov, PeyoSt. Petersburg State University1978
According to our current on-line database, Stanislav Vinogradov has 4 students and 4 descendants.
We welcome any additional information.
If you have additional information or corrections regarding this mathematician, please use the update form. To submit students of this mathematician, please use the new data form, noting this mathematician's MGP ID of 116059 for the advisor ID.
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2022-05-19 18:36:24
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https://gomathanswerkey.com/go-math-grade-2-answer-key-chapter-9/
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# Go Math Grade 2 Chapter 9 Answer Key Pdf Length in Metric Units
Go Math Grade 2 Chapter 9 Answer Key Pdf: Hey guys!!! Are you searching for the grade 2 answer key on various websites? If yes then you can stop your search now. Here you can get detailed explanations for all the questions on this page. As per your convenience, we have provided the Go Math Grade 2 Answer Key Chapter 9 Length in Metric Units in pdf format. Hence Download HMH Go Math Grade 2 Chapter 9 Solution Key Length in Metric Units Concepts for free of cost.
## Length in Metric Units Go Math Grade 2 Chapter 9 Answer Key Pdf
Check out the topics before starting your preparation for the exams. We have provided solutions for all the questions as per the textbook. Test yourself by solving the problems given in the mid-chapter checkpoint and review test. By solving the review test problems you can know how much you have learned from this chapter. Click on the below-given links and kickstart your preparation.
Chapter: 9 – Length in Metric Units
Lesson: 1 Measure with a Centimeter Model
Lesson: 2 Estimate Lengths in Centimeters
Lesson: 3 Measure with a Centimeter Ruler
Lesson: 4 Problem Solving • Add and Subtract Lengths
Mid-Chapter Checkpoint
Lesson: 5 Centimeters and Meters
Lesson: 6 Estimate Lengths in Meters
Lesson: 7 Measure and Compare Lengths
Review/Test
### Length in Metric Units Show What You Know
Compare Lengths
Question 1.
Order the strings from shortest to longest.
Write 1, 2, 3.
By seeing the above figure we can order the strings from shortest to longest.
Use Nonstandard Units to Measure Length
Question 2.
Explanation:
Use the cube to measure the length of the pencil. The estimated measure of a pencil is 7
Question 3.
Explanation:
Use the cube to measure the length of the pen. The estimated measure of a pen is 7.
Measure Length Twice: Nonstandard Units
Question 4.
Question 5.
### Length in Metric Units Vocabulary Builder
Visualize It
Fill in the graphic organizer. Think of an object and write about how you can measure the length of that object.
Understand Vocabulary
Use the color tiles to estimate the length of each straw.
Question 1.
By using the color tiles we can measure the length of the straw.
Question 2.
By using the color tiles we can measure the length of the straw.
### Length in Metric Units Game Estimating Length
Materials
Play with a partner.
1. Take turns choosing a picture. Find the real object.
2. Each player estimates the length of the object in cubes and then makes a cube train for his or her estimate.
3. Compare the cube trains to the length of the object. The player with the closer estimate puts a counter on the picture. If there is a tie, both players put a counter on the picture.
4. Repeat until all pictures are covered. The player with more counters on the board wins.
### Length in Metric Units Vocabulary Game
Go Fish
For 3 players
Materials
• 4 sets of word cards
How to Play
1. Every player is dealt 5 cards. Put the rest face-down in a draw pile.
2. Ask another player for a word card to match a word card you have.
• If the player has the word card, he or she gives it to you. Put both cards in front of you. Take another turn.
• If the player does not have the word, he or she answers, “Go fish.” Take a card from the pile. If the word you get matches one you are holding, put both cards in front of you. Take another turn. If it does not match, your turn is over.
3. The game is over when one player has no cards left. The player with the most pairs wins.
The Write Way
Reflect
Choose one idea. Write about it in the space below.
• Compare a centimeter to a meter. Explain how they are alike and how they are different.
• Explain how you would find the length of this crayon in centimeters.
How would you compare the length of a door to the length of a window in meters? Draw pictures and write to explain. Use another piece of paper for your drawing.
### Lesson 9.1 Measure with a Centimeter Model
Essential Question How do you use a centimeter model to measure the lengths of objects?
Listen and Draw
Use to measure the length.
______ unit cubes
______ unit cubes
______ unit cubes
HOME CONNECTION • Your child used unit cubes as an introduction to measurement of length before using metric measurement tools.
MATHEMATICAL PRACTICES
Use Tools
Describe how to use unit cubes to measure an object’s length.
We can measure the length of the object by locating the unit cubes in a sequence.
Share and Show
Use a unit cube. Measure the length in centimeters.
Question 1.
Answer: about 22 centimeters as one cube is equal to one centimeter
Question 2.
Answer: about 13 centimeters as one cube is equal to one centimeter
Question 3.
Answer: about 13 centimeters as one cube is equal to one centimeter
Use a unit cube. Measure the length in centimeters.
Question 4.
Answer: 7 centimeters as one cube is equal to one centimeter
Question 5.
Answer: about 18 centimeters as one cube is equal to one centimeter
Question 6.
Answer: about 14 centimeters as one cube is equal to one centimeter
Question 7.
Answer: about 10 centimeters as one cube is equal to one centimeter
Question 8.
Answer: about 17 centimeters as one cube is equal to one centimeter
Problem Solving • Applications
Solve. Write or draw to explain.
Question 9.
THINK SMARTER
Mrs. Duncan measured the lengths of a crayon and a pencil. The pencil is double the length of the crayon. The sum of their lengths is 24 centimeters. What are their lengths?
crayon : _____________________
pencil : _____________
Let the length of the crayon be x
Length of a pencil = 2x
As according to the question that the sum of their length is 24 centimeters.
So we have
x + 2x = 24
3x = 24
x = 8
Thus the length of the crayon = 8 centimeters
Length of pencil = 2x
Substitute the value of x
2 × 8 = 16
Thus the length of pencil is 16 centimeters and
length of crayons is 8 centimeters.
Question 13.
THINK SMARTER +
Marita uses unit cubes to measure the length of a straw. Circle the number in the box that makes the sentence true.
Answer: The straw is 7 centimeters long.
TAKE HOME ACTIVITY • Have your child compare the lengths of other objects to those in this lesson.
### Measure with a Centimeter Model Homework & Practice 9.1
Use a unit cube. Measure the length in centimeters.
Question 1.
Question 2.
Question 3.
Problem Solving
Solve. Write or draw to explain.
Question 4.
Susan has a pencil that is 3 centimeters shorter than this string. How long is the pencil?
Question 5.
WRITE
Write about using a unit cube to measure lengths in this lesson.
______________
_____________
Let us say your cube has the dimensions of 3 × 4 × 5 to find the area of the cube in the cubic measure you would multiply the three dimensions together and divide by the area of the cubes.
Lesson Check
Question 1.
Sarah used unit cubes to measure the length of a ribbon. Each unit cube is about 1 centimeter long. What is a good estimate for the length of ribbon?
_________ centimeters
Spiral Review
Question 2.
What is the time on this clock?
Question 3.
What is the time on this clock?
Question 4.
Dan has a paper strip that is 28 inches long. He tears 6 inches off the strip. How long is the paper strip now?
_________ inches
Explanation:
Given Dan has a paper strip that is 28 inches long
He tears 6 inches off the strip
so 28 inches – 6 inches = 22 inches
22 inches long is the paper strip.
Question 5.
Rita has 1 quarter, 1 dime, and 2 pennies. What is the total value of Rita’s coins?
### Lesson 9.3 Measure with a Centimeter Ruler
Essential Question How do you use a centimeter ruler to measure lengths?
Listen and Draw
Find three small objects in the classroom.
Use unit cubes to measure their lengths.
Draw the objects and write their lengths.
HOME CONNECTION • Your child used unit cubes to measure the lengths of some classroom objects as an introduction to measuring lengths in centimeters.
MATHEMATICAL PRACTICES
Apply. Compare
Describe how the three lengths compare. Which object is shortest?
Answer: pencil with 7 centimeters is shortest
Share and Show
Measure the length to the nearest centimeter.
Question 1.
_______ centimeters
Question 2.
_______ centimeters
Question 3.
________ centimeters
Measure the length to the nearest centimeter.
Question 4.
________ centimeters
Question 5.
_______ centimeters
Question 6.
________ centimeters
Question 7.
________ centimeters
Question 8.
GO DEEPER
A marker is almost 13 centimeters long. This length ends between which two centimeter-marks on a ruler?
____________________
_____________________
It ends between 13 cm and 14 cm mark.
Explanation:
Since this marker in the above question is almost 13 centimeters long and definitely ends between two centimeter marks, the two centimeters that the length of this marker ends between on a ruler should be 13cm and 14 cm marks on the ruler.
Problem Solving • Applications
Question 9.
THINK SMARTER
The crayon was on the table next to the centimeter ruler. The left edge of the crayon was not lined up with the zero mark on the ruler.
What is the length of the crayon?
____________________
____________________
Explanation:
Shown in the given figure that length of the crayon is between 3 to 11 cms
hence 11-3 we get 8 centimeters
Question 10.
THINK SMARTER
This is Lee’s string. Hana’s string is 7 centimeters long. Whose string is longer? Use a centimeter ruler to find out. Explain.
____________________
___________________
Answer: Hana’s string is longer than Lee’s string.
TAKE HOME ACTIVITY • Have your child measure the lengths of some objects using a centimeter ruler.
### Measure with a Centimeter Ruler Homework & Practice 9.3
Measure the length to the nearest centimeter.
Question 1.
_________ centimeters
Question 2.
__________ centimeters
Problem Solving
Question 3.
Draw a string that is about 8 centimeters long.
Then use a centimeter ruler to check the length.
Question 4.
WRITE
Measure the length of the top of your desk in centimeters.
Describe how you found the length.
___________________
___________________
by using rulers found the length of the desk
Lesson Check
Question 1.
Use a centimeter ruler. What is the length of this pencil to the nearest centimeter?
_______ centimeters
Spiral Review
Question 2.
What is the time on this clock?
Question 3.
What is the total value of this group of coins?
$______ or ______ cents Answer: 41c Question 4. Use the line plot. How many pencils are 5 inches long? ______ pencils Answer: 2 ### Lesson 9.4 Problem Solving • Add and Subtract Lengths Essential Question How can drawing a diagram help when solving problems about lengths? Nate had 23 centimeters of string. He gave 9 centimeters of string to Myra. How much string does Nate have now? Unlock the Problem What information do I need to use? Nate had _________ centimeters of string. He gave _______ centimeters of string to Myra. Show how to solve the problem. Nate has ________ centimeters of string now. Answer: Nate has 14 centimeters of string now. HOME CONNECTION • Your child drew a diagram to represent a problem about lengths. The diagram can be used to choose the operation for solving the problem. Try Another Problem Draw a diagram. Write a number sentence using a ☐ for the missing number. Then solve. Question 1. Ellie has a ribbon that is 12 centimeters long. Gwen has a ribbon that is 9 centimeters long. How many centimeters of ribbon do they have? ______________ They have _________ centimeters of ribbon. Answer: 12 + 9 = 21 cm Thus They have 21 centimeters of ribbon. Question 2. A string is 24 centimeters long. Justin cuts 8 centimeters off. How long is the string now? ___________ Now the string is _________ centimeters long. Answer: 16 cm 24 – 8 = 16 cm Now the string is 16 centimeters long. MATHEMATICAL PRACTICES Explain how your diagram shows what happened in the first problem. Answer: Share and Show Draw a diagram. Write a number sentence using a ☐ for the missing number. Then solve. Question 3. A chain of paper clips is 18 centimeters long. Sondra adds 6 centimeters of paper clips to the chain. How long is the chain now? _____________ The chain is _______ centimeters long now. Answer: 18 + 6 = 24 cm Thus The chain is 24 centimeters long now. Question 4. THINK SMARTER A ribbon was 22 centimeters long. Then Martha cut a piece off to give to Tao. Now the ribbon is 5 centimeters long. How many centimeters of ribbon did Martha give to Tao? Martha gave _________ centimeters of ribbon to Tao. Answer: Given, A ribbon was 22 centimeters long. Then Martha cut a piece off to give to Tao. Now the ribbon is 5 centimeters long. 22 – 5 = 17 cm Thus Martha gave 17 centimeters of ribbon to Tao. TAKE HOME ACTIVITY • Have your child explain how he or she used a diagram to solve one problem in this lesson. ### Problem Solving • Add and Subtract Lengths Homework & Practice 9.4 Draw a diagram. Write a number sentence using a ☐ for the missing number. Then solve. Question 1. A straw is 20 centimeters long. Mr. Jones cuts off 8 centimeters of the straw. How long is the straw now? ________________ The straw is _______ centimeters long now. Answer: Given, A straw is 20 centimeters long. Mr. Jones cuts off 8 centimeters of the straw. 20 – 8 = 12 cm The straw is 12 centimeters long now. Question 2. WRITE Describe a diagram for a problem about the total length of two ribbons, 13 centimeters long and 5 centimeters long. Answer: 13 + 5 = 18 centimeters Lesson Check Question 1. Tina has a paper clip chain that is 25 centimeters long. She takes off 8 centimeters of the chain. How long is the chain now? ________ centimeters Answer: Given, Tina has a paper clip chain that is 25 centimeters long. She takes off 8 centimeters of the chain. 25 – 8 = 17 centimeters The chain is 17 centimeters long. Spiral Review Question 2. What is the sum? Answer: 472 Question 3. What is another way to write the time half past 7? _____ : ______ Answer: 07 : 30 Question 4. Molly has these coins in her pocket. How much money does she have in her pocket?$ ______ or ______ cents
1 quarter = 25 cents
1 dime = 10 cents
1 dime = 10 cents
1 nickel = 5 cents
25c + 10c + 10c + 5c + 50c
= 100 cents
### Length in Metric Units Mid-Chapter Checkpoint
Concepts and Skills
Use a unit cube. Measure the length in centimeters.
Question 1.
Question 2.
Question 3.
The pencil is about 11 centimeters long. Circle the best estimate for the length of the string.
Question 4.
THINK SMARTER
Use a centimeter ruler. What is the length of this ribbon to the nearest centimeter?
_______ centimeters
### Lesson 9.5 Centimeters and Meters
Essential Question How is measuring in meters different from measuring in centimeters?
Listen and Draw
Draw or write to describe how you did each measurement.
1st measurement
2nd measurement
MATHEMATICAL PRACTICES
Describe how the lengths of the yarn and the sheet of paper are different.
Share and Show
Measure to the nearest centimeter.
Then measure to the nearest meter.
Question 1.
______ centimeters
______ meters
100 centimeters
1 meters
Question 2.
______ centimeters
______ meters
250 centimeters
2.5 meters
Question 3.
______ centimeters
______ meters
400 centimeters
4 meters
Measure to the nearest centimeter.
Then measure to the nearest meter.
Question 4.
______ centimeters
______ meters
200 centimeters
2 meters
Question 5.
______ centimeters
______ meters
150 centimeters
1.5 meters
Question 6.
______ centimeters
______ meters
200 centimeters
2 meters
Question 7.
GO DEEPER
Write these lengths in order from shortest to longest.
200 centimeters
1 meter
10 meters
Problem Solving • Applications
Question 8.
THINK SMARTER
Mr. Ryan walked next to a barn. He wants to measure the length of the barn. Would the length be a greater number of centimeters or a greater number of meters? Explain your answer.
Meters as it is a realistic figure, centimeters would be far too small.
Question 9.
THINK SMARTER
Write the word on the tile that makes the sentence true.
A bench is 2 _______ long.
A pencil is 15 _______ long.
A paper clip is 3 ______ long.
A bed is 3 ______ long.
A bench is 2 meters long.
A pencil is 15 centimeters long.
A paper clip is 3 centimeters long.
A bed is 3 meters long.
TAKE HOME ACTIVITY • Have your child describe how centimeters and meters are different.
### Centimeters and Meters Homework & Practice 9.5
Measure to the nearest centimeter.
Then measure to the nearest meter.
Question 1.
______ centimeters
______ meters
150 centimeters
1.5 meters
Question 2.
______ centimeters
______ meters
100 centimeters
1 meter
Problem Solving
Question 3.
Sally will measure the length of a wall in both centimeters and meters. Will there be fewer centimeters or fewer meters? Explain.
_________________
_________________
There would be a fewer meter. Since 1 m=100 cm.
Fewer centimeters because there will be fewer centimeters
Question 4.
WRITE
Would you measure the length of a bench in centimeters or in meters? Explain your choice.
Step-by-step explanation:
Reason: Usually, the bench length is bigger than the length of the chair and the table.
So it will be measured using meters. If we use a meter it will easy to measure and accurate because it somewhat bigger in length than a chair and table.
1 meter = 100 centimeters.
The standard length of a bench will 1.5 meters to 2 meters. Usually, 3 to 4 people will sit on the bench.
Lesson Check
Question 1.
Use a centimeter ruler. What is the length of the toothbrush to the nearest centimeter?
______ centimeters
Spiral Review
Question 2.
List a group of coins that equals 65 cents.
____________
_____________
Question 3.
Janet has a poster that is about 3 feet long. Fill in the blanks with the word inches or feet to make the statement true.
3 ______ is longer than 12 ________.
The measurement of feets is longer than inches.
1 feet = 12 inches
So, 3 feet is longer than 12 inches.
Question 4.
Last week, 483 children checked books out from the library. This week, only 162 children checked books out from the library. How many children checked out library books in the last two weeks?
Given,
Last week, 483 children checked books out from the library.
This week, only 162 children checked books out from the library.
483
+162
642
Question 5.
List a group of coins with a value of $1.00? ______________ _____________ Answer: 100 pennies, 20 nickels, 10 dimes, or 4 quarters ### Lesson 9.6 Estimate Lengths in Meters Essential Question How do you estimate the lengths of objects in meters? Listen and Draw Find an object that is about 10 centimeters long. Draw and label it. Answer: pencil Is there a classroom object that is about 50 centimeters long? Draw and label it. Answer: No MATHEMATICAL PRACTICES Describe how the lengths of the two real objects compare. Answer: Measure the length of an object twice, using length units of different lengths for the two measurements Model and Draw Estimate. About how many meter sticks will match the width of a door? about ______ meters Answer: about 1 meter Share and Show Find the real object. Estimate its length in meters. Question 1. bookshelf about _______ meters Answer: 2 meters Question 2. bulletin board about _______ meters Answer: 1 meter On Your Own Find the real object. Estimate its length in meters. Question 3. teacher’s desk about _______ meters Answer: 2.5 meters Question 4. wall about _______ meters Answer: 4 meters Question 5. window about _______ meters Answer: 2 meters Question 6. chalkboard about _______ meters Answer: 2 meters Problem Solving • Applications Question 7. THINK SMARTER R In meters, estimate the distance from your teacher’s desk to the door of your classroom. about ________ meters Explain how you made your estimate. ___________________ ___________________ Answer: 3 meters Explanation: by estimating the distance from your teacher’s desk to the door of your classroom in meters we will get 3 meters by using measuring tape Question 8. THINK SMARTER Estimate the length of an adult’s bicycle. Fill in the bubble next to all the sentences that are true. Answer: The bicycle is about 200 centimeters long. The bicycle is less than 1 meter long. TAKE HOME ACTIVITY • With your child, estimate the lengths of some objects in meters. ### Estimate Lengths in Meters Homework & Practice 9.6 Find the real object. Estimate its length in meters. Question 1. poster about ________ meters Answer: 2 meters Question 2. chalkboard about ________ meters Answer: 2 meters Problem Solving Question 3. Barbara and Luke each placed 2 meter sticks end-to-end along the length of a large table. About how long is the table? about ________ meters Answer: 4 meters Question 4. WRITE Choose one object from above. Describe how you estimated its length. __________________ __________________ Answer: Table By comparing with the real world we can estimate the length of the table. Lesson Check Question 1. What is the best estimate for the length of a real baseball bat? _______ meter Answer: 1 meter Question 2. What is the best estimate for the length of a real couch? _______ meters Answer: 2 meter Spiral Review Question 3. Sara has two$1 bills, 3 quarters, and 1 dime. How much money does she have?
$____ . _____ Answer:$1 + $0.25 +$0.10 = $1.35 Question 4. Use an inch ruler. What is the length of this straw to the nearest inch? _______ inches Answer: 3 inches Question 5. Scott has this money in his pocket. What is the total value of this money?$ _____ . _____
Answer: $1.15 ### Lesson 9.7 Measure and Compare Lengths Essential Question How do you find the difference between the lengths of two objects? Listen and Draw Measure and record each length. _______ centimeters Answer: 14 centimeters _______ centimeters Answer: 8 centimeters HOME CONNECTION • Your child measured these lengths as an introduction to measuring and then comparing lengths. MATHEMATICAL PRACTICES Name a classroom object that is longer than the paintbrush. Explain how you know. Answer: Chalk board as it is 2 meters long which is more than a paint brush Model and Draw How much longer is the pencil than the crayon? The pencil is _______ centimeters longer than the crayon. Answer: 3 centimeters Share and Show Measure the length of each object. Complete the number sentence to find the difference between the lengths. Question 1. The string is ________ centimeters longer than the straw. Answer: 13 – 9 = 4 centimeters Question 2. The paintbrush is ________ centimeters longer than the toothpick. Answer: 15 – 5 = 10 centimeters On Your Own Measure the length of each object. Complete the number sentence to find the difference between the lengths. Question 3. The yarn is ________ centimeters longer than the crayon. Answer: 10 – 8 = 2 Question 4. The string is ________ centimeters longer than the paper clip. Answer: 8 – 4 = 4 centimeters Question 5. THINK SMARTER Use a centimeter ruler. Measure the length of your desk and the length of a book. desk: ______ centimeters book: ______ centimeters Which is shorter? __________ How much shorter is it? __________ Answer: book is shorter than desk. Problem Solving • Applications Analyze Relationships Question 6. Mark has a rope that is 23 centimeters long. He cuts 15 centimeters off. What is the length of the rope now? _________ centimeters Answer: 8 centimeters Question 7. The yellow ribbon is 15 centimeters longer than the green ribbon. The green ribbon is 29 centimeters long. What is the length of the yellow ribbon? _________ centimeters Answer: 44 centimeters Question 8. THINK SMARTER + Measure the length of each object. Which object is longer? How much longer? Explain. ________________ __________________ Answer: Marker is 11 centimeters and the string is 6 centimeters hence the marker is longer than the string TAKE HOME ACTIVITY • Have your child tell you how he or she solved one of the problems in this lesson. ### Measure and Compare Lengths Homework & Practice 9.7 Measure the length of each object. Write a number sentence to find the difference between the lengths. Question 1. The craft stick is ________ centimeters longer than the chalk. Answer: 3 centimeters Explanation: The length of the craft stick is 8 centimeters The length of chalk is 6 centimeters 11 – 8 = 3 Thus the craft stick is 3 centimeters longer than the chalk. Problem Solving Solve. Write or draw to explain. Question 2. A string is 11 centimeters long, a ribbon is 24 centimeters long, and a large paper clip is 5 centimeters long. How much longer is the ribbon than the string? ________ centimeters longer Answer: 19 centimeters longer Question 3. WRITE Suppose the lengths of two strings are 10 centimeters and 17 centimeters. Describe how the lengths of these two strings compare. _____________________ ______________________ Answer: The length of the second string is 7 centimeters longer than the first string. Lesson Check Question 1. How much longer is the marker than the paper clip? Circle the correct answer. 11 centimeters longer 8 centimeters longer 10 centimeters longer 5 centimeters longer Answer: 11 centimeters Spiral Review Question 2. What is the total value of these coins?$ _______ or ______ cents
Answer: 100 cents and 1 \$
Question 3.
What is a reasonable estimate for the length of a real chalkboard?
_______ feet
Question 4.
Cindy leaves at half past 2. At what time does Cindy leave?
___ : ___
### Length in Metric Units Review/Test
Question 1.
Michael uses unit cubes to measure the length of the yarn. Circle the number in the box that makes the sentence true.
Answer: The yarn is 4 centimeters long.
Question 2.
The paper clip is about 4 centimeters long. Robin says the string is about 7 centimeters long. Gale says the string is about 20 centimeters long.
Which girl has the better estimate? Explain.
_______________
_______________
Answer: Robin has a better estimation.
Question 3.
GO DEEPER
Sandy’s paper chain is 14 centimeters long. Tim’s paper chain is 6 centimeters long. How many centimeters of the paper chain do they have? Draw a diagram. Write a number sentence using a ☐ for the missing number. Then solve.
____________
The paper chain is _________ centimeters long now.
Given,
Sandy’s paper chain is 14 centimeters long. Tim’s paper chain is 6 centimeters long.
14 + 6 = 20
The paper chain is 20 centimeters long now.
Question 4.
Write the word on the tile that makes the sentence true.
A hallway is 4 ______ long.
A marker is 15 ______ long.
A toothpick is 5 ______ long.
A sofa is 2 _______ long.
A hallway is 4 meters long.
A marker is 15 centimeters long.
A toothpick is 5 centimeters long.
A sofa is 2 meters long.
Question 5.
Estimate the length of a real car. Fill in the bubble next to all the sentences that are true.
The car is less than 6 meters long
Question 6.
THINK SMARTER
Measure the length of each object. Does the sentence describe the objects? Choose Yes or No.
Answer: By seeing the above figure we can say that the crayon is 4 centimeters shorter than the marker.
Question 7.
Ethan’s rope is 25 centimeters long. Ethan cuts the rope and gives a piece to Hank. Ethan’s rope is now 16 centimeters long. How many centimeters of rope does Hank have?
Draw a diagram. Write a number sentence using a ☐ for the missing number. Then solve.
______________
Hank has ________ centimeters of rope.
Given,
Ethan’s rope is 25 centimeters long.
Ethan cuts the rope and gives a piece to Hank.
Ethan’s rope is now 16 centimeters long.
25 – 16 = 9 centimeters
Thus Hank has 9 centimeters of rope.
Question 8.
Measure the length of the paintbrush to the nearest centimeter. Circle the number in the box that makes the sentence true.
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2022-08-18 22:10:18
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http://mathoverflow.net/revisions/110181/list
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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
Complicated (the special case $f: X \to \mathbf{F}_q$ proper smooth is Weil I!): Let $\mathcal{F}$ be mixed of weight $i$. \leq i$. Then$R^q\pi_!\mathcal{F}$is mixed of weight$\leq q+i$(see Deligne, Weil II, Théorème 1 (3.3.1) or Kiehl-Weissauer, Theorem I.7.1, strengthened in I.9.3) 1 Complicated: Let$\mathcal{F}$be mixed of weight$i$. Then$R^q\pi_!\mathcal{F}$is mixed of weight$q+i\$ (see Deligne, Weil II, or Kiehl-Weissauer, Theorem I.7.1, strengthened in I.9.3)
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2013-06-20 07:16:21
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http://mathhelpforum.com/calculus/38451-riemann-integrable-problem.html
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3. For the first question consider $f(x) = \left\{ \begin{array}{c}1\mbox{ if }x\in \mathbb{Q} \\ -1\mbox{ if }x\not \in \mathbb{Q} \end{array} \right.$.
For second question use $\max (a,b) = \frac{1}{2}(a+b) - \frac{1}{2}|a-b|$.
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2017-11-21 01:19:15
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https://physics.aps.org/synopsis-for/10.1103/PhysRevLett.106.083002
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Synopsis: A double whammy of x rays
Scientists have for the first time observed two-photon absorption with x rays.
Photons can pair up to ionize an atom when individually they’re not energetic enough to do the job. However, this requires an intense light source so that the atom can absorb two photons in rapid succession. Using the world’s most powerful x-ray laser, scientists have for the first time observed two-photon absorption in the x ray part of the spectrum. In the experiment, which is presented in Physical Review Letters, x rays from the Linac Coherent Light Source (LCLS) stripped the innermost electrons of a neon atom at a rate over $100$ times higher than predicted by theoretical models.
For fifty years now, physicists have been exploring two-photon absorption at energies corresponding to infrared, visible, and ultraviolet wavelengths. The technique has been used to study electronic properties of atoms and molecules, to image human tissues, and to improve the resolution for photolithography. Despite scientific interest in expanding this nonlinear optical effect to higher energies, x-ray sources with adequate intensity have been lacking until now.
Completed in 2009, the LCLS at the SLAC National Accelerator Laboratory produces ultrashort pulses of $10$ trillion x-ray photons. In their experiments, Gilles Doumy, of The Ohio State University, and his colleagues placed a neon target in the beam path and tuned the energy of the x rays to just above and just below the energy threshold ($1196\phantom{\rule{0.333em}{0ex}}\text{eV}$) for ionizing electrons in the $1s$ shell of neon. By analyzing the ions that emerged from the target, the team found evidence of two-photon absorption. The authors expect their results will be an important step towards probing the interiors of solid samples with x rays. – Michael Schirber
More Features »
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2019-02-16 10:33:38
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https://crypto.stackexchange.com/questions/35073/what-is-the-branch-number-of-this-matrix
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# What is the branch number of this matrix?
We have the following matrix: $$\begin{pmatrix}0&1&1&1\\ 1&0&1&1\\ 1&1&0&1\\ 1&1&1&0\end{pmatrix}$$
What is the branch number?
Is this a MDS marix?
• Can any matrix with a 0 entry be MDS? – poncho May 4 '16 at 18:02
• @poncho I would not ask, if I would know. But I suspect it can't be. – LightBit May 4 '16 at 18:04
• Hint: what's the definition of MDS? – poncho May 4 '16 at 18:49
• Also, the answer depends on the ring that the matrix is based on. This matrix is singular if the matrix elements are $GF(3)$ – poncho May 4 '16 at 21:44
• @poncho, you're right that a zero in $A$ constrains the rank. Having all nonzero entries over $GF(2)$ (thus all 1's) results in a singular $A$ and a (relatively) bad code. – kodlu May 5 '16 at 0:28
The matrix is not MDS over $GF(2)$; No binary MDS codes exist and non nonbinary (over $GF(2^n)$ MDS codes would have this generator whose scalar entries are in the field $GF(2)$). Over $GF(2^n)$ The branch number, which is the minimum weight of the corresponding linear code is 4, in $GF(2^n)$ for all $n$. This covers all possible fields of interest for crypto.
Note that you just need to take the given matrix, call it $A,$ form $G=[I | A]$ over the relevant field and determine the minimum weight of the resulting code. Magma (and GAP, Pari, Python) can do this for you easily. This code turns out to be a quasi-cyclic code over the fields $GF(2^n)$ with minimum distance 4.
Edit: It seems that 4 is the limit for the minimum distance of this type of code. In particular a $16\times16$ version, still has minimum distance only 4, for the fields $GF(2^n)$ where $n=1,2,\ldots,8.$ I assume $n=8$ is of interest to the OP, since it corresponds to having bytes as code symbols, as in AES.
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2020-01-28 00:19:28
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https://testbook.com/question-answer/what-is-the-boolean-expression-for-the-two-input-e--60359874976ecb3001c765c6
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# What is the Boolean expression for the two-input Exclusive -OR gate?
This question was previously asked in
UPPCL JE Previous Paper 8 (Held On: 27 November 2019 Shift 2 )
View all UPPCL JE Papers >
1. A̅B + AB̅
2. AB + A̅ B̅
3. AB
4. A + B
Option 1 : A̅B + AB̅
Free
ST 1: Logical Reasoning
1168
20 Questions 20 Marks 20 Mins
## Detailed Solution
XOR GATE
Symbol:
Truth Table:
Input A Input B Output Y = A ⊕ B 0 0 0 0 1 1 1 0 1 1 1 0
Output Equation: $$Y = {\bf{A}} \oplus {\bf{B}} = \bar AB + A\bar B$$
Key Points:
1) The output is low when both the inputs are the same
2) The output is high when both the inputs are different
XNOR GATE
Symbol:
Truth Table:
Input A Input B Output Y = AB 0 0 1 0 1 0 1 0 0 1 1 1
Output Equation: $$Y = \overline {{\bf{A}} \oplus {\bf{B}}} = A \odot B = AB + \bar A\bar B$$
Key Points:
1) The output is high when both the inputs are the same
2) The output is low when both the inputs are different
Important Points
NOT GATE
Symbol:
Truth Table:
Input (A) Output (A̅) 0 (Low) 1 (High) 1 (High) 0 (low)
Output Equation: Y = A̅
Key Points: The output of NOT gate is an invert of the input
AND GATE
Symbol:
Truth Table:
Input A Input B Output Y = A.B 0 0 0 0 1 0 1 0 0 1 1 1
Output Equation: Y = A.B
Key Points: The output is high only when both the inputs are high
OR GATE
Symbol:
Truth Table:
Input A Input B Output Y = A + B 0 0 0 0 1 1 1 0 1 1 1 1
Output Equation: Y = A + B
Key Points: The output is low only when both the inputs are low
NAND GATE
Symbol:
Truth Table:
Input A Input B Output $$Y = \overline {AB}$$ 0 0 1 0 1 1 1 0 1 1 1 0
Output Equation: $$Y = \overline {A.B} = \bar A + \bar B$$
Key Points:
1) The output is low only when both the inputs are high
2) It is a universal gate
NOR GATE
Symbol:
Truth Table:
Input A Input B Output $$Y = \overline {\left( {A + B} \right)}$$ 0 0 1 0 1 0 1 0 0 1 1 0
Output Equation: $$Y = \overline {A + B} = \bar A.\bar B$$
Key Points:
1) The output is high only when both the inputs are low
2) It is a universal gate
|
2021-09-19 07:14:59
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https://www.physicsforums.com/threads/equation-help.50739/
|
# Equation help
1. Oct 31, 2004
### zandra_z
If (x + a)² + b = x² - 6x + 13
find the values of a and b
2. Oct 31, 2004
### shmoe
You have a polynomial on the left hand side and a polynomial on the right hand side. Two polynomials are equal if and only if their coefficients match.
eg.if $$2x^2-3x+4=ax^2+bx+c$$ then we must have $$a=2,b=-3,c=4$$
For your question, try expanding the $$(x+a)^2$$ part and equating coefficients.
3. Oct 31, 2004
### zandra_z
x² + 2ax + a² + b = x² - 6x + 13
the x² cancel out
a² + 2ax + b = 13 - 6x
then I get stuck
4. Oct 31, 2004
### shmoe
Collect like terms (according to power of "x"). Your equation becomes:
$$(2a)x+(a^2+b)=-6x+13$$
Switching the order of the left hand side was just to make things match up nicer. Remember what I said about the coefficients of equal polynomials.
$$2a=??$$
$$a^2+b=??$$
Can you fill in the ??
5. Oct 31, 2004
### arildno
(2a+6)x=13-a^2-b
Since this equation shall hold for ALL choices of x, in particular it must hold for x=0
which means you get the equations:
0=13-a^2-b
And:
2a+6=0
6. Oct 31, 2004
### zandra_z
so:
a=-3
and:
0=16^2-b
b=2
7. Oct 31, 2004
### arildno
How do you get that flawed b-value?????
8. Oct 31, 2004
### zandra_z
I see, I did:
16^(2-b)
I should do:
13-(3^2)-b
b=4
|
2018-02-24 04:34:49
|
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|
https://socratic.org/questions/how-do-you-calculate-the-vapor-pressure-of-a-solution
|
How do you calculate the vapor pressure of a solution?
Dec 24, 2014
This is done by using Raoult's law, which states that, for an ideal solution, the partial vapor pressure of a component in that solution is equal to the mole fraction of that component multiplied by its vapor pressure when pure.
If you're dealing with a solution that has a non-volatile solute (a solute that does not have the tendency to form vapor at the temperature of the solution), then the vapor pressure of that solution is
${P}_{s o l u t i o n} = {\chi}_{s o l v e n t} \cdot {P}_{s o l v e n t}^{0}$, where
${P}_{s o l u t i o n}$ - the vapor pressure of the solution;
${\chi}_{s o l v e n t}$ - the mole fraction of the solvent;
${P}_{s o l v e n t}^{0}$ - the vapor pressure of pure solvent.
The mole fraction simply refers to ratio between solvent moles and the total number of moles in the solution.
If however you're dealing with a solution that contains a volatile solute, the vapor pressure of that solution is
${P}_{s o l u t i o n} = {\chi}_{s o l v e n t} \cdot {P}_{s o l v e n t}^{0} + {\chi}_{s o l u t e} \cdot {P}_{s o l u t e}^{0}$, where
${P}_{s o l u t e}^{0}$ - the vapor pressure of the pure solute;
${\chi}_{s o l u t e}$ = the mole fraction of the solute.
This equation can be extended to solutions containing three or more components.
|
2020-01-22 04:45:44
|
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|
http://nrich.maths.org/public/leg.php?code=12&cl=2&cldcmpid=7800
|
Search by Topic
Resources tagged with Factors and multiples similar to Going for Gold:
Filter by: Content type:
Stage:
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There are 143 results
Broad Topics > Numbers and the Number System > Factors and multiples
X Marks the Spot
Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
Hot Pursuit
Stage: 3 Challenge Level:
The sum of the first 'n' natural numbers is a 3 digit number in which all the digits are the same. How many numbers have been summed?
Which Numbers? (1)
Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. They are the red set, the green set and the blue set. Can you find all the numbers in the sets from these clues?
Three Spinners
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These red, yellow and blue spinners were each spun 45 times in total. Can you work out which numbers are on each spinner?
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Eminit
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The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
Diggits
Stage: 3 Challenge Level:
Can you find what the last two digits of the number $4^{1999}$ are?
Which Numbers? (2)
Stage: 2 Challenge Level:
I am thinking of three sets of numbers less than 101. Can you find all the numbers in each set from these clues?
Table Patterns Go Wild!
Stage: 2 Challenge Level:
Nearly all of us have made table patterns on hundred squares, that is 10 by 10 grids. This problem looks at the patterns on differently sized square grids.
Remainders
Stage: 3 Challenge Level:
I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
Factors and Multiple Challenges
Stage: 3 Challenge Level:
This package contains a collection of problems from the NRICH website that could be suitable for students who have a good understanding of Factors and Multiples and who feel ready to take on some. . . .
Three Neighbours
Stage: 2 Challenge Level:
Look at three 'next door neighbours' amongst the counting numbers. Add them together. What do you notice?
AB Search
Stage: 3 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
What Do You Need?
Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
A First Product Sudoku
Stage: 3 Challenge Level:
Given the products of adjacent cells, can you complete this Sudoku?
In the Money
Stage: 2 Challenge Level:
There are a number of coins on a table. One quarter of the coins show heads. If I turn over 2 coins, then one third show heads. How many coins are there altogether?
Biscuit Decorations
Stage: 1 and 2 Challenge Level:
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
Spelling Circle
Stage: 2 Challenge Level:
Find the words hidden inside each of the circles by counting around a certain number of spaces to find each letter in turn.
What's in the Box?
Stage: 2 Challenge Level:
This big box multiplies anything that goes inside it by the same number. If you know the numbers that come out, what multiplication might be going on in the box?
Red Balloons, Blue Balloons
Stage: 2 Challenge Level:
Katie and Will have some balloons. Will's balloon burst at exactly the same size as Katie's at the beginning of a puff. How many puffs had Will done before his balloon burst?
Tom's Number
Stage: 2 Challenge Level:
Work out Tom's number from the answers he gives his friend. He will only answer 'yes' or 'no'.
Flashing Lights
Stage: 2 Challenge Level:
Norrie sees two lights flash at the same time, then one of them flashes every 4th second, and the other flashes every 5th second. How many times do they flash together during a whole minute?
Becky's Number Plumber
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Becky created a number plumber which multiplies by 5 and subtracts 4. What do you notice about the numbers that it produces? Can you explain your findings?
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Stage: 2 and 3 Challenge Level:
Which pairs of cogs let the coloured tooth touch every tooth on the other cog? Which pairs do not let this happen? Why?
Factor Track
Stage: 2 and 3 Challenge Level:
Factor track is not a race but a game of skill. The idea is to go round the track in as few moves as possible, keeping to the rules.
Sieve of Eratosthenes
Stage: 3 Challenge Level:
Follow this recipe for sieving numbers and see what interesting patterns emerge.
Give Me Four Clues
Stage: 2 Challenge Level:
Four of these clues are needed to find the chosen number on this grid and four are true but do nothing to help in finding the number. Can you sort out the clues and find the number?
Mathematical Swimmer
Stage: 3 Challenge Level:
Twice a week I go swimming and swim the same number of lengths of the pool each time. As I swim, I count the lengths I've done so far, and make it into a fraction of the whole number of lengths I. . . .
Multiplication Series: Number Arrays
Stage: 1 and 2
This article for teachers describes how number arrays can be a useful reprentation for many number concepts.
Which Is Quicker?
Stage: 2 Challenge Level:
Which is quicker, counting up to 30 in ones or counting up to 300 in tens? Why?
Inclusion Exclusion
Stage: 3 Challenge Level:
How many integers between 1 and 1200 are NOT multiples of any of the numbers 2, 3 or 5?
Path to the Stars
Stage: 2 Challenge Level:
Is it possible to draw a 5-pointed star without taking your pencil off the paper? Is it possible to draw a 6-pointed star in the same way without taking your pen off?
Money Measure
Stage: 2 Challenge Level:
How can you use just one weighing to find out which box contains the lighter ten coins out of the ten boxes?
Factoring Factorials
Stage: 3 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
Thirty Six Exactly
Stage: 3 Challenge Level:
The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?
Divide it Out
Stage: 2 Challenge Level:
What is the lowest number which always leaves a remainder of 1 when divided by each of the numbers from 2 to 10?
Remainder
Stage: 3 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
Ewa's Eggs
Stage: 3 Challenge Level:
I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?
GOT IT Now
Stage: 2 and 3 Challenge Level:
For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target?
Digat
Stage: 3 Challenge Level:
What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A
Oh! Hidden Inside?
Stage: 3 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
Helen's Conjecture
Stage: 3 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
Graphing Number Patterns
Stage: 2 Challenge Level:
Does a graph of the triangular numbers cross a graph of the six times table? If so, where? Will a graph of the square numbers cross the times table too?
Powerful Factorial
Stage: 3 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
Number Detective
Stage: 2 Challenge Level:
Follow the clues to find the mystery number.
One to Eight
Stage: 3 Challenge Level:
Complete the following expressions so that each one gives a four digit number as the product of two two digit numbers and uses the digits 1 to 8 once and only once.
Two Much
Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
How Old Are the Children?
Stage: 3 Challenge Level:
A student in a maths class was trying to get some information from her teacher. She was given some clues and then the teacher ended by saying, "Well, how old are they?"
What Is Ziffle?
Stage: 2 Challenge Level:
Can you work out what a ziffle is on the planet Zargon?
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2015-12-01 07:29:23
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http://orinanobworld.blogspot.fr/2013/06/object-constraints-sequel.html
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## Friday, June 7, 2013
### Objective Constraints: The Sequel
It belatedly occurred to me that I could have done a much better job explaining my previous post with a picture than I did with (copious) algebra. To recap the previous post, I claimed that adding a constraint to a mixed-integer programming model that puts an upper or lower bound on the objective value can cause solver performance to degrade of any of a several reasons, the least obvious of which is creating either dual degeneracy or multiple optimal solutions to the dual in the node linear program (LP) relaxation. We start with a model of the form$\begin{array}{lrclr} \mathrm{minimize} & c'x\\ \mathrm{s.t.} & Ax & \ge & b\\ & x & \ge & 0\\ & x_{i} & \in & \mathbb{Z} & (i\in I) \end{array}$(where $\mathbb{Z}$ is the set of integers), relax integrality and add a constraint $c'x\ge d$ to obtain the modified node LP$\begin{array}{lrclr} \mathrm{minimize} & c'x\\ \mathrm{s.t.} & Ax & \ge & b\\ & c'x & \ge & d\\ & x & \ge & 0. \end{array}$The following picture depicts the scene at a typical node.
Vertex A represents the optimal solution in the absence of the added constraint (which runs through B and C), with optimal objective value $d^*$. As I mentioned in the previous post, at nodes where $d^*>d$ the added constraint will be nonbinding and just "extra baggage". Adding it can only benefit the user if $d>d^*$ at some nodes, as depicted here.
The shaded region is the feasible region including the objective constraint. The optimal value of the modified node LP is $d$, and B, C and every point on the line segment between them are all optimal solutions. So the node LP now has multiple optima, which will make the dual problem degenerate. The dual degeneracy can actually be seen in the picture. Slight increases or decreases in $d$ will shift the B-C edge up or down a bit, but that edge will clearly continue to be optimal. If we change $d$ to $d\pm \epsilon$ for small $\epsilon\gt 0$, the optimal objective value changes equally (to $d\pm \epsilon$), and so the dual variable $z$ corresponding to the objective constraint takes the value $z^*=1$.
Now observe what happens if we make small perturbations in the constraints that run through A and B or A and C. Either B or C will slide a bit in or out along the B-C edge (say, to B' or C' respectively), but the segment [B', C] or [B, C'] will continue to be optimal with the objective value $d$ unchanged. So the dual variables corresponding to those constraints must take value zero. Variables dual to the nonbinding constraints take value zero by virtue of the complementary slackness theorem of linear programming. We end up with all dual variables equal to zero except the one ($z$) connected to the added constraint. As I mentioned in the previous post, that is likely to have adverse consequences for pricing decisions, presolving, etc.
An upper bound constraint on the objective ($c'x\le d$) would switch the reduced feasible region to the triangle A-B-C, and A would be the sole optimum. This illustrates why the upper bound constraint is less likely to create algorithmic problems (other than drag). If it cuts through the feasible region (as depicted), it will be nonbinding, and the dual solution will be unaffected. If it misses the feasible region, the modified problem will be infeasible (and the node will be pruned). Only if the added constraint is tangent at A ($d=d^*$) do we run into misadventures.
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2015-05-28 19:50:29
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https://brilliant.org/discussions/thread/finance-quant-interview-iv/
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# Finance Quant Interview - IV
For what values of $$n$$ does there exist complex $$n \times n$$ matrices $$A, B$$ such that
$A B - BA = Id?$
Note by Calvin Lin
3 years, 1 month ago
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If d=0, the condition holds trivially for all n by taking A=B=I. If d is different from zero, taking trace of both sides and using the fact that tr(AB)=tr(BA), we conclude that none such n exists.
- 3 years, 1 month ago
Id means identity matrix.
Staff - 3 years, 1 month ago
Anyways, I took it as identity matrix times the scalar d.
- 3 years, 1 month ago
No issues. That's a more general case :)
Notation for identity matrix isn't standardized anyway. I was debating between I and Id.
Staff - 3 years, 1 month ago
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2018-07-20 07:15:15
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https://math.stackexchange.com/questions/1369846/substitute-for-finding-the-hypotenuse-of-a-right-triangle
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# Substitute for finding the hypotenuse of a right triangle?
All of us know the way to calculate the hypotenuse of a right triangle: Using the Pythagorean Theorem.
I came up with a substitute to this. Let the shortest leg of the right triangle be '$a$' units, and the relatively bigger leg be '$b$' units, where $a < b < \text{Hypotenuse}$.
My theorem is that in such a case, there will always be a variable '$x$' such that:
$a^2 - x^2 =2bx$.
As for the applications of this statement, we can find the hypotenuse of the triangle by finding the positive root of this equation, and then added to the biggest leg to obtain the value of the hypotenuse. I derived a few basic conclusions from The Dickson's method of generating Pythagorean Triples, to prove this.
Let us give it a try:
Let $a = 5$, $b = 12$.
$a^2-x^2=2bx$.
$25-x^2=24x$
$x^2+24x-25=0$
$x=1,-5$.
Therefore hypotenuse $= 12+1=13$, which satisfies Pythagorean Theorem. This satisfies only the triangles which have unequal sides. Obviously, we don't need a substitute for isosceles right triangles since given the legs '$a$', the hypotenuse is $a\sqrt{2}$.
My questions are:
1. Can the difficulty level of application of this theorem be compared to that of Pythagoras? If so, in the general case, is this difficult or easy to apply?
2. Are there any more possible practical applications of this theorem (like in geometric proofs)?
3. (Please forgive my imperiousness) Can I get this published somewhere?
I would like a detailed answer on how better or more tedious this is than the Pythagorean Theorem, and one practical example (like a geometric proof) involving this theorem.
Thanks,
Sandeep
• So, instead of the hypotenuse h, you prefer to look at h - b, where b is the longer of the two legs. Not harder than finding h, clearly. But of course you could also look at h + b or 3h - 79.8b or a + b + h and so on. To make it interesting, I'd say you had to find a strong geometric theorem that used your expression. – lulu Jul 22 '15 at 11:54
• If I'm right, that's what I'm asking... Is there any practical purpose of this theorem other than being able to find the hypotenuse? – Sandeep Jul 22 '15 at 12:00
• Well, I don't see one but of course there might be something. As a rule, though, I'd start with the Geometry (or maybe with the Number Theory) instead of just algebraically manipulating the definitions. You might, for example, take a look at the form a + b - h . That one has Geometry in it! It is the diameter of the inscribed circle. – lulu Jul 22 '15 at 12:52
1. Yes, in the general case it is applicable but difficult to apply as compared to Pythagorean theorem. Because the Pythagorean theorem is straight forward to apply while your theorem involves root finding of quadratic equation which is not simple in each case.
2. It can be applied in simple cases & geometric proofs but in general cases it may create lengthy calculations.
3. Yes, but for publishing this article, first you will have to submit manuscript for review by peers of journal if it fulfills the standards of particular journal, it can be published. Because journals have their own rules/terms/standards to publish articles if some reject this article then some other may accept this up-to the defined level.
• But the Pythagorean Theorem will make us Ind the squares of two numbers, adding them, and finding their positive square root. Wouldn't both of these be approximately the same difficulty? – Sandeep Jul 22 '15 at 11:54
• Alright, but finding the roots of a quadratic equation is more difficult than simply taking sq. root of sum of squares of two known numbers i.e. legs of right triangle. – Harish Chandra Rajpoot Jul 22 '15 at 13:24
• You stated this can be applied in simple proofs. Can you give one case where it makes things easy and one case where it creates complications? – Sandeep Jul 25 '15 at 9:25
• Alright, this can be used to find out the hypotenuse if legs are known, one of unknown legs if hypotenuse & other leg are known & other acute angles in a triangle instead of using Pythagorean theorem – Harish Chandra Rajpoot Jul 25 '15 at 9:31
Your equation: $$a^2-x^2=2bx\Longrightarrow x^2+2bx-a^2=0$$ So, $$x=-b\pm\sqrt{b^2+a^2},$$ and $x_2=-b+\sqrt{a^2+b^2}$ is a positive root. By your statement, $$c=b+x_2=\sqrt{a^2+b^2}.$$ And what? If you want to find hypotenuse, you need evaluate square root in any way.
• Isn't factoring the equation better than using the general formula? That way, you don't need to evaluate square root. And it always ends up with the equation being factorisable. – Sandeep Jul 26 '15 at 16:49
• @Sandeep, factoring? Ok, how do you factor $a^2 - x^2 - 2bx = 0$ in a general case? – Michael Galuza Jul 26 '15 at 16:51
• But 'a' and 'b' are given. And speaking from a statistical point of view, it is always factorisable. – Sandeep Jul 26 '15 at 16:53
• @Sandeep, yes, it is. By finding roots ))) Anyway, your method is correct, but useless. If I want to find hypotenuse, I evaluate square root. All I need is Pythagorean Theorem. Continue your studying and research! – Michael Galuza Jul 26 '15 at 16:58
• Thanks a lot for your 'valuable' advice. Just as you prefer the Pythagorean Theorem, I prefer my own. Let's not carry on the argument any further. As a second thought, you might guess for who the bounty is going to be... – Sandeep Jul 26 '15 at 17:14
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2019-08-20 07:14:34
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https://poc-library.readthedocs.io/en/release/IPCores/misc/sync/sync_Strobe.html
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PoC.misc.sync.Strobe¶
This module synchronizes multiple high-active bits from clock-domain Clock1 to clock-domain Clock2. The clock-domain boundary crossing is done by a T-FF, two synchronizer D-FFs and a reconstructive XOR. A busy flag is additionally calculated and can be used to block new inputs. All bits are independent from each other. Multiple consecutive strobes are suppressed by a rising edge detection.
Attention
Use this synchronizer only for one-cycle high-active signals (strobes).
Constraints:
This module uses sub modules which need to be constrained. Please attend to the notes of the instantiated sub modules.
Entity Declaration:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 entity sync_Strobe is generic ( BITS : positive := 1; -- number of bit to be synchronized GATED_INPUT_BY_BUSY : boolean := TRUE; -- use gated input (by busy signal) SYNC_DEPTH : T_MISC_SYNC_DEPTH := T_MISC_SYNC_DEPTH'low -- generate SYNC_DEPTH many stages, at least 2 ); port ( Clock1 : in std_logic; -- input clock domain Clock2 : in std_logic; -- output clock domain Input : in std_logic_vector(BITS - 1 downto 0); -- @Clock1: input bits Output : out std_logic_vector(BITS - 1 downto 0); -- @Clock2: output bits Busy : out std_logic_vector(BITS - 1 downto 0) -- @Clock1: busy bits ); end entity;
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2021-04-22 11:47:41
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http://gmatclub.com/forum/profiles-admitted-to-top-b-schools-w-low-gmat-or-low-gpa-73999-40.html?kudos=1
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# Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA
Author Message
TAGS:
Intern
Joined: 28 Jan 2010
Posts: 43
Schools: [color=red][CBS Class of 2013]
WE 1: 2 years investment banking (middle-market)
WE 2: 3.5 years transaction modeling / consulting (big 4)
Followers: 0
Kudos [?]: 9 [1] , given: 1
Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 01 Mar 2010, 12:09
1
KUDOS
aa5786 wrote:
Had a chat with a staff member at Wharton recently and discussed my 50 percentile quant score. Before I could jump into my standard explanation, she looked at my resume and said......"your work experience and undergrad econ major are all the examples I need of your quant ability, don't worry about it."
Guys - Don't let these +700'ers psych you out if you have other components of your background which can speak to ability.
thanks for the information. I had/have a similar issue. I have an undergrad in finance and econ, worked as an i-banking analyst for 2 years and i am currently a financial model anlayst for a big 4 transaction advisory practice. Despite my background, I have struggled with the quant section of the GMAt.
I got a 640 the first time i took the test: 49 percentile quant. Retook the test a week after submitting my last app on jan 15th. My new score was much better: 700, altho the quant was still fairly poor= 67 percentile. I was sure that Booth would ding me considering its analytical bent and because I sent the new score to the school as an update to my app, 3 weeks after the round 2 deadline. Miraculously, I received an interview invite a couple of weeks back and just had my interview this past saturday (which I think went fairly well).
So I am curious how you setup this appointment to discuss ur app. So you met with an adcom member in person and discussed different parts of ur app at length? was he/she willing to discuss the other strenghts and weaknesses? do you know if other schools (e.g. Booth and Kellogg) are receptive to this kind of discussion?
Altho I havent made an appointment with the add office, most of the questions that I've asked have gotten fairly standard responses and got the impression that they would/could not comment on the specifics of my application.
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Senior Manager
Joined: 20 Jan 2010
Posts: 277
Schools: HBS, Stanford, Haas, Ross, Cornell, LBS, INSEAD, Oxford, IESE/IE
Followers: 15
Kudos [?]: 183 [1] , given: 117
Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 14 Mar 2010, 19:52
1
KUDOS
WoW! Congrats man. I am also in a same boat you were before but my transcript is all messed up. once I got 2.51 and then it's been low . Your post did boost me and I am trying to take some courses like Calculus,Statistics and Accounting. Problem is we don't have community colleges but Linda from Accepted.com suggested online course so will try that and also have to ace GMAT. I am aiming Haas, Ross, and UCLA or Cornell so do pray for me guys. And best of luck with your future ahead.
_________________
"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so."
Target=780
http://challengemba.blogspot.com
Kudos??
Manager
Joined: 25 Jan 2010
Posts: 58
Location: London
Schools: HBS, Booth, Stern, Kellogg
Followers: 0
Kudos [?]: 19 [1] , given: 3
Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 25 Mar 2010, 23:27
1
KUDOS
Just an update after finding out about schools - Got into Chicago Booth and NYU Stern with a low GPA (2.85 unweighted, 3.01 weighted), an average GMAT (730), and average work experiences (less than 4 years). So, there is hope!
I feel like adcoms know there is very little you can do about your undergrad record - you just have to focus on things you CAN have an impact on, like your GMAT score and your work experience, and demonstrate that you have plans for the future. I didn't fill in the optional essay in the application to explain my poor grades, but during my interviews, I openly acknowledged my sub-par undergrad performance and didn't try to excuse my lack of focus. Instead, I just noted that in retrospect, I should have studied more in school, but I've learned from my mistakes and can only move forward.
Intern
Joined: 10 Jun 2010
Posts: 35
Followers: 0
Kudos [?]: 5 [1] , given: 0
Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 10 Jun 2010, 11:06
1
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I'm applying this fall to a number of schools and am in the process of choosing which ones to pursue. I've included my information below:
GMAT: 670 (lower quant than I expected)
University attended: Large state school
Major(s): Econ w/ Business Minor (took extra Accounting courses as electives)
GPA / Honors: 3.79 (graduated with distinction and Phi Beta Kappa)
Age: 26
Demographic / Gender: American Male
Work Experience: 3.5 years currently (4.5 years when I matriculate) in commercial real estate group of top 3 financial institution in US (analyst for 2.5 years > 6 month training program for associates > associate / relationship manager)
Extra-curricular Activities: Many including playing guitar in a band I started, skiing, other outdoor activities, etc.
Community Services: board of directors for city chapter of a well-known national charity organization (also on finance committee), young leader committee member for commercial real estate organization (also on membership committee)
Last year I applied to MIT Sloan as a reach and was dinged. This year, I'm contemplating the following schools:
Darden
UT-McCombs
HBS (reach)
Wharton (reach)
MIT (reach)
UNC
Tuck (reach)
Does anyone have suggestions for me in selecting which schools to pursue given my profile? Feel free to suggest ones that aren't on my list.
I realize I need to pick schools that are a "fit" for me, but that's a given. I'm thinking about adding Kellogg to the list and possibly Stanford, which would both be reaches.
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Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 22 Jun 2010, 16:59
1
KUDOS
Via wrote:
Hey Guys/Gals,
I have a low undergrad GPA (2.99) and a low GMAT (650/44Q/35V/5.5AWA).
All in all, I applied to 8 schools (Kellogg, Tuck, Yale, Darden, McDonough, Kelley, Owen, Olin) and I was accepted to all of them. I also received significant scholarships (half tuition or full tuition) from 4 of them (Yale, Darden, Owen, McDonough).
My success, I believe, was a product of my research efforts and application strategy. Using my research on the GMAT Club and several other sources, I realized that I had several attributes that were appealing to admissions committees and I worked hard to highlight every one of them.
My approach:
Two and a half years ago, I realized that I wanted to attend graduate school. Specifically, I wanted to attend a competitive b-school program but I realized that my undergrad GPA was low. Therefore, I took graduate level classes within an MA in leadership and management program. I only took four classes within a year and a half period but I received a 3.75 GPA. I was able to highlight this GPA on my application to counter my undergrad GPA. Although I only took four classes, I argued that I earned good grades despite working full time as a general manager, dealing with several life changing events, and volunteering/etc...
About a year ago this month, I started researching b-schools in depth. I was addicted to the GMAT CLUB, BW Forums, the GMAC website to see GMAT statistics, books ("How to get into top B-school", and 15 b-school websites. I also attended MBA Fairs and discovered groups like The Consortium. Basically, I was a research-a-holic last year.
Using me research, I decided to apply to the 8 schools I listed above. Given my low GPA and GMAT, I didn't know where I'd lie on the competitive ladder among b-school applicants. Therefore, I decided to apply to 1 top-5 school, 1 top-10, 2 top-15, and a couple of other programs that lied in the 15-35 range according to US News and BW rankings.
I then developed a very specific essay strategy that highlighted any and every attribute that I thought gave me a competitive advantage. My goal was to show that I was a success story, that I could succeed in any rigorous academic program, and that I would contribute to any b-school as a student and alum.
In my opinion, my positive attributes include: I'm a first generation hispanic/ I'm the first member of my family to pursue a graduate degree/ I'm in the Army and have deployed twice/ While I was at my undergrad institution- I was heavily involved in my student government as an elected officer/ I served as a tutor, Div 1 athlete, and as an elected and appointed club leader throughout my time in college / I continue to serve as a Class Officer for my undergrad alumni organization/ I earned a 3.75 GPA in graduate level courses despite losing two family members and working full time as a general manager. As a general manager, I received formal ratings that eventually placed me in the top 2% of my peer group. There were a couple of other small data points I added to my essays but you get the point.
When I was writing my essays, I tried to answer the unique questions of each b-school while inserting these attributes to develop a unique story of enduring success. After endless revisions to my b-school essays, I submitted all of my applications by Round 2 for each school. I applied to 5 schools through the Consortium and 3 schools independently.
Some other things that I believe helped me:
I visited every school to which I applied and I highlighted that fact during my interviews. I searched for "key" terms each admissions committee was likely looking for. In my opinion, some of these include: Olin: Research driven thinking/Kelley: family, community, involvement / Kellogg: collaboration, Look at their 4 pillars ie. Intellectual Depth, Diversity of thought.../ Tuck: loyalty, community, friendship, advantage of a b-school education in a rural setting (ie. less distractions, community, focus, etc.)/ Darden: Case Method, working on a team to solve cases, interactive classroom environment.
Every school has a unique community and academic experience that they want to sustain. Figure out what that is and show how you could contribute to it.
Many of you have pointed out that, unless you are an underrepresented minority, female, or have unique work experience, you may not be able to overcome a low GPA or GMAT. Overall, that may be true. However, there are no absolutes in this world. I think everyone has positive attributes that can make them look very competitive despite poor stats. I could have easily applied to less prestigious schools and have saved a lot of time & money in dropping my Kellogg, Tuck, Darden, and Yale applications. Also, had I failed to highlight my strengths to the admissions committee, I could have experienced multiple rejections that some applicants with very high GPAs and GMATs have experienced. In the end, it never hurts to fight for your candidacy and apply to the b-school programs you love.
Furthermore, I believe that, even as a minority, I still overcame major weaknesses in my application. I am certain that over 80% of the minority applicant pool at Kellogg had better stats than I did. It is rare to have both a low GPA AND low GMAT and get into a top 5 program.
Now that my application journey is over, I hope that this post will help some of you. I feel indebted to all of you who have contributed to GMAT Club and have educated me for the last couple of years.
-Via
FANTASTIC MAN!!! Guess I have one more hero from GC
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Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 30 Sep 2010, 12:06
1
KUDOS
3.2 Ivy (econ). Worked 15-20 hrs/ week to pay for education
730 GMAT (50 Q, 38 V), 6.0 AWA
Applied to CBS ED August 23rd, Admitted September 21st.
Good luck guys. It can happen
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Schools: UNC, DUKE
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Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 20 Nov 2010, 16:08
1
KUDOS
nhype wrote:
Howdy just finished my first test and found this string interesting.
GMAT: 710 (Q:46, V:41, AWA:??)
University attended: Large State university (Top 50ish)
Major(s): History
GPA / Honours: 2.92
Nationality: White American
Age: (at matriculation): 28
Gender: Male
Work Experience: No. of Years / Industry / Role : 4.5/ Energy/Oil & Gas / Consultant & Project Management
Extra-curricular Activities: Fraternity President in college and volunteered for a few local charities
Community Services:
Basically right now have the mentality that the 2.92 precludes applying to top 10 and I don't want to waste time with "saftey schools" so the plan as of now is to apply January round to:
UVA
Duke
UNC
UT-Austin
Georgetown
4 questions:
1. Thoughts on schools above UVA that are somewhat low gpa friendly?
2. Does energy qualify as unusual work experience?
3. Is the GMAT worth another testing, meaning can i make up for the 2.92 with a higher GMAT ?
4.$cholariships: I've read there are GMAT thresholds, but will the 2.92 keep me from getting scholarships? Hi, I can only help with information about DUKE and UNC (schools that I applied and got into) 2.- Energy in both school does not qualify as unusual work experience I think, quite a few people of the 2012 class actually have that same background. 3.- I feel that the higher your GMAT the better you will look to admissions but with that said, when they say they look at the whole package they really mean that. I got into both school with a very low GMAT and I know of people with 750 that got dings. Applications goes beyond GPA and GMAT, your story has to make sense, your essays have to convey a coherent and pretty good story, your interview has to be darn good. 4.- Most likely it will, with the way the economy still is hurting, way too many qualified candidates are applying (and getting accepted) to good schools. For scholarships they look at the whole package, one thing doesn't make up for another. With that said, if they want you bad enough, they might offer some money but it is not going to be just because your GMAT. I hope that helps, and good luck with the application Current Student Joined: 05 Aug 2008 Posts: 1262 Schools: Ross 2012 WE 1: 5 Years at Fortune 50 Company in Manufacturing Followers: 21 Kudos [?]: 168 [1] , given: 20 Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 24 Oct 2011, 19:13 1 This post received KUDOS 1 This post was BOOKMARKED Sorry for reviving a old thread but I would just like to give a follow up on the Low GPA when it comes to recruiting. So all and all, most companies care very little to 0 on GPA. I have been asked twice, both this year, both consulting, one during the first round which resulted in an offer, the other during a second round which is pending. Outside of consulting, no one has asked me about GPA during an interview. But here's the results Interviewed with 2 of the M/B/B, [*]closed listed (Moved to 2nd round) [*]got a call which the recruiter asked to bid (dinged after first round), [*]Screening call (no interview) As for the other consulting firms, I was closelisted my first and/or second year from most top-20 consulting firms that recruited at Ross. I currently two offers from vault's top 10, one is a returning offer from my summer internship. If you have questions outside of Consulting, I can do a debrief of those also, but in general your prior experience matters most, I received most interviews that I either fit my background or I tried to networked. Alternatively, I went 1 for 4 in tech. I hope this helps all you low GPA people! _________________ Intern Joined: 10 Aug 2012 Posts: 19 Location: India Concentration: General Management, Technology GPA: 3.96 Followers: 0 Kudos [?]: 6 [1] , given: 15 Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 21 Feb 2013, 07:43 1 This post received KUDOS Hi, I have friend having a GMAT score of 670, average ECs but good profile...he got call from ISB this year and also got selected... So apart from GMAT score, profile matters a lot.. Manager Joined: 12 Dec 2012 Posts: 230 Concentration: Leadership, Marketing GMAT 1: 540 Q36 V28 GMAT 2: 550 Q39 V27 GMAT 3: 620 Q42 V33 GPA: 2.82 WE: Human Resources (Health Care) Followers: 3 Kudos [?]: 63 [1] , given: 181 Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 21 Feb 2013, 14:24 1 This post received KUDOS Any stories from Desautels? _________________ My RC Recipe http://gmatclub.com/forum/the-rc-recipe-149577.html My Problem Takeaway Template http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html Director Joined: 26 May 2010 Posts: 719 Location: United States (MA) Concentration: Strategy Schools: MIT Sloan - Class of 2015 WE: Consulting (Mutual Funds and Brokerage) Followers: 18 Kudos [?]: 204 [1] , given: 642 Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 21 Mar 2013, 20:42 1 This post received KUDOS panama, it's easy to see why you were accepted at Kellogg and interviewed at Wharton with writing skills like that. Kudos again! Intern Joined: 12 Oct 2012 Posts: 4 Followers: 0 Kudos [?]: 1 [1] , given: 0 Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 25 Mar 2013, 21:35 1 This post received KUDOS I think a high GMAT and good work experience can cover up for the weak GPA... not sure how the adcoms view it. Maybe need to explain in those extra essays. Manager Joined: 12 Dec 2012 Posts: 230 Concentration: Leadership, Marketing GMAT 1: 540 Q36 V28 GMAT 2: 550 Q39 V27 GMAT 3: 620 Q42 V33 GPA: 2.82 WE: Human Resources (Health Care) Followers: 3 Kudos [?]: 63 [1] , given: 181 Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 22 Apr 2013, 07:07 1 This post received KUDOS This topic simply inspires me .I hope I can post here one day _________________ My RC Recipe http://gmatclub.com/forum/the-rc-recipe-149577.html My Problem Takeaway Template http://gmatclub.com/forum/the-simplest-problem-takeaway-template-150646.html MBA Section Director Affiliations: GMAT Club Joined: 21 Feb 2012 Posts: 2383 Location: India City: Pune GPA: 3.4 WE: Business Development (Manufacturing) Followers: 289 Kudos [?]: 2039 [1] , given: 1629 Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 05 Mar 2014, 11:11 1 This post received KUDOS Expert's post Bump!!! _________________ Current Student Joined: 15 Jun 2012 Posts: 114 Location: United States Concentration: General Management, Strategy GMAT 1: 760 Q48 V48 GPA: 3.1 WE: Engineering (Manufacturing) Followers: 4 Kudos [?]: 46 [1] , given: 17 Re: Profiles - Admitted to Top B-Schools w/ Low GMAT or Low GPA [#permalink] 05 Jun 2014, 11:10 1 This post received KUDOS 2 This post was BOOKMARKED Program: MBA Area of interest in MBA: General Management, Strategy GMAT: 760 Q48 V48 GPA: University of Michigan, Engineering, 3.1 (ok ok, 3.06) Work experience(WE): ~5 years at matriculation, Manufacturing Nationality: US, White Age: 26 Gender: Male Extra-curriculars/community: Heavy involvement (including chair of board) in local chapter of professional society. Schools applied to: Ross, Round 1, Accepted Kellogg, Round 1, Accepted and Matriculating Booth, Round 1, Waitlisted with interview Tepper, Round 1, Accepted (with$)
Fuqua, Round 1, Accepted (with \$)
I used the optional essay in each app to briefly explain the grading curve in my technical classes. I also said that while I wish my GPA had been higher, that I was proud to have graduated from a competitive program in four years while showing improvement and working part time. I used 118 words, which I think was enough to show that I recognized the weakness, but not so much that I irritated the adcom with another lengthy essay.
For someone with a similar situation (high GMAT, low GPA), just know that you can get into a really good school.
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Re: Profiles - Admitted to Top B-Schools with Low GMAT Score [#permalink] 18 Dec 2008, 08:49
This is a great idea. Thanks, yhc.
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Re: Profiles - Admitted to Top B-Schools with Low GMAT Score [#permalink] 18 Dec 2008, 10:26
Good one to see the stats.
_________________
To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton
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Re: Profiles - Admitted to Top B-Schools with Low GMAT Score [#permalink] 20 Dec 2008, 22:01
This thread has been here for several days, but no profiles have been posted. Does that mean people with a low GMAT score are destined to be denied?
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Re: Profiles - Admitted to Top B-Schools with Low GMAT Score [#permalink] 20 Dec 2008, 22:13
yhc25 wrote:
This thread has been here for several days, but no profiles have been posted. Does that mean people with a low GMAT score are destined to be denied?
_________________
The one who flies is worthy. The one who is worthy flies. The one who doesn't fly isn't worthy
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Status: Burning mid-night oil....daily
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Re: Profiles - Admitted to Top B-Schools with Low GMAT Score [#permalink] 23 Dec 2008, 07:00
pelihu wrote:
I've said it many times, and received negative comments many many times, but I'll say it again. People admitted with GMAT scores more than 20-30 points below the median for a given school are either underrepresented minorities or have some other easily identifiable unique factors that support their candidacy. People without such factors and not from such underrepresented groups can rely on little more than pure luck if their scores are more than 30 points below the average.
I have to agree with this assessment.
We're talking about top programs here. It's difficult enough for many of us to even get admit to one!
Sure, there are exceptions, but they are outliers.
One admissions consultant I've talked to once said:
"The 80 percentile GMAT range posted in the class profile? Unless you are an african american, native american, hispanic, or female - don't bother look at the lower end of that range."
I didn't believe that person at first, but more and more that consultant sounds right.
_________________
Re: Profiles - Admitted to Top B-Schools with Low GMAT Score [#permalink] 23 Dec 2008, 07:00
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Help... Low GPA, Low GMAT 3 29 May 2013, 09:38
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2016-02-12 14:08:22
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https://brilliant.org/practice/geometric-progression-sum/
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Algebra
# Geometric Progression Sum
Evaluate
$1 + 2 + 2^2 + 2^3 + \ldots + 2^{ 24 }.$
A bookshelf has $6$ shelves. The top shelf has $2$ books, and each other shelf has $3$ times as many books as the shelf directly above it. If you want to read a book from the shelves, how many choices do you have?
Details and assumptions
All of the books are different.
The series $a_1, a_2, a_3, a_4, a_5, a_6, a_7$ is a geometric progression with $a_4 = 48$ and $a_5 = 96$. What is the sum of the these $7$ terms?
Details and assumptions
A geometric progression is a sequence of numbers, in which each subsequent number is obtained by multiplying the previous number by a common ratio / multiple. For example, $1, 2, 4, 8, 16, 32, 64, \ldots$ is a geometric progression with initial term 1 and common ratio 2.
Consider a geometric progression with initial term $4$ and common ratio $5$. What is the smallest value of $n$ for which the sum of the first $n$ terms is greater than $800$?
Evaluate the sum to infinity:
$\frac{ 14 } { 2^0} + \frac{ 14 } { 2^1 } + \frac{ 14 } { 2^2 } + \ldots .$
×
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2020-02-23 22:15:01
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https://asmedigitalcollection.asme.org/biomechanical/article/139/10/101003/371335/Experimental-Study-of-Anisotropic-Stress-Strain
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Homografts and synthetic grafts are used in surgery for congenital heart disease (CHD). Determining these materials' mechanical properties will aid in understanding tissue behavior when subjected to abnormal CHD hemodynamics. Homograft tissue samples from anterior/posterior aspects, of ascending/descending aorta (AA, DA), innominate artery (IA), left subclavian artery (LScA), left common carotid artery (LCCA), main/left/right pulmonary artery (MPA, LPA, RPA), and synthetic vascular grafts, were obtained in three orientations: circumferential, diagonal (45 deg relative to circumferential direction), and longitudinal. Samples were subjected to uniaxial tensile testing (UTT). True strain-Cauchy stress curves were individually fitted for each orientation to calibrate Fung model. Then, they were used to calibrate anisotropic Holzapfel–Gasser model (R2 > 0.95). Most samples demonstrated a nonlinear hyperelastic strain–stress response to UTT. Stiffness (measured by tangent modulus at different strains) in all orientations were compared and shown as contour plots. For each vessel segment at all strain levels, stiffness was not significantly different among aspects and orientations. For synthetic grafts, stiffness was significantly different among orientations (p < 0.042). Aorta is significantly stiffer than pulmonary artery at 10% strain, comparing all orientations, aspects, and regions (p = 0.0001). Synthetic grafts are significantly stiffer than aortic and pulmonary homografts at all strain levels (p < 0.046). Aortic, pulmonary artery, and synthetic grafts exhibit hyperelastic biomechanical behavior with anisotropic effect. Differences in mechanical properties among vascular grafts may affect native tissue behavior and ventricular/arterial mechanical coupling, and increase the risk of deformation due to abnormal CHD hemodynamics.
## Introduction
Surgical cardiovascular reconstruction of congenital heart disease (CHD) is frequently accomplished by using biological or synthetic materials. Cardiovascular mechanical properties are determined by tissue structure which is integrally related to its function [1]. Adult aorta and pulmonary artery homografts, as well as grafts made of dacron (polyethylene terephthalate) or Teflon (expanded polytetra-fluoroethylene (ePTFE)), have been used clinically in cardiovascular surgeries [24]. These materials have very different structures than that of the native tissue of a neonate [5,6].
The abnormal blood flow patterns of CHD may have adverse effects on the implanted material, the native vessel, and the mechanical load on the heart [1,7,8]. These include changes in effective arterial impedance that could impair ventricular efficiency; stress changes on the adjacent native vessel; and deformation of both the native and implanted material with the risk of either luminal narrowing or aneurysm formation. Mechanical properties of homograft tissue and synthetic vascular grafts are currently not available in the literature. Determining these properties will aid in understanding tissue behavior when subjected to the abnormal hemodynamics of CHD.
We have previously reported the anisotropic stress/strain relationship of the piglet great vessels using uniaxial tension testing (UTT) [9]. The purpose of this study was to determine and compare biomechanical characteristics among adult aorta and pulmonary artery homografts, and ePTFE grafts. UTT was performed in three orientations on several vessels and graft segments, and then used to calibrate coefficients of the Holzapfel–Gasser anisotropic constitutive model. These data will be useful in future computational fluid dynamics (CFD) modeling of blood flow incorporating vessel compliance [10]. Furthermore, it will strengthen CFD use in planning surgical treatment of CHD.
## Material and Methods
### Material Preparation.
Recently expired clinical-grade homografts, two aortic from CryoLife, Inc., Kennesaw, GA (donor's age Ao1 = 17 and Ao2 = 21 yr old), and two pulmonary from LifeNet Health, Inc., Virginia Beach, VA (donor's age PA1 = 44 and PA2 = 47 yr old), and two synthetic vascular grafts, one GORE-TEX (made of ePTFE and fluorinated ethylenepropylene, SVG-GORE-TEX) from W. L. Gore & Associates, Inc., Flagstaff, AZ, and the other IMPRA (made only of ePTFE, EVG-IMPRA) from BARD Peripheral Vascular, Inc., Tempe, AZ were donated to our lab. The homografts were kept frozen at −140 °C in Dulbecco's Modified Eagle's Medium containing 10% dimethylsulfoxide and 10% fetal bovine serum and thawed to room temperature on the day of mechanical testing. Homograft segments were excised and labeled for vessel origin and orientation as ascending aorta (AA), aortic arch (Arch), innominate artery (IA), left subclavian artery (LScA), left common carotid artery (LCCA), descending aorta (DA), main pulmonary artery (MPA), left pulmonary artery (LPA), and right pulmonary artery (RPA) in circumferential (C), diagonal (D), and longitudinal (L) orientations, and anterior and posterior aspects (Fig. 1). It is mentioned that the direction of angle bisector between circumferential and longitudinal directions was considered as the diagonal direction, for some geometrically complicated regions like the Arch. Dimensions of tissue samples were acquired immediately prior to mechanical testing given in Table 1. Mechanical testing was subsequently performed within 6 h of tissue defrosting. Homograph samples were moistened until tests start by the same solution as it was stored (Dulbecco's modified Eagle's medium containing 10% dimethylsulfoxide and 10% fetal bovine serum).
The SVG-GORE-TEX graft was 40 cm long with 12 mm inner diameter, and the EVG-IMPRA graft was 10 cm long with 13 mm inner diameter. Both of them were 1.0 mm thick. The cylindrical grafts were longitudinally cut and extended into a rectangular shape, allowing us to obtain full-width samples along C, D, and L orientations (referring to Fig. 1(d)). SVG-GORE-TEX comes with a reinforced layer attached to the outer surface which was kept during mechanical testing (Fig. 1(e)).
### Mechanical Properties Testing.
The details of the testing technique have been reported elsewhere [9]. Briefly, opposite edges of samples of the homografts segments defined earlier, and synthetic grafts, were clamped with a biomedical tissue grip (G341 mechanical vice grip, Testresources, Inc., Shakopee, MN) and loaded onto a MTS Tytron 250 microforce computer-controlled high-precision tensile testing machine (MTS Systems Corporation, Eden Prairie, MN) for UTT. A 5 N load cell was used for small homograft tissue samples [11], and a 250 N load cell was utilized for the synthetic grafts samples. In all cases, UTT was conducted at a velocity of 0.1 mm/s, until tissue fracture. Strain rate effect on the tissue was ignored [12]. Tensile force data were obtained from the loading frame, and the samples' horizontal displacement during UTT was obtained by digital imaging correlation (DIC). DIC system consists of a Tokina AT-X Pro macro 100 mm-f/2.8-d 2448 × 2048 resolution lens connected to a desktop computer with VIC-2D 2009 software (Correlated Solutions, Inc., Columbia, SC). Images for DIC were obtained one per second. A figure demonstrating the entire experimental setup can be found in our previous report [9].
### Data Collection.
We collected, processed the data, and applied the constitutive model in the same manner as described in our previous report [9]. Briefly, recorded force–displacement datasets were used to calculate Cauchy stress ($σ$) and true strain ($ε$) which is the logarithm of the tissue stretch under UTT [9]. The principal true strain [13], ε, is
$ε=lnλ$
(1)
where $λ=l/l0$ is the stretch, representing the ratio of the deformed length to the original tissue length under uniaxial tension. The Cauchy stress along tensile direction is related to the original cross-sectional area of the tissue defined as the specific width of the sample multiplied by the average thickness of that sample as
$σ=FAλ=FAeε$
(2)
where F is the tensile force, and A is original cross-sectional area of the tissue defined as the specific width of the sample multiplied by the average thickness of that sample (Table 1). The tissue stiffness $E0$ (tangent value in stress–strain curve) is the first derivative of stress–strain curve at a given strain level $ε0$
$E0=dσdε|ε=ε0$
(3)
was computed and compared at $ε0$ = 10%, 20%, and 30% strain to assess fiber engagement to maintain stiffness (from elastin-dominated to partial fiber engagement), among different tissue samples [12]. A schematic of this method can be found in our previous report [9].
### Constitutive Modeling.
Arterial tissue typically exhibits anisotropic hyperelastic behavior and incompressibility [10,14]. First, a unidirectional Fung exponential stress–strain relationship [12] was utilized for data processing and to perform individual least-square curve fitting for all vessel segments individually along C, D, and L orientations (denoted by C: $θ=0 deg$, D: $θ=45 deg$, and L: $θ=90 deg$) utilizing a two-coefficient relation
$σfitting=C(eQ ε−1)$
(4)
where the C and Q fitting coefficients are obtained to describe the mechanical behavior for each orientation individually.
Subsequently, the Holzapfel–Gasser constitutive model was applied to fully capture the hyperelastic anisotropic (orientations dependent) relation for each vessel segment. This model assumes: (a) arterial tissues are treated as fiber-reinforced composite and (b) two families of collagen fibers are embedded in an isotropic ground matrix [10]. The modeled Cauchy stress–strain relationship, dependent on the sample orientation, reads
$σ(θ)=λ2[C0(1−1λ3)+2k1∑i=12Ei¯∂Ei¯∂C11ek2Ei¯2]$
where
$∂Ei¯∂C11=κ(1−1λ3)+(1−3κ)[ cos2(θ+αi)−12λ3 sin2(θ+αi)]Ei¯=κ(I1−3)+(1−3κ)(I4i−1)$
(5)
Here, $Ei¯$ is a Green strainlike quantity characterizing the strain in the collagen fiber mean direction $αi$ of the $i$ th family of fibers [10], with respect to the C tissue orientation ($θ=0 deg$). Parameter $(0≤κ≤1/3$) describes the level of dispersion in fiber directions, which directly controls degree of material anisotropy, $I1=C11+C22+C33$ is the first invariant of right Cauchy–Green strain tensor, and $I4i=C11 cos2(θ+αi)+C22 sin2(θ+αi)$ is a pseudoinvariant dependent on specimen orientation $θ$ and collagen fiber direction $αi$. $C11=λ12$, $C22=λ22$, and $C33=λ32$ are the components of the right Cauchy–Green strain tensor, respectively. $λ1=λ$ and $λ2=λ3=1/λ$ are principal stretches, respectively, based on UTT and incompressibility ($λ1λ2λ3=1$). The subscripts 11, 22, and 33 indicate direction along tension, lateral, and thickness of the UTT sample.
The modeled Cauchy stress–strain relationship is calibrated using the measured data to determine the five model parameters $C0$, $k1$, $k2$, $αi$, and $κ$ as previously described (i = 1,2 for all variables with subscript i) [9]. These parameters were obtained using a generalized reduced gradient (GRG) algorithm to correlate the Holzapfel–Gasser model with the unidirectional fitted Fung stress–strain curves in the three tested orientations [15].
The first step of Fung model least-squares curve-fitting helps to remove scatter in the original test data for each orientation. A combined fit was applied for all available experimental curves under each case. The Pearson R2-value was set as the objective function to be maximized iteratively by the GRG algorithm by adjusting the model parameters in curve-fitting first to the experimental and Fung models and then Fung and Holzapfel Gasser models.
## Statistical Analysis
To check the data normality, Kruskal–Wallis tests have been conducted for all the tissue sample stiffnesses investigated. A value of $α=0.05$ is taken as the significance level of the Kruskal–Wallis test. For all aortic and pulmonary artery homografts segments, probability distributions are the same for anterior and posterior aspects, when comparing all orientations for each region under true strain of 10%, 20%, and 30% (p > 0.11 in all regions). For all aortic and pulmonary artery homografts segments, probability distributions are the same for different orientations (C, D, and L), when comparing both aspects for each region under true strain of 10%, 20%, and 30% (p > 0.10 in all regions). For SVG-GORE-TEX, when comparing all the orientations, probability distributions are the same under true strain of 10% (p = 0.051), but not the same under true strain of 20% (p = 0.027), and 30% (p = 0.027). For EVG-IMPRA, when comparing all the orientations, probability distributions are the same under true strain of 10%, 20%, and 30% (p > 0.10).
Individual paired t-tests were applied to compare tissue stiffness of different sample orientations, aspects and vessel segments of homografts and synthetic grafts, using Microsoft Excel (version 2013). A p < 0.05 was considered statistically significant. Pearson R2 was calculated to show the strength of association between experimental and Fung models, and Holzapfel–Gasser and Fung models. Reported values are quoted as mean value ± standard deviation.
## Results
Sample length and width were measured with calipers, and the thickness was measured using DIC. Samples of IA, LScA, and LCCA were too small to be excised in more than one orientation of their anterior and posterior aspects. Therefore, these three sets of the vessel segments were obtained only in C orientation from both aspects. The general characteristics of the tissue sample are provided in Table 1.
### Constitutive Modeling.
Figures 2 and 3 illustrate raw experimental data with least-squared fitting curves (solid line) from the Fung model with associated Pearson R2 correlation values. All tested samples demonstrated nonlinear hyperelastic behavior under UTT except for SVG-GORE-TEX in C orientation and EVG-IMPRA in L orientation in the true stress–true strain curves. This prevents the application of Holzapfel–Gasser model for SVG-GORE-TEX and EVG-IMPRA. Therefore, in Fig. 3, the Fung model and raw experimental data for SVG-GORE-TEX and EVG-IMPRA are provided, but SVG-GORE-TEX in C orientation and EVG-IMPRA in L orientation only display the raw experimental data.
Figure 4 provides the GRG-calibrated Holzapfel–Gasser model parameters and displays the predicted true stress–true strain curves with fitted Fung model curves for all orientations of homograft samples. The calibrated anisotropic Holzapfel–Gasser model is capable of predicting the fitted Fung model curves with excellent correlation for all orientations (R2-values > 0.97), capturing full anisotropic mechanical behavior.
### Stiffness.
Average stiffness for all samples, orientations, and aspects obtained at the true strain levels of 10%, 20%, and 30% are exhibited in Tables 2 and 3 and Fig. 5. Figures 6 and 7 show the entire aortic and pulmonary artery homografts stiffness contours at true strain levels of 10% and 30%, respectively. A series of comparisons were made among different regions, orientations, and aspects, as follows:
1. (1)
For all aortic and pulmonary artery homografts segments, no significant differences were observed between anterior and posterior aspects, when comparing all orientations for each region under true strain of 10%, 20%, and 30% (p > 0.05 in all regions).
2. (2)
For most vessel segments, no significant differences were observed among C, D, and L orientations, when comparing both aspects for each region at three strain levels (p > 0.05 in all regions except for: DA-10% between C and L, p = 0.042; DA-30% strain between D and L, p = 0.011; and LPA-30% between D and L, p = 0.041).
3. (3)
Stiffness was significantly different among orientations for most of SVG-GORE-TEX and EVG-IMPRA samples. SVG-GORE-TEX is significantly stiffer along C than D and L, respectively, at all strain levels (p < 0.035); and D is stiffer than L at 30% strain (p = 0.002). EVG-IMPRA in L is stiffer than D orientation at all strain levels (p < 0.042); and L is stiffer than C at 20% and 30% strain (p < 0.024).
4. (4)
Stiffness was significantly different among the following regions when comparing all orientations and aspects between each two regions: Arch > AA at 20% strain (p = 0.026); Arch > DA at 10% (p = 0.040) and 20% strain (p = 0.011); AA, DA, and Arch > MPA at 10% strain (p = 0.003, 0.028, 0.004, respectively); AA > RPA at 20% strain (p = 0.043); Arch > LPA and RPA at 10% strain (p = 0.011, 0.007, respectively); Arch > RPA at 20% strain (p = 0.006); and LPA > RPA, at 30% strain (p = 0.024).
5. (5)
Stiffness is significantly greater in aortic than pulmonary artery homograft at 10% strain, when comparing all orientations, aspects, and regions (p = 0.0001). This difference in stiffness is not observed at higher strain levels of 20% (p = 0.080) and 30% strain (p = 0.588), nonetheless the aortic homograft is still slightly stiffer.
6. (6)
SVG-GORE-TEX samples along C and D orientations, and EVG-IMPRA samples along all orientations are significantly stiffer than aortic and pulmonary artery homografts at all strain levels (p < 0.046).
The limited number of samples of IA, LScA, and LCCA preclude statistical analyses of observed differences. Obtained measurements are however included for qualitative reference. The stiffnesses of IA at 10%, 20%, and 30% strain are 1078.72 ± 1197.46 kPa, 1246.70 ± 1101.51 kPa, and 1525.63 ± 1092.66 kPa, respectively. The stiffnesses of LCCA at 10%, 20%, and 30% strain are 586.52 ± 369.84 kPa, 985.84 ± 323.73 kPa, and 1100.51 ± 610.31 kPa, respectively. The stiffnesses of LScA at 10%, 20%, and 30% strain are 438.00 ± 109.47 kPa, 624.06 ± 12.58 kPa, and 980.72 ± 132.67 kPa. In contrast to stiffness at these specific points (10%, 20%, and 30% strains), the entire stress–strain behavior exhibited strong dependence among different orientations and anisotropy is captured in individual Fung and comprehensive Holzapfel–Gasser model curve fits.
## Comment and Discussion
Mechanical limitations of materials currently used by surgeons in cardiovascular operations, mainly homografts and either Dacron or polytetrafluoroethylene grafts, are sometimes evidenced by complications such as graft stenosis or dilation, noted particularly after aortic arch reconstruction [1618]. These complications could be mitigated by utilizing the graft material that best matches the mechanical properties of the native tissue being repaired. Graft stiffness and variation of its properties with orientation are important considerations. Moreover, surgical planning aids combining imaging techniques such as magnetic resonance, and computed tomography together with catheterization and CFD, will substantially benefit and become more reliable by accurately accounting for the mechanical response of graft materials in their models [1622]. This enables more sophisticated studies that may possibly predict phenomena such as local tissue strain, aneurysm formation, ventriculoarterial and arterioarterial impedance mismatch, and ventricular efficiency [23].
Our results show that stiffness of synthetic graft materials varies significantly with orientation and that all orientations of SVG-GORE-TEX and EVG-IMPRA except for L-orientation of SVG-GORE-TEX are significantly stiffer than aortic and pulmonary artery homografts. Furthermore, the L-orientation of SVG-GORE-TEX shows a good fit to the homografts in regards to stiffness at physiologic stress levels ($ε0∼10−20%$). This suggests that an implantation of the graft may provide a better mechanical match between the graft and native vasculature in the longitudinal direction.
In our previous study, we performed UTT on piglet aorta and pulmonary artery. Piglet tissue approximates the biomechanical properties of tissue from a human neonate as the anatomy of the porcine heart and vasculature is the most similar research model to that of a human [9,24].
Comparison of the aortic and pulmonary artery homografts data with that of the piglet great vessels shows that homografts from human adults are about 100 times larger in stress value when subjected to the same strain condition [9]. The increase in stress and stiffness as age increases, and tissue matures from neonatal into adult tissue, is due to the structural tissue changes occurring during growth [2527]. The age-related vascular structural changes that have been proposed to be associated with increased stiffness of the aorta include reduced collagen and elastin contents/mm2 of the aortic wall, increased fenestration of elastic laminae, and accumulation of fluorescent material in collagen and elastin [26,28]. This phenomenon results in an overall decrease in compliance as age increases. A histological study could reveal the structural difference between samples from different locations, ages, and so on. The elastic lamina and collagen bundle configuration from a histological analysis could better explain the anisotropy and mechanical variation of arteries.
A limitation of this study is that the constitutive model calibration was determined by the strain–stress response of a single loading condition (uniaxial tension, fully exhibiting the anisotropic behavior). However, hyperelastic mechanical behavior is dependent on loading conditions (uniaxial tension, biaxial tension, etc.) [8,14]. Thus, the anisotropic hyperelastic model could be better calibrated to biaxial or equibiaxial loading conditions [29]. Additionally, in order to achieve statistical power, it was necessary to pool sufficient data from the same location of the arteries.
In conclusion, aortic and pulmonary artery homografts exhibit hyperelastic mechanical characteristic with anisotropic effect. The calibrated Holzapfel–Gasser constitutive model can be applied in surgical preparation and related simulation to predict anisotropic biomechanical behavior. The synthetic vascular grafts are significantly stiffer than homografts. The mechanical properties of SVG-GORE-TEX in L orientation are comparable to those of the homografts, and this characteristic should be considered by surgeons when performing reconstructive cardiovascular surgery with this material. These differences between aortic and pulmonary artery homografts and synthetic vascular grafts may play a role in native neonatal tissue behavior, ventricular/arterial mechanical coupling, and risk of deformation due to abnormal hemodynamics of CHD.
## Acknowledgment
Professor Quanfang Chen provided access to the microloading frame. Mr. Shenghong Zhang assisted with mechanical testing. Thanks also due to partial financial support from the University of Central Florida.
## Funding Data
• Army Research Laboratory (Grant No. W911NF-15-2-0011).
## Nomenclature
### Abbreviations
Abbreviations
• AA =
ascending aorta
•
• Arch =
aortic arch
•
• C =
circumferential
•
• CHD =
congenital heart disease
•
• D =
diagonal
•
• DA =
descending aorta
•
• DIC =
digital imaging correlation
•
• ePTFE =
expanded polytetra-fluoroethylene
•
• GRG =
•
• IA =
innominate artery
•
• L =
longitudinal
•
• LCCA =
left common carotid artery
•
• LPA =
left pulmonary artery
•
• LScA =
left subclavian artery
•
• MPA =
main pulmonary artery
•
• RPA =
right pulmonary artery
•
• UTT =
uniaxial tension testing
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2019-08-24 05:31:24
|
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|
https://cs.stackexchange.com/questions/19207/find-least-probable-path-in-graph
|
# Find least probable path in graph
I am working on a special case of the longest path problem. For a cyclic directed graph $G=(V, E)$, where the edge-weights are probability values (i.e., $P(\_) = w(s, q)$ with $s,q \in V$), my aim is to find the least 'probable' path between two vertices.
My initial approach is to generate an graph $G'$ where the weights are the complementary probabilities $1- w(s, q)$ (with strictly positive values), and compute Dijkstra's shortest path on $G'$. Is this reasoning sound? Or am I getting myself into an NP-hard disaster?
• If you want the least probable path, wouldn't that just be the one with the smallest weight, not the largest weight? – G. Bach Dec 23 '13 at 14:01
• You mean the longest path on $G'$? If so, then what does is have do to with Dijkstra's shortest path? However, even if longest path wasn't NP-hard, this reasoning would still be wrong. – Parham Dec 23 '13 at 15:15
• This doesn't work at all. If the product of the weights along a path is the probability that the given path is taken, the probability that the path is not taken is the sum of the probabilities of all other paths, not the probability you describe. But I'm not even sure your original problem is well-defined: how do you know that all the "probabilities" sum to one, for example? – David Richerby Dec 24 '13 at 13:55
Your approach doesn't work. Presumably, you want to define the probability of a path to be the product of the probabilities on its edges. It sounds like you want to define the weight $w(e)$ on an edge $e$ to be $w(e)=1-p(e)$ (one minus its probability). However, doesn't work. It doesn't do what you want: you want $w(e)+w(e')$ to correspond to $p(e) \times p(e')$, but it doesn't.
Instead, you should be taking logarithms. In particular, you should define the weight on edge $e$ by $w(e) = -\log p(e)$. Now addition of weights corresponds to multiplication of probabilities:
$$w(e) + w(e') = -(\log p(e) + \log p(e')) = - \log(p(e) \times p(e')),$$
as desired. At this point all of the weights in $G'$ will be non-negative (do you see why?), so you can use Dijkstra's algorithm to find the shortest path in $G'$, and that will correspond to the path in $G$ with highest probability. As you can see, this is not a longest-path problem at all; it is just a straightforward shortest-paths problem.
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2020-09-24 11:50:07
|
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|
https://www.cfm.brown.edu/people/sg/SSPpage/sspsite/erk_lin.html
|
# Optimal Methods
• For nonlinear/nonautonomous systems:
• Explicit methods
• Implicit methods
• For linear, autonomous systems:
• Explicit methods
• Implicit methods
# Maple Scripts
• First Order
Optimal first order explicit SSP Runge-Kutta methods consist simply of repeated forward Euler steps. These methods all have effective SSP coefficient equal to unity and are equivalent to simply using the forward Euler method.
• Second Order
• The $m$ stage, second order SSP methods: \begin{eqnarray} u^{(0)} &=& u^n \\ u^{(i)} &=& \left( 1+ \frac{\Delta t}{m-1} L \right) u^{(i-1)} , i=1 ,..., m-1 \\ u^{m} & = & \frac{1}{m} u^{(0)} + \frac{m-1}{m} \left( 1+ \frac{\Delta t}{m-1} L \right) u^{(m-1)} \\ u^{n+1} & = & u^{(m)} \end{eqnarray} have an optimal SSP coefficient $c= m-1$ among all methods with nonnegative coefficients. Although these methods were designed for linear problems, they are also nonlinearly second order Each such method uses $m$ stages to attain the order usually obtained by a $2$-stage method, but has SSP coefficient $c=m-1$, thus the effective SSP coefficient here is $c_{\text{eff}} = \frac{m-1}{m}$.
• $m$-Stage, Order $m$
• The class of $m$ stage schemes given by: \begin{eqnarray} u^{(i)} & = & u^{(i-1)} + \Delta t L u^{(i-1)}, \qquad i=1, . . . , m-1 \\ u^{(m)} & = & \sum_{k=0}^{m-2} \alpha_{m,k} u^{(k)} + \alpha_{m,m-1} \left(u^{(m-1)}+\Delta t L u^{(m-1)} \right), \end{eqnarray} where $\alpha_{1,0} = 1$ and \begin{eqnarray} \alpha_{m,k} &=& \frac{1}{k} \alpha_{m-1,k-1}, \qquad k=1, . . . , m-2 \\ \alpha_{m,m-1} & = & \frac{1}{m!}, \qquad \alpha_{m,0} = 1- \sum_{k=1}^{m-1} \alpha_{m,k} \end{eqnarray} is an $m$-order linear Runge-Kutta method which is SSP with SSP coefficient $c=1$, which is optimal among all $m$ stage, $p=m$ order SSPRK methods with nonnegative coefficients. The effective SSP coefficient is $c_{\text{eff}} = \frac{1}{m}$.
• $m$-Stage, Order $m-1$
• The $m$ stage, order $p=m-1$ method: \begin{eqnarray} u^{(0)} &=& u^n \\ u^{(i)} & = & u^{(i-1)} + \frac{1}{2} \Delta t L u^{(i-1)}, \qquad i=1, . . . , m-1 \\ u^{(m)} & = & \sum_{k=0}^{m-2} \alpha_{m,k} u^{(k)} + \alpha_{m,m-1} \left( u^{(m-1)}+ \frac{1}{2} \Delta t L u^{(m-1)} \right), \\ u^{n+1} & = & u^{(m)} \end{eqnarray} Where the coefficients are given by: \begin{eqnarray} \alpha_{2,0} & = & 0 \qquad \alpha_{2,1}=1 \\ \alpha_{m,k} &=& \frac{2}{k} \alpha_{m-1,k-1}, \qquad k=1, . . . , m-2 \\ \alpha_{m,m-1} & = & \frac{2}{m} \alpha_{m-1,m-2} , \qquad \alpha_{m,0} = 1- \sum_{k=1}^{m-1} \alpha_{m,k} \end{eqnarray} is SSP with optimal (for methods with nonnegative coefficients) SSP coefficient $c=2$. The effective SSP coefficient for these methods is $c_{\text{eff}} = \frac{2}{m}$.
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2022-08-12 20:48:28
|
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|
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Composition_of_Mappings
|
# Equivalence of Definitions of Composition of Mappings
## Theorem
The following definitions of the concept of Composition of Mappings are equivalent:
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that the domain of $f_2$ is the same set as the codomain of $f_1$.
### Definition 1
The composite mapping $f_2 \circ f_1$ is defined as:
$\forall x \in S_1: \map {\paren {f_2 \circ f_1} } x := \map {f_2} {\map {f_1} x}$
### Definition 2
The composite of $f_1$ and $f_2$ is defined and denoted as:
$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
### Definition 3
The composite of $f_1$ and $f_2$ is defined and denoted as:
$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z}$
## Proof
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that:
$\Dom {f_2} = \Cdm {f_1}$
### $(1)$ implies $(2)$
Let $f_2 \circ f_1$ be a composite mapping by definition 1.
Then by definition:
$\forall x \in S_1: \map {\paren {f_2 \circ f_1} } x := \map {f_2} {\map {f_1} x}$
We have that $f_1$ is a mapping, and so:
$\forall x \in S_1: \exists \map {f_1} x \in S_2$
We have that $f_2$ is a mapping, and so:
$\forall \map {f_1} x \in S_2: \exists \map {f_2} {\map {f_1} x} \in S_3$
Then:
$\displaystyle z$ $=$ $\displaystyle \map {f_2} {\map {f_1} x}$ $\displaystyle \leadsto \ \$ $\displaystyle \tuple {\map {f_1} x, z}$ $\in$ $\displaystyle f_2$ Definition of Mapping as a Relation
and so:
$f_2 \circ f_1 = \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
Thus $f_2 \circ f_1$ is a composite mapping by definition 2.
$\Box$
### $(2)$ implies $(1)$
Let $f_2 \circ f_1$ be a composite mapping by definition 2.
Then by definition:
$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
We have that $f_1$ is a mapping, and so:
$\forall x \in S_1: \exists \map {f_1} x \in S_2$
and so by definition of a Definition of Mapping as a Relation:
$\, \displaystyle \forall x \in S_1: \,$ $\displaystyle \tuple {\map {f_1} x, z}$ $\in$ $\displaystyle f_2$ $\displaystyle \leadsto \ \$ $\displaystyle z$ $=$ $\displaystyle \map {f_2} {\map {f_1} x}$
Thus $f_2 \circ f_1$ is a composite mapping by definition 1.
$\Box$
### $(2)$ implies $(3)$
Let $f_2 \circ f_1$ be a composite mapping by definition 2.
Then by definition:
$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
Because $f_1$ is a mapping, it follows that:
$\forall x \in S_1: \exists y \in S_2: \map {f_1} x = y$
Similarly:
$\forall y \in S_2: \exists z \in S_3: \map {f_2} y = z$
Hence:
$\tuple {\map {f_1} x, z} \in f_2 \implies \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z$
and so:
$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z}$
Thus $f_2 \circ f_1$ is a composite mapping by definition 3.
$\Box$
### $(3)$ implies $(2)$
Let $f_2 \circ f_1$ be a composite mapping by definition 3.
Then by definition:
$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z}$
We have that:
$\forall x \in S_1: \exists y \in S_2: \map {f_1} x = y$
and that:
$\forall y \in S_2: \exists z \in S_3: \map {f_2} y = z$
Hence:
$\forall x \in S_1: \exists z \in S_3: \map {f_1} x = y \land \map {f_2} y = z$
Thus:
$\forall x \in S_1: \exists z \in S_3: \tuple {\map {f_1} x, y} \in f_2$
and so:
$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$
Thus $f_2 \circ f_1$ is a composite mapping by definition 2.
$\blacksquare$
|
2019-12-16 12:33:29
|
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|
http://by.tc/tag/information-theory/
|
#### information theory
(Credit to xkcd for the comic, of course.)
Near as I can figure, the go-to framework for mathematics these days is Zermelo-Fraenkel set theory, with or without the Axiom of Choice (denoted as ZF or ZFC, respectively). The Axiom of Choice (AC) was contentious in its day, but the majority of mathematicians now accept it as a valid tool. It is independent of ZF proper, and many useful results require that you either adopt or reject it in your proof, but you're allowed to do either.
The simplified version of AC is that it allows you to take an element from any number of non-empty sets, thereby forming a new set. Where it diverges from ZF is that it allows for rigorous handling of infinite sets, something that is impossible within ZF. Imagine you have an infinite number of buckets, each one containing an infinite number of identical marbles; AC finds it acceptable for you to posit, in the course of a proof, that you somehow select one or more marbles from each of those buckets into a new infinite pile.
Unsurprisingly, this means that AC is equivalent to a number of other famous results, notably the well-ordering principle, which counter-intuitively states that there is a way to totally order any set given the right binary operator, even e.g. \RR. Worse yet, it lets you (via the Banach-Tarski paradox) disassemble a tennis ball into five or six pieces and reassemble them into something the size of the sun. In other words, AC is meant to be an abstract mathematical tool rather than a reflection of anything in reality.
Now, I could spend forever and a day discussing this sort of stuff, but let's move along to my thought: what it has to do with the Big Bang.
As you all know, the Big Bang ostensibly kicked off our universe some 13.8 billion years ago. However, some of the details are still kind of hazy, especially the further back you look. Past the chronological blinder of the Planck Epoch—the first ~10^-43 seconds of our universe—there is essentially a giant question mark. All of our physics models and equations break down past that point. Theory says that the big bang, at its moment of inception, was an infinitely dense zero-dimensional point, the mother of all singularities. One moment it's chillin', the next it explodes, and over time becomes our universe.
I don't love this theory, but it's the best fit we've got so far for what we've observed, particularly with red-shifting galaxies and the cosmic microwave background radiation, so we'll run with it for the moment. So where did all of these planets and stars actually come from, originally? There is a long and detailed chronology getting from there to here, but let's concern ourselves with looking at it from an information-theoretic point of view, or rather, 'How did all of this order and structure come out of an infinitesimal point?'
It's unclear, but the leading (though disputed) explanation is inflation, referring to an extremely rapid and sizable burst of expansion in the universe's first moments. There is apparently observational data corroborating this phenomenon, but a lot of the explanation sounds hand-wavy to me, and as though it were made to fit the evidence. The actual large structure of the universe is supposed to have arisen out of scale-invariant quantum fluctuations during this inflation phase, which is a cute notion.
Note, by the way, that entropy was also rapidly increasing during this step. In fact, my gut feeling on the matter is that since entropy is expected to strictly increase until the end of time (maximum entropy), it makes plenty of sense that the Big Bang kernel would have had zero entropy—hell, it's already breaking all the rules. While thermodynamic and information-theoretic entropy are ostensibly two different properties, they're one and the same principle underneath. Unless I'm gravely mistaken, no entropy would translate to complete order, or put another way, absolute uniformity.
If that was indeed the case, its information content may have been nothing more than one infinitely-charged bit (or bits, if you like); and if that were the case, there must have been something between that first nascent moment and the subsequent arrival of complex structure that tore that perfect node of null information asunder. Whether it was indeed quantum fluctuations or some other phenomenon, this is an important point which we will shortly circle back to.
It's far from settled, but a lot of folks in the know believe the universe to be spatially infinite. Our observable universe is currently limited to something just shy of 100 billion light years across; however, necessary but not sufficient for the Big Bang theory is the cosmological principle, which states that we should expect the universe to be overwhelmingly isotropic and homogeneous, particularly on a large scale (300+ mil. light years). This is apparently the case, with statistical variance shrinking down to a couple percent or much less, depending on what you're examining.
That last bit is icing on the cake for us. The real victory took place during whatever that moment was where a uniform singularity became perturbed. (Worth noting that if the uniformity thing is true, that mandates that it really was an outside agency that affected it in order to spawn today's superstructure, which of course makes about as little sense as anything else, what with there presumably being no 'outside' from which to act.)
So here's the punchline. If you assume an infinite universe, that means that the energy at one time trapped featureless in that dimensionless point has since been split apart into an infinitude of pieces. "But it's still one universe!" you say. Yes, but I think it's equally valid to recognize that our observable universe is finite, and as such, could be represented by NN (if discrete) numbers, or RR (if not), or \aleph_n if things are crazier than we know. Regardless, it could be described mathematically, as could any other of the infinitely many light cones which probably exist, cozy in their own observable (but creepily similar) universe.
Likewise, we could view each observable universe as a subset of the original Big Bang kernel, since that is from whence everything came. It must be representable as a set of equal or larger cardinality to the power set of all observable universe pockets, and therefore the act of splitting it into these subsets was a physical demonstration of the Axiom of Choice in reality!
I'm not sure what exactly that implies, but I find it awfully spooky. I feel like it either means that some things thought impossible are possible, or that this violation happened when the Big Bang kernel was still in its pre-Planck state; but if that's the case, not only do our physical models break down, our fundamental math would be shown not to apply in that realm either, which means it could have been an anti-logical ineffable maelstrom of quasi-reality which we have no hope of ever reasoning about in a meaningful way.
This is something I worked on a year ago, so I'll keep it (relatively) brief.
##### Inspiration
There's a keypad on my apartment building which accepts 5-digit codes. One day on the way in, I started thinking about how long it would take to guess a working code by brute force. The most obvious answer is that with five digits, you have 10^5 = 100,000 possible codes; since each one of these is 5 digits long, you're looking at 500,000 button pushes to guarantee finding an arbitrary working code.
##### Realization
But then I realized you could be a little smarter about it. If you assume the keypad only keeps track of the most recent five digits, you can do a rolling approach that cuts down your workload. For example, say I hit 12345 67890. If the keypad works as described, you're not just checking two codes there, you're checking 6: 12345, 23456, 34567, 45678, 56789, 67890.
##### Foundation
The next natural question was how many button-pushes this could actually save. After doing some work, I satisfied myself that you could cut it down by a factor of ~D, where D is the number of digits in a code. So if you typed the right digits in the right order, instead of the 500,000 before, you're only looking at 100,004 (you need an extra 4 to wrap the last few codes up).
##### Experimentation
The next natural question was: how do you actually come up with that string of digits? It has to be perfect, in that it contains every possible 5-digit combination without repeating any of them. As with almost any exploratory problem, the best approach is to simplify as much as possible. For instance, consider a binary code three digits long, which only has 2^3 = 8 different codes: {000, 001, 010, 011, 100, 101, 110, 111}.
My formula suggested you should be able to hit all eight of those in a 2^3 + 3 - 1 = 10 digit string, and it's easy enough to put one together by a little trial and error: 0 0 0 1 1 1 0 1. I found it was easiest to treat these strings as cyclical, so the 0 1 at the end wrap around to give you 0 1 0 and 1 0 0. As a bonus, any rotation of this string will work just as well.
##### Perspiration
As I scaled the problem up, however, more and more things became clear. First, above a certain point, you start getting multiple viable optimal strings that are not simple transformations of one another. Second, finding an elegant way to generate these strings was not turning out to be easy.
I found one mechanical way of generating a valid string that worked, but I didn't love it. If you list all the combinations you have to cover, and then slot each combination into a buffer greedily, meaning the earliest spot where it can fit (potentially with overlap), it works out.
E.g.:
##### Beautification
At some point I realized the generative process could be viewed as a directed graph, the nodes representing an N-length code, its successors delineating alternatives for continuing the string. After a few attempts, I got a pretty clear-looking one down (the node labels 0-7 are standing in for their binary counterparts):
As it turns out, you can start on any node and if you trace a Hamiltonian cycle—touching each vertex only once and ending back at the start—the numbers you hit along the way form a valid optimal string. This approach also scales with the parameters of the problem, but requires a messier or multi-dimensional graph.
##### Confirmation
Whenever I stumble into open-ended problems like this, I avoid Googling for the answer at first, because what's the fun in that? I was pretty sure this would be a solved problem, though, and after spending a while working this through myself, I looked for and found Wikipedia's page on De Bruijn sequences. As usual, I was beaten to the punch by somebody a hundred years earlier. Hilariously, however, in this case the results matched up better than expected. Check out the Wiki page, notably a) the graph on the right, and b) the first line of the section "Uses".
##### Notation
If you want to see the wild scratch pad which brought me from point A to point B, by all means, enjoy.
This is a quick story about today's thing that I discovered that is already in Wiki. I must be up in the triple digits at this point. That said, this is one of the more obvious ones.
I was reading a thread about which sorting algorithm could be considered "best," and someone mentioned a couple of algorithms which allegedly run in O(n log log n) time. This came as a shock, since I'd thought n log n was the brick wall for an honest sort, and I wondered if these other sorts were for real, whether that would imply the existence of a linear time sort.
I tried to imagine what the lower bound would be, and figured there must be a minimum number of comparisons required for some number of elements. Didn't take long to get from there to realizing that n integers can be arranged in n! different permutations, which I reasoned meant that one must gather enough information (read: take comparisons) from the data to uniquely identify which of n! different arrangements the sort is in.
That, in turn, screams out for a log, specifically log_2(n!). If the sort permutation is truly random, then on average, we should expect to be able to identify it from log_2(n!) bits (read: comparisons, again.) To be a little more precise, I guess it'd be more like $\frac{\lceil{n \log_2 (n!)}\rceil}{n} .$
I cheated a little here and plugged lim_{n->\infty} [log_2(n!)] into Wolfram Alpha, and it was clear the dominating factor is, surprise surprise, n log n. As for those mystery n log log n algorithms, they were tough to track down too, and there seemed to be a lack of final consensus on the issue. Due to the math described herein if nothing else, they must operate with some limitation or another on domain or input, assuming they do work.
Later, seeing the bottom of the Wikipedia page on sorting algorithms, I saw all of this done similarly, with the weird surprise that once you get to n=12, the maximum number of comparisons required suddenly goes up by one inexplicably, requiring 30 comparisons where we predict 29. Sadly, the footnote links explaining those aberrations were in journals behind paywalls, but the gist seemed to be that it was a bit of an open (or at least highly non-trivial) question.
So here's how I went down the information theory rabbit hole this morning.
How much information is in a bit?
You might say '1 bit.' I am inclined to agree.
How much information is in two bits?
Disregarding any other considerations, and on average, I would agree that two bits stores two bits of information. However, let's say that the first bit you receive from some black box is a 0. (Or it could be a 1, and after inverting the rest of the following argument, it would yield identical results, which perversely almost makes it seems as though that first bit carries no information after all.)
But I digress. So, you have your 0. Now, out comes another 0 as the second bit.
A word on this black box. You don't know what's inside it. It could be a friend connected through the internet, it could be a random number generator, it could be sequential bits from a vacation picture on your hard drive, it could be aliens trying to make contact, you name it.
Now, the specific details of the type of distribution and manner of bit-generating agents that might be in there is actually an absolutely critical point which is not to be glossed over, and so I will cheerfully completely disregard it, except to say that for our purposes, we will allow in a hand-wavy way that all the possibilities listed each represent valid non-negligible possibilities, as that is the spirit of this thought experiment.
Moving on. A priori, the second bit (b_2) could have been a 0 or a 1. To be able to estimate which was more likely, without knowing the details about what the probability distribution of possible bit-generating entities actually is, is arguably impossible, but with the barest of assumptions, this is no longer the case. I maintain that a second 0 was indeed the more likely bit to arrive, and by a significant margin, too.
Consider it through the lens of entropy. Whatever is in the box, it will be emitting bits following some kind of overarching guidelines, which you can consider ranging from "perfectly ordered" to "perfectly entropic", the latter being more or less synonymous with "completely random." If it is a true RNG in there, then your odds are 50/50 for each bit that comes out, including b_2.
However, if it is something highly ordered, say a simple computer program, it may well be printing an endless stream of zeros. As was strongly indicated by my previous excursion into the cartography of computation space, repeated streams of bits are the single easiest thing to generate with any kind of reasonable computational framework; note that computational framework here can refer to anything from a programming language to the behavior of an ant colony on down to physics itself. It simply takes less work to generate some numbers than others. So, all things being roughly equal, I submit that you'll find a second 0 more often than not. This represents the more ordered possibility, while b_2=1 would be the more entropic possibility.
Because of these probabilities, the underlying information content of that second bit is changed. After all, if something very likely happens, you've gained very little information; if the sun rises tomorrow, you can't conclude much from it. If the sun were not to rise, well, that contains a LOT of information, information you'd get to enjoy for maybe a month or two before the atmosphere freezes over.
Let's try a more mathematical but equally stark demonstration of the same essential argument. Consider: you watch the box for days, and all it does it spit out 0s, one after another. Millions of them. Each new zero says even less than the one before it (or, arguably, devalues all bits received retroactively—probably moot.)
And then, one day, you wake up and see a single 1 came through. Being a man or woman of science, you eventually think to count exactly how many zeros came out before the 1, and to your amazement, there are 3,141,592,654 zeros, which you naturally immediately recognize as \pi \times 10^9. Even setting aside any deeper implications about what that might portend, it is now obvious that the odds are extremely good that there is an agent or process which is highly structured at work in the box.
All that from a single bit. Clearly that 3,141,592,655^(th) bit contained more information than the millions before it. You might say it shares its information content among all the bits before it, but it has still, by itself, made an enormous net change to the Shannon entropy and Kolmogorov complexity of the data. As a lower bound, you might say that in such a pattern, where you have n 0s followed by a single 1, that 1 contains at least a ballpark figure of log_2 n bits of actual information, that being the amount of information inherent in the number generated. E.g., 0000 0000 1 gives you 8 zeros, and log_2 8 = 3, which is the number of bits you'd require to generate a number between 1 and 8 inclusive.
So what?
I don't know. Stuff is neat.
(Maybe I'll show how primes fit into all this tomorrow!)
### The birth of Skynet
A couple of years back, I wrote a little program named Skynet. It uses a configurable set of assembly-like instructions to spawn tiny programs, cobbled together completely at random. The original idea was to be able to feed it a set of input and output integer lists representing some function you were trying to find, and have it then blindly fumble its way through millions or billions of those micro-programs until it hit on one that satisfied your conditions.
For example, if you wanted some code that would exponentiate, you might feed it a configuration file signifying
``````Input 1: 2 3
Output 1: 8
Input 2: 4 3
Output 2: 256
``````
and it would run forever until it stumbled upon a solution, or you killed it.
And it worked, and it was fun to screw around with, if you like that sort of thing. Feel free to grab it from the link above, but it was a quicky that I wrote for myself, so it might be rough to figure out. It's Turing-complete, but it doesn't do anything very clever—there's no fancy machine learning algorithms here, just straight random goodness—but you can specify a few parameters to try to guide its efforts, such as minimum and maximum program length to try, or how many instructions to execute before giving up on a microprogram as a runaway and moving on. I think it also tries to spot and discard overtly obvious dead ends and NOOPs.
##### Limitations
So yes, it works as intended, but the problem is that the realm of all possible computable functions is one of those seriously immense virtual universes that tend to pop up when dealing with combinatorics or math in general. You might have 16 instructions enabled at one time, but instructions at the assembly level don't accomplish a hell of a lot, and when you consider a branching factor that big, all of a sudden you're looking at 16^10 possible microprograms for even a modest 10-instruction snippet of code.
That's just over a trillion, many of which are likely to be non-halting and thus run several hundred instructions before bailing. The upshot is that to exhaustively try all of them, you're probably talking days or even months, depending on how optimized it all is. After quickly discovering the futility of systematically enumerating them, I said "fuck it" and resdesigned things with its current random approach, perfectly happy to try out 50-instruction programs if the pRNG requests it. Of course, if there were actually some program out in the ether that needed to be 50 instructions long, the odds of running into it through chance are almost nothing—you could expect to hit it around the same time the last protons and neutrons are decaying.
But I digress. A lot of simpler algorithms can be squeezed into a tractable space.
###### Things it's figured out:
• Fibonacci sequence
• factorials
• doubling, cubing
• for loops
• adding pairs of input numbers on the stack
• finding the mean of input numbers
• determining divisibility of input numbers by 7
• reversing order of stack
###### Not so much:
• primality detection
• prime generation
• misc. other NP problems
• bootstrapping into sentience
##### Details
Some of those successes were much easier than others. Fibonacci, for example, seems to be about the next easiest thing for it to do besides count. The other day I got GitLab CI going which I'm abusing mercilessly by having it spawn Skynet jobs instead of actually testing anything. But manually, this is how it rolls:
``````[12:00] ~/progs/Skynet> time ./4bit --file fib.skynet -r
fscanfs return 6111111111
low=4 up=8 exact=0 stopAfter=1 maxSteps=50
Input 1:
Output 1: 1 1 2 3 5 8 13
Randomizing instructions...
[ INS SWAB MULT PUSH POP INC ADD JNE NAND DEC PEEK SKIP IFF NOT LOCO_1 MOD ]
Satisficing program (length 5): INC PUSH ADD SWAB JNE -=- Tracing operation -=-
[init a=0 b=0 c=0 pc=0 sp=1] ( __ )
INC [1 a=1 b=0 c=0 pc=1 sp=1] ( __ )
PUSH [2 a=1 b=0 c=0 pc=2 sp=2] ( __ 1 )
ADD [3 a=1 b=0 c=0 pc=3 sp=2] ( __ 1 )
SWAB [4 a=0 b=1 c=0 pc=4 sp=2] ( __ 1 )
JNE [5 a=0 b=1 c=0 pc=0 sp=2] ( __ 1 )
INC [6 a=1 b=1 c=0 pc=1 sp=2] ( __ 1 )
PUSH [7 a=1 b=1 c=0 pc=2 sp=3] ( __ 1 1 )
ADD [8 a=2 b=1 c=0 pc=3 sp=3] ( __ 1 1 )
SWAB [9 a=1 b=2 c=0 pc=4 sp=3] ( __ 1 1 )
JNE [10 a=1 b=2 c=0 pc=0 sp=3] ( __ 1 1 )
INC [11 a=2 b=2 c=0 pc=1 sp=3] ( __ 1 1 )
PUSH [12 a=2 b=2 c=0 pc=2 sp=4] ( __ 1 1 2 )
ADD [13 a=4 b=2 c=0 pc=3 sp=4] ( __ 1 1 2 )
SWAB [14 a=2 b=4 c=0 pc=4 sp=4] ( __ 1 1 2 )
JNE [15 a=2 b=4 c=0 pc=0 sp=4] ( __ 1 1 2 )
INC [16 a=3 b=4 c=0 pc=1 sp=4] ( __ 1 1 2 )
PUSH [17 a=3 b=4 c=0 pc=2 sp=5] ( __ 1 1 2 3 )
ADD [18 a=7 b=4 c=0 pc=3 sp=5] ( __ 1 1 2 3 )
SWAB [19 a=4 b=7 c=0 pc=4 sp=5] ( __ 1 1 2 3 )
JNE [20 a=4 b=7 c=0 pc=0 sp=5] ( __ 1 1 2 3 )
INC [21 a=5 b=7 c=0 pc=1 sp=5] ( __ 1 1 2 3 )
PUSH [22 a=5 b=7 c=0 pc=2 sp=6] ( __ 1 1 2 3 5 )
ADD [23 a=12 b=7 c=0 pc=3 sp=6] ( __ 1 1 2 3 5 )
SWAB [24 a=7 b=12 c=0 pc=4 sp=6] ( __ 1 1 2 3 5 )
JNE [25 a=7 b=12 c=0 pc=0 sp=6] ( __ 1 1 2 3 5 )
INC [26 a=8 b=12 c=0 pc=1 sp=6] ( __ 1 1 2 3 5 )
PUSH [27 a=8 b=12 c=0 pc=2 sp=7] ( __ 1 1 2 3 5 8 )
ADD [28 a=20 b=12 c=0 pc=3 sp=7] ( __ 1 1 2 3 5 8 )
SWAB [29 a=12 b=20 c=0 pc=4 sp=7] ( __ 1 1 2 3 5 8 )
JNE [30 a=12 b=20 c=0 pc=0 sp=7] ( __ 1 1 2 3 5 8 )
INC [31 a=13 b=20 c=0 pc=1 sp=7] ( __ 1 1 2 3 5 8 )
PUSH [32 a=13 b=20 c=0 pc=2 sp=8] ( __ 1 1 2 3 5 8 13 )
ADD [33 a=33 b=20 c=0 pc=3 sp=8] ( __ 1 1 2 3 5 8 13 )
SWAB [34 a=20 b=33 c=0 pc=4 sp=8] ( __ 1 1 2 3 5 8 13 )
JNE [35 a=20 b=33 c=0 pc=0 sp=8] ( __ 1 1 2 3 5 8 13 )
INC [36 a=21 b=33 c=0 pc=1 sp=8] ( __ 1 1 2 3 5 8 13 )
PUSH [37 a=21 b=33 c=0 pc=2 sp=9] ( __ 1 1 2 3 5 8 13 21 )
ADD [38 a=54 b=33 c=0 pc=3 sp=9] ( __ 1 1 2 3 5 8 13 21 )
SWAB [39 a=33 b=54 c=0 pc=4 sp=9] ( __ 1 1 2 3 5 8 13 21 )
JNE [40 a=33 b=54 c=0 pc=0 sp=9] ( __ 1 1 2 3 5 8 13 21 )
INC [41 a=34 b=54 c=0 pc=1 sp=9] ( __ 1 1 2 3 5 8 13 21 )
PUSH [42 a=34 b=54 c=0 pc=2 sp=10] ( __ 1 1 2 3 5 8 13 21 34 )
...
Total=357252 Timeouts=206816 Redundant=6694
real 0m0.130s
user 0m0.127s
sys 0m0.002s
[12:00] ~/progs/Skynet>
``````
As may or may not be clear, it spits out its working program at the top, and then walks you through a trace execution.
One of my discoveries was that program length matters a lot less than I would have guessed. You can set your program length bounds to look for stuff 30-50 instructions long, and it'll still pop out a Fibonacci program after a second. This is presumably because when you throw shit randomly together, you end up with an abundance of NOOPs (at least with my instruction set.)
To put that in slightly more concrete perspective, there are one or two 5-length versions of Fibonacci that exist, out of something like a million possible programs; put another way, you could say ~0.0002 of 5-lengthers are Fibonacci programs. But as it turns out, if you go out to 20 instructions, that percentage doesn't drop by much—in fact, for some things it climbs. The point I'm trying to make, which surprised me, is that regardless of the program length you're searching for, you'll hit anything provided it's simple enough to fit in that length or simpler, and these types of algorithms seem to occupy a pretty constant market share at any depth.
### Present day
(Well, yesterday. )
I discovered PyPy and Cython, things which I should have known about a long time ago but didn't. I wrote Skynet in C because it's second to none for this sort of thing, probably a 30x speed increase over the same thing in vanilla Python. But Cython lets you statically type variables, embed C code, and other wily stuff, while PyPy appears to be a magical JIT compiler that makes your pure Python run 5-10x faster (if it's this sort of project).
Anyway, I wanted to try them out, so last night I ported Skynet back to Python, yielding skython, a name which I'm half proud of but made me die a little inside to use. After I got the basics working, I started to adapt it for Cython, but realized I would effectively be porting it awkwardly back to C-ish if I continued, so I've stuck with Python and am using PyPy, which seems to run it at maybe 25 of the speed of the C version. In other words, close enough to be an acceptable trade-off. It's not up to snuff yet, but it is working, and it's certainly faster to add features than it was with C.
To wit: I was thinking about the program-market-share concept I mentioned, and it led me to add a method which creates programs randomly without looking for any input or output, but rather keeping track of everything it churns out on its own. And this is where we actually rejoin the title and original thematic intent of this post, because the idea there was to explore the distribution of algorithmic information in computation space, by which I mean seeing just how often it opts for some sequences over others when selecting out of the vast set of all computable functions.
#### And?
And it was interesting. I gave it some pretty neutral parameters and had it run a billion microprograms. There's lots I could have done (and perhaps will do) to clean up and solidify the results, but there was enough to get a sense of things.
###### A few results
• Half the programs do nothing. Give or take.
• Most of the rest, percentage-wise, print out endless strings of 0s or 1s.
• The last few percent, though, starts to get interesting.
• When you get down to 0.5 or so, you start to get 0101010... and 123456... sequences.
• They're followed close behind by the 22222s and the 110110s and the 012012s... etc.
The first eye-catching result is down at 0.00272 computation-space-share (but this is still in the top percent of frequency):
``````0,1,3,7,15,31,63,127,255,511,1023,2047,4095,8191,16383,32767,65535,131071,262143,524287,1048575,2097151,4194303,8388607,16777215
``````
The pattern's clear enough. A little ways on, down at 0.00045, I spot a 1, 4, 7, 10, ... along with the first Fibonacci. Down further, I recognize a pattern that came up in some math problem I worked on months ago:
``````0,1,2,5,10,21,42,85,170,341,682,1365,2730
``````
But as I went deeper, I found sequences that clearly had structure but were foreign to me:
``````1,3,8,20,49,119,288,696,1681,4059,9800,23660,57121,137903
``````
I started looking them up on OEIS, and found that most of the interesting-looking patterns were indeed in there, and at least frickin' half of them were related in some way to the Fibonacci sequence. Not that this is especially surprising or perplexing in hindsight, I'd just never given much thought to how fundamental the principle is upon which it's built. No wonder that it pops up in nature everywhere; God must have noticed the same thing. So yeah, there are some interesting patterns when you get deep, but I didn't find anything earth-shattering.
#### Zip 'em up!
When I got tired of that, I moved on to a test of algorithmic information theory. The expectation is that the more frequently a program appears, the less complexity it is apt to have. A quick'n'dirty test for complexity is compressibility—the more complex a thing, the less compressible, to the point where any data sufficiently close to random (which is 'perfectly' complex, in this sense) will only get bigger.
I had my big honkin' 88mb data file, sorted by output frequency, so first I split it into equal parts, going from least to most common outputs:
``````[13:17] ~/progs/skython> split -da 2 -n 25 1bil.txt
[13:18] ~/progs/skython> ls -l
-rw-r--r-- 1 tc tc 88516009 Aug 1 09:46 1bil.txt
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x00
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x01
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x02
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x03
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x04
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x05
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x06
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x07
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x08
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x09
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x10
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x11
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x12
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x13
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x14
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x15
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x16
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x17
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x18
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x19
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x20
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x21
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x22
-rw-r--r-- 1 tc tc 3540640 Aug 1 13:18 x23
-rw-r--r-- 1 tc tc 3540649 Aug 1 13:18 x24
``````
And then I zipped 'em:
``````[13:18] ~/progs/skython> gzip -9 x??; ll
-rw-r--r-- 1 tc tc 88516009 Aug 1 09:46 1bil.txt
-rw-r--r-- 1 tc tc 851286 Aug 1 13:18 x00.gz
-rw-r--r-- 1 tc tc 850941 Aug 1 13:18 x01.gz
-rw-r--r-- 1 tc tc 851373 Aug 1 13:18 x02.gz
-rw-r--r-- 1 tc tc 852635 Aug 1 13:18 x03.gz
-rw-r--r-- 1 tc tc 849508 Aug 1 13:18 x04.gz
-rw-r--r-- 1 tc tc 857673 Aug 1 13:18 x05.gz
-rw-r--r-- 1 tc tc 853534 Aug 1 13:18 x06.gz
-rw-r--r-- 1 tc tc 852645 Aug 1 13:18 x07.gz
-rw-r--r-- 1 tc tc 854754 Aug 1 13:18 x08.gz
-rw-r--r-- 1 tc tc 854989 Aug 1 13:18 x09.gz
-rw-r--r-- 1 tc tc 855178 Aug 1 13:18 x10.gz
-rw-r--r-- 1 tc tc 848454 Aug 1 13:18 x11.gz
-rw-r--r-- 1 tc tc 846578 Aug 1 13:18 x12.gz
-rw-r--r-- 1 tc tc 844529 Aug 1 13:18 x13.gz
-rw-r--r-- 1 tc tc 849767 Aug 1 13:18 x14.gz
-rw-r--r-- 1 tc tc 849013 Aug 1 13:18 x15.gz
-rw-r--r-- 1 tc tc 846908 Aug 1 13:18 x16.gz
-rw-r--r-- 1 tc tc 852737 Aug 1 13:18 x17.gz
-rw-r--r-- 1 tc tc 847214 Aug 1 13:18 x18.gz
-rw-r--r-- 1 tc tc 821187 Aug 1 13:18 x19.gz
-rw-r--r-- 1 tc tc 760842 Aug 1 13:18 x20.gz
-rw-r--r-- 1 tc tc 752602 Aug 1 13:18 x21.gz
-rw-r--r-- 1 tc tc 727900 Aug 1 13:18 x22.gz
-rw-r--r-- 1 tc tc 692860 Aug 1 13:18 x23.gz
-rw-r--r-- 1 tc tc 605098 Aug 1 13:18 x24.gz
``````
...and as expected, they all got smaller, because these were text dumps which are always gonna be highly compressible. But that aside, it's evident that gzip had a much easier time with the more common stuff. Again, no surprise, but cool. Note the implicit curve that drops off sharply until you get to the top 75 or 80 of them, where it levels off and stays effectively flat.
#### One last thing to check
I'm beginning to suspect that primes are inherently complex, not just from number theory but in a Kolmogorov-y, robust sense. More Linux tricks:
``````[10:24] ~/progs/skython> for i in {1..30}; do cut -s --delimiter=, -f \$i 1bil.txt >> 1bil_vals.txt; done
[10:27] ~/progs/skython> sed 's/$\s\|\-$//g' 1bil_vals.txt | uniq -c | sort -gr > 1bil_sorted_abs_final.txt
``````
which gives a sorted text file of how many times every individual term came up overall, ignoring patterns:
``````[10:30] ~/progs/skython> head -20 1bil_sorted_abs_final.txt
16245701
11338986 1
8543839 0
5732203 2
3037257 3
1661122 4
980248 5
838860 6
582802 7
457519 8
338796 9
289842 10
226633 12
208586 11
161242 13
154787 14
149209 15
140625 16
90908 18
90094 17
``````
It's my suspicion that primes will be disproportionately low on this list—as they should be, given how computers work—and a first glance confirmed that somewhat, but then I stopped to write all this.
Okay. That's it. Here's the big ugly data file if you want it. Maybe I'll write up my thoughts on an approach to P=^{?}NP via transfinite induction tomorrow. Peace out.
|
2019-02-20 02:27:43
|
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|
https://minorfree.github.io/tree-shortcutting/
|
Rambling on Graphs
# Shortcutting Trees
Over the past two years or so, I have been thinking about a cute problem, which turned out to be much more useful to my research than I initially thought. Here it is:
Tree Shortcutting Problem: Given an edge-weighted tree $T$, add (weighted) edges to $T$, called shortcuts, to get a graph $K$ such that:
1. $d_K(u,v) = d_T(u,v)~\quad \forall u,v\in V(T)$. That is, $K$ preserves distances in $T$.
2. For every $u,v\in V(T)$, there exists a shortest path from $u$ to $v$ in $K$ containing at most $k$ edges for some $k\geq 2$.
The goal is to minimize the product $k \cdot \mathrm{tw}(K)$, where $\mathrm{tw}(K)$ is the treewidth of $K$.
Figure 1: An emulator $K$ obtained by adding one edge to the tree $T$ has hop bound $3$ and treewidth $2$. Compared to $T$, the hop bound decreases by 1 while the treewidth increases by 1.
Such a graph $K$ is called a low-hop and low-treewidth emulator of the tree $T$. The parameter $k$ is called the hop bound of $K$. It is expected that there will be some trade-off between the hop bound and the treewidth. We are interested in minimizing $k \cdot \mathrm{tw}(K)$. This product directly affects parameters in our application; see the conclusion section for more details.
For readers who are not familar with treeewidth, see here and here for an excellent introduction, and why treewidth is an interesting graph parameter.
Remark 1: The tree shortcutting problem is already non-trivial for unweighted trees; the shortcuts must be weighted though. Furthermore, for each edge $(u,v)$ added to $K$, the weight of the edge will be $d_T(u,v)$. Thus, in the construction below, we do not explicitly assign weights to the added edges.
A version of a tree shortcutting problem where one seeks to minimize the number of edges of $K$, given a hop bound $k$, was studied extensively (see 1,2,3,4, including my own work with others), arising in the context of spanners and minimum spanning tree problem. What is interesting there IMO is that we see all kinds of crazy slowly growing functions in computer science: for $k = 2$, the number of edges of $K$ is $\Theta(n \log n)$; for $k = 3$, the number of edges of $K$ is $\Theta(n \log\log n)$; for $k = 4$, the number of edges is $\Theta(n \log^* n)$; $\ldots$ [too difficult to describe]; and for $k = \alpha(n)$, the number of edges is $\Theta(n)$. Here, as you might guess, $\alpha(\cdot)$ is the notorious (one parameter) inverse Ackermann function. (The $\Theta$ notation in the number of edges means there exist matching lower bounds.) I hope to cover this problem in a future blog post.
Now back to our tree shortcutting problem. Let $n = \lvert V(T) \rvert$. There are two extreme regimes that I am aware of:
1. Hop bound $k=2$ and treewidth $\mathrm{tw}(K) = O(\log n)$. This is a relatively simple exercise.
2. Hop bound $k=O(\log n)$ and treewidth $\mathrm{tw}(K) = O(1)$. This regime is harder to prove; trying to show this for a path graph will be an insightful exercise. It follows from a well-known fact [1] that any tree decomposition of width $t$ can be turned into a tree decomposition of width $O(t)$ and depth $O(\log n)$.
The two regimes might suggest that a lower bound $k\cdot \mathrm{tw}(K) = \Omega(\log n)$ for any $k$. In our recent paper [2], we show that this is not the case:
Theorem 1: There exists an emulator $K$ for any $n$-vertex tree $T$ such that $h(K) = O(\log \log n)$ and $\mathrm{tw}(K) = O(\log \log n)$.
Theorem 1 implies that one can get an emulator with $k\cdot \mathrm{tw}(K) = O((\log \log n)^2)$, which is exponentially smaller than $O(\log(n))$. The goal of this post is to discuss the proof of Theorem 1. See the conclusion section for a more thorough discussion on other aspects of Theorem 1, in particular, the construction time and application.
For the tree shortcutting problem, it is often insightful to look into the path graph with $n$ vertices, which is a special case. Once we solve the path graph, extending the ideas to trees is not that difficult.
# 1. The Path Graph
The (unweighted )path graph $P_n$ is a path of $n$ vertices. To simplify the presentation, assume that $\sqrt{n}$ is an integer. The construction is recursive and described in the pseudo-code below. First we divide $P_n$ into $\sqrt{n}$ sub paths of size $\sqrt{n}$ each. Denote the endpoints of these subpaths by $b_i = i\sqrt{n}$ for $1\leq i\leq\sqrt{n}$. We call ${b_i}_{i}$ boundary vertices. We have two types of recursions: (1) top level recursion – lines 2 and 3 – and (2) subpath recursion – lines 5 to 9. The intuition of the top level recursion is a bit tricky and we will get to that later. See Figure 2.
Figure 2: (a) The recursive construction applied to the path $P_n$. (b) Gluing $\mathcal T_B$ and all $\{\mathcal T_i\}$ to obtain a tree decomposition $\mathcal{T}$ of $K$ via red edges.
The subpath recursion is natural: recursively shortcut each subpath $P[b_i,b_{i+1}]$ (line 6). The subpath recursion returns the shortcut graph $K_i$ and its tree decomposition $\mathcal T_i$. Next, we add edges from each boundary vertex $b_i, b_{i+1}$ of the subpath to all other vertices on the subpath (lines 7 and 8). This step guarantees that each vertex of the subpath can “jump” to the boundary vertices using only one edge. In terms of tree decomposition, it means that we add both $b_i$ and $b_{i+1}$ to every bag of $\mathcal T_i$ (line 9).
The top level recursion serves two purposes: (i) creating a low hop emulator for boundary vertices (recall that each vertex can jump to a boundary vertex in the same subpath using one edge) and (ii) gluing ${\mathcal T_i}$ together. More precisely, let $P_{\sqrt{n}}$ be a path of boundary vertices, i.e., $b_i$ is adjacent to $b_{i+1}$ in $P_{\sqrt{n}}$. We shortcut $P_{\sqrt{n}}$ recursively, getting the shortcut graph $K_B$ and its tree decomposition $\mathcal T_B$ (line 3). Since $(b_i,b_{i+1})$ is an edge in $P_{\sqrt{n}}$, there must be a bag in $\mathcal T_B$ containing both $b_i,b_{i+1}$; that is, the bag $X$ in line 11 exists. See Figure 2(b). Recall that in line 9, every bag in $\mathcal T_i$ contains both $b_i,b_{i+1}$, and that $K_B$ and $K_i$ only share two boundary vertices $b_i,b_{i+1}$. Thus, we can connect $X$ to an arbitrary bag of $\mathcal T_i$ as done in line 12. This completes the shortcutting algorithm.
PathShortcutting$(P_n)$
$1.$ $B \leftarrow {0,\sqrt{n}, 2\sqrt{n}, \ldots, n}$ and $b_i \leftarrow i\sqrt{n}$ for every $0\leq i \leq \sqrt{n}$
$2.$ $P_{\sqrt{n}} \leftarrow$ unweighted path graph with vertex set $B$.
$3.$ $(K_B,\mathcal T_B) \leftarrow$PathShortcutting$(P_{\sqrt{n}})$
$4.$ $K\leftarrow K_B,\quad \mathcal{T}\leftarrow \mathcal T_B$
$5.$ for $i\leftarrow 0$ to $\sqrt{n}-1$
$6.$ $(K_i,\mathcal T_i) \leftarrow$PathShortcutting$(P_{n}[b_i, b_{i+1}])$
$7.$ for each $v\in P_{n}[b_i, b_{i+1}]$
$8.$ $E(K_i)\leftarrow {(v,b_i), (v,b_{i+1})}$
$9.$ add both ${v_i,v_{i+1}}$ to every bag of $\mathcal T_i$
$10.$ $K\leftarrow K \cup K_i$
$11.$ Let $X$ be a bag in $\mathcal{T}$ containing both $b_i,b_{i+1}$
$12.$ Add $\mathcal T_i$ to $\mathcal{T}$ by connecting $X$ to an arbitrary bag of $\mathcal T_i$
$13.$ return $(K,\mathcal{T})$
It is not difficult to show that $\mathcal{T}$ indeed is a tree decomposition of $K$. Thus, we focus on analyzing the hop bound and the treewidth.
Remark 2: For notational convenience, we include $0$ in the set $B$ though $0\not\in P_n$. When calling the recursion, one could simply drop 0.
Figure 3: A low-hop path from $u$ to $v$.
Analyzing the hop bound $h(K)$. Let $u$ and $v$ be any two vertices of $P_{n}$; w.l.o.g, assume that $u \leq v$, and $h(n)$ be the hop bound. Let $b_{u}$ and $b_v$ be two boundary vertices of the subpaths containing $u$ and $v$, respectively, such that $b_u,b_v \in P[u,v]$. See Figure 3. As mentioned above, line 8 of the algorithms guarantees that there are two edges $(u,b_u)$ and $(b_v,v)$ in $K$, and the top level recursion (line 3) guarantees that there is a shortest path of hop length $h(\sqrt{n})$ between $b_u$ and $b_v$ in $K_B$. Thus we have:
$h(n) \leq h(\sqrt{n}) + 2$
which solves to $h(n)= O(\log\log n)$.
Analyzing the treewidth $\mathrm{tw}(K)$. Note that the treewidth of $\mathcal T_B$ and all ${\mathcal{T_i}}$ (before adding boundary vertices in line 9) is bounded by $\mathrm{tw}(\sqrt{n})$. Line 9 increases the treewidth of ${\mathcal{T_i}}$ by at most $2$. Since the treewidth of $K$ is the maximum treewidth $\mathcal T_B$ and all ${\mathcal{T_i}}$, we have:
$\mathrm{tw}(n) \leq \mathrm{tw}(\sqrt{n}) + 2$
which solves to $\mathrm{tw}(n)= O(\log\log n)$.
This completes the proof of Theorem 1 for the path graph $P_n$.
# 2. Trees
What is needed to extend the construction of a path graph to a general tree? Two properties we exploited in the construction of $P_n$:
1. $P_n$ can be decomposed into $\sqrt{n}$ (connected) subpaths.
2. Each subpath in the decomposition has at most two boundary vertices.
Property 2 implies that the total number of boundary vertices is about $\sqrt{n}$, which plays a key role in analyzing the top level recursion.
It is well known that we can obtain a somewhat similar but weaker decomposition for trees: one can decompose a tree of $n$ vertices to roughly $\sqrt{n}$ connected subtrees such that the number of boundary vertices is $\sqrt{n}$. (A vertex is a boundary vertex of a subtree if it is incident to an edge not in the subtree.) This decomposition is weaker in the sense that a subtree could have more than 2, and indeed up to $\Omega(\sqrt{n})$, boundary vertices. Is this enough?
Not quite. To glue the tree decomposition $\mathcal T_i$ to $\mathcal{T}$ (and effectively to $\mathcal T_B$), we rely on the fact that there is a bag $X\in \mathcal T_B$ containing both boundary vertices in line 11. The analogy for trees would be: there exists a bag $X$ containing all boundary vertices of each subtree. This is problematic if a subtree has $\Omega(\sqrt{n})$ boundary vertices.
Then how abound guaranteeing that each subtree has $O(1)$ vertices, say 3 vertices. Will this be enough? The answer is pathetically no. To guarantee 3 vertices in the same bag, one has to add a clique of size 3 between the boundary vertices in the top level recursion. What it means is that, the graph between boundary vertices on which we recursively call the shortcuting procedure is no longer a tree. Thus, we really need a decomposition where every subtree has at most 2 boundary vertices.
Lemma 1: Let $T$ be any tree of $n$ vertices, one can decompose $T$ into a collection $\mathcal{D}$ of $O(\sqrt{n})$ subtrees such that every tree $T’\in \mathcal{D}$ has $\lvert V(T’)\rvert \leq \sqrt{n}$ and at most 2 boundary vertices.
Lemma 1 is all we need to prove Theorem 1, following exactly the same construction for the path graph $P_n$; the details are left to readers.
Remark 3: In developing our shortcutting tree result, we were unaware that Lemma 1 was already known in the literature. A reviewer later pointed out that one can get Lemma 1 from a weaker decomposition using least common ancestor closure [3], which I reproduce below.
Proof of Lemma 1: First, decompose $T$ into a collection $\mathcal{D}’$ of $O(\sqrt{n})$ subtrees such that each tree in $\mathcal{D}’$ has size at most $\sqrt{n}$ and that the total number of boundary vertices is $O(\sqrt{n})$. As mentioned above, this decomposition is well known; see Claim 1 in our paper [2] for a proof.
Let $A_1$ be the set of boundary vertices; $\lvert A_1\rvert = O(\sqrt{n})$. Root $T$ at an arbitrary vertex. Let $A_2$ be the set containing the ancestor of every pair of vertices in $A_1$. Let $B = A_1\cup A_2$.
It is not hard to see that $\lvert A_2\rvert \leq \lvert A_1\rvert - 1 = O(\sqrt{n})$. Thus, $\lvert B\rvert = O(\sqrt{n})$. Furthermore, every connected component of $T\setminus B$ has at most $\sqrt{n}$ vertices and has edges to at most 2 vertices in $B$. The set $B$ induces a decomposition $\mathcal{D}$ of $T$ claimed in Lemma 1.
# 3. Conclusion
The emulator in Theorem 1 can be constructed in time $O(n \log \log n)$; see Theorem 10 our paper [2] for more details. The major open problem is:
Open problem: Is $O((\log \log n)^2)$ the best possible bound for the product $\mathrm{tw}(K)\cdot h(K)$?
This open problem is intimately connected to another problem: embedding planar graphs of diameter $D$ into low treewidth graphs with additive distortion at most $+\epsilon D$ for any $\epsilon \in (0,1)$. More precisely, though not explicitly stated [2], one of our main results is:
Theorem 2: If one can construct an emulator $K$ of treewidth $\mathrm{tw}(n)$ and hop bound $h(n)$ for any tree of $n$ vertices, then one can embed any planar graphs with $n$ vertices and diameter $D$ into a graph of treewidth $O(h(n)\cdot \mathrm{tw}(n)/\epsilon)$ and additive distortion $+\epsilon D$ for any given $\epsilon \in (0,1)$.
That is, the product of treewdith and hop bound directly bounds the treewdith of the embedding. Theorem 1 give us an embedding with treewidth $O((\log\log n)^2/\epsilon)$, which has various algorithmic applications [2]. My belief is that the bound $O((\log \log n)^2)$ is the best possible.
# 4. References
[1] Bodlaender, H.L. and Hagerup, T. (1995). Parallel algorithms with optimal speedup for bounded treewidth. In ICALP ‘95, 268-279.
[2] Filtser, A. and Hung, L. (2022). Low Treewidth Embeddings of Planar and Minor-Free Metrics. ArXiv preprint arXiv:2203.15627.
[3] Fomin, F.V., Lokshtanov, D., Saurabh, S. and Zehavi, M. (2019). Kernelization: theory of parameterized preprocessing. Cambridge University Press.
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2023-03-26 05:57:04
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https://math.meta.stackexchange.com/questions/34745/short-answer-turned-into-a-comment-by-moderator
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# Short answer turned into a comment by moderator
I'd like to update my poor knowledge about the policy of this site:
The hints, which are usually very short answers, have become unwanted (as answers, I mean)?
I'm asking this since my my answer was turned into a comment by the moderator Alexander Gruber. As one can see, the answer had one upvote and helped the OP to post his own answer. So why delete it?
• A starting point is the FAQ "Why and how are some answers deleted?". I'd be happy to expand on the issue of not answering the Question if you wish, in the form of an Answer below. Apr 27 at 13:20
• The current state of your question here on meta is okay. The previous version, which you keep rolling back to, is inappropriate. Please do not roll it back again.
– Xander Henderson Mod
May 19 at 22:01
• @XanderHenderson Please elaborate on "inappropriate". Which part you found unacceptable and why? Moreover, can you tell me why was that part a reason for my 15 days suspension? Your collegue into moderation said that he didn't feel offended, but considered it as a personal attack. Since when making public what is already public, that is, someone (lack of) activity on M.SE is a personal attack? May 21 at 8:06
• @XanderHenderson Since I noticed that you are very kind and reply to my questions, please make me clear why I was suspended if I accepted quid's edit which removed my "snide" remark? Just to show me what's going on if I mess with a mod as the suspension message implies? (I know, the suspension reasons are not discussed in public. But wait, I sent to the MODERATION TEAM a message containing, among others, the same.question and yoh didn't bother to reply. Who am.I to.deserve a reply from you!) May 21 at 19:07
This meta post is tagged "specific answer", so I'll frame my remarks accordingly.
The posting of very short "hint" answers has come up a few dozen times here, for example "Is it acceptable to leave hints as answers?" and its linked posts.
Hint. Consider $$\frac{1}{s}\in R_S$$.
The Question Exercise about localization and monic polynomial to which this responds is about proving $$R = R_S$$ when $$R$$ is an integral domain, $$S$$ a multiplicative subset, and every element of the localization $$R_S$$ is integral (satisfies a monic polynomial) over $$R$$. The body of the Question includes de minimis context, namely the application of the hypothesis to a typical element $$r/s \in R_S$$
So your hint might be understood by the user-who-asked that it suffices to consider just the case $$1/s \in R_S$$. This hint evidently was useful as the user subsequently (within the day) posted a self-answer (one that you edited lightly for grammar, etc.).
I also consider the self-answer as evidence that your hint did not answer the Question. In such cases deleting the Answer and reposting it as a Comment is a possible moderator action. I have flagged such posts for moderator attention (in the past, not in this case) with a request that it be changed to a Comment.
Given your long participation on the site, it is possible that this policy is known to you. Perhaps you even intend to test the boundaries of brevity by posting this way.
In some cases the rush to post a "hint" as an Answer doesn't succeed in producing a viable approach. You recently posted such a failure regarding "Isomorphism between two field extensions":
Hint. Notice that $$f(x+1)=g(x)$$.
I've often left a Comment (where a "hint" ought to be expanded to at least a sketch of the solution) to "be wary of posting an Answer that is significantly shorter than the Question" to which it responds. The wariness I recommend to posters of "hint" Answers is intended to benefit you as well as the Community. Our mission is to collect and curate content useful to learners of math at all levels, so even when a hint is useful to the original poster, fuller explanations can have greater value to future Readers.
• Please notice that the polynomials are over $\mathbb F_2$. Apr 27 at 17:31
• @noname: That's a useful insight. The user who asked that second Question said, "I can't tell what is the exact field extension and how to approach such question." If that were enough context (I'm unpersuaded), then Answers need to address it. For what it's worth, I voted to close the second Question (for lack of context) without flagging or downvoting the Answer. Apr 27 at 19:26
Hello,
We're writing in reference to your Mathematics Stack Exchange account:
https://math.stackexchange.com/users/121097/user26857
Moderation on this site is not perfect, we know. We, the moderators, are far from perfect, we also know that. Sometimes we make decisions that you may disagree with, sometimes we are wrong, and sometimes we just can't reveal the whole picture because of user privacy rules.
It is true that when you disagree with the actions of moderators, complaining on meta is the right way to go (another route is to contact SE directly and file a complaint with them). Having said that, your meta post Short answer turned into a comment by moderator is absolutely not the way to complain about moderation.
Getting angry is natural, venting your anger is reasonable, but aiming it, publicly, at the moderators is absolutely unacceptable.
Your account has been temporarily suspended for 15 days. While you’re suspended, your reputation will show as 1 but will be restored once the suspension ends.
Regards, Mathematics Stack Exchange Moderation Team
If the moderators felt offended by my post why don't delete it and eventually send me a warning?
Well, when you have the suspension tool at hand maybe it's easier to do things like this.
Let me say that I also felt offended by the way you proceeded, and by your constant disrespect for high rated users who dare to question your methods.
• I don't care about 10 reputation points. All that matter to me was that some guy turned my answer into a comment and didn't have the minimum respect to explain his action. May 19 at 11:20
• I didn't realize you wanted to hear specifically from me. If it will help, I'll oblige. I hadn't answered because hardmath pretty much hit it on the head. I think that hints are great, but that they make more sense as comments than answers, particularly for short "prompt" style hints which neatly fit into comment length. That's especially true when there are other complete answers on the question. Your post was flagged as not an answer, and I agreed that it was more appropriate as a comment, so I converted it. (I didn't downvote you, by the way.)
– Alexander Gruber Mod
May 20 at 2:19
• Regarding how your suspension happened, it's not because I or any other mod was offended by your post. Personally, I wasn't. (Glad to hear I'm your favorite moderator-- aren't I everyone's?) But that doesn't matter, because SE doesn't operate on a system of restorative justice. We need to enforce a certain standard of civility, and personal attacks violate that. Keeping your post as-is would signal that that type of callout is within the social norms tolerated here, and it's not. It's fine in its current form-- there's nothing wrong with debating policy or (politely) questioning decisions.
– Alexander Gruber Mod
May 20 at 2:41
• @AlexanderGruber It's a personal attack to say that your contribution to M.SE is at a very low level? What was wrong in saying this? You posted only one answer in 2021, and this is public. Is it forbidden to say this cause you are a moderator and can suspend me? Moreover, in the time when we saw more acurately when someone has visited this site for the last time, I remember that I pointed out that you were missing for months. So why this obstination to be a moderator if you are so bussy with other things? Last but not least, please let me decide if I post a hint as a comment or as an answer. May 21 at 8:01
• In context, you clearly intended it out to be an insult, much like how here you're using inflammatory language like "obstination." You've been here long enough to know that hostile language isn't acceptable. I'm afraid you are no more exempt from the rules of civility here than you are from comment conversions.
– Alexander Gruber Mod
May 21 at 14:57
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2022-06-26 20:45:57
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http://eul.ink/complex-analysis/Complex-Step%20Differentiation/
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Introduction
You may already have used numerical differentiation to estimate the derivative of a function, using for example Newton’s finite difference approximation $f'(x) \approx \frac{f(x+h) - f(x)}{h}.$ The implementation of this scheme in Python is straightforward:
def FD(f, x, h):
return (f(x + h) - f(x)) / h
However, the relationship between the value of the step $$h$$ and the accuracy of the numerical derivative is more complex. Consider the following sample data:
Expression Value
$$\exp'(0)$$ $$1$$
FD(exp, 0, 1e-4) 1.000050001667141
FD(exp, 0, 1e-8) 0.99999999392252903
FD(exp, 0, 1e-12) 1.000088900582341
The most accurate value of the numerical derivative is obtained for $$h=10^{-8}$$ and only 8 digits of the result are significant. For the larger value of $$h=10^{-4},$$ the accuracy is limited by the quality of the Taylor development of $$\exp$$ at the first order; this truncation error decreases linearly with the step size. For the smaller value of $$h=10^{-12},$$ the accuracy is essentially undermined by round-off errors in computations.
In this document, we show that complex-step differentiation may be used to get rid of the influence of the round-off error for the computation of the first derivative. For higher-order derivatives, we introduce a spectral method, a fast algorithm with an error that decreases exponentially with the number of function evaluations.
Computer Arithmetic
You may skip this section if you are already familiar with the representation of real numbers as “doubles” on computers and with their basic properties. At the opposite, if you wish to have more details on this subject, it is probably a good idea to have a look at the classic “What every computer scientist should know about computer arithmetic” (Goldberg 1991).
In the sequel, the examples are provided as snippets of Python code that often use the Numerical Python (NumPy) library; first of all, let’s make sure that all NumPy symbols are available:
>>> from numpy import *
Floating-Point Numbers: First Contact
The most obvious way to display a number is to print it:
>>> print pi
3.14159265359
This is a lie of course: print is not supposed to display an accurate information about its argument, but something readable. To get something unambiguous instead, we can do:
>>> pi
3.141592653589793
When we say “unambiguous”, we mean that there is enough information in this sequence of digits to compute the original floating-point number; and indeed:
>>> pi == eval("3.141592653589793")
True
Actually, this representation is also a lie: it is not an exact decimal representation of the number pi stored in the computer memory. To get an exact representation of pi, we can request the display of a large number of the decimal digits:
>>> def all_digits(number):
... print "{0:.100g}".format(number)
>>> all_digits(pi)
3.141592653589793115997963468544185161590576171875
Asking for 100 digits was actually good enough: only 49 of them are displayed anyway, as the extra digits are all zeros.
Note that we obtained an exact representation of the floating-point number pi with 49 digits. That does not mean that all – or even most – of these digits are significant in the representation the real number of $$\pi.$$ Indeed, if we use the Python library for multiprecision floating-point arithmetic mpmath, we see that
>>> import mpmath
>>> mpmath.mp.dps = 49; mpmath.mp.pretty = True
>>> +mpmath.pi
3.141592653589793238462643383279502884197169399375
and both representations are identical only up to the 16th digit.
Binary Floating-Point Numbers
Representation of floating-point numbers appears to be complex so far, but it’s only because we insist on using a decimal representation when these numbers are actually stored as binary numbers. In other words, instead of using a sequence of (decimal) digits $$f_i \in \{0,1,\dots,9\}$$ to represent a real number $$x$$ as $x = \pm (f_0.f_1f_2 \dots f_i \dots) \times 10^{e}$ we should use binary digits – aka bits$$f_i \in \{0,1\}$$ to write: $x = \pm (f_0.f_1f_2 \dots f_i \dots) \times 2^{e}.$ These representations are normalized if the leading digit of the significand $$(f_0.f_1f_2 \dots f_i \dots)$$ is non-zero; for example, with this convention, the rational number $$999/1000$$ would be represented in base 10 as $$9.99 \times 10^{-1}$$ and not as $$0.999 \times 10^{0}.$$ In base 2, the only non-zero digit is 1, hence the significand of a normalized representation is always $$(1.f_1f_2\dots f_i \dots).$$
In scientific computing, real numbers are usually approximated to fit into a 64-bit layout named “double”1. In Python standard library, doubles are available as instances of float – or alternatively as float64 in NumPy.
A triple of
• sign bit $$s \in \{0,1\},$$
• biased exponent $$e\in\{1,\dots, 2046\}$$ (11-bit),
• fraction $$f=(f_1,\dots,f_{52}) \in \{0,1\}^{52}.$$
represents a normalized double $x = (-1)^s \times 2^{e-1023} \times (1.f_1f_2 \dots f_{52}).$
The doubles that are not normalized are not-a-number (nan), infinity (inf) and zero (0.0) (actually signed infinities and zeros), and denormalized numbers. In the sequel, we will never consider such numbers.
Accuracy
Almost all real numbers cannot be represented exactly as doubles. It makes sense to associate to a real number $$x$$ the nearest double $$[x].$$ A “round-to-nearest” method that does this is fully specified in the IEE754 standard (see IEEE Task P754 1985), together with alternate (“directed rounding”) methods.
To have any kind of confidence in our computations with doubles, we need to be able to estimate the error in the representation of $$x$$ by $$[x].$$ The machine epsilon, denoted $$\epsilon$$ in the sequel, is a key number in this respect. It is defined as the gap between $$1.0$$ – that can be represented exactly as a double – and the next double in the direction $$+\infty.$$
>>> after_one = nextafter(1.0, +inf)
>>> after_one
1.0000000000000002
>>> all_digits(after_one)
1.0000000000000002220446049250313080847263336181640625
>>> eps = after_one - 1.0
>>> all_digits(eps)
2.220446049250313080847263336181640625e-16
This number is also available as an attribute of the finfo class of NumPy that gathers machine limits for floating-point data types:
>>> all_digits(finfo(float).eps)
2.220446049250313080847263336181640625e-16
Alternatively, the examination of the structure of normalized doubles yields directly the value of $$\epsilon$$: the fraction of the number after $$1.0$$ is $$(f_1, f_2, \dots, f_{51}, f_{52}) = (0,0,\dots,0,1),$$ hence $$\epsilon =2^{-52},$$ a result confirmed by:
>>> all_digits(2**-52)
2.220446049250313080847263336181640625e-16
The machine epsilon matters so much because it provides a simple bound on the relative error of the representation of a real number as a double. Indeed, for any sensible rounding method, the structure of normalized doubles yields $\frac{|[x] - x|}{|x|} \leq \epsilon.$ If the “round-to-nearest” method is used, you can actually derive a tighter bound: the inequality above still holds with $$\epsilon / 2$$ instead of $$\epsilon.$$
Significant Digits
This relative error translates directly into how many significant decimal digits there are in the best approximation of a real number by a double. Consider the exact representation of $$[x]$$ in the scientific notation: $[x] = \pm (f_0.f_1 \dots f_{p-1} \dots) \times 10^{e}.$ We say that it is significant up to the $$p$$-th digit if $|x - [x]| \leq \frac{10^{e-(p-1)}}{2}.$ On the other hand, the error bound on $$[x]$$ yields $|x - [x]| \leq \frac{\epsilon}{2} |x| \leq \frac{\epsilon}{2} \times 10^{e+1}.$ Hence, the desired precision is achieved as long as $p \leq - \log_{10} \epsilon/2 = 52 \log_{10} 2 \approx 15.7.$ Consequently, doubles provide a 15-th digit approximation of real numbers.
Functions
Most real numbers cannot be represented exactly as doubles; accordingly, most real functions of real variables cannot be represented exactly as functions operating on doubles either. The best we can hope for are correctly rounded approximations. An approximation $$[f]$$ of a function $$f$$ of $$n$$ variables is correctly rounded if for any $$n$$-uple $$(x_1,\dots,x_n),$$ we have $[f](x_1,\dots,x_n) = [f([x_1], \dots, [x_n])].$ The IEEE 754 standard (see IEEE Task P754 1985) mandates that some functions have a correctly rounded implementation; they are:
add, substract, multiply, divide, remainder and square root.
Other standard elementary functions – such as sine, cosine, exponential, logarithm, etc. – are usually not correctly rounded; the design of computation algorithms that have a decent performance and are provably correctly rounded is a complex problem (see for example the documentation of the Correctly Rounded mathematical library).
Complex Step Differentiation
Forward Difference
Let $$f$$ be a real-valued function defined in some open interval. In many concrete use cases, we can make the assumption that the function is actually analytic and never have to worry about the existence of derivatives. As a bonus, for any real number $$x$$ in the domain of the function, the (truncated) Taylor expansion $f(x+h) = f(x) + f'(x) h + \frac{f''(x)}{2} h^2 + \dots +\frac{f^{(n)}}{n!} h^n + \mathcal{O}(h^{n+1})$ is locally valid2. A straighforward computation shows that $f'(x) = \frac{f(x+h) - f(x)}{h} + \mathcal{O}(h)$ The asymptotic behavior of this forward difference scheme – controlled by the term $$\mathcal{O}(h^1)$$ – is said to be of order 1. An implementation of this scheme is defined for doubles $$x$$ and $$h$$ as $\mathrm{FD}(f, x, h) = \left[\frac{[[f] ( [x] + [h]) - [f] (x)]}{[h]} \right].$ or equivalently, in Python as:
def FD(f, x, h):
return (f(x + h) - f(x)) / h
Round-Off Error
We consider again the function $$f(x) = \exp(x)$$ used in the introduction and compute the numerical derivative based on the forward difference at $$x=0$$ for several values of $$h.$$ The graph of $$h \mapsto \mathrm{FD}(\exp, 0, h)$$ shows that for values of $$h$$ near or below the machine epsilon $$\epsilon,$$ the difference between the numerical derivative and the exact value of the derivative is not explained by the classic asymptotic analysis.
If we take into account the representation of real numbers as doubles however, we can explain and quantify the phenomenon. To focus only on the effect of the round-off errors, we’d like to get rid of the truncation error. To achieve this, in the following computations, instead of $$\exp,$$ we use $$\exp_0,$$ the Taylor expansion of $$\exp$$ of order $$1$$ at $$x=0;$$ we have $$\exp_0 (x) = 1 + x.$$
Assume that the rounding scheme is “round-to-nearest”; select a floating-point number $$h>0$$ and compare it to the machine epsilon:
• If $$h \ll \epsilon,$$ then $$1 + h$$ is close to $$1,$$ actually, closer to $$1$$ than from the next binary floating-point value, which is $$1 + \epsilon.$$ Hence, the value is rounded to $$[\exp_0](h) = 1,$$ and a catastrophic cancellation happens: $\mathrm{FD}(\exp_0, 0, h) = \left[\frac{\left[ [\exp_0](h) - 1 \right]}{h}\right] = 0.$
• If $$h \approx \epsilon,$$ then $$1+h$$ is closer from $$1+\epsilon$$ than it is from $$1,$$ hence we have $$[\exp_0](h) = 1+\epsilon$$ and $\mathrm{FD}(\exp_0, 0, h) = \left[\frac{\left[ [\exp_0](h) - 1 \right]}{h}\right] = \left[ \frac{\epsilon}{h} \right].$
• If $$\epsilon \ll h \ll 1,$$ then $$[1+h] = 1+ h \pm \epsilon(1+h)$$ (the symbol $$\pm$$ is used here to define a confidence interval3). Hence $[[\exp_0](h) - 1] = h \pm \epsilon \pm \epsilon(2h + \epsilon + \epsilon h)$ and $\left[ \frac{[[\exp_0](h) - 1]}{h} \right] = 1 \pm \frac{\epsilon}{h} + \frac{\epsilon}{h}(3h + 2\epsilon + 3h \epsilon +\epsilon^2 + \epsilon^2 h)$ therefore $\mathrm{FD}(\exp_0, 0, h) = \exp_0'(0) \pm \frac{\epsilon}{h} \pm \epsilon', \; \epsilon' \ll \frac{\epsilon}{h}.$
Going back to $$\mathrm{FD}(\exp, 0, h)$$ and using a log-log scale to display the total error, we can clearly distinguish the region where the error is dominated by the round-off error – the curve envelope is $$\log(\epsilon/h)$$ – and where it is dominated by the truncation error – a slope of $$1$$ being characteristic of schemes of order 1.
Higher-Order Scheme
The theoretical asymptotic behavior of the forward difference scheme can be improved, for example if instead of the forward difference quotient we use a central difference quotient. Consider the Taylor expansion at the order 2 of $$f(x+h)$$ and $$f(x-h)$$: $f(x+h) = f(x) + f'(x) (+h)+ \frac{f''(x)}{2} (+h)^2 + \mathcal{O}\left(h^3\right)$ and $f(x-h) = f(x) + f'(x) (-h) + \frac{f''(x)}{2} (-h)^2 + \mathcal{O}\left(h^3\right).$ We have $f'(x) = \frac{f(x+h) - f(x-h)}{2h} + \mathcal{O}(h^2),$ hence, the central difference scheme is a scheme of order 2, with the
implementation: $\mathrm{CD}(f, x, h) = \left[\frac{[[f] ( [x] + [h]) - [f] ([x]-[h])]}{[2 \times [h]]} \right].$ or equivalently, in Python:
def CD(f, x, h):
return 0.5 * (f(x + h) - f(x - h)) / h
The error graph for the central difference scheme confirms that a truncation error of order two may be used to improve the accuracy. However, it also shows that a higher-order actually increases the region dominated by the round-off error, making the problem of selection of a correct step size $$h$$ even more difficult.
Complex Step Differentiation
If the function $$f$$ is analytic at $$x,$$ the Taylor expansion is also valid for (small values of) complex numbers $$h.$$ In particular, if we replace $$h$$ by a pure imaginary number $$ih,$$ we end up with
$f(x + ih) = f(x) + f'(x) i h + \frac{f''(x)}{2} (ih)^2 + \mathcal{O}(h^3)$ If $$f$$ is real-valued, using the imaginary part yields: $\mathrm{Im} \,\left( \frac{f(x + ih)}{h} \right) = f'(x) + \mathcal{O}(h^2).$ This is a method of order 2. The straightforward implementation of the complex-step differentiation is $\mathrm{CSD}(f, x, h) = \left[ \frac{ \mathrm{Im}( [f]([x] + i [h])) }{[h]} \right].$ or equivalently, in Python:
def CSD(f, x, h):
return imag(f(x + 1j * h)) / h
The distinguishing feature of this scheme: it almost totally gets rid of the truncation error. Indeed, let’s consider again $$\exp_0;$$ when $$x$$ and $$y$$ are floating-point real numbers, the sum $$x + i y$$ can be computed with any round-off, hence, if $$h$$ is a floating-point number, $$[\exp_0](ih) = [1+ih]= 1 + ih$$ and consequently, $$\mathrm{Im}( [\exp_0](ih)) = h,$$ which yields $\mathrm{CSD}(\exp_0, 0, h) = \left[\frac{h}{h}\right]= 1 = \exp_0'(0).$
Spectral Method
The complex step differentiation is a powerful method but it also has limits. We can use it to compute the first derivative of a real analytic function $$f,$$ but not its second derivative because our estimate $$x\mapsto \mathrm{CSD}(f,x,h)$$ of the first derivative is only available for real values of $$x,$$ hence the method cannot be iterated. We cannot use it either if we know that $$f$$ is analytic but not real-valued.
We introduce in this section an alternate method to compute first, second and higher-order derivatives of – real or complex-valued – analytic functions. More details may be found in (Fornberg 2006) and (Trefethen 2000).
Computation Method
Let $$f$$ be a function that is holomorphic in an open neighbourhood of $$x \in \mathbb{R}$$ that contains the closed disk with center $$x$$ and radius $$r.$$ In this disk, the values of $$f$$ can be computed by the Taylor series $f(z) = \sum_{n=0}^{+\infty} a_n (z-x)^n, \; \; a_n = \frac{f^{(n)}(x)}{n!}.$ The open disk of convergence of the series has a radius that is larger than $$r$$, thus the growth bound of the sequence $$a_n$$ is smaller than $$1/r$$. Hence, $\exists \, \kappa >0, \, \forall \, n \in \mathbb{N}, \; |a_n| \leq \kappa \, r^{-n}.$
Let $$h \in (0,r)$$ and $$N$$ be a positive integer; let $$f_k$$ be the sequence of $$N$$ values of $$f$$ on the circle with center $$x$$ and radius $$h$$ defined by $f_k = f(x + h w^{k}), \; w = e^{-i2\pi /N}, \; k =0, \dots ,N-1.$
Estimate and Accuracy
The values $$f_k$$ can be computed as $f_k = \sum_{n=0}^{+\infty} a_n (hw^k)^n = \sum_{n=0}^{N-1} \left[\sum_{m=0}^{+\infty} a_{n+mN} h^{n+mN} \right] w^{k(n+mN)}.$ Notice that we have $$w^{k(n+mN)} = w^{kn} (w^N)^{km} = w^{kn}.$$ Hence, if we define $c_n = a_n h^n + a_{n+N} h^{n+N} + \dots = \sum_{m=0}^{+\infty} a_{n+mN} h^{n+mN},$ we end up with the following relationship between the values $$f_k$$ and $$c_n$$: $f_k = \sum_{n=0}^{N-1} w^{kn} c_n.$ It is useful because the coefficients $$c_n/h^n$$ provide an approximation of $$a_n$$: $\left|a_n - \frac{c_n}{h^n} \right| \leq \kappa r^{-n} \sum_{m=1}^{+\infty} (h/r)^{mN} = \kappa r^{-n} \frac{(h/r)^N}{1 - (h/r)^N}$ There are two ways to look at this approximation: if we freeze $$N$$ and consider the behavior of the error when the radius $$h$$ approaches $$0,$$ we derive $a_n = \frac{c_n}{h^n} + \mathcal{O}(h^N)$ and conclude that the approximation of $$a_n$$ is of order $$N$$ with respect to $$h;$$ on the other hand, if we freeze $$h$$ and let $$N$$ grow to $$+\infty,$$ we obtain instead $a_n = \frac{c_n}{h^n} + \mathcal{O}(e^{-\alpha N}) \; \mbox{ with } \; \alpha = -\log(h/r) > 0,$ in other words, the approximation is exponential with respect to $$N.$$
Computation of the Estimate
The right-hand side of the equation $f_k = \sum_{n=0}^{N-1} w^{kn}c_n.$ can be interpreted as a classic matrix-vector product; the mapping from the $$c_n$$ to the $$f_k$$ is known has the discrete Fourier transform (DFT). The inverse mapping – the inverse discrete Fourier transform – is given by $c_n = \frac{1}{N}\sum_{n=0}^{N-1}w^{-kn} f_k.$ Both the discrete Fourier transform and it inverse can be computed by algorithms having a $$\mathcal{O}(N\log N)$$ complexity (instead of the $$\mathcal{O}(N^2)$$ of the obvious method), the aptly named fast Fourier transform (FFT) and inverse fast Fourier transform (IFFT). Some of these algorithms actually deliver the minimal complexity only when $$N$$ is a power of two, so it is safer to pick only such numbers if you don’t know exactly what algorithm you are actually using.
The implementation of this scheme is simple:
from numpy.fft import ifft
from scipy.misc import factorial
def SM(f, x, h, N):
w = exp(-1j * 2 * pi / N)
k = n = arange(N)
f_k = f(x + h * w**k)
c_n = ifft(f_k)
a_n = c_n / h ** n
return a_n * factorial(n)
Error Analysis
The algorithm introduced in the previous section provides approximation methods with an arbitrary large order for $$n$$-th order derivatives. However, the region in which the round-off error dominates the truncation error is large and actually increases when the integer $$n$$ grows. A specific analysis has to be made to control both kind of errors.
We conduct the detailled error analysis for the function $f(z) = \frac{1}{1-z}$ at $$x=0$$ and attempt to estimate the derivatives up to the fourth order. We have selected this example because $$a_n=1$$ for every $$n,$$ hence the computation of the relative errors of the results are simple.
Round-off error
We assume that the main source of round-off errors is in the computations of the $$f_k.$$ The distance between $$f_k$$ and its approximation $$[f_k]$$ is bounded by $$|f_k|\times \epsilon/2;$$ the coefficients in the IFFT matrix are of modulus $$1/N,$$ hence, if the sum is exact, we end up with an absolute error on $$c_n$$ bounded by $$M(h) \times {\epsilon}/{2}$$ with $M(h)=\max_{|z-x|=h} |f(z)|.$
Hence the absolute error on $$a_n = c_n / h^n$$ is bounded by $$M(h) \epsilon / (2 h^n).$$ Using the rough estimate $$|a_n| \simeq \kappa r^{-n},$$ we end up with a relative error for $$a_n$$ controlled by $\left(\frac{M(h)}{\kappa}\frac{\epsilon}{2}\right) \left( \frac{h}{r} \right)^{-n}$ On our example, we can pick $$M(h) = 1/(1-h),$$ $$\kappa=1,$$ and $$r=1,$$ hence the best error bound we can hope for is obtained for the value of $$h$$ that minimizes $${1}/{((1-h)h^n)} {\epsilon/}{2};$$ the best $$h$$ and round-off error bound are actually $h=\frac{n}{n+1} \; \mbox{ and } \; \mbox{round-off}(a_n) \leq \frac{(n+1)^{n+1}}{n^n} \frac{\epsilon}{2}.$
The error bound is always bigger than the structural relative error $$\epsilon/2$$ and increases with $$n,$$ hence the worst case is obtained for the highest derivative order that we want to compute, that is $$n=4.$$ If for example we settle on a round-off relative error of $$1000$$ times $$\epsilon/2,$$ we can select $$h=0.2.$$
Truncation error
We have already estimated the difference between $$a_n$$ and $$c_n / h^n;$$ if we again model $$|a_n|$$ as $$\kappa r^{-n},$$ the relative error of this estimate is bounded by $\frac{(h/r)^N}{1 - (h/r)^N} \simeq \left( \frac{h}{r}\right)^N,$ hence to obtain a truncation error of the same magnitude than the truncation error – that is $$1000 \times \epsilon /2,$$ we may select $$N$$ such that $$0.2^N \leq 1000 \times {\epsilon}/{2},$$ that is
$N \geq \left \lceil \frac{\log (1000 \times {\epsilon}/{2})}{\log 0.2} \right \rceil = 19.$
We pick for $$N$$ the next power of two after $$19;$$ the choices $$h=0.2$$ and $$N=32$$ yield the following estimates of the first 8 $$n$$-th order derivatives of $$f.$$
$$n$$ $$f^{(n)}(0)$$ estimate relative error
$$0$$ $$1.0000000000000000$$ $$0.0$$
$$1$$ $$0.9999999999999998$$ $$2.2 \times 10^{-16}$$
$$2$$ $$1.9999999999999984$$ $$7.8 \times 10^{-16}$$
$$3$$ $$6.0000000000000284$$ $$4.7 \times 10^{-15}$$
$$4$$ $$23.999999999999996$$ $$1.1 \times 10^{-16}$$
$$5$$ $$120.00000000001297$$ $$1.1 \times 10^{-13}$$
$$6$$ $$720.00000000016007$$ $$2.2 \times 10^{-13}$$
$$7$$ $$5040.0000000075588$$ $$1.5 \times 10^{-12}$$
Appendix
Augustin-Louis is the proud author of a very simple Python CSD code fragment that people cut-and-paste in their numerical code:
from numpy import imag
def CSD(f, x, h=1e-100):
return imag(f(x + 1j * h)) / h
One day, he receives the following mail:
Dear Augustin-Louis,
We are afraid that your Python CSD code is defective; we used it to compute the derivative of $$f(x) = \sqrt{|x|}$$ at $$x=1$$ and got $$0.$$ We’re pretty sure that it should be $$0.5$$ instead.
Yours truly,
Isaac & Gottfried Wilhelm.
1. Should the complex-step differentiation method work in this case?
2. How do you think that Isaac and Gottfried Wilhelm have implemented the function $$f$$? Would that explain the value of CSD(f, x=1.0) that they report?
3. Can you modify the CSD code fragment to “make it work” with the kind of function and implementation that Isaac and Gottried Wilhelm are using? Of course, you cannot change their code, only yours.
Bibliography
1. Numerical Differentiation of Analytic Functions
Bengt Fornberg, 2006.
JSON: / URL:
2. What Every Computer Scientist Should Know about Floating-Point Arithmetic
David Goldberg, 1991.
JSON: / URL:
3. ANSI/IEEE 754-1985, Standard for Binary Floating-Point Arithmetic
JSON:
4. Spectral Methods in Matlab
Lloyd N. Trefethen, 2000.
JSON: / URL:
Notes
1. “Double”
is a shortcut for “double-precision floating-point format”, defined in the IEEE 754 standard, see (IEEE Task P754 1985). A single-precision format is also defined, that uses only 32 bits. NumPy provides it under the name float32.
2. Bachmann-Landau notation.
For a real or complex variable $$h,$$ we write $$\psi(h) = \mathcal{O}(\phi(h))$$ if there is a suitable deleted neighbourhood of $$h=0$$ where the functions $$\psi$$ and $$\phi$$ are defined and the inequality $$|\psi(h)| \leq \kappa |\phi(h)|$$ holds for some $$\kappa > 0.$$ When $$N$$ is a natural number, we write $$\psi(N) = \mathcal{O}(\phi(N))$$ if there is a $$n$$ such that $$\psi$$ and $$\phi$$ are defined for $$N\geq n$$ and for any such $$N,$$ the inequality $$|\psi(N)| \leq \kappa |\phi(N)|$$ holds for some $$\kappa > 0.$$
3. Plus-minus sign and confidence interval. The equation $$a = b \pm c$$ should be interpreted as the inequality $$|a - b| \leq |c|.$$
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2020-02-25 00:04:40
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https://stats.stackexchange.com/questions/424765/multi-test-correction-and-hypothesis-testing
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# Multi test correction and hypothesis testing
I have 9 datasets with one predictor and one target attribute. For each of the dataset, I am testing for a single hypothesis - whether the attributes are associated. I have got the following based on the test-statistic:
• Uncorrected p-values: 8 out of 9 p-values are significant ($$p\le\alpha$$)
• Bonferroni correction (FWER): 3 out of 9 p-values are significant ($$p\le\alpha_{corrected}$$)
• Benjamini–Hochberg correction (FDR): 6 out of 9 p-values are significant ($$p\le\alpha_{B\&H}$$)
I could combine 9 datasets but I am testing for each dataset separately because the context of the data in each dataset is important.
Question: Based on these findings, should I accept or reject the null hypothesis (the 2 attributes are not correlated?) and what could be the formal reasoning behind that?
The model is expected to produce few FP/FN but we are not sure to which extent. So we can allow a few errors from the model.
• Well, the two attributes are significantly different in 6 out of 9 data sets, is what I would say. – user2974951 Sep 4 '19 at 6:27
If you are interested in an association between the same predictor & target in all the datasets, you are using your dataset inefficiently by doing independent tests in each of them. Instead, consider using a (a.k.a. hierarchical model) with dataset as a random effect (random intercept or random intercept + slope).
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2021-03-08 00:36:17
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http://dictionnaire.sensagent.leparisien.fr/SEICHE/en-en/
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Publicité ▼
# définition - SEICHE
seiche (n.)
1.a wave on the surface of a lake or landlocked bay; caused by atmospheric or seismic disturbances
## définition (complément)
voir la définition de Wikipedia
Publicité ▼
physics[Domaine]
Motion[Domaine]
seiche (n.)
Publicité ▼
Wikipedia
# Seiche
"Seiche" is also a French term for a cuttlefish or Bobtail squid.
A seiche ( SAYSH) is a standing wave in an enclosed or partially enclosed body of water. Seiches and seiche-related phenomena have been observed on lakes, reservoirs, swimming pools, bays, harbors and seas. The key requirement for formation of a seiche is that the body of water be at least partially bounded, allowing the formation of the standing wave.
The term was promoted by the Swiss hydrologist François-Alphonse Forel in 1890, who was the first to make scientific observations of the effect in Lake Geneva, Switzerland.[1] The word originates in a Swiss French dialect word that means "to sway back and forth", which had apparently long been used in the region to describe oscillations in alpine lakes.
## Causes and nature of seiches
A standing wave (red) depicted as a sum of two propagating waves traveling in opposite directions (green and blue).
Seiches are often imperceptible to the naked eye, and observers in boats on the surface may not notice that a seiche is occurring due to the extremely long wavelengths. The effect is caused by resonances in a body of water that has been disturbed by one or more of a number of factors, most often meteorological effects (wind and atmospheric pressure variations), seismic activity or by tsunamis.[2] Gravity always seeks to restore the horizontal surface of a body of liquid water, as this represents the configuration in which the water is in hydrostatic equilibrium. Vertical harmonic motion results, producing an impulse that travels the length of the basin at a velocity that depends on the depth of the water. The impulse is reflected back from the end of the basin, generating interference. Repeated reflections produce standing waves with one or more nodes, or points, that experience no vertical motion. The frequency of the oscillation is determined by the size of the basin, its depth and contours, and the water temperature.
The longest natural period[3] for a seiche in an enclosed rectangular body of water is usually represented by the Merian formula:
$\mbox{Period}(T) = \frac{2L}{\sqrt{g h}}$
where L is the length, h the average depth of the body of water, and g the acceleration of gravity.[4]
Higher order harmonics are also observed. The period of the second harmonic will be half the natural period, the period of the third harmonic will be a third of the natural period, and so forth.
## Seiches around the world
Seiches have been observed on both lakes and seas. The key requirement is that the body of water be partially constrained to allow formation of standing waves. Regularity of geometry is not required, even harbors with exceedingly irregular shapes are routinely observed to oscillate with very stable frequencies.
### Lake seiches
Small rhythmic seiches are almost always present on larger lakes. On the North American Great Lakes, seiche is often called slosh. It is always present, but is usually unnoticeable, except during periods of unusual calm. Harbours, bays, and estuaries are often prone to small seiches with amplitudes of a few centimeters and periods of a few minutes. Seiches can also form in semi-enclosed seas; the North Sea often experiences a lengthwise seiche with a period of about 36 hours.
Differences in water level caused by a seiche on Lake Erie, recorded between Buffalo, New York (red) and Toledo, Ohio (blue) on November 14, 2003
The National Weather Service issues low water advisories for portions of the Great Lakes when seiches of 2 feet or greater are likely to occur.[5] Lake Erie is particularly prone to wind-caused seiches because of its shallowness and elongation. These can lead to extreme seiches of up to 5 m (16 feet) between the ends of the lake. The effect is similar to a storm surge like that caused by hurricanes along ocean coasts, but the seiche effect can cause oscillation back and forth across the lake for some time. In 1954, Hurricane Hazel piled up water along the northwestern Lake Ontario shoreline near Toronto, causing extensive flooding, and established a seiche that subsequently caused flooding along the south shore.
Lake seiches can occur very quickly: on July 13, 1995, a big seiche on Lake Superior caused the water level to fall and then rise again by three feet (one meter) within fifteen minutes, leaving some boats hanging from the docks on their mooring lines when the water retreated.[6] The same storm system that caused the 1995 seiche on Lake Superior produced a similar effect in Lake Huron, in which the water level at Port Huron changed by six feet (1.8 m) over two hours.[7] On Lake Michigan, eight fishermen were swept away and drowned when a 10-foot seiche hit the Chicago waterfront on June 26, 1954.[8]
Lakes in seismically active areas, such as Lake Tahoe in California/Nevada, are significantly at risk from seiches. Geological evidence indicates that the shores of Lake Tahoe may have been hit by seiches and tsunamis as much as 10 m (33 feet) high in prehistoric times, and local researchers have called for the risk to be factored into emergency plans for the region.[9]
Earthquake-generated seiches can be observed thousands of miles away from the epicentre of a quake. Swimming pools are especially prone to seiches caused by earthquakes, as the ground tremors often match the resonant frequencies of small bodies of water. The 1994 Northridge earthquake in California caused swimming pools to overflow across southern California. The massive Good Friday Earthquake that hit Alaska in 1964 caused seiches in swimming pools as far away as Puerto Rico. The earthquake that hit Lisbon, Portugal in 1755 caused seiches 2,000 miles (3,000 km) away in Loch Lomond, Loch Long, Loch Katrine and Loch Ness in Scotland [1] and in canals in Sweden. The 2004 Indian Ocean earthquake caused seiches in standing water bodies in many Indian states as well as in Bangladesh, Nepal and northern Thailand.[10] Seiches were again observed in Uttar Pradesh, Tamil Nadu and West Bengal in India as well as in many locations in Bangladesh during the 2005 Kashmir earthquake.[11] The 1950 Chayu-Upper Assam earthquake is known to have generated seiches as far as Norway and southern England. Other earthquakes in the Indian sub-continent known to have generated seiches include the 1803 Kumaon-Barahat, 1819 Allah Bund, 1842 Central Bengal, 1905 Kangra, 1930 Dhubri, 1934 Nepal-Bihar, 2001 Bhuj, 2005 Nias, 2005 Teresa Island earthquakes. The February 27, 2010 Chile earthquake produced a seiche on Lake Pontchartrain, Louisiana with a height of around 0.5 feet. The 2010 Sierra El Mayor earthquake produced large seiches that quickly became an internet phenomenon.[12] Seiches up to at least 1.8 m (6 feet) were observed in Sognefjorden, Norway during the 2011 Tōhoku earthquake.[13]
### Sea and bay seiches
Seiches have been observed in seas such as the Adriatic Sea and the Baltic Sea, resulting in flooding of Venice and St. Petersburg respectively. The latter is constructed on drained marshlands at the mouth of the Neva river. Seiche-induced flooding is common along the Neva river in the autumn. The seiche is driven by a low pressure region in the North Atlantic moving onshore, giving rise to cyclonic lows on the Baltic Sea. The low pressure of the cyclone draws greater-than-normal quantities of water into the virtually land-locked Baltic. As the cyclone continues inland, long, low-frequency seiche waves with wavelengths up to several hundred kilometers are established in the Baltic. When the waves reach the narrow and shallow Neva Bay, they become much higher — ultimately flooding the Neva embankments.[14] Similar phenomena are observed at Venice, resulting in the MOSE Project, a system of 79 mobile barriers designed to protect the three entrances to the Venetian Lagoon.
Nagasaki Bay is a typical area in Japan where seiches have been observed from time to time, most often in the spring — especially in March. On 31 March 1979, the Nagasaki tide station recorded a maximum water-level displacement of 2.78 metres (9.1 ft), at that location and due to the seiche. The maximum water-level displacement in the whole bay during this seiche event is assumed to have reached 4.70 metres (15.4 ft), at the bottom of the bay. Seiches in Western Kyushu — including Nagasaki Bay — are often induced by a low in the atmospheric pressure passing South of Kyushu island.[15] Seiches in Nagasaki Bay have a period of about 30 to 40 minutes. Locally, seiche is called Abiki. The word of Abiki is considered to have been derived from Amibiki, which literally means: the dragging-away (biki) of a fishing net (ami). Seiches not only cause damage to the local fishery but also may result in flooding of the coast around the bay, as well as in the destruction of port facilities.
Seiches can also be induced by tsunami, a wave train (series of waves) generated in a body of water by a pulsating or abrupt disturbance that vertically displaces the water column. On occasion, tsunamis can produce seiches as a result of local geographic peculiarities. For instance, the tsunami that hit Hawaii in 1946 had a fifteen-minute interval between wave fronts. The natural resonant period of Hilo Bay is about thirty minutes. That meant that every second wave was in phase with the motion of Hilo Bay, creating a seiche in the bay. As a result, Hilo suffered worse damage than any other place in Hawaii, with the tsunami/seiche reaching a height of 26 feet along the Hilo Bayfront, killing 96 people in the city alone. Seiche waves may continue for several days after a tsunami.
### Underwater (internal) waves
Although the bulk of the technical literature addresses surface seiches, which are readily observed, seiches are also observed beneath the lake surface acting along the thermocline[16] in constrained bodies of water.
## Engineering for seiche protection
Engineers consider seiche phenomena in the design of flood protection works (e.g., Saint Petersburg Dam), reservoirs and dams (e.g., Grand Coulee Dam), potable water storage basins, harbours and even spent nuclear fuel storage basins.
## Notes
1. ^ Darwin, G. H. (1898). The Tides and Kindred Phenomena in the Solar System. London: John Murray. . See pp. 21–31.
2. ^ Tsunamis are normally associated with earthquakes, but landslides, volcanic eruptions and meteorite impacts all have the potential to generate a tsunami.
3. ^ The longest natural period is the period associated with the fundamental resonance for the body of water—corresponding to the longest standing wave.
4. ^ As an example, the period for a seiche wave in a body of water 10 meters deep and 5 kilometers long would be 1000 seconds or about 17 minutes, while a body about 300 km long (such as the Gulf of Finland) and somewhat deeper has a period closer to 12 hours.
5. ^ National Weather Service. National Weather Service Instruction 10-301. Retrieved on 2008-01-31.
6. ^ Ben Korgen. Bonanza for Lake Superior: Seiches Do More Than Move Water. Retrieved on 2008-01-31
7. ^ "Lake Huron Storm Surge July 13, 1995". NOAA. Retrieved 2009-03-13.
8. ^ Illinois State Geological Survey. Seiches: Sudden, Large Waves a Lake Michigan Danger. Retrieved on 2008-01-31.
9. ^ Kathryn Brown. Tsunami! At Lake Tahoe? Retrieved on 2008-01-31.
10. ^ In fact, one person was drowned after being swept away in a particularly energetic seiches in the Jalangi River in the Nadia district to the north of Kolkata in West Bengal (see also Sumatra-Andaman Earthquake)
11. ^ Kashmir earthquake
12. ^ Arizona Geology: Video of seiche in Devils Hole pupfish pond. (Posted: April 27, 2010)
13. ^ Fjorden svinga av skjelvet Retrieved on 2011-03-17.
14. ^ This behaves in a fashion similar to a tidal bore where incoming tides are funneled into a shallow, narrowing river via a broad bay. The funnel-like shape increases the height of the tide above normal, and the flood appears as a relatively rapid increase in the water level.
15. ^ Hibiya, Toshiyuki; Kinjiro Kajiura (1982). "Origin of the Abiki Phenomenon (a kind of Seiche) in Nagasaki Bay" (PDF). Journal of Oceanographical Society of Japan 38 (3): 172–182. DOI:10.1007/BF02110288. Retrieved 2009-02-26.
16. ^ The thermocline is the boundary between colder lower layer (hypolimnion) and warmer upper layer (epilimnion).
• Jackson, J. R. (1833). "On the Seiches of Lakes". Journal of the Royal Geographical Society of London 3: 271–275. DOI:10.2307/1797612.
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2021-01-18 20:12:18
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https://papers.nips.cc/paper/2019/hash/9308b0d6e5898366a4a986bc33f3d3e7-Abstract.html
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#### Authors
Xiaobo Xia, Tongliang Liu, Nannan Wang, Bo Han, Chen Gong, Gang Niu, Masashi Sugiyama
#### Abstract
In label-noise learning, the \textit{noise transition matrix}, denoting the probabilities that clean labels flip into noisy labels, plays a central role in building \textit{statistically consistent classifiers}. Existing theories have shown that the transition matrix can be learned by exploiting \textit{anchor points} (i.e., data points that belong to a specific class almost surely). However, when there are no anchor points, the transition matrix will be poorly learned, and those previously consistent classifiers will significantly degenerate. In this paper, without employing anchor points, we propose a \textit{transition-revision} ($T$-Revision) method to effectively learn transition matrices, leading to better classifiers. Specifically, to learn a transition matrix, we first initialize it by exploiting data points that are similar to anchor points, having high \textit{noisy class posterior probabilities}. Then, we modify the initialized matrix by adding a \textit{slack variable}, which can be learned and validated together with the classifier by using noisy data. Empirical results on benchmark-simulated and real-world label-noise datasets demonstrate that without using exact anchor points, the proposed method is superior to state-of-the-art label-noise learning methods.
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2022-07-01 23:13:27
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http://crumbs.mypopescu.com/since/642115527?mode=own
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You are at the newest post.
## January282018
### Handball and management: from success to failure and back up
On January 29th, 2017 France was winning the World Handball Championship. Yesterday, after 6 wins in the European Handball Championship—the only team with only wins until the semifinal, France was eliminated by Spain after a game in which Les Bleus looked very different.
I’m trying to understand what led to this unexpected result given the previous results obtained in the tournament and also the fact that the French team this year is very similar to the one in 2017—fact that I take as an advantage of unity and team efficiency, more than a sign of a team getting old. To be very clear, I have no intention to minimize Spain’s success which was won with a great strategy, game plan, and effort.
My curiousity is mostly related to what mechanics can change so radically the way a very experienced and well greased team with a perfect track record performs in a critical moment. And next, how do you regain your confidence to play the next game.
Imagine working with a very experienced team that delivers consistently high quality releases. Now with no visible changes in the objectives or area of work, we not only fail to ship in time, but during the whole iteration we are almost clueless of how to address the growing delays and issues.
Update: next day France won the bronze medal in a game with Denmark. When I wrote the title the other day, I had no intention to make it sound like a prediction. I was rather thinking that in sports, you need to be able to refocus very, very fast after a win or a lost.
## September052017
### SteerMouse is indeed the ultimate tool for configuring mice on macOS
It looks like it didn’t take me long to get to SteerMouse.
I continued to ask questions about the Evoluent Mouse Manager and learned that there are no plans to support application-based mappings in the macOS Evoluent Mouse Manager.
Next I tried BetterTouchTool which is my tool of choise when it comes to configuring the touchpad, or an Apple mouse, or even the touchbar. While BetterTouchTool supports the idea of multiple configurations, switching between them is a manual process. What this means is there isn’t an easy way to maintain and switch configurations for multiple mice.
These attempts finally brought me to try SteerMouse. And so far it provides exactly the functionality I’ve been looking for:
1. recognizing different mice and being able to switch configurations accordingly
2. recognize all the extra buttons in my mice (e.g. Evoluent VerticalMouse, Anker vertical mouse, Logitech M705, and Logitech MX Performance)
3. ability to configure different behaviors for the mouse buttons based on the active application.
Based on my initial experience, SteerMouse seems indeed to be the BetterTouchTool for mice. And I haven’t even tried yet its chording operations.
## August312017
### Evoluent Mouse Manager supports the advanced per-app button mappings. But not on macOS
I just learned that the Evoluent Mouse Manager for Windows supports the feature that I was looking for that allows defining per app mappings. The button in the screenshot is pretty clear: “Customize functions for different programs”.
To make things even more disappointing, I have found a screenshot of an older version (3.3) in which this feature already existed:
I can only hope that at some point this feature will also be supported by the macOS Evoluent Mouse Manager tool.
## August272017
### SteerMouse - BetterTouchTool but for mice
In researching how to best configure the Evoluent vertical mouse , I’ve run into SteerMouse. I haven’t had a chance to play with it, but it looks like BetterTouchTool but for mice. I’m bookmarking it here, just in case I’ll need it later.
### Mapping the Evoluent vertical mouse buttons on macOS: BetterTouchTool or Evoluent’s tool?
After trying an Anker vertical mouse for a few months, I finally decided to try the original vertical mouse from Evoluent 1. As with other advanced multi-button mice (e.g. Logitech MX, etc.), Evoluent offers a tool to customize the extra-buttons. The configuration tool offers pretty much all the mapping options I’d expect; or at least I didn’t run into any limitations. I could map buttons to simple actions like Back or Forward, other predefined actions like Mission control, and even keyboard shortcuts.
What I found missing though is the ability of defining per application mappings. In case you haven’t used such mappings, they offer a mechanism to define differfent button behaviors depending on the active application. Such a feature truly maximizes the value of the extra buttons these advanced mice are coming with. Logitech’s application for configuring the mouse supports application-based profiles. Evoluent’s doesn’t.
This is when I started to wonder if using BetterTouchTool for defining custom behaviors for Evoluent’s extra buttons would give the flexibility I’m looking for. At the first glance, BetterTouchTool seem to recognize all Evoluent’s buttons and that means that this migth work nicely2.
As a side note, I couldn’t find much information out there about advanced mappings being used with the Evoluent vertical mouse. Any link or hints will be appreciated.
1. as far as I can tell, there’s a huge difference in the position of the hand while using these 2 mice. But this post won’t focus on this.
2. there is one default button mapping that I don’t know yet how to deal with
## June132017
### Reading "How will you measure your life?" by Clayton M. Christensen
The other day I’ve started to listen the “How will you measure your life?” audiobook by Clayton M. Christensen.
The 2nd chapter introduces Frederick Herzberg’s two-factor theory or motivation theory. I have found it very educational. As a leader, you might discover or realize these aspects—what the theory calls hygiene and motivation factors—by yourself over time. It’s reassuring though to learn there’s a theory and research behind.
I’m also planning to watch the homonymous TEDx talk:
## May292017
### What's the right way to organize my Go project code?
I’ve only worked with Go code a couple of times (the Go wrapper of the DataStax C++ driver for Apache Cassandra is one of them). But my familiarity with the rules and practices in the Go land is very limited. So when I looked to start a toy project I have found myself asking again: what’s the right way to organize my Go project code?
I have found very little online. The first, is Ben Johnson’s Structuring Applications in Go:
I used to place my main.go file in the root of my project so that when someone runs “go get” then my application would be automagically installed. However, combining the main.go file and my application logic in the same package has two consequences:
1. It makes my application unusable as a library.
2. I can only have one application binary.
The best way I’ve found to fix this is to simply use a “cmd” directory in my project where each of its subdirectories is an application binary.
There are some additional hints in the article about Go project code org, and some
I had very high hopes that I’ll find the final answer in David Crawshaw’s slides Organizing Go code. The slides are really interesting, but unfortunately they don’t provide the answer I was looking for.
Finally, I’ve run into Dave Cheney’s Five suggestions for setting up a Go projec. This article provides suggestions for the following types of projects:
1. a single package
2. multiple packages
3. a command
4. a command and a package
5. multiple commands and multiple packages
All I read in this post makes sense to me. So until I hear other recommendations, this will be what I’ll be using.
## December242016
### Dealing with the leap second at AWS
AWS Adjusted Time – We will spread the extra second over the 24 hours surrounding the leap second (11:59:59 on December 31, 2016 to 12:00:00 on January 1, 2017). AWS Adjusted Time and Coordinated Universal time will be in sync at the end of this time period.
It sounds like AWS is using a similar approach for dealing with the leap second as Google does, by making the time slightly slower for a window of time. What seems to differ is the lenght and when this window of time is scheduled. For Google, the interval is 20 hours with the leap second right in the middle of it. For AWS, the interval is 24 hours, but finishes at the same time with the lead second.
I confess that I’m curious about any pros and cons of each of these approaches.
## December232016
### Dealing with the leap second at Google
No commonly used operating system is able to handle a minute with 61 seconds, and trying to special-case the leap second has caused many problems in the past. Instead of adding a single extra second to the end of the day, we’ll run the clocks 0.0014% slower across the ten hours before and ten hours after the leap second, and “smear” the extra second across these twenty hours. For timekeeping purposes, December 31 will seem like any other day.
I don’t know if this a widely used approach for the leap second to spread it across 20 hours.
## December162016
### Make sure you're using the correct USB-C charge cable
• If the first three characters of the serial number are C4M or FL4, the cable is for use with the Apple 29W USB-C Power Adapter.
• If the first three characters of the serial number are DLC or CTC, the cable is for use with the Apple 61W or 87W USB-C Power Adapter.
• If the cable says “Designed by Apple in California. Assembled in China” but has no serial number, you might be eligible for a replacement USB-C charge cable.
I wish it was a joke.
## December142016
### Apple AirPods shipping... in 4 weeks
The Apple AirPods re-announced through a press release are shipping in 4 weeks and are not available for pick up from stores. This looks like missing a deadline to me.
I was more interested in the BeatsX Earphones. They are using the same Apple W1 chip as the AirPods, but are less futuristic. The status of these headphones changed from “Coming this fall” to “Currently unavailable”.
2. the “Thunderbolt 3 (USB-C) to Thunderbolt 2”: $29 The “USB-C Digital AV Multiport Adapter” option is the one that worked. The first time I plugged in the dongle, I was prompted to perform a driver update. Then everything worked as expected. Please note that the Thunderbolt 3 to Thunderbolt 2 does not work. There is a note somewhere in the Apple pages about this dongle not working/supporting the Mini DisplayPort capabilities. Side note: besides HDMI, the AV Multiport adapter also has a charging USB-C port and USB-A port. As I don’t need the USB-A port, and having the charging USB-C is nice, but definitely not critical, I’ll be looking to find a cheaper adapter that only support HDMI. ### Connecting the Thunderbolt 3/USB-C MacBook Pro to my Monoprice IPS monitor The Monoprice IPS monitor I’ve been using for the last couple of years requires a Dual-link DVI adapter1. I confess that I didn’t have very high hopes. Even if my research hasn’t revealed any potential solutions, I wasn’t completely pessimistic when I walked into the Apple store. But things changed very fast after starting to describe how I use to connect to this monitor. I could tell by the grimace of the sale person that I’m touching a pain point2. We consulted with other technical people in the Apple store and in the end, we decided that the only adapter that seemed like having a slight chance to work is the “Thunderbolt 3 (USB-C) to Thunderbolt 2”. Judging by what we read online — none of the people we’ve talked to could tell with a good level of confidence what can be expected to work or not from this dongle — I didn’t expect much. And I was right, as this adapter didn’t work3. At this time, there is no way to use my Monoprice IPS monitor with the Touch Bar MacBook Pro. To make things more complicated and unpleasant for me: 1. the 4k LG monitor is not on display in the Apple store. Apple wants you to buy it blindly. I’m wondering why? 2. the 5k LG monitor is “coming this fall” (today is Dec. 11th, but astronomically(?) fall’s ending on the Winter solstice day, Dec. 21st) 3. I’m almost sure that neither of these will work with my 2013 retina MBP I got from work. I haven’t made up my mind yet what to do regarding the monitor. If you are in a similar situation, I’d love to hear your thoughts and/or decision. 1. Apple’s dongles have never been cheap. I had paid$99 for this Mini DisplayPort to Dual-Link DVI adapter. Compare it with the about $400 I paid for the monitor. But at that time, it was still a good deal considering the Dells were around$700-800.
2. we ended up chatting about the state of dongles and he confessed that 1) he postponed upgrading his own notebook; 2) he avoids recommending the Touch Bar MacBook Pros to non-geeks, and 3) it’s a nightmare.
3. right now, I have no idea what the “Thunderbolt 3 (USB-C) to Thunderbolt 2” adapter is good for. It definitely doesn’t support the DisplayPort features. But I’ve also tried chaining it with the Ethernet-to-Thunderbolt and it did nothing.
## December112016
### Dibya Chakravorty: My search for a MacBook Pro alternative
8 laptops reviewed to correspond to the following requirements:
• must have good build quality comparable to Apple products.
• must have powerful and upgradeable hardware (at least up to 16 GB RAM, 6th or 7th gen processors and 512 GB SSD).
• must have a good form factor, be light and portable.
• must be Linux compatible out of the box.
• must be cheaper than/as costly as the MacBook Pro with similar configuration.
What she finds is not looking very encouraging. Slightly less disappointing than The best MacBook Pro alternatives, now that Apple ruined everything
## December082016
### What happened to my Touch Bar?
This is how DayOne displays my movie journal in which each entry has the movie poster. That’s super cool!
## December022016
Looking at this I’m scratching my head wondering why I haven’t saved the models I had owned. I’m sure there are plently that had them all; I started with 3GS and I’ll most probably skip the 7.
### Console app: how many bugs can you count?
How many bugs can you count in the screenshot below?
This is not about some third party app, but the Console app itself on macOS 10.12.1.
## November272016
### Tree and Ranger: 2 utilities for the terminal
Just a quick note about 2 utilities for the terminal:
1. Tree: a recursive directory listing command that produces a depth indented listing of files, which is colorized
2. ranger: a console file manager with VI key bindings
Both can be installed on macOS through Homebrew. And they are great!
## November212016
### Reddit: I just tried spacemacs for two weeks. Here are my impressions
Great things:
• The ability to roll back package updates easily
• Pre-configured layers that can auto-install if you open a ruby (or whatever) file
• Really good defaults (whoever came up with Emacs’ default scrolling needs a stern talking to in my book)
• I really like the idea of the mnemonic keybindings. I ended up disliking how Spacemacs does it, more below
I learned about Spacemacs a while ago. Before that I’ve tried to get comfortable with Emacs a couple of times and ended up having a decent working configuration; I’ve built it from scratch following the advice of adding to it only the bits and pieces that I understood.
Today I still have Spacemacs on all my machines. I don’t use it very frequently though, but I keep it up to date and, now and then, I check how random things are done.
Then I go back to my Vim environment. There are a few things that I wish I had in Vim too:
1. a central listing of Vim plugins (download numbers, last update info would be super useful)
2. feature packages. Not having to figure out what plugins to combine to get an end to end experience would make things so much easier. I do wonder if Vim’s 8.0 packages will start be using for this.
3. Consistent mappings. Even more important, a keybindings hint system that could suggest mappings that are available. Spacemacs uses which-key which provides a help buffer listing the available key bindings and their associated commands. It is amazing!
## November202016
A question on Twitter made me quite curious about AsciiDoc. I’ve heard about it in the past and as someone writting almost everything in Markdown I was naturally curious to see what improvements could it bring to my environment.
What follows is the series of remarks, comments, head scratchers, and complains I accumulated over a couple of hours of learning AsciiDoc:
1. searching anything about AsciiDoc led me invariably to either AsciiDoc or Asciidoctor. That was confusing as hell. Even if the 2 projects seem to be related.
2. after doing quite a bit of reading, I still cannot summarize the main advantages of AsciiDoc when compared with Markdown. I did noticed some interesting features, but I had to read the docs twice. Now when writting this, I do realize that I should’ve create that summary myself.
3. I didn’t find a way to simply generate an HTML fragment; AsciiDoc supports multiple output formats, but not HTML fragments (see update at the end)
4. it feels to me that AsciiDoc has a bit too much ceremony. While I did look at an example file, I failed the test of creating a simple document right after going through the docs.
5. I have found an AsciiDoc plugin for Vim that is quite good. Spacemacs also has a layer for AsciiDoc, but the formatting used goes too far and I could hardly use it.
6. weirdly enough, while AsciiDoc has better support for standardizing quotations (inline and block), I have discovered that when generating HTML, it doesn’t use either of the specialized tags cite or blockquote. I have found a corresponding ticket from 2013.
In the times of myNoSQL, I have developed my own toolkit and extensions around various Python libraries, starting with Markdown and all the way to custom scripts for resizing and uploading images, and an end-to-end publishing workflow. I don’t need such a complete toolkit these days, but if I’d actually see the significant benefits of AsciiDoc, I’d probably be tempted to address some of the above points myself.
Update: thanks to Dan Allen, I’ve learned that it’s possible to generate HTML fragments by using the -s option.
Older posts are this way If this message doesn't go away, click anywhere on the page to continue loading posts.
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2019-12-15 04:49:41
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https://projecteuclid.org/euclid.nmj/1114631499
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## Nagoya Mathematical Journal
### Twistor theory of manifolds with Grassmannian structures
#### Abstract
As a generalization of the conformal structure of type $(2, 2)$, we study Grassmannian structures of type $(n, m)$ for $n, m \geq 2$. We develop their twistor theory by considerin the complete integrability of the associated null distributions. The integrability corresponds to global solutions of the geometric structures.
A Grassmannian structure of type $(n, m)$ on a manifold $M$ is, by definition, an isomorphism from the tangent bundle $TM$ of $M$ to the tensor product $V \otimes W$ of two vector bundles $V$ and $W$ with rank $n$ and $m$ over $M$ respectively. Because of the tensor product structure, we have two null plane bundles with fibres $P^{m-1}(\mathbb{R})$ and $P^{n-1}(\mathbb{R})$ over $M$. The tautological distribution is defined on each two bundles by a connection. We relate the integrability condition to the half flatness of the Grassmannian structures. Tanaka's normal Cartan connections are fully used and the Spencer cohomology groups of graded Lie algebras play a fundamental role.
Besides the integrability conditions corr[e]sponding to the twistor theory, the lifting theorems and the reduction theorems are derived. We also study twistor diagrams under Weyl connections.
#### Article information
Source
Nagoya Math. J., Volume 160 (2000), 17-102.
Dates
First available in Project Euclid: 27 April 2005
https://projecteuclid.org/euclid.nmj/1114631499
Mathematical Reviews number (MathSciNet)
MR1804138
Zentralblatt MATH identifier
1039.53055
Subjects
Secondary: 53C10: $G$-structures
#### Citation
Machida, Yoshinori; Sato, Hajime. Twistor theory of manifolds with Grassmannian structures. Nagoya Math. J. 160 (2000), 17--102. https://projecteuclid.org/euclid.nmj/1114631499
#### References
• M. A. Akivis and V. V. Goldberg, Conformal differential geometry and its generalizations, John Wiley and Sons, Inc., New York (1996).
• ––––, On the theory of almost Grassmann structures , in New developments in differential geometry (J. Szenthe, ed.), Kluwer Academic Publishers, Dordrecht, Boston, London (1998), 1–37.
• ––––, Conformal and Grassmann structures , Differential Geom. Appl., 8 (1998), 177–203.
• ––––, Semiintegrable almost Grassmann structures , Differential Geom. Appl., 10 (1999), 257–294.
• A. L. Besse, Manifolds all of whose geodesics are closed, Springer-Verlag, Berlin, Heidelberg, New York (1978).
• T. N. Bailey and M. G. Eastwood, Complex paraconformal manifolds –- their differential geometry and twistor theory , Forum Math., 3 (1991), 61–103.
• A. B. Goncharov, Generalized conformal structures on manifolds , Selecta Math. Sov., 6 (1987), 307–340.
• T. Hangan, Geómétrie différentielle Grassmannienne , Rev. Roum. Math. Pures Appl., 11 (1966), 519–531.
• S. Helgason, Differential geometry, Lie groups, and symmetric spaces, Academic Press, New York, London (1978).
• N. J. Hitchin, Complex manifolds and Einstein's equations , in Twistor geometry and non-linear systems (H. D. Doebner, T. D. Palev, eds.), Lecture Notes in Math. 970, Springer-Verlag, Berlin, Heidelberg, New York (1982), 73–99.
• T. Ishihara, On tensor-product structures and Grassmannian structures , J. Math. Tokushima Univ., 4 (1970), 1–17.
• P. E. Jones and K. P. Tod, Minitwistor spaces and Einstein-Weyl spaces , Class. Quantum. Grav., 2 (1985), 565–577.
• S. Kaneyuki, On the subalgebras ${\smallfrak g}_{0}$ and ${\smallfrak g}_{\it ev}$ of semisimple graded Lie algebras , J. Math. Soc. Japan., 45 (1993), 1–19.
• S. Kobayashi, Transformation groups in differential geometry, Springer-Verlag, Berlin, Heidelberg, New York (1972).
• H. Kamada and Y. Machida, Self-duality of metrics of type $(2, 2)$ on four-dimensional manifolds , Tohoku Math. J., 49 (1997), 259–275.
• S. Kobayashi and T. Nagano, On filterd Lie algebras and geometric structures I , J. Math. Mech., 13 (1964), 875–907.
• S. Kobayashi and K. Nomizu, Foundations of differential geometry I, Interscience Publishers, John Wiley and Sons, New York, London (1963).
• Y. I. Manin, Gauge field theory and complex geometry, Springer-Verlag, Berlin, Heidelberg, New York (1988).
• Y. I. Mikhailov, On the structure of almost Grassmannian manifolds , Soviet Maht., 22 (1978), 54–63.
• Y. Machida and H. Sato, Twistor spaces for real four-dimensional Lorentzian manifolds , Nagoya Math. J., 134 (1994), 107–135.
• J. Milnor and J.D. Stasheff, Characteristic classes, Princeton Univ. (1974).
• T. Ochiai, Geometry associated with semisimple flat homogeneous spaces , Trans. Amer. Math. Soc., 152 (1970), 159–193.
• H. Pedersen and A. Swann, Riemannian submersions, four-manifolds and Einstein-Weyl geometry , Proc. London Math. Soc., 66 (1993), 381–399.
• H. Pedersen and K. P. Tod, Three-dimensional Einstein-Weyl geometry , Adv. in Math., 97 (1993), 74–109.
• S. Sternberg, Lectures on differential geometry, Prentice-Hall, Inc., New Jersey (1964).
• H. Sato and K. Yamaguchi, Lie contact manifods , in Geometry of manifolds (K. Shiohama, ed.), Academic Press, Boston (1989), 191–238.
• ––––, Lie contact manifolds II , Math. Ann., 297 (1993), 33–57.
• M. Takeuchi, A remark on Lie contact structures , Science Report Coll. Ge. Ed. Osaka Univ., 42 (1993), 29–37.
• K. P. Tod, Compact $3$-dimensional Einstein-Weyl structures , J. London Math. Soc., 45 (1992), 341–351.
• N. Tanaka, On the equivalence problems associated with simple graded Lie algebras , Hokkaido Math. J., 8 (1979), 23–84.
• ––––, On affine symmetric spaces and the automorphism groups of product manifolds , Hokkaido Math. J., 14 (1985), 277–351.
• ––––, On geometric theory of systems of ordinary differential equations, Lectures in Colloq. on Diff. Geom. at Sendai, August (1989).
• R. O. Wells, Jr., Complex geometry in mathematical physics, Presses de l'Université de Montréal, Montréal (1982).
• R. S. Ward and R. O. Wells, Jr., Twistor geometry and field theory, Cambridge Univ. Press (1990).
• K. Yamaguchi, Differential systems associated with simple graded Lie algebras , Advanced Studies in Pure Math., 22 (1993), 413–494.
• K. Yano and S. Ishihara, Tangent and cotangent bundles, Marcel Dekker, New york (1973).
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2020-07-09 06:12:42
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https://math.stackexchange.com/questions/2921883/is-there-any-isomorphism-between-the-non-zero-complex-numbers-under-multiplicati/2921901
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# Is there any isomorphism between the non-zero complex numbers under multiplication and the complex numbers under addition? [duplicate]
It seems like there is no trivial isomorphism between them and both are abelian and non-cyclic, so, it makes it even harder to conclude anything. I need some hints.
If I had to make a guess, I'd probably say there is no isomorphism between them.
## marked as duplicate by Arnaud D., Cam McLeman, Delta-u, Mostafa Ayaz, StrantsSep 19 '18 at 16:02
• There is one and only one solution to $nx=a$ for $a\in\mathbb{C}$, $n>0$; however, there are $n$ solutions to $x^n=a$ for $a\in\mathbb{C}$. – Arturo Magidin Sep 18 '18 at 19:07
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2019-06-18 04:40:57
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https://kids.kiddle.co/Special_functions
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# Special functions facts for kids
Kids Encyclopedia Facts
Special functions are some mathematical functions used in mathematical analysis or physics. Most of them appear in higher education. Some experts are studying numerical methods for them.
## Definition
In mathematics, most functions are defined as a solution of a differential equation. For example, the exponential function $\exp(x)$ is the solution of the ordinary differential equation $y^\prime=y$. Due to this relation, some mathematicians are studying the connection between ODEs and special functions.
## Examples
For more examples, find textbooks named "special functions".
Special functions Facts for Kids. Kiddle Encyclopedia.
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2021-02-26 13:12:24
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https://pos.sissa.it/390/397/
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Volume 390 - 40th International Conference on High Energy physics (ICHEP2020) - Parallel: Quark and Lepton Flavour Physics
Angular analysis of modes with $b \to c \ell \bar \nu$ transition and new physics
R. Mandal
Full text: Not available
Abstract
Starting with the most general dimension six beyond the standard model Hamiltonian, we derive the four fold angular distributions for $B \to D \pi \ell \bar{\nu}$ mode where the $D\pi$ final state arises from a vector $D^*$ or a tensor meson $D_2^*$ onshell decay. Distinguishable features in the angular distributions of these two modes are explored. A Plenty of observables are constructed on which the experimental information would help to disentangle the dynamical origin of the observed anomalies in $b \to c \tau \bar{\nu}$ transition.
How to cite
Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.
Open Access
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2020-11-29 10:02:22
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https://injurystats.wordpress.com/tag/chi-square-test/
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# Transcribing Problems with cycle-helmets.com Analysis
I recently discussed problems replicating the results found in an assessment of mandatory helmet legislation in Australia published in Accident Analysis and Prevention (Robinson, 1996). This issue was introduced to me by Linda Ward who has pointed to a related issue.
The anti-helmet website http://www.cycle-helmets.com has a page titled “Spinning” helmet law statistics. Under the heading Measuring changes in cycle use, the webpage states
Similarly, in South Australia a telephone survey found no significant decline in the amount people said they cycled but there was a large, significant drop in how much they had actually cycled in the past week 24. In 1990 (pre-law), 17.5% of males aged at least 15 years reported cycling in the past week (210 out of 1201), compared to 13.2% (165 out of 1236) post-law in 1993. For females, 8.1% (102 out of 1357) had cycled in the past week in 1990 compared to 5.9% (98 out 1768) in 1993 24.
These reductions (24% for males, 26% for females aged at least 15 years) are statistically significant (P < 0.005 for males, P = 0.025 for females).
The citation given is a technical report that evaluated the introduction of helmet legislation in South Australia.[1] Table 1 of the SA report gives frequencies of bicycle riding from two surveys, one in 1990 and the other in 1993, for those aged 15 years or older separated by gender. In this survey, the amount of cycling split into four categories: “At Least Once A Week”, “At Least Once A Month”, “At Least Once Every 3 Months” and “Less Often or Never”. The SA helmet law went into effect on 1 July 1991.
The main problem here is the numbers in the above quote don’t match up to the data in the original report. Here is a screenshot of the table.
When these numbers are corrected and a comparison is made for those cycling at least once a week versus everyone else, the p-values are 0.279 and 0.450 for males and females respectively. Additionally, the relative risks are 0.90 (95% CI: 0.76,1.08) and 0.91 (95% CI: 0.71, 1.17) for males and females respectively. The point estimates for changes in the proportion cycling in the past week are much less than those reported on the webpage.
In addition to using the wrong data, I don’t agree with the analysis. There are four cycling categories which have been collapsed into two — those who cycle at least once a week and those who don’t. A lot of information is needlessly removed from the data. Instead, a chi-square test for independence could’ve been performed and individual changes could be assessed through an investigation of the residuals.
The Pearson residuals for an individual cell from a chi-square test are
$r=\dfrac{O-E}{\sqrt{E}}$
where $O$ are the observed frequencies and $E$ is the expected frequency under an assumption of independence, i.e., no relationship between helmet legislation and the amount of cycling. These residuals are asymptotically normal, so residuals with absolute value greater 1.96 may be considered “statistically significant”. The sign would indicate observing more than expected (if positive) or less than expected (if negative).
When analyses are performed on the full tables, the chi-square tests give p-values of 0.20 and 0.85 for males and females respectively. None of the residuals have absolute value anywhere near 1.96. The largest residual pair is for males cycling “at least once every 3 months”. The signs of the residuals indicate there is less cycling than expected in 1990 (r=-1.04) and more cycling than expected in 1993 (r=1.02) if there is no relationship between helmet legislation and amount of cycling. Here is some R code to do those analyses.
males=matrix(c(204,190,66,83,58,77,871,886),nrow=2)
males
females=matrix(c(104,123,59,74,52,64,1141,1507),nrow=2)
females
chisq.test(males,correct=F)
chisq.test(females,correct=F)
chisq.test(males,correct=F)$residuals chisq.test(females,correct=F)$residuals
The analyses above are stratified by gender and we could perform a unified analysis using Poisson regression. This model is essentially
$log(\mu)=\beta_0+\beta_1YEAR+\beta_2FREQ+\beta_3GENDER+\beta_4YEAR*FREQ+\beta_5YEAR*GENDER+\beta_6FREQ*GENDER+\beta_7YEAR*FREQ*GENDER$
I’ve simplified things a bit here because the variable $FREQ$ has four categories and therefore gets estimated by three dummy variables.
The important comparison here is the interaction between $YEAR$ and $FREQ$. If significant, this would indicate helmet legislation and amount of cycling are associated. Using the given South Australian data, the three-way interaction was non-signficant, so was removed from the model. The p-value of the interaction between $YEAR$ and $FREQ$ is not statistically significant (p=0.41).
No analysis I’ve performed indicates a significant relationship between helmet legislation and amount of cycling in South Australia among those 15 years or older when using the correct data.
Note: The anti-helmet website http://www.cycle-helmets.com is maintained by Chris Gillham. I previously discussed problems with this website here. If you download the PDF version of this report, the author is listed as “Dorre” who I believe is Dorothy Robinson. Both Gillham and Robinson are editorial board members of the anti-helmet organisation Bicycle Helmet Research Foundation.
1. Marshall, J. and M. White, Evaluation of the compulsory helmet wearing legislation for bicyclists in South Australia Report 8/94, 1994, South Australian Department of Transport, Walkerville, South Australia.
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2018-02-20 13:41:57
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https://stats.stackexchange.com/questions/338928/what-is-the-loss-function-for-a-probabilistic-decoder-in-the-variational-autoenc
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# What is the loss function for a probabilistic decoder in the Variational Autoencoder?
In a VAE with Gaussian output the loss function is usually:$$\sum{(\hat x - x)^2} + KL,$$ so the sum of squared errors plus KL divergence. When I also want to predict the variance of the reconstructed input I need 2 outputs for each dimension of x: mean and variance. But in this case how should I calculate the reconstruction error? Should I sample from a normal distribution with the mean and variance I have as current output and calculate sum of squared errors for that sample?
If you're predicting a mean and a variance with the decoder, you can use a log likelihood loss for the reconstruction loss. So if you predict $\mu_x$ and $\sigma_x$, then the reconstruction loss is:
$$2(\mu_x - x) - 2\sigma_x^2 + \operatorname{log}2\sigma_x + C$$
• Thank you! Just to be sure, so in TensorFlow it would be: 2 * (mu - input) - 2 * tf.square(sigma) + tf.log(2 * sigma) where input, mu and sigma are rank 2 tensors with shape (number of samples, input dimensions)? – Márton György Apr 9 '18 at 7:35
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2021-04-11 10:52:37
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https://www.physicsforums.com/threads/stumped-de-problem.91093/
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# Stumped- DE problem
1. Sep 27, 2005
### hola
I am stumped..... here is the problem:
Solve the DE using the following:
L and R are constants
$$L\frac{di}{dt} + Ri = E(t)$$
$$i(0) = i_0$$
$$E(t) = E_0*sin(wt)$$
Here is my work so far:
I got the integrating factor to become $$e^{Rt/L}$$. But now:
$$\frac{d(e^{\frac{Rt}{L}}*i)}{dt} = e^{\frac{Rt}{L}}\frac{E_0}{L}*sin(wt)$$
But I am stuck from there. Help would be appreciated.
2. Sep 27, 2005
### Corneo
Since you have
$$\mu(t) = e^{\frac {R}{L}t}$$
and you also have
$$(\mu(t)i)' = \mu(t) \frac {E_0}{L} \sin (\omega t)$$
You can integrate both sides and divide by $\mu(t)$
Hence
$$i(t) = \frac {\int \mu(s) \frac {E_0}{L} \sin (\omega s) ds}{\mu(t)}$$
I switched the t to a s in the numerator to avoid confusion. After you integrate the numerator, you can replace the s with a t.
3. Sep 27, 2005
### hola
That's the problem..... I can't integrate it.
4. Sep 27, 2005
### Tom Mattson
Staff Emeritus
Integrate it by parts. Let $u=\sin(\omega t)$ and let $dv=exp\left(\frac{Rt}{L})$.
You'll have to integrate by parts twice and then algebraically solve for the integral. This integral actually pops up all the time in second order dynamic systems.
5. Sep 27, 2005
### Tide
You can either integrate by parts or, if you're comfortable with complex analysis, you can note that
$$\int e^{at} \sin \omega t dt = Im \int e^{(a + i \omega) t} dt$$
and extract the imaginary part after performing the integration.
6. Sep 27, 2005
### hola
I got some really messy answer.... is that ok?
7. Sep 27, 2005
### Tom Mattson
Staff Emeritus
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2017-11-17 17:52:33
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https://www.physicsforums.com/threads/driving-me-batty-csc-4x-cot-4x-simplify.345086/
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Homework Help: Driving me batty! CSC^4x -COT^4x; simplify.
1. Oct 12, 2009
Agent M27
1. The problem statement, all variables and given/known data
The problem is to simplify the following trig equation:
csc^4x - cot^4x
2. Relevant equations
There really isn't any I can think of.
3. The attempt at a solution
When I attempt to simplify csc^4x - cot^4x I arrive at the result, from factoring, (csc^2x + cot^2x)(csc^2x - cot^2x). The book shows just the addition portion of my factoring. I have asked my professor and he said that I need not worry as the exam will be multiple guess?? I however am not satisfied with guessing, and would like to know if there is some fundamental rule or operation which I have overlooked. Thanks in advance and I apologize for the eq not being in a Tex format, I'm on my phone. Take care.
Joe
2. Oct 12, 2009
Bohrok
You can use a pythagorean trig identity with one of the factors you have.
3. Oct 12, 2009
Agent M27
Well that's embarassing... Thanks buddy. I knew it had to be something simple but I have a tendency to make things way more difficult than they are. I am assuming you were refering to csc^2x - cot^2x = 1 ?
Joe
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2018-05-27 20:01:39
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http://www.statemaster.com/encyclopedia/Catastrophe-theory
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Encyclopedia > Catastrophe theory
In mathematics, catastrophe theory is a branch of bifurcation theory in the study of dynamical systems; it is also a particular special case of more general singularity theory in geometry. For other meanings of mathematics or uses of math and maths, see Mathematics (disambiguation) and Math (disambiguation). ... In mathematics, specifically in the study of dynamical systems, a bifurcation occurs when a small smooth change made to the parameter values (the bifurcation parameters) of a system causes a sudden qualitative or topological change in the systems long-term dynamical behaviour. ... The Lorenz attractor is an example of a non-linear dynamical system. ... For non-mathematical singularity theories, see singularity. ... For other uses, see Geometry (disambiguation). ...
Bifurcation theory studies and classifies phenomena characterized by sudden shifts in behavior arising from small changes in circumstances, analysing how the qualitative nature of equation solutions depends on the parameters that appear in the equation. This may lead to sudden and dramatic changes, for example the unpredictable timing and magnitude of a landslide. Qualitative is an important qualifier in the following subject titles: Qualitative identity Qualitative marketing research Qualitative method Qualitative research THE BIG J This is a disambiguation page — a list of pages that otherwise might share the same title. ... The magnitude of a mathematical object is its size: a property by which it can be larger or smaller than other objects of the same kind; in technical terms, an ordering of the class of objects to which it belongs. ... This article is about geological phenomenon. ...
Catastrophe theory, which was originated with the work of the French mathematician René Thom in the 1960s, and became very popular due to the efforts of Christopher Zeeman in the 1970s, considers the special case where the long-run stable equilibrium can be identified with the minimum of a smooth, well-defined potential function (Lyapunov function). René Thom (September 2, 1923 - October 25, 2002) was a French mathematician and founder of the catastrophe theory. ... Sir Erik Christopher Zeeman (born February 4, 1925), is a mathematician known for work in geometric topology and singularity theory. ... It has been suggested that this article or section be merged with Potential. ... Lyapunov functions are of interest in mathematics, especially in stability theory. ...
Small changes in certain parameters of a nonlinear system can cause equilibria to appear or disappear, or to change from attracting to repelling and vice versa, leading to large and sudden changes of the behaviour of the system. However, examined in a larger parameter space, catastrophe theory reveals that such bifurcation points tend to occur as part of well-defined qualitative geometrical structures.
Catastrophe theory analyses degenerate critical points of the potential function — points where not just the first derivative, but one or more higher derivatives of the potential function are also zero. These are called the germs of the catastrophe geometries. The degeneracy of these critical points can be unfolded by expanding the potential function as a Taylor series in small perturbations of the parameters. In mathematics, a germ is an equivalence class of continuous functions from one topological space to another (often from the real line to itself), in which one point x0 in the domain has been singled out as privileged. ... Series expansion redirects here. ...
When the degenerate points are not merely accidental, but are structurally stable, the degenerate points exist as organising centres for particular geometric structures of lower degeneracy, with critical features in the parameter space around them. If the potential function depends on two or fewer active variables, and four or fewer active parameters, then there are only seven generic structures for these bifurcation geometries, with corresponding standard forms into which the Taylor series around the catastrophe germs can be transformed by diffeomorphism (a smooth transformation whose inverse is also smooth). These seven fundamental types are now presented, with the names that Thom gave them. In mathematics, structural stability is an aspect of stability theory concerning whether a given function is sensitive to a small perturbation. ... In mathematics, a diffeomorphism is a kind of isomorphism of smooth manifolds. ...
## Potential functions of one active variable
### Fold catastrophe
Stable and unstable pair of extrema disappear at a fold bifurcation
$V = x^3 + ax,$
At negative values of a, the potential has two extrema - one stable, and one unstable. If the parameter a is slowly increased, the system can follow the stable minimum point. But at a=0 the stable and unstable extrema meet, and annihilate. This is the bifurcation point. At a>0 there is no longer a stable solution. If a physical system is followed through a fold bifurcation, one therefore finds that as a reaches 0, the stability of the a<0 solution is suddenly lost, and the system will make a sudden transition to a new, very different behaviour. This bifurcation value of the parameter a is sometimes called the tipping point.
Image File history File links Fold_bifurcation. ... The phrase tipping point or angle of repose is a sociology term that refers to that dramatic moment when something unique becomes common. ...
### Cusp catastrophe
Diagram of cusp catastrophe, showing curves (brown, red) of x satisfying dV / dx = 0 for parameters (a,b), drawn for parameter b continuously varied, for several values of parameter a. Outside the cusp locus of bifurcations (blue), for each point (a,b) in parameter space there is only one extremising value of x. Inside the cusp, there are two different values of x giving local minima of V(x) for each (a,b), separated by a value of x giving a local maximum. Cusp shape in parameter space (a,b) near the catastrophe point, showing the locus of fold bifurcations separating the region with two stable solutions from the region with one. Pitchfork bifurcation at a=0 on the surface b=0
The cusp geometry is very common, when one explores what happens to a fold bifurcation if a second parameter, b, is added to the control space. Varying the parameters, one finds that there is now a curve (blue) of points in (a, b) space where stability is lost, where the stable solution will suddenly jump to an alternate outcome. Image File history File links Cusp_catastrophe. ... Image File history File links Cusp_shape. ... Image File history File links Pitchfork_bifurcation_left. ...
But in a cusp geometry the bifurcation curve loops back on itself, giving a second branch where this alternate solution itself loses stability, and will make a jump back to the original solution set. By repeatedly increasing b and then decreasing it, one can therefore observe hysteresis loops, as the system alternately follows one solution, jumps to the other, follows the other back, then jumps back to the first. A system with hysteresis can be summarised as a system that may be in any number of states, independent of the inputs to the system. ...
However, this is only possible in the region of parameter space a<0. As a is increased, the hysteresis loops become smaller and smaller, until above a=0 they disappear altogether (the cusp catastrophe), and there is only one stable solution.
One can also consider what happens if one holds b constant and varies a. In the symmetrical case b=0, one observes a pitchfork bifurcation as a is reduced, with one stable solution suddenly splitting into two stable solutions and one unstable solution as the physical system passes to a<0 through the cusp point a=0, b=0 (an example of spontaneous symmetry breaking). Away from the cusp point, there is no sudden change in a physical solution being followed: when passing through the curve of fold bifurcations, all that happens is an alternate second solution becomes available. In bifurcation theory, a field within mathematics, a pitchfork bifurcation is a special case zero-eigenvalue bifurcation. ... Spontaneous symmetry breaking in physics takes place when a system that is symmetric with respect to some symmetry group goes into a vacuum state that is not symmetric. ...
A famous suggestion is that the cusp catastrophe can be used to model the behaviour of a stressed dog, which may respond by becoming cowed or becoming angry. The suggestion is that at moderate stress (a>0), the dog will exhibit a smooth transition of response from cowed to angry, depending on how it is provoked. But higher stress levels correspond to moving to the region (a<0). Then, if the dog starts cowed, it will remain cowed as it is irritated more and more, until it reaches the 'fold' point, when it will suddenly, discontinuously snap through to angry mode. Once in 'angry' mode, it will remain angry, even if the direct irritation parameter is considerably reduced.
Another application example is for the outer sphere electron transfer frequently encountered in chemical and biological systems (Xu, F. Application of catastrophe theory to the ∆G to -∆G relationship in electron transfer reactions. Zeitschrift für Physikalische Chemie Neue Folge 166, 79-91 (1990)). Outer sphere electron transfer is an inorganic reaction mechanism defined as electron transfer (or redox) in which the coordination shells of the two metal centers remain intact during the electron transfer event. ...
Fold bifurcations and the cusp geometry are by far the most important practical consequences of catastrophe theory. They are patterns which reoccur again and again in physics, engineering and mathematical modelling.
The remaining simple catastrophe geometries are very specialised in comparison, and presented here only for curiosity value.
### Swallowtail catastrophe
$V = x^5 + ax^3 + bx^2 + cx ,$
The control parameter space is three dimensional. The bifurcation set in parameter space is made up of three surfaces of fold bifurcations, which meet in two lines of cusp bifurcations, which in turn meet at a single swallowtail bifurcation point.
As the parameters go through the surface of fold bifurcations, one minimum and one maximum of the potential function disappear. At the cusp bifurcations, two minima and one maximum are replaced by one minimum; beyond them the fold bifurcations disappear. At the swallowtail point, two minima and two maxima all meet at a single value of x. For values of a>0, beyond the swallowtail, there is either one maximum-minimum pair, or none at all, depending on the values of b and c. Two of the surfaces of fold bifurcations, and the two lines of cusp bifurcations where they meet for a<0, therefore disappear at the swallowtail point, to be replaced with only a single surface of fold bifurcations remaining. Salvador Dalí's last painting, The Swallow's Tail, was based on this catastrophe. Salvador Domingo Felipe Jacinto Dalà i Domènech, 1st Marquis of Púbol (May 11, 1904 – January 23, 1989), was a Spanish surrealist painter of Catalan descent born in Figueres, Catalonia (Spain). ... La queue daronde - Série des catastrophes (The Swallows Tail - Series on Catastrophes) was the last painting of Salvador DalÃ, done in May 1983. ...
### Butterfly catastrophe
$V = x^6 + ax^4 + bx^3 + cx^2 + dx ,$
Depending on the parameter values, the potential function may have three, two, or one different local minima, separated by the loci of fold bifurcations. At the butterfly point, the different 3-surfaces of fold bifurcations, the 2-surfaces of cusp bifurcations, and the lines of swallowtail bifurcations all meet up and disappear, leaving a single cusp structure remaining when a>0
## Potential functions of two active variables
Umbilic catastrophes are examples of corank 2 catastrophes. They can be observed in optics in the focal surfaces created by light reflecting off a surface in three dimensions and are intimately connected with the geometry of nearly spherical surfaces. Thom proposed that the Hyperbolic umbilic catastrophe modeled the breaking of a wave and the elliptical umbilic modeled the creation of hair like structures. For the book by Sir Isaac Newton, see Opticks. ...
### Hyperbolic umbilic catastrophe
$V = x^3 + y^3 + axy + bx + cy ,$
### Parabolic umbilic catastrophe
$V = x^2y + y^4 + ax^2 + by^2 + cx + dy ,$
## Arnold's notation
Vladimir Arnol'd gave the catastrophes the ADE classification, due to a deep connection with simple Lie groups. Vladimir Igorevich Arnold (ВладиÌмир ИÌгоревич ÐрноÌльд, born June 12, 1937 in Odessa, USSR) is one of the worlds most prolific mathematicians. ... In mathematics, the ADE classification is the complete list of simply laced groups or other mathematical objects satisfying analogous axioms. ... In mathematics, a simple Lie group is a Lie group which is also a simple group. ...
• A0 - a non singular point: V = x.
• A1 - a local extrema, either a stable minimum or unstable maximum $V = pm x^2 + a x$.
• A2 - the fold
• A3 - the cusp
• A4 - the swallowtail
• A5 - the butterfly
• Ak - an infinite sequence of one variable forms $V=x^{k+1}+cdots$
• D4- - the elliptical umbilic
• D4+ - the hyperbolic umbilic
• D5 - the parabolic umbilic
• Dk - an infinite sequence of further umbilic forms
• E6 - the symbolic umbilic V = x3 + y4 + axy2 + bxy + cx + dy + ey2
• E7
• E8
There are objects in singularity theory which correspond to most of the other simple Lie groups.
Broken symmetry is a concept used in mathematics and physics when an object breaks either rotational symmetry or translational symmetry. ... The phrase tipping point or angle of repose is a sociology term that refers to that dramatic moment when something unique becomes common. ... This diagram shows the nomenclature for the different phase transitions. ... The domino effect refers to a small change which will cause a similar change nearby, which then will cause another similar change, and so on in linear sequence, by analogy to a falling row of dominoes standing on end. ... Production Order Snowball Effect is a SpongeBob SquarePants episode from season three. ... Point attractors in 2D phase space. ... Spontaneous symmetry breaking in physics takes place when a system that is symmetric with respect to some symmetry group goes into a vacuum state that is not symmetric. ... For other uses, see Chaos Theory (disambiguation). ...
## References
• Arnol'd, Vladimir Igorevich. Catastrophe Theory, 3rd ed. Berlin: Springer-Verlag, 1992.
• Gilmore, Robert. Catastrophe Theory for Scientists and Engineers. New York: Dover, 1993.
• Postle, Denis. Catastrophe Theory – Predict and avoid personal disasters. Fontana Paperbacks 1980. ISBN 0-00-635559-5
• Poston, T. and Stewart, Ian. Catastrophe: Theory and Its Applications. New York: Dover, 1998. ISBN 0-486-69271-X.
• Sanns, Werner. Catastrophe Theory with Mathematica: A Geometric Approach. Germany: DAV, 2000.
• Saunders, Peter Timothy. An Introduction to Catastrophe Theory. Cambridge, England: Cambridge University Press, 1980.
• Thom, René. Structural Stability and Morphogenesis: An Outline of a General Theory of Models. Reading, MA: Addison-Wesley, 1989. ISBN 0-201-09419-3.
• Thompson, J. Michael T. Instabilities and Catastrophes in Science and Engineering. New York: Wiley, 1982.
• Woodcock, Alexander Edward Richard and Davis, Monte. Catastrophe Theory. New York: E. P. Dutton, 1978. ISBN 0525078126.
Vladimir Igorevich Arnold (Russian: ВладиÌмир ИÌгоревич ÐрноÌльд, born June 12, 1937 in Odessa, USSR) is one of the worlds most prolific mathematicians. ... Ian Stewart, FRS (b. ... René Thom (September 2, 1923 - October 25, 2002) was a French mathematician and founder of the catastrophe theory. ... Sir Erik Christopher Zeeman (born February 4, 1925), is a mathematician known for work in geometric topology and singularity theory. ...
Results from FactBites:
Catastrophe Theory (134 words) Originated by the French mathematician Rene Thom in the 1960s, catastrophe theory is a special branch of dynamical systems theory. Catastrophes are bifurcations between different equilibria, or fixed point attractors. Catastrophe theory has been applied to a number of different phenomena, such as the stability of ships at sea and their capsizing, bridge collapse, and, with some less convincing success,
More results at FactBites »
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2020-07-12 04:13:39
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https://math.stackexchange.com/questions/3919459/generalizing-topological-spaces
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# Generalizing topological spaces
I was studying topological spaces having studied metric spaces. I guess the properties of open sets in metric space led into (in terms of open set) definition of Topological spaces. Although there are many equivalent definitions such as Kuratowski closure axioms, the most used definition is in terms of open sets.
I know that we can extend these topological spaces into vector spaces or any other algebraic structures by defining appropriate operations in the given topological space. I'm not aware of any generalization of topological spaces which further broadens the definition of topological spaces.
For example, if we replace union and intersection by any other two operations (let's call it op1 and op2) such that arbitrary op1 and finite op2 are closed in the given class of subsets of a set and conventionally we can call them as open$$^*$$ sets(I'm calling them open$$^*$$ to distinguish from the open sets in the normal sense). This idea is itself is worth reflecting because if we consider op1 as intersection and op2 as union then closed sets(in the normal sense) are open$$^*$$ sets and open sets(in the normal sense) are closed$$^*$$ sets. If we consider some other operations then the properties will differ entirely. In general, it may not lead us into metric spaces or may not have pleasant properties which always is the aim of mathematicians.
Does this kind of generalization of topological space exist? Did anyone work in this direction?
• Do you know Stone duality? It deals with something similar, namely finding algebraic structures which behave as if they were the sets of open/closed sets of a topological space. Nov 23 '20 at 13:16
• I don't know about "revising" topological notions by using set operations to replace union and intersection, but if one were to venture down this road I'd recommend looking at Hausdorff operations (see also this search and this search). For the usual generalizations of topology, see my answers to this question, especially the references. Nov 23 '20 at 14:01
• Regarding generalizations and extensions of Hausdorff operations, see Kolmogorov's ideas in the theory of operations on sets by Vladimir Grigorevich Kanovei (1988) and Memoir on the analytical operations and projective sets (I) (II) by Kantorovitch/Livenson (1932, 1933). Nov 23 '20 at 14:12
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2021-11-30 02:45:44
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https://number.subwiki.org/wiki/Von_Mangoldt_function
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# Von Mangoldt function
This article defines an arithmetic function or number-theoretic function: a function from the natural numbers to a ring (usually, the ring of integers, rational numbers, real numbers, or complex numbers).
View a complete list of arithmetic functions
## Definition
### Hands-on definition
The von Mangoldt function, denoted , is an arithmetic function defined as follows:
• .
• , for a prime and a positive integer.
• if has more than one prime divisor.
### Definition in terms of Dirichlet product
The von Mangoldt function is the unique function such that:
where is the logarithm, is the all ones function, and denotes Dirichlet product.
By the Mobius inversion formula, this is equivalent to defining:
,
where is the Mobius function.
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2021-02-26 04:39:45
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https://stoneswww.academickids.com/encyclopedia/index.php/Hypergeometric_series
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# Hypergeometric series
In mathematics, a hypergeometric series is the sum of a sequence of terms in which the ratios of successive coefficients k is a rational function of k. The series, if convergent, will define a hypergeometric function which may then be defined over a wider domain of the argument by analytic continuation. The hypergeometric series is generally written:
[itex]\,_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;x)=\sum_{k=0}^\infty c_k x^k[itex]
where [itex]c_0[itex]=1 and
[itex]\frac{c_{k+1}}{c_k}=\frac{(k+a_1)(k+a_2)\cdots(k+a_p)}{(k+b_1)(k+b_2)\cdots(k+b_q)}\,\frac{1}{k+1}[itex]
The series may also be written:
[itex]\,_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;x)=\sum_{k=0}^\infty
\frac{(a_1)_k(a_2)_k\ldots(a_p)_k}{(b_1)_k(b_1)_k\ldots(b_q)_k}\,\frac{x^k}{k!}[itex]
where [itex](a)_k=a(a+1)\ldots(a+k-1)[itex] is the rising factorial or Pochhammer symbol
Contents
## Introduction
A hypergeometric series could in principle be any formal power series in which the ratio of successive coefficients [itex]\alpha_n/\alpha_{n-1}[itex] is a rational function of n. That is,
[itex]\frac{\alpha_n}{\alpha_{n-1}} = \frac{\tilde P(n)}{\tilde Q(n)}[itex]
for some polynomials [itex]\tilde P(n)[itex] and [itex]\tilde Q(n)[itex]. Thus, for example, in the case of a geometric series, this ratio is a constant. Another example is the series for the exponential function, for which
[itex]\alpha_n/\alpha_{n-1}=z/n\,.[itex]
In practice the series is written as an exponential generating function, modifying the coefficients so that a general term of the series takes the form
[itex]\alpha_n = \beta_n z^n /n!\,[itex]
and [itex]\beta_0=1[itex]. Notice that z corresponds to a constant in the ratio [itex]\tilde P/\tilde Q[itex]. One uses the exponential function as a 'baseline' for discussion.
Many interesting series in mathematics have the property that the ratio of successive terms is a rational function. However, when expressed as an exponential generating function, such series have a non-zero radius of convergence only under restricted conditions. Thus, by convention, the use of the term hypergeometric series is usually restricted to the case where the series defines an actual analytic function with a non-zero radius of convergence. Such a function, and its analytic continuations, is called the hypergeometric function.
Convergence conditions were given by Carl Friedrich Gauss, who examined the case of
[itex]\frac{\beta_n}{\beta_{n-1}} = \frac{(n+a)(n+b)}{(n+c)}[itex],
leading to the classical standard hypergeometric series
[itex]\,_2F_1(a,b,c;z).[itex]
## Notation
The standard notation for the general hypergeometric series is
[itex]\,_mF_p.[itex]
Here, the integers m and p refer to the degree of the polynomials P and Q, respectively, referring to the ratio
[itex]\frac{\beta_n}{\beta_{n-1}} = \frac{P(n)}{Q(n)}.[itex]
If m>p+1, the radius of convergence is zero and so there is no analytic function. The series naturally terminates in case P(n) is ever 0 for n a natural number. If Q(n) were ever zero, the coefficients would be undefined.
The full notation for F assumes that P and Q are monic and factorised, so that the notation for F includes an m-tuple that is the list of the zeroes of P and a p-tuple of the zeroes of Q. Note that this is not much restriction: the fundamental theorem of algebra applies, and we can also absorb a leading coefficient of P or Q by redefining z. As a result of the factorisation, a general term in the series then takes the form of a ratio of products of Pochhammer symbols. Since Pochhammer notation for rising factorials is traditional it is neater to write F with the negatives of the zeros. Thus, to complete the notational example, one has
[itex] \,_2F_1 (a,b;c;z) = \sum_{n=0}^\infty
\frac{(a)_n(b)_n}{(c)_n} \, \frac {z^n} {n!} [itex] where [itex](a)_n=a(a+1)(a+2)...(a+n-1)[itex] is the rising factorial or Pochhammer symbol. Here, the zeros of P were −a and −b, while the zero of Q was −c.
## Special cases and applications
The classic orthogonal polynomials can all be expressed as special cases of [itex]{\;}_2F_1[itex] with one or both a and b being (negative) integers. Similarly, the Legendre functions are a special case as well.
Applications of hypergeometric series includes the inversion of elliptic integrals; these are constructed by taking the ratio of the two linearly independent solutions of the hypergeometric differential equation to form Schwarz-Christoffel maps of the fundamental domain to the complex projective line or Riemann sphere.
The Kummer function 1F1(a,b;x) is known as the confluent hypergeometric function.
The function 2F1 has several integral representations, including the Euler hypergeometric integral.
## History and generalizations
Studies in the nineteenth century included those of Ernst Kummer, and the fundamental characterisation by Bernhard Riemann of the F-function by means of the differential equation it satisfies. Riemann showed that the second-order differential equation (in z) for F, examined in the complex plane, could be characterised (on the Riemann sphere) by its three regular singularities: that effectively the entire algorithmic side of the theory was a consequence of basic facts and the use of Möbius transformations as a symmetry group.
Subsequently the hypergeometric series were generalised to several variables, for example by Appell; but a comparable general theory took long to emerge. Many identities were found, some quite remarkable. What are called q-series analogues were found. During the twentieth century this was a fruitful area of combinatorial mathematics, with numerous connections to other fields. There are a number of new definitions of hypergeometric series, by Aomoto, Gel'fand and others; and applications for example to the combinatorics of arranging a number of hyperplanes in complex N-space (see arrangement of hyperplanes).
Hypergeometric series can be developed on Riemannian symmetric spaces and semi-simple Lie groups. Their importance and role can be understood through a special case: the hypergeometric series 2F1 is closely related to the Legendre polynomials, and when used in the form of spherical harmonics, it expresses, in a certain sense, the symmetry properties of the two-sphere or equivalently the rotations given by the Lie group SO(3). Concrete representations are analogous to the Clebsch-Gordan coefficients.
## References
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy
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2021-06-20 00:01:19
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https://projecteuclid.org/euclid.ade/1366896316
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On the solutions to some elliptic equations with nonlinear Neumann boundary conditions
Abstract
We describe all nontrivial nonnegative solutions to the problem $$\begin{cases} -\Delta u = a u^{{n+2}\over{n-2}} \quad &\hbox{in }\ H, \cr {{\partial u }\over {\partial \nu }} = bu^{{n}\over{n-2}} &\hbox{on } \ \partial H, \end{cases}$$ where $H$ is the half space of $\mathbb{R}^n (n\ge3)$.
Article information
Source
Adv. Differential Equations Volume 1, Number 1 (1996), 91-110.
Dates
First available in Project Euclid: 25 April 2013
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2018-01-21 04:45:19
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http://mathhelpforum.com/geometry/219982-if-cube-has-volume-1000-then-what-smallest-sphere-could-circu.html
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# Thread: If a cube has the volume of 1000 then what is the smallest sphere that could be circu
1. ## If a cube has the volume of 1000 then what is the smallest sphere that could be circu
If a cube has the volume of 1000 then what is the smallest sphere that could be circumscribed around it?
Is it 500√2 or 500√3
how did you do it
2. ## Re: If a cube has the volume of 1000 then what is the smallest sphere that could be c
Are you looking for the radius of the smallest spehere? First determine the length of each side of the cube. Then the spehere will have a radius that reaches from the center of the cube to each corner, so use Pythagorus to determine that length. By the way, neither of your answers are correct - perhaps the volume isn't 1000 but rather 1,000,000,000?
3. ## Re: If a cube has the volume of 1000 then what is the smallest sphere that could be c
Equivalently, the length of a diagonal of a cube of side s has length $s\sqrt{3}$ and that is the diameter of the circumscribing sphere.
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2016-08-29 06:12:39
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http://talkstats.com/threads/im-not-sure-how-to-use-egen-sum-by-rows.25669/
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# I'm not sure how to use "egen sum by rows"
#### jfca283
##### Member
Hello
i have the following problem. I need to sum by rows. For example:
Code:
v1 v2 v3
1 1 1
2 5 12
2 6 12
2 1 12
3 2 9
3 7 9
1 2 21
1 9 21
1 8 21
1 2 21
So, i need to sum variable v2 by grouping variable 1 and therefore i obtain v3.
I ran egen v3=sum(v2), by(v1), but stata summed all rows from the database. What i did wrong? Thanks for your time and help.
#### bukharin
##### RoboStataRaptor
Generally speaking you should use -egen, total- rather than -egen, sum- which is not documented. The reason that -egen, sum- is not documented is to prevent confusion with the function sum() which generates a running sum.
Why shouldn't v3 be 22 for v1==1? Does the order of v1 matter? If not, just use -egen, total- and you should get the correct result. If this doesn't work, you need to post an example (like the above) with your output.
If the order does matter then you could create a new grouping variable prior to running -egen-:
Code:
gen group=sum(v1!=v1[_n-1])
egen v3=total(v2), by(group)
list, clean noobs
v1 v2 group v3
1 1 1 1
2 5 2 12
2 6 2 12
2 1 2 12
3 2 3 9
3 7 3 9
1 2 4 21
1 9 4 21
1 8 4 21
1 2 4 21
This is a "trick" in the sense that it's a running sum of the true/false (1/0) comparison of v1 with the preceding v1 - so it goes up by 1 whenever v1 changes, creating your groups.
#### jfca283
##### Member
It worked flawless. That's the reply one is looking for! Thanks a lot.
#### jfca283
##### Member
Now it emerged another problem. Here is my data
Code:
v1 v2 v3
1 150 150
3 137 350
3 103 350
3 110 350
3 117 364
3 127 364
3 110 364
2 139 289
2 150 289
Now i need to obtain v3, which is the sum of v2 according to the rows indicated by v1. Look when v1 is 3. I must count from that row 2 rows down and sum the total. If appears another 3, that's other sum, like i did in the example posted above. i don't have any variable to differentiate than going down arrays and jump depending on the first value found in the row. Did i myself clear? Again, thanks for your time and patience.
#### bukharin
##### RoboStataRaptor
That's a bit trickier. Here's one solution (not very well tested):
Code:
clear
input v1 v2
1 150
3 137
3 103
3 110
3 117
3 127
3 110
2 139
2 150
end
gen n=_n
gen group1=sum(v1!=v1[_n-1])
bysort group1 (n): gen byte group2=mod(_n-1, v1[1])==0
gen group3=sum(group2)
egen v3=sum(v2), by(group3)
Result:
Code:
v1 v2 n group1 group2 group3 v3
1 150 1 1 1 1 150
3 137 2 2 1 2 350
3 103 3 2 0 2 350
3 110 4 2 0 2 350
3 117 5 2 1 3 354
3 127 6 2 0 3 354
3 110 7 2 0 3 354
2 139 8 3 1 4 289
2 150 9 3 0 4 289
Last edited:
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2020-07-15 19:07:17
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https://brilliant.org/problems/fatorial-4/
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# Fatorial #4
Algebra Level 1
If ($$\frac{15! + 14! + 13!}{14! + 13!}$$ = a) and ((5b - 7)! = 1),
find the sum of all possible values of "a", and add it to the sum of all possible values of "b".
×
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2018-06-19 08:32:51
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https://academiccommons.columbia.edu/doi/10.7916/D8ZG6SGM
|
Theses Doctoral
# Essays on Approximation Algorithms for Robust Linear Optimization Problems
Lu, Brian Yin
Solving optimization problems under uncertainty has been an important topic since the appearance of mathematical optimization in the mid 19th century. George Dantzig’s 1955 paper, “Linear Programming under Uncertainty” is considered one of the ten most influential papers in Management Science [26]. The methodology introduced in Dantzig’s paper is named stochastic programming, since it assumes an underlying probability distribution of the uncertain input parameters. However, stochastic programming suffers from the “curse of dimensionality”, and knowing the exact distribution of the input parameter may not be realistic. On the other hand, robust optimization models the uncertainty using a deterministic uncertainty set. The goal is to optimize the worst-case scenario from the uncertainty set. In recent years, many studies in robust optimization have been conducted and we refer the reader to Ben-Tal and Nemirovski [4–6], El Ghaoui and Lebret [19], Bertsimas and
Sim [15, 16], Goldfarb and Iyengar [23], Bertsimas et al. [8] for a review of robust optimization. Computing an optimal adjustable (or dynamic) solution to a robust optimization problem is generally hard. This motivates us to study the hardness of approximation of the problem and provide efficient approximation algorithms. In this dissertation, we consider adjustable robust linear optimization problems with packing and covering formulations and their approximation algorithms. In particular, we study the performances of static solution and affine solution as approximations for the adjustable robust problem.
Chapter 2 and 3 consider two-stage adjustable robust linear packing problem with uncertain second-stage constraint coefficients. For general convex, compact and down-monotone uncertainty sets, the problem is often intractable since it requires to compute a solution for all possible realizations of uncertain parameters [22]. In particular, for a fairly general class of uncertainty sets, we show that the two-stage adjustable robust problem is NP-hard to approximate within a factor that is better than Ω(logn), where n is the number of columns of the uncertain coefficient matrix. On the other hand, a static solution is a single (here and now) solution that is feasible for all possible realizations of the uncertain parameters and can be computed efficiently. We study the performance of static solutions an approximation for the adjustable robust problem and relate its optimality to a transformation of the uncertain set. With this transformation, we show that for a fairly general class of uncertainty sets, static solution is optimal for the adjustable robust problem. This is surprising since the static solution is widely perceived as highly conservative. Moreover, when the static solution is not optimal, we provide an instance-based tight approximation bound that is related to a measure of non-convexity of the transformation of the uncertain set. We also show that for two-stage problems, our bound is at least as good (and in many case significantly better) as the bound given by the symmetry of the uncertainty set [11, 12]. Moreover, our results can be generalized to the case where the objective coefficients and right-hand-side are also uncertainty.
In Chapter 3, we focus on the two-stage problems with a family of column-wise and constraint-wise uncertainty sets where any constraint describing the set involves entries of only a single column or a single row. This is a fairly general class of uncertainty sets to model constraint coefficient uncertainty. Moreover, it is the family of uncertainty sets that gives the previous hardness result. On the positive side, we show that a static solution is an
O(\log n · min(\log \Gamma, \log(m+n))-approximation for the two-stage adjustable robust problem where m and n denote the numbers of rows and columns of the constraint matrix and \Gamma is the maximum possible ratio of upper bounds of the uncertain constraint coefficients. Therefore, for constant \Gamma, surprisingly the performance bound for static solutions matches
the hardness of approximation for the adjustable problem. Furthermore, in general the static solution provides nearly the best efficient approximation for the two-stage adjustable robust problem.
In Chapter 4, we extend our result in Chapter 2 to a multi-stage adjustable robust linear optimization problem. In particular, we consider the case where the choice of the uncertain constraint coefficient matrix for each stage is independent of the others. In real world applications, decision problems are often of multiple stages and a iterative implementation of two-stage solution may result in a suboptimal solution for multi-stage problem. We consider the static solution for the adjustable robust problem and the transformation of the uncertainty sets introduced in Chapter 2. We show that the static solution is optimal for the adjustable robust problem when the transformation of the uncertainty set for each stage is convex.
Chapters 5 considers a two-stage adjustable robust linear covering problem with uncertain right-hand-side parameter. As mentioned earlier, such problems are often intractable due to astronomically many extreme points of the uncertainty set. We introduce a new approximation framework where we consider a “simple” set that is “close” to the original uncertainty set. Moreover, the adjustable robust problem can be solved efficiently over the extended set. We show that the approximation bound is related to a geometric factor that represents the Banach-Mazur distance between the two sets. Using this framework, we provide approximation bounds that are better than the bounds given by an affine policy in [7] for a large class of interesting uncertainty sets. For instance, we provide an approximation solution that gives a m^{1/4}-approximation for the two-stage adjustable robust problem with hypersphere uncertainty set, while the affine policy has an approximation ratio of O(\sqrt{m}).
Moreover, our bound for general p-norm ball is m^{\frac{p-1}{p^2}} as opposed to m^{1/p} as given by an affine policy.
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2022-08-08 06:50:08
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https://codegolf.stackexchange.com/questions/109043/count-my-change/109054
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# Count My Change
Your task is to sort an array containing the strings "quarter", "dime", "nickel", and "penny" any number of times in no specific order and sort them so that they are in this order: quarter dime nickel penny (in other words, greatest to least monetary value).
## Rules
1. Your program must take an array as input containing the names of U.S coins and sort them from greatest to least by monetary value.
• For those who are not from the U.S or don't use change, the values of U.S coins, from greatest to least, are:
• Quarter: 25 cents
• Dime: 10 cents
• Nickel: 5 cents
• Penny: 1 cent
2. You may sort this array in any way you wish, as long as the output is ordered by the monetary values shown above.
3. Input can be taken in any way, be it command-line arguments or STDIN.
4. An input array would be all lowercase strings, something like this:
• quarter dime nickel nickel quarter dime penny penny
5. The actual format of input and output is up to you.
## Test Cases
"penny nickel dime quarter"
-> "quarter dime nickel penny"
"nickel penny penny quarter quarter quarter dime dime dime dime"
-> "quarter quarter quarter dime dime dime dime nickel penny penny"
"quarter dime nickel nickel quarter dime penny penny"
-> "quarter quarter dime dime nickel nickel penny penny"
This is , so standard rules & loopholes apply.
• – ckjbgames Feb 3 '17 at 19:24
• All test cases should include output. In the mobile the second and third cases are shown in two lines, so it looks as if the second line is the output – Luis Mendo Feb 3 '17 at 19:33
• i'm canadian, can i assume the input has no pennies? ;) – undergroundmonorail Feb 3 '17 at 21:48
• @undergroundmonorail Sadly, no. – ckjbgames Feb 5 '17 at 13:46
• What happened to loonies and half-dollars? – Adám Feb 6 '17 at 6:29
# Japt, 5 3 bytes
ñg9
Test it online!
### Explanation
I, too, have added a sorting function to my language in the last few weeks :-) ñ takes in an array and a function and sorts the array as if each item had been mapped through that function.
The g function on a string takes in a number n and returns the nth char in the string, wrapping if n is negative or past the end of the string. The strings can thus be aligned as follows:
quarterquarter...
dimedimedimedi...
nickelnickelni...
pennypennypenn...
The 9th char (0-indexed) of each string has been highlighted in bold. These are in the correct order, so all we have to do is ñg9. (Though now that I look back on it, ñg5 would work as well...)
• It should also work with 5, i think. – FlipTack Feb 3 '17 at 19:08
• @FlipTack Yeah, I just noticed that myself. Not that it makes a difference ;-) – ETHproductions Feb 3 '17 at 19:09
• This. Cannot. Be. Defeated. – ckjbgames Mar 19 '17 at 21:28
• @ckjbgames Dennis didn't post any answers ;-) – ETHproductions Apr 26 '17 at 16:55
• @ETHproductions He probably will. Just show him this question. – ckjbgames Apr 26 '17 at 17:03
# V, 7 bytes
ú/¨qu©¿
Try it online!
This uses the spiffy new sort command I added to V around a week ago (ú). Sweet timing!
The way this works is by sorting every line by default sorting (ASCII values) but ignoring the first match of a certain regex. In this case, the regex is (qu)?, although it has some gross non-ASCII stuff to avoid using backslashes. If you ignore the first two letters of "quarter", it starts with 'a', and then all of the coins are already in alphabetical order.
# Non-competing version, 4 bytes
ú!/.
This feature was already implemented, but I hadn't tested it extensively yet so it had a bug that I only realized because of this challenge. There is no TIO link because TIO is slightly behind.
This works by reverse sorting every line but ignoring the first character on each line.
# Python, 36 bytes
lambda a:a.sort(key=lambda x:x[-5:])
Unnamed function that sorts the list in-place by the key function given.
The slices of each coin name are then, arter, dime, ickel, and penny - which are in alphabetical (or more importantly, ordinal) order.
• Oh oops - if I don't get the el the wrong way around I miss the c :p – Jonathan Allan Feb 3 '17 at 19:52
# Bash + coreutils, 18
tr q b|sort|tr b q
• Transliterate q to b
• Sort
• Transliterate b back to q
# Python 3, 4241 38 bytes
An unnamed lambda function which takes input as a list of strings, sorts in place.
(Outgolfed by Jonathan Allan)
lambda x:x.sort(key=lambda s:(s*2)[5])
Try it online!
Other solutions I messed around with:
lambda x:x.sort(key=lambda s:s*(s<'q'))
lambda x:x.sort(key=lambda s:(s+'pi')[5])
lambda x:x.sort(key=lambda s:ord(s[3])%16)
# PowerShell, 21 bytes
$args|sort{($_*3)[9]}
Try it online!
### Explanation
Shamelessly stole the algorithm in ETHproductions's answer (basically). I multiply each string by 3, then sort based on the 9th character of the resulting string.
• What is $_ in PowerShell? – ckjbgames Feb 3 '17 at 19:15 • @ckjbgames In a pipeline, within a scriptblock, it refers to the current item. So something like 1,2,3,4 | ForEach-Object {$_*2 } will output each number times 2; the script block is run once per input item. – briantist Feb 3 '17 at 19:18
• That makes sense. – ckjbgames Feb 3 '17 at 19:31
# Jelly, 4 bytes
6ịµÞ
Try it online! (the footer, ÇY, joins the resulting list with line feeds for a prettier print out.)
### How?
6ịµÞ - Main link: list of strings
Þ - sort the list of strings by the monadic function:
6ị - the sixth index - Jelly indexing is modular and 1-based
The Nth index of a list in Jelly is the Nth item starting at the left, counting from 1, and looping back to the start when need be. (The 0th is at the right, the -1th one left of that and so on too).
So the sixth character of ['d','i','m','e'] is 'i' since six is congruent to two modulo four.
The sixth character of the four coins in order are quarter, dime, nickel, penny . These are in alphabetical (or more importantly, ordinal) order.
Another way to achieve the same thing would be to sort by the rotated strings with ṙ5µÞ, where ṙ rotates to the right, making the strings erquart, imed, lnicke, and penny.
# Python, 32 bytes
lambda s:s.sort(key="npr".strip)
Try it online! Sorts the list in place.
The idea is to use a sorting key function without a lambda. A good candidate was x.strip, which takes the string x and removes its the left and right edges all characters in the input. For example, "abcdef".strip("faces") == "bcd".
The method "npr".strip takes:
quarter -> np
dime -> npr
nickel -> pr
penny -> r
which are lexicographically sorted. I found the string npr by brute force. npu and npt also work, and there are none shorter.
# Bash (+coreutils) 11 bytes
Golfed
sort -rk1.2
How It Works
Reverse sort, with the "sort key" from the second character of the first field (word) till the end of line, i.e.:
uarter
ime
ickel
enny
Test
>echo penny nickel dime quarter|tr ' ' '\n'|sort -rk1.2
quarter
dime
nickel
penny
Try It Online !
# CJam, 8 bytes
q~{5=}$p Try it online! Explanation q~ Get and eval all input (array of coin names) {5=}$ Sort the array by the 5th index of each element (array indices loop in CJam)
p Print the result
## Pyke, 97 5 bytes
.#5R@
Try it here!
.#5R@ - sort_by:
5R@ - i[5]
# Retina, 10
• 6 10 bytes saved thanks to @ETHproductions
q
b
O
b
q
• Substitute q to b
• Sort
• Substitute b back to q
• @ETHproductions Great - thanks! – Digital Trauma Feb 3 '17 at 19:57
# V, 8 7 bytes
1 byte saved thanks to @DJMcMayhem
Úçq/:m0
[Try it online!]
See @DJMcMayhem's answer in V (1 0 bytes shorter than mine)
Try it online!
Ú " sort all lines "
ç " globally "
q/ " where there a q is matched, "
:m0 " move it to the top of the buffer "
Here is an older solution at 1 byte larger, but I really like it.
### V, 8 bytes
Ú/q
dGHP
[Try it online!]
Try it online!
### Explanation
Ú " sorts the lines
Now the buffer will be in this format:
dimes
nickels
pennies
quarters
The only thing left to do now is to move the quarters to the top.
/q " search for a q "
dG " delete everything from the first quarter to the end of buffer "
HP " and paste it at the top of the buffer
• You can do :m0 on your alternate solution to save a byte (and tie me) Úçq/:m0 – DJMcMayhem Feb 3 '17 at 21:38
• @DJMcMayhem Thanks, TIL about :move – Kritixi Lithos Feb 4 '17 at 8:30
# Japt, 3 bytes
ñé5
Try it online!
## A couple other 3-byte solutions:
üÅw
Try it online!
ü é
Try it onlin!
# T-SQL, 4136 34 bytes
select*from @ order by right(a,5)
### Explanation
Assume the input is pre-loaded in a table variable named @, with a single column named a, where each value is one coin to be sorted.
The select * from @ part is boiler-plate 'get all values to return'. The real magic happens in the order by clause.
Using the same strategy as Johnathan Allan, I sort by the last five characters (SQL will return the entire string if it's too short): arter, dime, ickel, penny.
• q is the next letter after p, so for a simple mod to result in q less than p the value needs to be a factor of q, which is prime. You could subtract 1 first and then a modulus of 7 would work, but that would presumably take at least as many bytes as 113. – Neil Feb 3 '17 at 19:46
• @Neil Yeah, I realized 113 being prime was wrecking my attempts to reduce the count. Doing -1 and then mod 7 is more bytes (including required parentheses. – Brian J Feb 3 '17 at 19:50
## JavaScript (ES6), 35 33 bytes
a=>a.sort(([,...a],[,...b])=>b>a)
### Test cases
let f =
a=>a.sort(([,...a],[,...b])=>b>a)
console.log(f(['penny','nickel','dime','quarter']).join );
console.log(f(['nickel','penny','penny','quarter','quarter','quarter','dime','dime','dime','dime']).join );
console.log(f(['quarter','dime','nickel','nickel','quarter','dime','penny','penny']).join );
# Befunge, 158 bytes
~:0v0v4<~_
9:%8_^>8*^1p9\+1g
$:!#@_1-0" ynnep">:#,_> 1-0" lekcin">:#,_>$:!#^_
" emid">:#,_>$:!#^_1-0 >:#,_$1>-:!#^_0" retrauq"
*84g9< ^*84g91-*84g94-*84g96-
Try it online!
String processing and sorting are not the sorts of things you'd typically want to attempt in Befunge, but this solution is taking advantage of John Kasunich's observation that we don't actually need to sort anything. We just count the number of occurrences of each coin (which can easily be determined from the first character), and then output that many of each coin name in the appropriate order.
It's still not at all competitive with other languages in terms of size, but this approach is at least better than it would have been if we'd tried to handle the challenge as a string sorting exercise.
# Pyth, 3 bytes
@D5
Demonstration
Based on ETHproductions's answer in Japt.
Explanation:
@D5
@D5Q Variable introduction
D Q Sort the input by
@ 5 Its fifth character
# APL (Dyalog APL), 11 bytes
Takes and returns list of strings.
{⍵[⍋↑5⌽¨⍵]}
Try it online!
{ anonymous function:
⍵[] the argument indexed by
⍋ the ascending-making indices of
↑ the matrix whose rows are the padded
5⌽ five-steps-rotated
¨⍵ items of the argument
}
# Brachylog, 3 bytes
↺₉ᵒ
Try it online!
Approach stolen from ETHproductions' Japt answer.
The input
ᵒ sorted by
↺ its elements rotated left
₉ by 9
is the output.
# Husk, 3 bytes
Öṙ9
Try it online!
Ported from my Brachylog answer, which rips off ETHproductions' Japt answer but isn't an exact translation since it uses rotation instead of simple accessing-the-nth-element. There, I did it because ∋ doesn't let you index past the end of the input (which is probably quite helpful in many circumstances due to the declarative nature of the language). In Husk, ! does let you index past the end of the input, with the same modular wrapping around that Japt ñg9 uses, but it's from 1 so this program in particular would end up being one byte longer: Ö!10.
Ö Sort by
ṙ rotation by
9 9.
## Batch, 82 bytes
@for %%c in (quarter dime nickel penny)do @for %%a in (%*)do @if %%a==%%c echo %%c
Takes input as command-line arguments and outputs to STDOUT. Works by concatenating the lists resulting from filtering the original list on each coin.
# Ruby, 34 bytes
->m{m.sort_by{|c|c[1..2]}.reverse}
input and output as an array of strings
## Ruby, 31 bytes
->s{s.sort_by{|i|i[1]}.reverse}
• This will not always sort "nickel" and "dime" correctly. – daniero Feb 5 '17 at 1:10
# Ruby, 30 bytes
->m{m.sort_by{|s|s[3].ord^16}}
Magical numbers found by trial and error. A bit clumsy, but shorter than using .reverse.
# Perl 6, 40 36 34 bytes
*.sort: {(<q d n p>Zxx 1..*).Bag{~m/./}}
Try it
*.sort: {%(<q d n p>Z=>1..*){~m/./}}
Try it
*.sort: {%(<q d n p>Z=>^4){~m/./}}
Try it
## Expanded:
*\ # WhateverCode lambda ( this is the parameter )
.sort: # sort using the following code block
{ # bare block lambda with implicit parameter ｢$_｣ %( # treat this list as a hash <q d n p> # list of first characters Z[=>] # zipped using pair constructor ^4 # Range from 0 up-to 4 ( excludes 4 ) ){ # get the value associated with this key ~m/./ # regex which gets the first character ( implicitly from ｢$_｣ )
}
}
# Mathematica, 50 bytes
Flatten@{Last@#,Most@#}&@Split@Sort@StringSplit@#&
# RProgN, 18 bytes
~{3mtt¢\3mtt¢>}$ ## Explained ~ # Zero Space Segment { } # Anonymous Function 3m # Repeat the inputted string 3 times tt¢ # And take the tenth character of that \3mtt¢ # Do the same for the entry underneith > # Compare the ascii value of the two$# Sort the input by the anonymous function.
Try it online!
# java 8, 128 112 bytes
This is a lambda expression for a java.util.function.Function<String[],String[]>
s->{String r="";for(String k:"q d c p".split(" "))for(String t:s)if(t.contains(k))r+=" "+t;return r.split(" ");}
Explantion: For each of the 4 coins in order, go through the input and append the coin's name to the result every time there is a match for that coin's unique character. Split the result into an array and return it.
## Ruby, 27 bytes
->s{s.sort_by{|x|(x*2)[5]}}
`
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2019-11-14 01:14:28
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http://openstudy.com/updates/5582ef22e4b091b59af0bab4
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## TrojanPoem one year ago Find the limit of : ( No L'hopital's rule)
1. TrojanPoem
$\lim_{x \rightarrow \frac{ \pi }{ 6 }} \frac{ 2\sin x -1 }{ \sqrt{3} - 2\cos x }$
2. TrojanPoem
Approaches ( pi/ 6 )
3. phi
I would use L'hopital's rule
4. TrojanPoem
Out of my curriculum.
5. phi
one trick that often works is to multiply by the conjugate of the denominator the idea is (a+b)(a-b) = a^2 - b^2
6. TrojanPoem
But what made us think of doing so. To me It's clear when there is square roots.
7. TrojanPoem
I even never so example when they multiply by the conjugate with sin , cos , tan
8. TrojanPoem
i have never even seen*
9. phi
I agree it is not an obvious tactic.
10. TrojanPoem
After talking to my teacher , he told me that It was included in exam in 1998 and showed me it , he even saw the answer and wasn't obvious to him.
11. TrojanPoem
was confused by it*
12. TrojanPoem
By the way , thanks phi.
13. TrojanPoem
If you ever though of trying to it and found alternative solution , Pm me !
14. phi
multiplying top and bottom by (2 sin x + 1)( sqr(3) + 2 cos x) doesn't seem to help (as far as I can see)
15. TrojanPoem
Give me a second , I will see the solution written in my notebook.
16. TrojanPoem
(4sin^2x - 1) (sqrt(3) + 2cosx) / (3 - 4 cos ^2 x) (2 sinx +1 )
17. TrojanPoem
(4( 1- cos^2x ) - 1 ) ( sqrt(3) + 2 cosx) / (3-4cos^2x) (2sinx + 1)
18. TrojanPoem
(4 - 4cos^2x - 1)(sqrt(3) + 2 cosx) / (3-4cos^2x) (2 sinx + 1)
19. TrojanPoem
we have 4 - 1 - 4cos^2x which is 3- 4cos^2x ( down)
20. TrojanPoem
so we are left with sqrt(3) + 2cosx / 2sinx + 1
21. TrojanPoem
sqrt3( + 2 cos 39 . 2 sin30 + 1 = sqrt(3)
22. phi
ok it does work, but it's definitely a case of "lucking out" we get $\frac{(4 \sin^2 -1)(\sqrt{3 }+ 2 \cos x)}{(3-4 \cos^2 x)(2 \sin x +1)}$ we can separate that into the limit of the product of two terms. the second term as a limit of sqr(3) $\lim_{x \rightarrow \pi/6} \frac{(\sqrt{3 }+ 2 \cos x)}{(2 \sin x +1)} = \sqrt{3}$ the other term needs some work, as it looks like 0/0 but notice $3 - 4 \cos^2 x= 3-4(1-\sin^2 x) \\ = 3 -4 +4 \sin^2 x\\ = 4 \sin^2 x -1$ and (magically) we have $\frac{ 4 \sin^2 x -1}{ 4 \sin^2 x -1}$ which has a limit of 1
23. TrojanPoem
xD
24. TrojanPoem
but how did the creator of it think of doing that ?
25. phi
In this case it is a lucky trick. You try things (or work with these trig expressions so much you become very good at seeing the way to the end).
26. TrojanPoem
I tried every trig formula the end was similar each time a more complicated expression. I am stuck with another limit , may you help ? I will tag you
27. phi
ok
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2016-10-29 00:18:21
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https://mathematica.stackexchange.com/questions/239361/periodicinterpolation-does-not-work-in-elementmeshinterpolation
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# PeriodicInterpolation does not work in ElementMeshInterpolation
I want to use ElementMeshInterpolation to generate interpolation function with periodic boundary condition.
I use below data as an example
data=Flatten[Table[{i,j,Sin[i+j]},{i,0,2\[Pi],2\[Pi]/50},{j,0,2\[Pi],2\[Pi]/50}],1];
ListContourPlot[data]
which gives
This data is periodic along x and y direction.
Using Interpolation
f = Interpolation[data, PeriodicInterpolation -> True];
{ContourPlot[f[x, y], {x, 0, 2 \[Pi]}, {y, 0, 2 \[Pi]}],
ContourPlot[f[x, y], {x, 0, 4 \[Pi]}, {y, 0, 4 \[Pi]}]}
gives
We can see the Interpolation function is fine with periodic condition as wanted.
using ElementMeshInterpolation
Though Interpolation works fine for this data set. But Interpolation has problem that it frequently run into "femimq" problem. So ElementMeshInterpolation on a refined mesh is necessary sometimes.
mesh = ToElementMesh[data[[;; , 1 ;; 2]]];
f = ElementMeshInterpolation[{mesh}, data[[;; , -1]],
PeriodicInterpolation -> {True, True}];
{ContourPlot[f[x, y], {x, 0, 2 \[Pi]}, {y, 0, 2 \[Pi]}],
ContourPlot[f[x, y], {x, 0, 4 \[Pi]}, {y, 0, 4 \[Pi]}]}
this gives
You see the generated Interpolation function has no periodicity.
using ListInterpolation
mesh can also be used in ListInterpolation
mesh = ToElementMesh[data[[;; , 1 ;; 2]]];
f = ListInterpolation[data[[;; , -1]], mesh,
PeriodicInterpolation -> {True, True}];
{ContourPlot[f[x, y], {x, 0, 2 \[Pi]}, {y, 0, 2 \[Pi]}],
ContourPlot[f[x, y], {x, 0, 4 \[Pi]}, {y, 0, 4 \[Pi]}]}
but this gives the same result as ElementMeshInterpolation.
So the question is how to correctly make periodic interpolation function using ElementMeshInterpolation.
You can not really. The fact that Interpolation can do this hinges on the data being structured. In other works what I am going to show next is not easily generally possible for meshes that represent a non rectangullar domain; which is the common case for FEM meshes.
You can hack it by using the ExtrapolationHandler option.
Needs["NDSolveFEM"]
data = Flatten[
Table[{i, j, Sin[i + j]}, {i, 0, 2 \[Pi], 2 \[Pi]/50}, {j, 0,
2 \[Pi], 2 \[Pi]/50}], 1];
mesh = ToElementMesh[data[[;; , 1 ;; 2]]];
f = ElementMeshInterpolation[{mesh}, data[[;; , -1]]];
Now, we can use f as a function in the extrapolation handler and map the coordinates outside the domain back onto f. This mapping back to the original domain is tricky to do generally. Here we use Mod.
f2 = ElementMeshInterpolation[{mesh}, data[[;; , -1]],
"ExtrapolationHandler" -> {Function[{x, y},
f[Mod[x, 2 \[Pi]], Mod[y, 2 \[Pi]]]]}]
{ContourPlot[f2[x, y], {x, 0, 2 \[Pi]}, {y, 0, 2 \[Pi]}],
ContourPlot[f2[x, y], {x, 0, 4 \[Pi]}, {y, 0, 4 \[Pi]}]}
If you want to switch of the warning message you can do so with:
f2 = ElementMeshInterpolation[{mesh}, data[[;; , -1]],
"ExtrapolationHandler" -> {Function[{x, y},
f[Mod[x, 2 \[Pi]], Mod[y, 2 \[Pi]]]], "WarningMessage" -> False}]
Perhaps an idea for a future implementation.
• Thank you so much. I am interested in the overhead of "ExtrapolationHandler". I found a peculiar thing. The ContourPlot of f2 is even faster than f. However, Table[Quiet@f2[x,y],{x,100.,150.,1},{y,100.,150.,1}];//AbsoluteTiming is much slower than Table[f[x,y],{x,100.,150.,1},{y,100.,150.,1}];//AbsoluteTiming. Why is that? – matheorem Feb 4 at 8:22
• And what do you mean by "Interpolation can do this hinges on the data being structured." – matheorem Feb 4 at 8:23
• @matheorem, for your second question see update – user21 Feb 4 at 8:29
• @matheorem, here is a guess, If you are querying f outside it will have to generate an extrapolation value and it has to generate the Message. That might be the cost. If f2 is queried outside then you have the expense of the finding that the value is outside the domain plus the evaluation of f to get the value. – user21 Feb 4 at 8:32
• @matheorem, I have added this example to the documentation. I hope you do not mind. – user21 Feb 4 at 8:33
Since currently ElementMeshInterpolation does not support PeriodicInterpolation and Interpolation only support PeriodicInterpolation on rectangular grid. Apart from user21's workaround, I developed a workaround for arbitrary parallel or parallelipiped grid periodic interpolation.
The idea is naive, just to pull back points outside region by base vectors. Below is helper function.
pullBack2Dcom=Compile[{x1,x2,y1,y2,x,y},Mod[{(-x2 y+x y2)/(-x2 y1+x1 y2),(x1 y-x y1)/(-x2 y1+x1 y2)},1].{{x1,y1},{x2,y2}}];
pullBack3Dcom=Compile[{x1,y1,z1,x2,y2,z2,x3,y3,z3,x,y,z},Mod[{(x3 y2 z-x2 y3 z-x3 y z2+x y3 z2+x2 y z3-x y2 z3)/(x3 y2 z1-x2 y3 z1-x3 y1 z2+x1 y3 z2+x2 y1 z3-x1 y2 z3),(x3 y1 z-x1 y3 z-x3 y z1+x y3 z1+x1 y z3-x y1 z3)/(-x3 y2 z1+x2 y3 z1+x3 y1 z2-x1 y3 z2-x2 y1 z3+x1 y2 z3),(x2 y1 z-x1 y2 z-x2 y z1+x y2 z1+x1 y z2-x y1 z2)/(x3 y2 z1-x2 y3 z1-x3 y1 z2+x1 y3 z2+x2 y1 z3-x1 y2 z3)},1].{{x1,y1,z1},{x2,y2,z2},{x3,y3,z3}}];
pullBack2D[{{x1_,y1_},{x2_,y2_}},{x_,y_}]:=pullBack2Dcom[x1,x2,y1,y2,x,y];
pullBack3D[{{x1_,y1_,z1_},{x2_,y2_,z2_},{x3_,y3_,z3_}},{x_,y_,z_}]:=pullBack3Dcom[x1,y1,z1,x2,y2,z2,x3,y3,z3,x,y,z];
The expressions above seems complicated, but they are just solution of LinearSolve plus using Mod function.
Now I prepare a parallel grid data set which has periodic boundary condition.
data=N@Flatten[Table[Append[i{1,0}+j*{1,1},Sin[Total[i{1,0}+j*{1,1}]*2*\[Pi]]],{i,0,1.,1/100},{j,0,1.,1/100}],1];
ListContourPlot[data,AspectRatio->Automatic]
which gives
and then do mesh interpolation like below
mesh=ToElementMesh[data[[;;,1;;2]]];
f=ElementMeshInterpolation[{mesh},data[[;;,-1]]];
ContourPlot[f[x,y],{x,y}\[Element]ConvexHullMesh[data[[;;,1;;2]]],AspectRatio->Automatic]
gives
which is good.
Now we can use pullBack2D to generate a periodic function using base vector bvecs
bvecs = {{1, 0}, {1, 1}}
g[x_?NumericQ, y_?NumericQ] := f @@ pullBack2D[bvecs, {x, y}]
plot it using
ContourPlot[g[x, y], {x, 0, 2}, {y, 0, 2}, AspectRatio -> Automatic]
we get
• You should be able to use a technique like the one here if you have the basis vectors of your "fundamental parallelogram". – J. M.'s torpor Feb 4 at 13:15
• @J.M. Thank you for the link. I see you use LinearSolve and that is exactly what I use except that I use the explict symbolic solution of LinearSolve to use Compile. I found Compile is much faster :) – matheorem Feb 4 at 13:19
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2021-07-28 11:20:37
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https://www.math4refugees.de/html/en/1.8.2/xcontent3.html
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#### Chapter 8 Integral Calculus
Section 8.2 Definite Integral
# 8.2.4 Properties of the Integral
For odd functions $f:\left[-c;c\right]\to ℝ$, the integral is zero. For example, consider the function $f$ on $\left[-2;2\right]$ with $f\left(x\right)={x}^{3}$ shown in the figure below.
The graph of $f$ is subdivided into two parts, namely in a part between $-2$ and $0$ and a part between $0$ and $2$, and the two regions bounded by the graph and the $x$-axis are investigated. The two regions can be transferred into each other by a point reflection (central inversion). Both regions are equal in size. However, forming the Riemann sum of both regions, one finds that the area of the region below the $x$-axis takes a negative value. Thus, if the integrals for the two regions are added to calculate the integral over the interval from $-2$ to $2$, the area of the region above the positive $x$-axis is positive. The area of the region below the negative $x$-axis is equal in size but has a negative sign. Thus, the sum of the two areas equals zero. Thus, for odd functions $f$, we have the following rule:
${\int }_{-c}^{c}f\left(x\right) dx=0 .$
For an even function $g:\left[-c;c\right]\to ℝ$, the graph is symmetric with respect to the $y$-axis.
The region between the graph of $g$ and the $x$-axis is here symmetric with respect to the $y$-axis. Thus, the region to the left of the $y$-axis is the mirror image of the region to the right. The sum of the areas of the two regions is
${\int }_{-c}^{c}g\left(x\right) dx=2·{\int }_{0}^{c}g\left(x\right) dx .$
This rule for the integral applies to every integrable function $g$ that is even, even if the function takes negative values. Due to the calculation rule above, it is then sufficient to calculate the integral for non-negative values of $x$ with the lower limit $0$ and the upper limit $c$.
Often, the calculation of an integral is easier if the integrand is first transformed into a known form. Examples of possible transformations shall be considered below. In the first example, power functions are investigated.
##### Example 8.2.13
Calculate the integral
${\int }_{1}^{4}\left(x-2\right)·\sqrt{x} dx .$
First, the integrand is transformed to simplify the calculation:
$\left(x-2\right)·\sqrt{x}=x\sqrt{x}-2\sqrt{x}={x}^{\frac{3}{2}}-2{x}^{\frac{1}{2}} .$
Now the integral can be calculated more easily:
$\begin{array}{ccc}\multicolumn{1}{c}{{\int }_{1}^{4}\left(x-2\right)·\sqrt{x} dx}& =\hfill & {\int }_{1}^{4}\left({x}^{\frac{3}{2}}-2{x}^{\frac{1}{2}}\right)dx={\left[\frac{2}{5}{x}^{\frac{5}{2}}-\frac{4}{3}{x}^{\frac{3}{2}}\right]}_{1}^{4}\hfill \\ \multicolumn{1}{c}{}& =\hfill & \left(\frac{2}{5}{\left(\sqrt{4}\right)}^{5}-\frac{4}{3}{\left(\sqrt{4}\right)}^{3}\right)-\left(\frac{2}{5}·1-\frac{4}{3}·1\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & \left(\frac{64}{5}-\frac{32}{3}\right)-\left(\frac{2}{5}-\frac{4}{3}\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & \frac{62}{5}-\frac{28}{3}\hfill \\ \multicolumn{1}{c}{}& =\hfill & 3+\frac{1}{15} .\hfill \end{array}$
The next example demonstrates a transformation of an integrand involving exponential functions.
##### Example 8.2.14
Calculate the integral
${\int }_{-2}^{3}\frac{8e{}^{3+x}-12e{}^{2x}}{2e{}^{x}} dx .$
According to the calculation rule for exponential functions one obtains
$\frac{8e{}^{3+x}-12e{}^{2x}}{2e{}^{x}}=\frac{8e{}^{3+x}}{2e{}^{x}}-\frac{12e{}^{2x}}{2e{}^{x}}=4e{}^{3+x-x}-6e{}^{2x-x}=4e{}^{3}-6e{}^{x} ,$
such that the integral can finally be calculated easily:
$\begin{array}{ccc}\multicolumn{1}{c}{{\int }_{-2}^{3}\frac{8e{}^{3+x}-12e{}^{2x}}{2e{}^{x}} dx={\int }_{-2}^{3}\left(4e{}^{3}-6e{}^{x}\right)dx}& =\hfill & {\left[4e{}^{3}·x-6e{}^{x}\right]}_{-2}^{3}\hfill \\ \multicolumn{1}{c}{}& =\hfill & \left(4e{}^{3}·3-6e{}^{3}\right)-\left(4e{}^{3}·\left(-2\right)-6e{}^{-2}\right)\hfill \\ \multicolumn{1}{c}{}& =\hfill & 14e{}^{3}+\frac{6}{e{}^{2}} .\hfill \end{array}$
Consider a rational function. If the degree of the numerator polynomial is greater or equal to the degree of the denominator polynomial, a polynomial long division is carried out first (see Module 6). Depending on the situation, further transformations (e.g. partial fraction decomposition) may be appropriate. These can be found in advanced textbooks and formularies. In the following example, a polynomial long division is carried out to integrate a rational function.
##### Example 8.2.15
Calculate the integral
${\int }_{-1}^{1}\frac{4{x}^{2}-x+4}{{x}^{2}+1} dx .$
First, we transform the integrand using polynomial long division:
$4{x}^{2}-x+4=\left({x}^{2}+1\right)·4-x$
so
$\frac{4{x}^{2}-x+4}{{x}^{2}+1}=4-\frac{x}{{x}^{2}+1} .$
Thus, we have
$\begin{array}{ccc}\multicolumn{1}{c}{{\int }_{-1}^{1}\frac{4{x}^{2}-x+4}{{x}^{2}+1} dx}& =\hfill & {\int }_{-1}^{1}\left(4-\frac{x}{{x}^{2}+1}\right)dx\hfill \\ \multicolumn{1}{c}{}& =\hfill & {\int }_{-1}^{1}4 dx-{\int }_{-1}^{1}\frac{x}{{x}^{2}+1} dx={\left[4x\right]}_{-1}^{1}-0=8 .\hfill \end{array}$
The integrand in the second integral is an odd function and centrally symmetric on the interval $\left[-1;1\right]$, so the second integral equals zero.
Here, a specific example was given to provide a first impression of the integration of rational functions. In advanced mathematics lectures and in the literature this approach is described in general terms.
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2023-01-26 23:18:45
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https://www.reddit.com/r/learnmath/comments/10537i/partial_derivatives_how_do_they_work/
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×
[–] 1 point2 points (0 children)
It's a derivative of a multivariable function, with respect to one of the variables, holding the others constant.
Say for instance, I have a rectangle with one side of length x and the other of length y. The area is xy. The rate of change of the area as you alter x while keeping y fixed is dA/dx=y. We're just treating y as a constant, temporarily.
And naturally you can then differentiate dA/dx with respect to y; or you could differentiate wrt x again to get a second derivative; or whatever.
Obviously in applying this, you have to be careful. Sometimes in a given problem, the variables actually depend on each other and it doesn't make sense to ask what happens if they were allowed to vary independently.
[–] 1 point2 points (0 children)
As has been said, a partial derivative is basically a way of differentiating a multivariable function in terms of only one variable. If you want more information here is the wikipedia article on it.
[–] 1 point2 points (1 child)
I give up. What are partial derivatives?
My turn:
What is the first derivative of a cow?
The prime rib!
[–][S] 0 points1 point (0 children)
I'm using this in class tomorrow and there's nothing you can do about it except enjoy that the joke is spreading.
[–] 0 points1 point (0 children)
If you have a function of several variables, e.g. f(x,y,z,t), then you can take the derivative to one of those variables. One such derivative is called a partial derivative. So in the example, df/dx, df/dy, df/dz and df/dt are partial derivatives.
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2018-03-21 21:07:48
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https://proofwiki.org/wiki/Equivalence_of_Axiom_Schemata_for_Groups/Warning
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# Equivalence of Axiom Schemata for Groups/Warning
## Theorem
Suppose we build an algebraic structure with the following axioms:
$(0)$ $:$ Closure Axiom $\displaystyle \forall a, b \in G:$ $\displaystyle a \circ b \in G$ $(1)$ $:$ Associativity Axiom $\displaystyle \forall a, b, c \in G:$ $\displaystyle a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ $(2)$ $:$ Right Identity Axiom $\displaystyle \exists e \in G: \forall a \in G:$ $\displaystyle a \circ e = a$ $(3)$ $:$ Left Inverse Axiom $\displaystyle \forall x \in G: \exists b \in G:$ $\displaystyle b \circ a = e$
Then this does not (necessarily) define a group (although clearly a group fulfils those axioms).
## Proof
Let $\struct {S, \circ}$ be the algebraic structure defined as:
$\forall x, y \in S: x \circ y = x$
That is, $\circ$ is the left operation.
From Element under Left Operation is Right Identity, every element serves as a right identity.
Then given any $a \in S$, we have that $x \circ a = x$ and as $x$ is an identity, axiom $(3)$ is fulfilled as well.
But from More than one Right Identity then no Left Identity, there is no left identity and therefore no identity element.
Hence $\struct {S, \circ}$ is not a group.
$\blacksquare$
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2019-07-24 09:45:39
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https://www.projecteuclid.org/euclid.ijm/1318598669
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## Illinois Journal of Mathematics
### Cohomology of decomposition and the multiplicity theorem with applications to dynamical systems
Leon A. Luxemburg
#### Abstract
The article obtains some lower bounds for the sectional category of a map based on the cohomology of the base space and the total space. We also obtain geometric results on the multiplicity of maps and show applications to equilibria on the boundary of stability regions (basins of attraction) of dynamical systems on differentiable manifolds. We consider a number of generalizations of Lusternik–Schnirelmann’s theorem which states that if a covering of an $n$ dimensional sphere consists of $n + 1$ closed sets, then at least one of the sets contains antipodal points. An elementary proof is given for a generalization of this result to packings of Euclidean spaces.
#### Article information
Source
Illinois J. Math., Volume 54, Number 2 (2010), 485-499.
Dates
First available in Project Euclid: 14 October 2011
https://projecteuclid.org/euclid.ijm/1318598669
Digital Object Identifier
doi:10.1215/ijm/1318598669
Mathematical Reviews number (MathSciNet)
MR2846470
Zentralblatt MATH identifier
1246.55002
#### Citation
Luxemburg, Leon A. Cohomology of decomposition and the multiplicity theorem with applications to dynamical systems. Illinois J. Math. 54 (2010), no. 2, 485--499. doi:10.1215/ijm/1318598669. https://projecteuclid.org/euclid.ijm/1318598669
#### References
• V. Hurewicz and H. Wallman, Dimension theory, Revised edition, Princeton University Press, Princeton, NJ, 1996.
• L. Lyusternik and L. Schnirelmann, Topological methods in variational problems, Issledowatelskiy Institut Mathematiki i Mechaniki pri M G U, Moscow, 1930 (Russian).
• E. Spanier, Algebraic topology, McGraw-Hill, NewYork, 1966.
• M. Arkowitz and J. Strom, The sectional category of a map, Proc. of Royal Soc. Edinburgh, 134A (2004), 639–650.
• I. James, On category in the sense of Lusternik–Schnirelmann, Topology 17 (1978), 331–348.
• I. Berstein and T. Ganea, The category of a map and a cohomology class, Fund. Math. 50 (1962), 265–279.
• S. Smale, Morse inequalities for dynamical systems, Bull. Amer. Math. Soc. 66 (1960), 43–49.
• R. H. Fox, On the Lusternik–Schnirelmann category, Annals of Mathematics 42 (1941), 333–370.
• S. Eilenberg and T. Ganea, On the Lusternik–Schnirelmann category of abstract groups, Annals of Mathematics, 2nd Ser., 65 (1957), 517–518.
• A. Schwarz, The genus of a fiber space, A.M.S. Trans. 55 (1966), 49–140.
• L. F. Suarez, P. Ghienne, T. Kahl and L. Vandembroucq, Joins of DGA modules and sectional category, Algebraic & Geometric Topology 6 (2006), 119–144.
• L. A. Luxemburg and G. Huang, Generalized Morse theory and its applications to control and stability analysis, Circuits Systems Signal Process. 10 (1991), 175–209.
• L. A. Luxemburg and G. Huang, Characterization of equilibrium points on the stability boundary via algebraic topology approach, Analysis and Control of Nonlinear Systems, (C. I. Byrnes, C. F. Martin and R. E. Saeks, eds.), Amsterdam, North Holland, 1988, pp. 71–76.
• M. Huber and W. Meier, Cohomology theories and infinite CW complexes, Comm. Math. Helvet. 53 (1978), 239–257.
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2019-10-16 03:44:51
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https://math.stackexchange.com/questions/3361589/proof-that-sum-limits-n-1-infty-1n1-sinn-overn-1-over2
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# Proof that $\sum\limits_{n=1}^{\infty}{(-1)^{n+1}\sin(n)\over{n}}={1\over2}$
While messing around with WolframAlpha, I came across this identity that
$${\sin{1}\over{1}}-{\sin{2}\over{2}}+{\sin{3}\over{3}}-{\sin{4}\over{4}}+{\sin{5}\over{5}}\cdots={1\over{2}}$$.
One would perhaps expect such a seemingly simple identity to have a clean / intuitive proof, however after I have had trouble finding one while looking around online.
What is the simplest/ most intuitive way to prove this?
• It follows from the Fourier series of the fractional part but that is not particularly elementary – Conrad Sep 19 '19 at 0:35
If you see https://math.stackexchange.com/a/13494/706414 the author (in an elementary way) shows that $$f(x) = \lim_{n\to\infty}\ \sum_{k=1}^{n}\frac{\sin kx}{k}=\frac{\pi-x}{2},\qquad x\in(0,2\pi).$$ Note that $$f(2) = 2\sum_{n=1}^\infty \frac{\sin(2k)}{2k}$$ Your desired sum then becomes $$f(1)-f(2) = \frac{1}{2}$$
• (+1) You might want to explain (in your answer) why the desired sum is $f(1)-f(2)$. – robjohn Sep 19 '19 at 2:57
$$S$$ is the imaginary part of $$-\sum_{r=1}^\infty\dfrac{(-1)^re^{ir}}r$$
which is $$=\ln(1-(-1)e^i)=\ln(e^{i/2}+e^{-i/2})+\ln(e^{i/2})=\ln\left(2\cos\dfrac12\right)+\dfrac i2$$(considering the principal branch)
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2021-03-08 12:21:21
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https://kerodon.net/tag/033U
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# Kerodon
$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$
Corollary 4.5.2.25. Suppose we are given a categorical pullback square of $\infty$-categories
$\xymatrix@R =50pt@C=50pt{ \widetilde{\operatorname{\mathcal{C}}} \ar [d]^-{U} \ar [r] & \widetilde{\operatorname{\mathcal{D}}} \ar [d]^{V} \\ \operatorname{\mathcal{C}}\ar [r]^-{F} & \operatorname{\mathcal{D}}, }$
where $U$ and $V$ are isofibrations. Let $C \in \operatorname{\mathcal{C}}$ be an object having image $D = F(C)$. Then the induced map
$\widetilde{\operatorname{\mathcal{C}}}_{C} = \{ C\} \times _{\operatorname{\mathcal{C}}} \widetilde{\operatorname{\mathcal{C}}} \rightarrow \{ D\} \times _{\operatorname{\mathcal{D}}} \widetilde{\operatorname{\mathcal{D}}} = \widetilde{\operatorname{\mathcal{D}}}_{D}$
is an equivalence of $\infty$-categories.
Proof. Apply Corollary 4.5.2.24 in the special case $\operatorname{\mathcal{C}}_1 = \{ C\}$ and $\operatorname{\mathcal{D}}_1 = \{ D\}$. $\square$
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2022-08-19 04:53:03
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https://www.nature.com/articles/s41467-019-08601-7?error=cookies_not_supported&code=99bdb15a-a563-4d1c-94ee-4131533c0dde
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## Introduction
The beauty of symmetry in nature has long been appreciated in both art and science. Although the mirror symmetry is ubiquitous in nature, other symmetries are rather unusual. For example, the sevenfold rotational symmetry is rarely observed, apart from few exceptions such as the seven-armed starfish, the arctic starflower and certain proteins. In crystallography, it is well established that three-dimensional (3D) periodic lattices can only have 2-, 3-, 4- and 6-fold rotational symmetries. However, electron diffraction patterns with fivefold rotational symmetry have been experimentally observed1. In 2011, the Nobel Prize in chemistry was awarded to Dan Shechtman “for the discovery of quasicrystals” highlighting the significance of matter with exceptional symmetry properties2. In this paper, we demonstrate the generation of free electron wave packets (FEWPs) with arbitrary rotational symmetry using polarization-tailored bichromatic femtosecond (fs) laser pulses.
In ultrafast optics, bichromatic fields with cycloidal polarization profiles have emerged as a powerful twist to control the time evolution and symmetry of coherent matter waves. For example, propeller-type fs laser pulses, composed of (ω:2ω) counterrotating circularly polarized (CRCP) fields, were employed to control the polarization of attosecond pulses from high-harmonic generation (HHG)3,4,5 and to generate FEWPs with threefold rotational symmetry by strong field ionization of argon atoms6,7,8.
The interaction of bichromatic fields with matter creates FEWPs inaccessible with single color pulse sequences. For example, temporally overlapping commensurable frequency bichromatic CRCP fields result in propeller-type polarization profiles, which maintain polarization characteristics of both colors. In contrast, single color CRCP pulse sequences lose their symmetry as the time delay vanishes and become linearly polarized. The advantage of bichromatic over single color fields has profound consequences in light-matter-interactions. For example, FEWPs from multiphoton ionization (MPI) with single color CRCP pulse sequences have always even-numbered rotational symmetry, whereas FEWPs created by commensurable bichromatic CRCP fields exhibit either even- or the exceptional odd-numbered symmetry, selectable via the frequency ratio. A detailed comparison of single color and bichromatic MPI is given in Supplementary Note 4.
Recently, we introduced an optical scheme for the generation of bichromatic carrier-envelope phase-stable polarization-tailored supercontinuum (BiCEPS) fields of variable frequency ratio, utilizing a polarization pulse shaper to tailor the spectral amplitude, phase and polarization of a CEP-stable white light supercontinuum (WLS)9. In this paper, we report on an application of BiCEPS pulses for multipath coherent control of bichromatic MPI. In the experiment we combine the BiCEPS setup with high resolution photoelectron tomography10. We devise a general scheme to generate FEWPs with arbitrary rotational symmetry. The scheme is exemplified on the experimental demonstration of 3D FEWPs with sevenfold rotational symmetry, created by MPI of sodium (Na) atoms using (3ω:4ω) BiCEPS pulses. Shaper-generated parallel linearly polarized (PLP) bichromatic fields are already used to exert CEP control on left/right asymmetries in the photoelectron momentum distribution (PMD)11. The full power of BiCEPS pulses unfolds when circularly polarized bichromatic fields are employed. By combination of bichromatic MPI with polarization-tailoring12, we exert full spatial control on the FEWP by design of specific energy and angular momentum superposition states, thus preparing matter waves with odd-numbered rotational symmetry.
To elucidate the physical mechanisms, we study the interplay between the symmetry properties of the driving BiCEPS fields and the resulting FEWPs. We examine how the optical properties, such as the polarization, CEP, relative phases and time delays, control the symmetry of the PMD. Our results reveal that in the multiphoton regime the symmetry of FEWPs is not fully determined by the field symmetry, but completely described by the quantum interference of states with different angular momenta.
## Results
### Polarization profile and wave packet
Recently, the intriguing properties of polarization-controlled bichromatic fields have been highlighted in the context of HHG3. In general, circularly polarized bichromatic fields exhibit cycloidal polarization profiles. Measured polarization profiles of propeller-type CRCP and heart-shaped corotating circularly polarized (COCP) bichromatic pulses are illustrated in Fig. 1. To discuss the interplay of phase effects in optics and quantum mechanics, we start by considering the properties of the polarization profile of commensurable BiCEPS fields with center frequency ratio ω1 : ω2 = N1 : N2. The rotational symmetry of the field is given by $${\cal S}_{{\mathrm{opt}}} = (N_2 \mp N_1){\mathrm{/}}gcd(N_1,N_2)$$9, where the minus (plus) sign corresponds to the COCP (CRCP) case and gcd denotes the greatest common divisor. By introducing the CEP φce and the relative phases $$\hat \varphi _1 = \varphi _1 - \omega _1\tau$$ (τ: time delay) and φ2 of the low- (red) and high-frequency (blue) field, respectively, an (N1ω:N2ω) BiCEPS pulse rotates in the polarization plane by an angle of
$$\begin{array}{l}\alpha _\tau ^{{\mathrm{cr}}} = \frac{{N_2 - N_1}}{{N_2 + N_1}}\varphi _{{\mathrm{ce}}} + \frac{{N_2}}{{N_2 + N_1}}\hat \varphi _1 - \frac{{N_1}}{{N_2 + N_1}}\varphi _2,\\ \alpha _\tau ^{{\mathrm{co}}} = \varphi _{{\mathrm{ce}}} + \frac{{N_2}}{{N_2 - N_1}}\hat \varphi _1 - \frac{{N_1}}{{N_2 - N_1}}\varphi _2,\end{array}$$
(1)
measured in ϕ-direction [cf. Fig. 1a and Supplementary Note 1]. The angle $$\alpha _\tau ^{{\mathrm{co}}}$$ applies to a pair of left-handed circularly polarized (LCP) pulses, whereas a right-handed circularly polarized (RCP) sequence rotates by $$- \alpha _\tau ^{{\mathrm{co}}}$$. An (N1ω:N2ω) CRCP field, consisting of an LCP (RCP) red and an RCP (LCP) blue field, rotates about $$\alpha _\tau ^{{\mathrm{cr}}}$$ $$\left( { - \alpha _\tau ^{{\mathrm{cr}}}} \right)$$. For extremely short few-cycle fields the pulse envelope needs to be taken into account as well. Equation (1) shows that the sense of rotation is opposite for φ1 and φ2. Moreover, the sensitivity of the rotation to φce, φ1 and φ2 is different for CRCP and COCP fields and generally more sensitive for COCP than for CRCP pulses. For example, the rotation of a (3ω:4ω) CRCP pulse, as used in the experiment, by the CEP is $$\alpha _\tau ^{{\mathrm{cr}}} = \varphi _{{\mathrm{ce}}}{\mathrm{/}}7$$, whereas the rotation of any COCP pulse is $$\alpha _\tau ^{{\mathrm{co}}} = \varphi _{{\mathrm{ce}}}$$. The CEP-dependent rotation of the field explains, why CEP stability is required in experiments using shaper-generated bichromatic fields.
The coarse structure of FEWPs created by MPI of Na atoms with BiCEPS pulses can be deduced from quantum mechanical selection rules for optical transitions. For excitation with σ+ (LCP) and σ (RCP) pulses, the selection rules $${\mathrm{\Delta }}\ell = 1$$ and Δm = ±1 apply. N-photon ionization prepares a quantum state with $$\ell = N$$ and m = ±N. We consider MPI induced by an (N1ω:N2ω) pulse sequence consisting of an initial LCP red pulse followed by an RCP (CRCP case) or an LCP (COCP case) blue pulse. The frequency ratio is chosen such that the red pulse induces N2-, while the blue pulse induces N1-photon ionization. To describe the created FEWP, we consider the azimuthal part of the wave function. Absorption of N2 σ+ photons yields an azimuthal phase of N2ϕ, a spectral phase of −N2(φ1 + φce) and an additional phase due to the time evolution of −ετ/ħ (ε: photoelectron kinetic energy). Analogously, absorption of N1 σ± photons from the blue pulse yields an azimuthal phase of ±N1ϕ and a spectral phase of −N1(φ2 + φce). In addition, N-th order perturbation theory yields a factor of iN for N-photon processes13. Hence, the photoelectron wave function is described by the superposition state $$\psi _{{\mathrm{co}}/{\mathrm{cr}}} = \tilde \psi _{N_1, \pm N_1} + \tilde \psi _{N_2,N_2}$$, with contributions
$$\begin{array}{l}\tilde \psi _{N_1, \pm N_1} \propto i^{N_1}{\kern 1pt} e^{ \pm iN_1\phi }e^{ - iN_1(\varphi _2 + \varphi _{{\mathrm{ce}}})},\\ \tilde \psi _{N_2,N_2} \propto i^{N_2}{\kern 1pt} e^{iN_2\phi }e^{ - iN_2(\varphi _1 + \varphi _{{\mathrm{ce}}})}e^{ - i\varepsilon \tau /\hbar }.\end{array}$$
(2)
The rotational symmetry of the FEWP is $${\cal S}_{{\mathrm{wp}}} = \left| {m_2 - m_1} \right| = N_2 \mp N_1$$, where the minus (plus) sign applies to COCP (CRCP) ionization. Specifically for the case N1 = 3 and N2 = 4 used in the experiment the electron density is given by
$$\left| {\psi _{{\mathrm{co}}/{\mathrm{cr}}}} \right|^2 \propto 1 \pm {\mathrm{sin}}\left[ {(4 \mp 3)\left( {\phi - \alpha _0^{{\mathrm{co}}/{\mathrm{cr}}}} \right) - \frac{{\varepsilon \tau }}{\hbar }} \right],$$
(3)
(see Supplementary Note 2). Equation (3) reveals that, for τ ≠ 0, energies of constant electron density form an Archimedean spiral (n: integer number)14,15,16
$$\varepsilon _\tau ^{{\mathrm{co}}/{\mathrm{cr}}}(\phi ) = \frac{{(4 \mp 3)\hbar }}{\tau }\left( {\phi - \alpha _0^{{\mathrm{co}}/{\mathrm{cr}}}} \right) - \frac{{\left( {2\pi n \pm \pi {\mathrm{/}}2} \right)\hbar }}{\tau }.$$
(4)
### Experiment
Here we combine bichromatic polarization pulse shaping9 with photoelectron tomography10, as illustrated in Fig. 1a. A 4f polarization pulse shaper9 is employed to sculpture (3ω:4ω) fields from a CEP-stable over-octave spanning WLS by spectral amplitude and phase modulation. The shaper provides access to all pulse parameters of both colors, including the relative phases φ1,2 and linear spectral phases φ1,2(ω) = τ1,2 · (ω − ω1,2) to introduce a time delay between the two colors. By application of custom composite polarizers in the Fourier plane and a λ/4 wave plate at the output, the bichromatic polarization state is controlled to generate PLP, COCP and CRCP pulses9. In addition, the shaper is utilized for pulse compression and characterization. The CEP of the BiCEPS pulses is actively stabilized and controlled by feeding a single-shot f-2f interferometer with an (ω:2ω) field extracted additionally from the wings of the WLS (see Methods). In the experiment, commensurable center wavelengths λ1 = 880 nm (red field) and $$\lambda _2 = \frac{3}{4}\lambda _1 = 660\,{\mathrm{nm}}$$ (blue field) are chosen. A measured spectral amplitude profile is depicted in Fig. 1b, along with an in situ cross correlation (CC) trace in Fig. 1c and measured polarization profiles of a CRCP and a COCP pulse in Fig. 1d. The (3ω:4ω) BiCEPS pulses are focused (intensity I ≈ 2 × 1012 W cm−2) into the interaction region of a velocity map imaging spectrometer (VMIS) loaded with Na vapor. FEWPs are imaged onto a 2D multi-channel-plate (MCP) detector. By rotation of a λ/2 wave plate, 2D projections of the PMD are recorded under different angles to reconstruct the 3D density using tomographic techniques10 (Methods).
The experiment is subdivided in three parts. First, we check the coherence properties of interfering FEWPs from bichromatic MPI with (3ω:4ω) PLP fields. Subsequently, we use temporally overlapping bichromatic CRCP and COCP pulses to generate sevenfold rotationally symmetric (c7) and asymmetric FEWPs and manipulate their orientation by the optical phases of the BiCEPS pulses. Both measurements exemplify our general approach to the generation and manipulation of odd-numbered FEWPs. Finally, we introduce a time delay within the (3ω:4ω) CRCP sequence to create spiral-shaped FEWPs with c7 symmetry, that is, electron densities in the shape of 7-armed Archimedean spirals14.
We start by investigating CEP-dependent asymmetries in the photoemission induced by (3ω:4ω) PLP fields, which arise due to the interference of continuum states with opposite parity11,17. The asymmetry in the PMD is utilized as sensitive probe of the temporal, spatial and energetic overlap of the released FEWPs. MPI with a (3ω:4ω) field creates an f-type FEWP (odd parity) via N1 = 3-photon ionization by the blue band and a g-type FEWP (even parity) via N2 = 4-photon ionization by the red band, depicted in Fig. 2a. Single color photoelectron spectra measured by MPI with either the red or the blue band are shown in Fig. 2b, c, respectively. Additional ionization pathways to s-, p-, and d-type continua are taken into account for the fits to the measured angular distributions. The spectra confirm the energetic overlap of the two FEWPs centered around ε ≈ 0.5 eV. Coherent superposition of both contributions gives rise to a directional photoelectron wave function
$$\psi _{{\mathrm{dir}}} \propto \psi _{3,0} + i{\kern 1pt} \psi _{4,0}{\kern 1pt} e^{ - i{\mathrm{\Delta }}\varphi },$$
(5)
with Δφ = 4φ1 − 3φ2 + φce determining the photoelectron asymmetry. Initially, we set φ1 = φ2 = 0 and study the φce-dependence of the PMD. The photoelectron spectra measured by MPI with the bichromatic field, shown in Fig. 2d, e, display a pronounced asymmetry along the laser polarization direction. For φce = 0, the global maximum of the photoemission is observed around θ = 3π/2 (negative y-direction). By switching the CEP to φce = π, the asymmetry is inverted and the photoemission localizes around θ = π/2.
While (3ω:4ω) PLP pulses enable us to control the directional photoemission along the laser polarization11, circularly polarized BiCEPS pulses provide full 3D control of the final state by creation of angular momentum superposition states. In the second part of the experiment, we investigate 3D FEWPs created by MPI with (3ω:4ω) propeller-type CRCP and heart-shaped COCP fields. The excitation scheme is depicted in Fig. 3a. In the CRCP case, with an LCP red and an RCP blue pulse, the final state wave function arises from interference of two counterrotating torus-shaped waves:
$$\psi _{{\mathrm{cr}}} \propto \psi _{3, - 3} + i{\kern 1pt} \psi _{4,4}{\kern 1pt} e^{ - i{\mathrm{\Delta }}\varphi }.$$
(6)
This superposition describes a spherical standing wave with $${\cal S}_{{\mathrm{wp}}} = N_2 + N_1 = 7$$ lobes in the laser polarization plane (x-y-plane), that is, with sevenfold rotational symmetry. The tomographically reconstructed FEWP presented in Fig. 3b, maps the c7 symmetry of the polarization profile of the field. In the COCP case, consisting of two LCP pulses, the final state wave function is a coherent superposition of two corotating waves:
$$\psi _{{\mathrm{co}}} \propto \psi _{3,3} + i{\kern 1pt} \psi _{4,4}{\kern 1pt} e^{ - i\Delta \varphi },$$
(7)
resulting in a standing wave with $${\cal S}_{{\mathrm{wp}}} = N_2 - N_1 = 1$$ lobe. The crescent-shaped reconstructed FEWP, shown in Fig. 3c, is reminiscent of the heart-shaped laser field. In contrast to the CRCP case, this electron distribution exhibits no rotational symmetry and is localized in a predefined half of the polarization plane.
The shape of BiCEPS pulses is highly sensitive to φce, φ1 and φ2. Each phase rotates the field according to Eq. (1). Quantum mechanically, the three phases manifest in the relative phase Δφ between the interfering FEWPs. Since Δφ adds to the azimuthal phase ϕ, the FEWP rotates about the laser propagation direction, following the optical rotation. On the one hand, this implies that CEP stability of the BiCEPS pulses is required for bichromatic control of quantum interferences, because otherwise the azimuthal interference pattern is averaged out. On the other hand, the CEP can be used to manipulate the spatial orientation of the FEWP. To demonstrate phase control, we investigate the phase dependence of FEWPs generated by (3ω:4ω) COCP and CRCP fields on two characteristic examples. We show that shifting the CEP by Δφce = π and the relative phase by Δφ2 = π/3 cause an equivalent rotation of the field and the FEWP, albeit in opposite directions. The experimental results for the CRCP case are illustrated in Fig. 4a–c. For the analysis, reconstructed sections through the FEWP, taken in the polarization plane, are presented in cartesian and polar representation. Figure 4a displays the results at Δφ = 0 for reference. By variation of the CEP from φce = 0 to π, the CRCP pulse is rotated clockwise by $$\alpha _0^{{\mathrm{cr}}} = \pi {\mathrm{/}}7$$, as indicated in the top inset to Fig. 4b. Quantum mechanically, this CEP variation translates into a relative phase of Δφ = φce = π, shifting the standing wave pattern by half a cycle. Accordingly, the FEWP in Fig. 4b is rotated clockwise by π/7, visible by the interchange of azimuthal lobes and nodes. The sense of rotation is verified in a measurement presented in Supplementary Note 3. To achieve an equivalent effect via the relative phase between the two colors, the phase φ2 of the blue component is varied from φ2 = 0 to π/3. Hence, the CRCP pulse is rotated counterclockwise by $$\alpha _0^{{\mathrm{cr}}} = - \pi {\mathrm{/}}7$$, which reproduces the pulse shape shown in (b). The corresponding quantum phase Δφ = −3φ2 = −π again induces a half cycle rotation of the FEWP. Therefore, the measured FEWP shown in Fig. 4c has the same orientation as the FEWP in (b), but arises from the FEWP in (a) by counterclockwise rotation (see also Supplementary Note 3). The experimental results for the COCP case are shown in Fig. 4d–f. According to Eq. (1), the rotation of bichromatic COCP laser fields is directly determined by the CEP. Comparison of the experimental results shown in Fig. 4d, e reveals that the crescent-shaped FEWP rotates by π upon corresponding variation of φce. Variation of the relative phase between the two colors from φ2 = 0 to π/3 rotates the COCP pulse counterclockwise by $$\alpha _0^{{\mathrm{co}}} = - \pi$$. Again, the FEWP follows the field rotation, as shown in Fig. 4f. The results presented in Fig. 4 demonstrate the use of BiCEPS pulses to accurately control both, the symmetry and the spatial orientation of the FEWP by the optical phases. In turn, the COCP results can be utilized as a CEP-clock18 to measure the CEP at comparatively low laser intensities without the use of an f-2f interferometer.
In the third part, we introduce a time delay between the two colors of a (3ω:4ω) CRCP field. Ionization with two time-delayed CRCP pulses creates FEWPs with an Archimedean spiral pattern in the laser polarization plane14,16. Inspired by the helical interference structures, this type of photoelectron momentum distribution was termed ‘electron vortex’ by Starace and coworkers14. This notion of an electron vortex needs to be distinguished from the traditional definition of vortex states in quantum systems19,20. The latter are derived from the hydrodynamic formulation of quantum mechanics21 and defined by vortex structures in the velocity field of the wave function. Thus, this vortex type manifests in the quantum mechanical phase. Those vortex states are subject of intense studies in collision physics20,22 and the generation of electron vortex beams23,24. In contrast, the free electron vortices discussed, e.g., in refs. 14,16,25,26 are spiral-shaped angular distributions of photoelectron wave packets and manifest in the electron density. So far, the creation of even-armed electron vortices was demonstrated experimentally16,27. Using time-delayed (3ω:4ω) CRCP pulse sequences, we create FEWPs characterized by a 7-armed Archimedean spiral in the polarization plane. The delay is implemented by applying a linear spectral phase φ1(ω) = τ1 · (ω − ω1) to the red band, which advances (τ1 > 0) or delays (τ1 < 0) the red pulse relative to the blue pulse in time. We start with an LCP red pulse preceding an RCP blue pulse by τ1 = −10 fs. During the time delay, the 4-photon FEWP acquires an additional energy-dependent quantum phase of ετ1/ħ, which induces a linear tilt to the lobes of the standing wave. The resulting wave function
$$\psi _{{\mathrm{spi}}} \propto \psi _{3, - 3} + i{\kern 1pt} \psi _{4,4}{\kern 1pt} e^{ - i{\mathrm{\Delta }}\varphi }e^{i\varepsilon \tau _1/\hbar }$$
(8)
describes a 7-armed spiral-shaped FEWP with clockwise sense of rotation in the polarization plane, as observed in the experimental results shown in Fig. 5a (upper panel). The linear tilt of the lobes is best discernible in the polar representation (lower panel). From the slope, we derive a time delay of τ1 (−10.0 ± 1.0) fs using principal component analysis (PCA) to determine the major axis of the elliptical distribution of each lobe in polar representation. This result is in good accordance with the pulse shaper settings. Reversal of the pulse ordering by changing the sign of τ1 inverts the rotational sense of the spiral-shaped FEWP, as illustrated in Fig. 5b. Increasing the time delay to τ1 = −20 fs results in temporally separated pulses with opposite circularity. The increased pulse separation leads to a stronger tilt of the lobes in the PMD, indicating a time delay of τ1 (−19.8 ± 0.5) fs again in good agreement with the shaper settings. The apparent circular symmetry of the sequence of separated circularly polarized pulses is depicted by the polarization profile displayed in the top inset to Fig. 5c. Although the field is circularly symmetric, the measured photoelectron distribution shown in Fig. 5c retains the sevenfold rotational symmetry. Along with our recent single color results on even-numbered electron vortices16, our findings presented in Fig. 5, demonstrate the power of the BiCEPS scheme to generate spiral-shaped FEWPs of arbitrary rotational symmetry.
## Discussion
We have introduced a general optical scheme to create, measure and manipulate FEWPs with arbitrary rotational symmetry by combining advanced spectral amplitude, phase and polarization shaping of a CEP-stable supercontinuum with high resolution photoelectron tomography. Due to their cycloidal polarization profiles, BiCEPS pulses are a versatile tool to create quantum superposition states with exceptional symmetry properties. As an example, we presented the first measurement of FEWPs with sevenfold rotational symmetry, along with a crescent-shaped photoelectron angular distribution by MPI of sodium atoms using shaper-generated (3ω:4ω) CRCP and COCP pulse sequences. By increasing the delay in the CRCP sequence, we found that the symmetry of the field changed from c7 to circular but the symmetry of the FEWP remained c7. In contrast to single color and interferometric (ω:2ω) realizations of CRCP and COCP sequences, the photoelectron angular distribution from shaper-generated BiCEPS pulse sequences is CEP-sensitive. Especially FEWPs from MPI with bichromatic COCP fields directly indicate the CEP by their orientation, which can be utilized as an in situ CEP-clock similar to the attoclock technique18,28. In general, the generated FEWPs are susceptible to additional quantum mechanical phases from intensity-dependent energy shifts, resonances or the propagation in the continuum. Therefore, refined FEWP measurements can serve as a sensitive tool for spectroscopic and holographic applications. Our results show that unprecedented control on matter waves was attained by BiCEPS pulse sequences, leading to promising perspectives for numerous applications in physics. For example, in HHG, BiCEPS pulses will enable enhanced possibilities for polarization control of the generated XUV light3. Also, application of BiCEPS pulses to advanced chiral recognition by photoelectron circular dichroism is foreseen. Eventually, tailored FEWPs may be used as a source for electron pulses in ultrafast electron diffraction or scattering experiments.
## Methods
### Shaper-generated CEP-stable bichromatic fields
We use a CEP-stabilized FEMTOLASERS multipass chirped pulse amplifier (Rainbow 500, Femtopower HR 3 kHz CEP, center wavelength λ0 ≈ 785 nm, pulse duration Δτ ≈ 20 fs, 0.8 mJ pulse energy) to seed a neon-filled hollow-core fiber (absolute gas pressure of 2.2 bar) for the generation of a CEP-stable over-octave-spanning WLS (pulse duration Δτ ≈ 5 fs, center wavelength λ0 ≈ 770 nm, wavelength range from 450 to 1100 nm, pulse energy of 0.6 mJ). The WLS pulses are spectrally modulated by employing a home-built 4f polarization pulse shaper29,30 specifically adapted to the ultra-broadband WLS9. Spectral phase and amplitude modulation of the WLS is realized by the combination of a dual-layer Liquid Crystal Spatial Light Modulator (LC-SLM; Jenoptik SLM-640d) positioned in the Fourier plane of the 4f setup and custom composite broadband polarizers (CODIXX colorPol)9. The composite polarizer (CP) is mounted behind the LC-SLM in order to sculpture spectrally disjoint OLP or PLP bands from the input WLS (cf. inset to Fig. 1). By optional use of a superachromatic λ/4 wave plate (Bernhard Halle Nachfl.) at the shaper output, the OLP and PLP bichromatic fields are converted to CRCP or COCP bichromatic fields, respectively9. To compress the BiCEPs pulses, residual spectral phases are compensated by shaper-based adaptive optimization of the second harmonic generation in a thin β-barium borate crystal (GWU-Lasertechnik, θ = 29.2°, 5 μm thickness) using an evolutionary algorithm31,32. In order to compensate for longterm CEP-drifts of the shaped output pulses, an additional (ω:2ω) field with center frequencies ω3 = 2.00 rad fs−1 and ω4 = 4.00 rad fs−1 is extracted from the wings of the input WLS. The (ω:2ω) field is split off the main beam by a dichroic mirror (DM) (Thorlabs DMLP567R) to be detected with a home-built single-shot f-2f interferometer. The interferometer output feeds the CEP control loop of the laser system resulting in a longterm CEP-stability of about 200 mrad root mean square (measured over 3 h)11.
### Photoelectron tomography
Photoelectron imaging techniques are used to measure angular and energy-resolved projections of the 3D PMD from MPI of Na atoms with BiCEPs pulses. The laser pulses are focused into the interaction region of a VMIS33 using a spherical focusing mirror (SFM; focal length f = 250 mm) with an intensity I ≈ 2 × 1012 W cm−2 in the laser focus. The Na vapor is supplied by a dispenser source (SAES Getters). The released FEWPs are imaged onto a position sensitive detector (Scientific Instruments S3075-10-I60-PS43-FM) consisting of a dual-layer multi-channel plate (MCP) in chevron configuration followed by a phosphor screen. The resulting 2D projections are detected by a charge coupled device (CCD) camera (Lumenera LW165M) using an exposure time of 250 ms. Each projection is acquired by accumulation of 150 images. The FEWPs are reconstructed employing tomographic techniques10. To this end, the pulse is rotated by 360° about the propagation axis by application of a λ/2 wave plate (Bernhard Halle Nachfl.) and various projections are recorded under 45 angles between ϕλ/2 = 0° … 176° with an angular step size of Δϕλ/2 = 4°. From the measured 2D projections, the 3D PMD is retrieved using the Fourier slice algorithm34. PMDs created by bichromatic PLP pulses are reconstructed by Abel inversion using the pBASEX algorithm35. Further details on the data processing procedure are provided in Supplementary Note 3.
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2023-03-31 04:36:57
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https://www.shaalaa.com/question-bank-solutions/form-the-differential-equation-representing-the-family-of-curves-y-e2x-a-bx-where-a-and-b-are-arbitrary-constants-procedure-form-differential-equation-that-will-represent-given-family-curves_103173
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# Form the Differential Equation Representing the Family of Curves Y = E2x (A + Bx), Where 'A' and 'B' Are Arbitrary Constants. - Mathematics
Sum
Form the differential equation representing the family of curves y = e2x (a + bx), where 'a' and 'b' are arbitrary constants.
#### Solution
Given: y = e2x (a + bx)
Differentiating the above equation, we get
(dy)/(dx) = be^(2x) + 2 (a + bx)e^(2x)
= (dy)/(dx) = be^(2x) + 2y ...("i") [∵ y = e^(2x) (a + bx)]
differentiating the above equation, we get
(d^2y)/(dx^2) = 2 be^(2x) + 2(dy)/(dx)
= (d^2y)/(dx^2) = 2 ((dy)/(dx) - 2y) + 2(dy)/(dx) ...[∵ "from" ("i") "we get", be^(2x) = (dy)/(dx) - 2y]
= (d^2y)/(dx^2) = 4(dy)/(dx)- 4y
Hence, the required differential equation is (d^2y)/(dx^2) - 4 (dy)/(dx) + 4y= 0.
Concept: Procedure to Form a Differential Equation that Will Represent a Given Family of Curves
Is there an error in this question or solution?
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2021-06-14 09:57:44
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https://codeforces.com/blog/entry/64928
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### jinlifu1999's blog
By jinlifu1999, history, 9 months ago, ,
Code
Code
Code
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• +85
» 9 months ago, # | ← Rev. 2 → +3 Problem E should have 512MB(or 1GB) of Memory limit :(
• » » 9 months ago, # ^ | +8 In the author's solution, DP state reduction is used.
• » » » 9 months ago, # ^ | 0 Yeah, state reduction.But in polygon, the system warns if the jury solution solved the problem with more than half of time limit. Why not for memory limit :( I am so sad.
• » » » 8 months ago, # ^ | ← Rev. 3 → 0 I am getting wrong answer on the 5th test case. Not sure what I am missing. Could you please help?I am using segment trees + lazy propagation for querying the maximum possible weight and d-value at an index and DP over these values. Here's my code: 50479788. Help is very much appreciated. Thanks.
» 9 months ago, # | +5 I believe there is a slight correction in the proof of C. bi and ci form a corresponding grouping only if the following condition holds:bi = cj if and only if bj = ci.The proof still holds because each grouping can be mapped to a bi, ci and the optimal bi, ci can be mapped to a grouping.
• » » 9 months ago, # ^ | 0 Fixed. I'm not sure if it is fine now. I'm a bit dizzy today. Wait a minute and the modification will be shown soon.
• » » » 9 months ago, # ^ | 0 can you explain why my solution is incorrect,please 49283839
• » » » » 9 months ago, # ^ | ← Rev. 2 → 0 DFS is not suitable here. See this comment
» 9 months ago, # | 0 Probably, in B, it should be O(n + m + MAXC) instead of O(n + m).
• » » 9 months ago, # ^ | 0 Did you use something called "bucket"? Actually it is not necessary to use that. A priority queue or a pointer is sufficient.
• » » » 9 months ago, # ^ | +8 Could you please explain how the pointer leads to linear solution? Since dishes are not ordered by cost, we need to sort them first, right? It is already .
• » » » » 9 months ago, # ^ | 0 Oh ... I did make a mistake. I will modify this. Sorry for that.
» 9 months ago, # | 0 In problem E , instead of the sweep line thing , i used a segment tree to cover the best red envelope in some range , and did the same Dp approach , however i'm still getting Wrong answer.Any help would be appreciated , if my approach is wrong please let me know.
• » » 9 months ago, # ^ | +1 10 0 5 4 9 10 1 7 10 10 1 3 8 9 1 6 7 10 1 6 6 7 1 This is pretest 22 and the answer should be 2. However, you output 1.
• » » » 9 months ago, # ^ | 0 Thanks for the test.I read over and over my code and it looks just okay , so im starting to wonder if the approach is just wrong ..
• » » » » 9 months ago, # ^ | +1 Well I don't know how to use segment tree with the operator max, except that there is a technique called Segment Tree Beats. Hope you figure out your problem. :)
• » » 9 months ago, # ^ | +2 Before assigning a value to your lazy array you should check that you don't have a lazy value in your current node or that the value you want to put is better that your current lazy.Something like this 49285040
• » » » 9 months ago, # ^ | 0 Tried it but still not working :\ But thanks !https://codeforces.com/contest/1106/submission/49289629 you meant me doing like that ?
• » » » » 9 months ago, # ^ | ← Rev. 2 → 0 UPD : I got it accepted just after reading this blog : https://codeforces.com/blog/entry/57319I highly recommend reading it , so beneficial !Thank you jiry_2 !Also thank you jinlifu1999 for mentioning the name of the trick !
» 9 months ago, # | ← Rev. 4 → 0 Any test case for this Problem:D https://codeforces.com/contest/1106/submission/49285955
• » » 9 months ago, # ^ | 0 Try playing around with the following test: 4 3 1 2 2 4 1 3 Your solution will print 1 2 4 3 which is not the correct answer.
• » » » 5 months ago, # ^ | 0 Can you tell the correct order
• » » » 4 months ago, # ^ | 0 hey .. can you please tell why sorting doesn't work in this case
• » » » » 5 weeks ago, # ^ | 0 Because it doesn't care the path. At node v, you can go back to parent(v) and go to other nodes with smaller number.
» 9 months ago, # | ← Rev. 2 → -30 In my opinion problem F is unreasonably hard for div2 (and not only to come up with a solution but also to implement)
» 9 months ago, # | ← Rev. 2 → 0 My WA approach for E-Sort all events Maximum Coins and then Maximum d.Iterate on all events. I'm assigning a specific event for each moment, which I will select If not blocked and I can select an event. Made dp states similar to editorial.Took minimum of -If blocked n-1(times),d-1(no of times blocked in i
• » » 9 months ago, # ^ | +13 After that, we use a set with a sweep line to deal with it. — How ??The transition is trivial and you can take a look at it in the code. — Not for me. Please elaborate exactly what you are doing. And why you are updating f[j & 1][a[i].d + 1]
» 9 months ago, # | ← Rev. 2 → 0 My solution for D is getting MLE. can someone help me? https://codeforces.com/contest/1106/submission/49291115
• » » 9 months ago, # ^ | ← Rev. 2 → 0 you shouldn't keep duplicates of vertices in a queue, so when vertex is visited don't push it to queuehttps://codeforces.com/contest/1106/submission/49292899
» 9 months ago, # | +15 Today I learned the big-step baby-step algorithm, and about primitive roots. Problems E and F are very interesting. I'm sorry that this round faced technical difficulties, but I enjoyed it nonetheless.
» 9 months ago, # | 0 Can someone explain why i am getting a "runtime error: constructor call on misaligned address" on problem E. Any help would be appreciated. https://codeforces.com/contest/1106/submission/49293426
• » » 9 months ago, # ^ | 0 The diagnostic message is not for your solution. Your solution gave a wrong answer that the checker wasn't prepared to handle, and the checker faced a runtime error. The diagnostic message is for the checker.Your verdict is simply WA.
• » » » 9 months ago, # ^ | 0 But the diagnostics mentions function, std::pair > &, std::pair, std::pair > &)> which i have used in my code.
• » » » » 9 months ago, # ^ | 0 I am very surprised to see that. I was certain that it might be due to the checker. But now I am not that sure.One of the things that is curious is that you asked for GNU C++ compilation, but the diagnostic message mentions clang++ and MSVC. So that's suspicious indeed. Please check once if your comparator is well defined.
• » » » » » 9 months ago, # ^ | 0 Checked the comparator.Looks alright.
• » » » » » » 9 months ago, # ^ | 0 I think there might be something wrong with your solution, I copied it and changed it to this and no more exception but still wrong answer on test 7. #include #define mp make_pair #define mt make_tuple #define x first #define y second #define all(v) (v).begin(),(v).end() #define rall(v) (v).rbegin(),(v).rend() #define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); #define ve vector #define forn(i,n) for(int i=0;i<(int)n;i++) #define for1(i,n) for(int i=1;i<=(int)(n);i++) #define ford(i,n) for(int i=(int)(n)-1;i>=0;i--) #define fore(i,a,b) for(int i=(int)(a);i<=(int)(b);i++) #define pb push_back #define print(v) for(int i:v) cout<>v[i]; #define sz(a) (int)(a.size()) using namespace std; typedef long long ll; typedef vector vi; typedef pair ii; typedef vector vii; typedef vector vvi; int main(){ IOS int n,m,k;cin>>n>>m>>k; vector> v(k); forn(i,k){ cin>>v[i].x.x>>v[i].x.y>>v[i].y.x>>v[i].y.y; v[i].x.x--;v[i].x.y--,v[i].y.x--; } sort(all(v)); // for(auto& p:v){ // cout<& a, const pair& b){ if(a.y.y==b.y.y) return a.y.x,ve>,decltype(com)> pq(com); for(int i=0;ii) break; // cout<<"ptr"<> DP(n+1,vector(m+1,-1)); // cout<<"s"< sol=[&](int i,int j){ // cout<=n) return 0LL; if(DP[i][j]!=-1) return DP[i][j]; DP[i][j]=sol(best[i].y+1,j)+(ll)best[i].x; if(j){ DP[i][j]=min(DP[i][j],sol(i+1,j-1)); } // cout<
• » » » » » » » 9 months ago, # ^ | 0 Figured already.I missed a condition in the DP part of code,which in no way could have caused byte misalignment.The diagnostic was incorrect. Thanks anyway.
• » » » » » » » 9 months ago, # ^ | +3 Your defines are awful.
» 9 months ago, # | ← Rev. 3 → +5 Can anyone explain the solution of problem E in detail,please ? The "SWEEP LINE" term mentioned in the editorial isn't clear to me.I understood the second part but i'm stuck with the first part. Thank you.
• » » 9 months ago, # ^ | ← Rev. 5 → +16 I'll try to explain, but my words might be unclear. Perhaps you could take a look at my submission. It might help. In my code dp[t][dist] = minimum possible money that can be earned by Bob when starting from time t, and dist disturbances are remaining with Alice. The answer that we are looking for is dp[1][m]. The recurrence for this would be as follows: If best letter at time t is (starti, finishi, delayi, rewardi), then the recurrence would be as follows: dp[t][dist] = min(dp[t + 1][dist - 1], dp[delayi + 1][dist] + rewardi) The first case corresponds to Alice using up a disturbance, and the second case is when she doesn't. If there is no letter that Bob can open at time t, then the dp recurrence is: dp[t][dist] = min(dp[t + 1][dist], dp[t + 1][dist - 1]) You can work out the base case of this dp yourself. A more fundamental question is that why does this DP work? Especially because we are not keeping any state of what are the envelopes used up or remaining at a particular time. One insight to have is this: If we could ignore the start and finish times of the envelopes, then there is a preference order amongst all the envelopes. (Basically prefer the ones that give more prize, and choose the ones that will delay more). However, at a time instant t, you can only choose amongst the envelopes that belong to that time instant, i.e. starti ≤ t ≤ finishi. You don't need to worry about any other envelopes. In particular you don't have to worry that these envelopes are already chosen, because if it had been chosen already, then in the recurrence, it will check only after the delay. So if we are iterating through time backwards and keeping a Priority Queue of the valid envelopes, then we need to insert and remove envelopes at the correct times (at finishi and starti) when they become valid and invalid. This is what is being referred to as SWEEP LINE. Maybe the phrase is used a bit incorrectly used in this case (but there are some strong similarities in the ideas).
• » » » 9 months ago, # ^ | +5 What is dp[delayi + 1][rewardi]?dp[delayi + 1][dist]?
• » » » » 9 months ago, # ^ | +10 Maybe dp[delayi + 1][dist] + rewardi?
• » » » » » 9 months ago, # ^ | +3 Apologies. Thanks. Fixed.
• » » » » 9 months ago, # ^ | 0 I've corrected the error. Thanks for pointing it out.
» 9 months ago, # | ← Rev. 2 → 0 I tried Problem E.I let dp[i][j] be the coins Bob can obtain in time j with Alice already disturbed i times.Let w[j] be the coin from the next envelope Bob can get if he is free at time j.Let d[j] be the next time Bob can be free after getting this envelope.then dp[m][j]=dp[m][d[j]]+w[j]dp[i][j]=min(dp[i+1][j+1],dp[i][d[j]]+w[j])But I got a Wrong Answer at test 5, which is very hard to find out the wrongs in my code.I'll keep trying...
• » » 9 months ago, # ^ | ← Rev. 2 → 0 Please notice that f[ * ][m] might not be the optimal policy. Alice may not use all m chances to disturb Bob, which might reach a smaller answer.
» 9 months ago, # | +5 In 1106A, scanf("%s", a[i] + 1);, why a[i]+1 is used? Why it is needed to add 1 there?
• » » 9 months ago, # ^ | 0 If you do not add 1, the variable j should start from 0.
» 9 months ago, # | +13 There's something worthy to note about problem F that I came up with while testing. After reducing the problem to finding an integer x such that xp = m using matrix expo, we can find x in O(log(n)) but if and only if gcd(p, mod - 1) = 1. Let inv be the modular inverse of p with respect to mod - 1:
• » » 9 months ago, # ^ | 0 It was this line of thinking that led me in quite a wrong direction in the contest.For the case where gcd(p, mod - 1) > 1, then we can reduce it to the case where p divides mod - 1 by the following steps: g = gcd(p, mod - 1) = ap + b(mod - 1)Then, Because mod - 1 has only three prime factors 2, 7 and 17, I was trying to find a special solution for those three powers, and entered into the number theory rabbit hole of quadratic residues.
• » » 8 months ago, # ^ | 0 Is there a way to solve the equation when gcd(p, mod-1) != 1 or this solution works only when gcd = 1?
• » » » 8 months ago, # ^ | +8
• » » » » 8 months ago, # ^ | 0 Great, Thank you!
» 9 months ago, # | ← Rev. 2 → 0 I also did the same thing mentioned in the editorial for B problem . Just used a set and simulated the process . I am getting a runtime error on the first case but in my local machine I am getting a right output . Can someone help me with the error ?Link for the submission is : 49266362
» 9 months ago, # | +5 At first I think D is very similar to NOIP2018 D2T1,but at last, it turns out that I am wrong.The code is same as the wrong code I wrote at NOIP2018.
• » » 9 months ago, # ^ | 0 same, took me a long time to find out that it is not NOIPD2T1 with extra edges :(
• » » » 8 months ago, # ^ | -8 真的是两个中国人说话讲缨语然后我也是D2T1写挂/px
» 9 months ago, # | 0 Why am I getting runtime error on pretest 2 in Problem D? It is running fine on other online ide like gfg. https://codeforces.com/contest/1106/submission/49308034 Pls help!
• » » 9 months ago, # ^ | 0 Do not use set, use priority_queue instead, it will work fine. It's with the erase function of set which is showing undefined behavior. I also faced same issue.
• » » » 9 months ago, # ^ | 0 Thanks buddy
» 9 months ago, # | 0 Is there anyone who can explain the problem E in details? I'm trying to figure it out, but it doesn't work. I hope someone do me a favor please.I understood what the problem requires, but I can't understand the way 1) calculating max coin 2) what structure you are using in calculating max coin (Why use map(set)?, what's the purpose of it?)Because of my understandings, I also don't know about DP definition. After understanding the way above, I will try to understand DP approach.Thanks in advance...
• » » 9 months ago, # ^ | ← Rev. 2 → 0 Perhaps you could take a look at my comment above. Comment Link
• » » 9 months ago, # ^ | 0 For DP part you can look at proofbycontradiction's comment above.We use map because we need to get the max event of each 'time' on timeline effectively.Map is a balanced binary search tree, which means elements in it is sorted.Therefore, we can get the max event(equals to the max coin) among all events in the tree by cur.begin()->first.In order to calculate the max coin of every 'time' on the timeline effectively, we introduce two vector array add[N] and remove[N]. At any 'time' i, if there're some events in add[i], just insert them into the tree(map). If there're some events in remove[i], remove every of them from the tree.By doing this, it is guaranteed that at all 'time' i, for all events in the tree, the event's range includes i, and there doesn't exist such an event in the tree such that this event's range doesn't include i. Also, all events whose range includes i are in the tree.When reading data from input, just add[s].push_back(event), remove[t + 1].push_back(event). We use t + 1 because s and t are inclusive, so you should remove it from tree at t + 1.You can calculate the max event of current 'time' during dp, or calculate them before dp. Both works.Sorry for bad English, hope this helps! :DPlease correct me if there's any mistake, thanks!
» 9 months ago, # | 0 I'd like to ask about C problem.I came to know that we want to minimize sum bi * ci.But, I can't know , why we sort ai(bi, ci), when we come to know we want to minimize sum bi * ci. Is there provement?
• » » 9 months ago, # ^ | +3 Here is the proof written by anayk 1: Groups must be formed as pairs of 2.Proof: As (a+b)^2 > a^2 + b^2 for positive a, b, we want to minimize group size. As size has to be greater than 1, and N is even, all group sizes must be 2.Now, we have to pair each number with another such that sum is least. We double this sum. This doubled sum can now be represented as summation (ai + bi)^2 where ai and bi are both permutations of the given numbers.2: ai and bi must be oppositely ordered sequencesProof: summation (ai + bi)^2 = summation (ai)^2 + summation (bi)^2 + summation (2*ai*bi); We want to minimize this sum. We observe that the first two terms are constants (sum of squares of all given numbers). So, we want to minimize the 3rd term. This is done when ai and bi are oppositely ordered, proof is a direct application of rearrangement inequality.Now, observe that this doubled sum corresponds to the pairing where ai is paired with a(N-i+1), with i <= N/2. Thus, we are done
• » » » 9 months ago, # ^ | 0 Thank you, ChevishaUchiha.but, I'd like to know, why when we want to minimize sum a_i * b_i , we sort?For example, a=(2, 3, 5, 7) we have to pair(2, 7), (3, 5). But I can't understand why is this correct?
• » » » » 9 months ago, # ^ | ← Rev. 4 → 0 Because you want to pair the largest one with the smallest one, next largest with next smallest, and so on. Since we are squaring each group having one group even slightly larger than it needs to be costs us much more than if we weren't squaring. e.g. (2+7)^2+(3+5)^2 == 9^2+8^2 == 145or try (7+3)^2+(5+2)^2 == 10^2+7^2 == 149See how we have an average of 8.5 in the first example terms and 8.5 in the second example terms, but because we are squaring its much better to have terms 8 and 9 to square as opposed to 7 and 10 to square. The terms have to be paired according to the question, makes no sense to make them groups of three or higher because every time you increase the square geometrically. So we know we are looking for pairs, and its best to keep the pair sums as close as possible. Think 10 and 1 vs 5 and 6, both sum to 11, but their sum of squares 5 and 6 is 61 and sum of squares 10 and 1 is 101. Do that for any sum you can think of, you will see the closer they are the better. Take 11:10^2 + 1^2 == 1019^2 + 2^2 == 858^2 + 3^2 == 737^2 + 4^2 == 656^2 + 5^2 == 61So how do you take an array of numbers and make their pairs as close as possible? You sort them, two pointers, one in front i and one in back j. Sum and square front plus back, increment front, decrement back, while i
• » » » » 9 months ago, # ^ | ← Rev. 2 → +3 Search for rearrangement inequality. It's simple to prove, intuitive and quite useful.Basically, it states that if ai and bi are permutations of 2 sequences of real numbers, then the sum ai*bi will be minimized when ai and bi are oppositely sorted and will be maximized when they are similarly sorted.Intuitively, if you want to purchase some items from a store, such that you want ai objects having price pi, then to minimize the total cost, you want to buy the costliest item as less times as possible and so on. So, if prices pi are increasing, then ai must be decreasing.
• » » » 9 months ago, # ^ | 0 I understood the second proof by reading about proof og Rearrangement Inequality. But couldn't understand your proof for 1st part — having only groups of size 2. Can you please elaborate why only groups 2 will lead to optimal solution?
» 9 months ago, # | ← Rev. 2 → 0 In the last part of problem F, there is no need to use extended euclidean algorithm to find a solution. Let's assume we want to find X such that Such a number X exists only if y is divisible by gcd(n, p - 1). If yes, then we can reduce the fraction to . Obviously now, is invertible mod(P - 1) since they are co-prime, and n - 1 = nphi(p - 1) - 1
» 9 months ago, # | 0 For problem E,which should be the times at which Alice will disturb for test 3? I can't find any combination in which he only gets 11 coins.
• » » 9 months ago, # ^ | 0 Bob collects 4 coins at time 1, collects 7 coins at time 6, Alice disturbs at time 11 and 12.
» 9 months ago, # | 0 Can anybody explain how is answer for testcase3 = 11 for Div2E?
• » » 9 months ago, # ^ | 0 Bob collects 4 coins at time 1, collects 7 coins at time 6, Alice disturbs at time 11 and 12.
• » » » 9 months ago, # ^ | 0 Thanks missed "after that" in statement : If Bob chooses to collect the coins in the i-th red envelope, he can do it only in an integer point of time between si and ti, inclusive, and he can't collect any more envelopes until time di (inclusive) after that.
» 9 months ago, # | 0 Can someone please help me fix my solution for problem E 49335045 I get WA on test 10
• » » 9 months ago, # ^ | ← Rev. 3 → +5 You shouldn't use LONG_MAX because LONG_MAX = 2 ^ 31 — 1. Try using 1e18.UPD AC with #define LONG_MAX 1e18 — 49360125
• » » » 9 months ago, # ^ | +1 Or use LLONG_MAX instead of LONG_MAX.
• » » » 9 months ago, # ^ | 0 Thank you very much! I am new to c++ so this helped me a lot.
» 9 months ago, # | ← Rev. 2 → +3 Hi, everybody! Please allow me to advertise my unofficial Chinese editorial of this round: Codeforces Round #536(Div.2)For problem C, I think author's proof may be incomplete (or just lack of detailed explanations):If bi and ci are settled, we can simply use Rearrangement Inequality to pair them.But why must bi and ci be {a1, a3, a5, ..., an - 1} and {a2, a4, a6, ..., an} ?My proof to this detail:Any such pairing ({bi}, {ci}) can be "doubled" and be come a complete pairing between {ai} and {ai} .For example: for 1 2 3 4 5 6, a pairing (3,1) (2,4) (5,6) can be seen as 1 2 3 4 5 6 3 4 1 2 6 5 so a pairing can be represent as A , a complete pairing can be represent as B = f(A) (note that B = f(A) always exists but not A = f - 1(B)).ThenThuscost(Ai) ≥ cost(A * )Q.E.D.
» 9 months ago, # | 0 For 1st part of problem E, how to include the condition for d(till when other envelopes can't be collected)?
» 9 months ago, # | 0 How to solve B ?
» 9 months ago, # | 0 In Problem F , Can anyone explain more about h_k ? and how can I obtain it from matrix exponentiation using linear recursive equation ?
» 9 months ago, # | ← Rev. 6 → 0 Does problem F have a simpler solution?Let's first say that f_i is a product of some powers of f1, f2,..., fk. This is true for all i. We can then consider the recurrence as an additive recurrence over k dimensional vectors f_i = b1*f_i-1 +... +bk*f_i-k. We are basically considering a "logarithmized" version of the recurrence.Then we can have a transition matrix M of size k x k to get from f_i, f_i+1,...,f_i+k to f_i+k+1,f_i+k+2,...,f_i+2k.We then use fast exponentiation with matrix multiplication to get from f1...fk to fn (mod p-1 for values). The result will be that we will have fn expressed as a product of powers of f1...fk. We will use again fast exponentiation on numbers now to compute the values for the powers of f1..fk-1. We multiply fn by the inverse of this product and we get fk^b = a (mod p). Finally we are left with just computing the b-th root of a (mod p). Final complexity should be O(k^3*log n + finding discrete b-th root). Does this make any sense to you? LATER EDIT: I just realized that this is more or less what the Tutorial also proposed.
» 9 months ago, # | 0 Hi,can you tell me the test of 7 on Problem B? i wrong answer on it long time
» 9 months ago, # | 0 Can Somebody help me in Problem E. I am getting TLE in testcase #13 I am using similar approach to this comment https://codeforces.com/blog/entry/64928?#comment-488863
» 9 months ago, # | 0 Solve this linear equation in or if you apply Fast Fourier Transform. In this equation, the mod number is 998244352 which doesn't have primitive root. You can't optimize it with NTT. I think the fastest solution should be instead of .
• » » 9 months ago, # ^ | ← Rev. 7 → 0 In fact, for arbitrary modulus (not very large) NTT, there is a trick called three-modulus-NTT (translated from Chinese). Suppose that we want to get . The intuition is that the coefficients for the answer polynomial C(x) = A(x)B(x) won't exceed nP2, where n is . Thus we choose 3 NTT-friendly modulus (whose multiplication is greater than nP2) to get the results and apply Chinese Reminder Theorem to retrieve the original answer back. The details can be found in some Chinese blogs (I'm sorry that I haven't found any in English).
» 9 months ago, # | ← Rev. 2 → +5 I think maybe the complexity of BSGS could be reduced to .In number-theory part, if we search for an x satisfying instead of , we have , so we don't need to get inversion modulo p which takes time.In data-structure part, maybe we can use unordered-map instead of map, then the total complexity is (although I have never implemented it before)
» 8 months ago, # | ← Rev. 2 → 0 .
» 6 months ago, # | 0 Could someone explain why in problem D sorting all the adjacency list and then using Dfs as it will always traverse the lowest nodes first gets a wrong answer? here is my code https://codeforces.com/contest/1106/submission/53541920
• » » 3 months ago, # ^ | ← Rev. 3 → 0 You will understand your mistake if you try this case 4 3 1 2 1 3 2 4 The mistake is we do not want to specifically use dfs or bfs, just a pq will do.
» 4 months ago, # | 0 Can anyone please explain how the state has been reduced in problem E.
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2019-10-21 21:26:33
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https://www.thejournal.club/c/paper/55655/
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#### An All-Around Near-Optimal Solution for the Classic Bin Packing Problem
##### Shahin Kamali, Alejandro López-Ortiz
In this paper we present the first algorithm with optimal average-case and close-to-best known worst-case performance for the classic on-line problem of bin packing. It has long been observed that known bin packing algorithms with optimal average-case performance were not optimal in the worst-case sense. In particular First Fit and Best Fit had optimal average-case ratio of 1 but a worst-case competitive ratio of 1.7. The wasted space of First Fit and Best Fit for a uniform random sequence of length $n$ is expected to be $\Theta(n^{2/3})$ and $\Theta(\sqrt{n} \log ^{3/4} n)$, respectively. The competitive ratio can be improved to 1.691 using the Harmonic algorithm; further variations of this algorithm can push down the competitive ratio to 1.588. However, Harmonic and its variations have poor performance on average; in particular, Harmonic has average-case ratio of around 1.27. In this paper, first we introduce a simple algorithm which we term Harmonic Match. This algorithm performs as well as Best Fit on average, i.e., it has an average-case ratio of 1 and expected wasted space of $\Theta(\sqrt{n} \log ^{3/4} n)$. Moreover, the competitive ratio of the algorithm is as good as Harmonic, i.e., it converges to $1.691$ which is an improvement over 1.7 of Best Fit and First Fit. We also introduce a different algorithm, termed as Refined Harmonic Match, which achieves an improved competitive ratio of $1.636$ while maintaining the good average-case performance of Harmonic Match and Best Fit. Finally, our extensive experimental evaluation of the studied bin packing algorithms shows that our proposed algorithms have comparable average-case performance with Best Fit and First Fit, and this holds also for sequences that follow distributions other than the uniform distribution.
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2022-08-14 22:06:30
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https://chemistry.stackexchange.com/questions/59368/work-done-in-reversible-isothermal-expansion
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# Work done in reversible isothermal expansion
A cylinder fitted with a frictionless piston is filled with $$10~\mathrm{mol}$$ of gaseous carbon tetrachloride and immersed in a large bath of water maintained at $$300~\mathrm{K}$$. The volume of the gas is initially at $$V_1~\mathrm{L}$$, and the external pressure on the piston is slowly decreased until the volume reaches $$V_2~\mathrm{L}$$. Calculate the work for this process, in $$\mathrm{kJ}$$, assuming the gas can be treated as ideal.
I think I have the general idea of how to solve this problem but I'm not sure how to find the external pressure. So far, I'm using the equation
$$\int \mathrm{d}w = \int_{V_2}^{V_1}-p_{\mathrm{ext}}\,\mathrm{d}V$$
since it is pressure-volume work and have tried calculating $$p_{\mathrm{ext}}$$ using $$pV=nRT$$ and the final volume of the system, but this doesn't seem to give the correct answer. How does it have to be solved?
• If you could show us your work, then someone here might be able to point out where exactly you went wrong. Cheers! – getafix Sep 20 '16 at 2:45
I agree with getafix, if you would like an answer that is more tailored to you, you should show us exactly what you've done.
However, I am going to make a (hopefully educated) guess that what you did was to pull $p_\mathrm{ext}$ out of the integral. That is incorrect, because $p_\mathrm{ext}$ is not a constant here.
This process is known as an isothermal expansion - isothermal because the temperature remains constant throughout - and expansion because volume is increased. In thermodynamics it is very important to note which variables are held constant, because then that lets you decide which formula is appropriate to use, or how to derive such formulae).
Since the process is reversible, the external pressure must always be equal to the pressure exerted by the gas, which can be calculated via the ideal gas law $pV = nRT$. Therefore, you have (where 1 and 2 denote the initial and final state respectively)
\begin{align} w &= -\int_1^2 p\,\mathrm{d}V \\ &= -\int_1^2 \frac{nRT}{V}\,\mathrm{d}V \end{align}
and now since $T$ is a constant, you can take it out of the integral (along with $n$ and $R$ which are also constants)
\begin{align} w = &= -nRT\int_1^2 \frac{1}{V}\,\mathrm{d}V \\ &= -nRT\ln\left(\frac{V_2}{V_1}\right) \end{align}
Since you weren't given numerical values for $V_2$ and $V_1$, you can't evaluate a numerical value for $w$. The closest you can get is therefore
$$w = (-24.9~\mathrm{kJ})\ln\left(\frac{V_2}{V_1}\right)$$
• It's worth noting that carbon tetrachloride has a boiling point (at atmospheric pressure) of well over 300 K, so we can at least infer that both V1 and V2 correspond to starting and ending pressures that are significant below one atmosphere. – Curt F. Jul 11 at 22:58
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2019-08-25 11:05:01
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http://www.physicspages.com/2016/07/30/relative-humidity-and-dew-point/
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# Relative humidity and dew point
Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.42.
The vapour pressure equation gives the phase boundary curve between the liquid and gas phases of a substance if we can assume that the gas is an ideal gas:
$\displaystyle P_{v}=Ke^{-L/RT} \ \ \ \ \ (1)$
where ${K}$ is a constant and ${L}$ is the latent heat of vapourization. If we’re interested in the vapour pressure of water for temperatures typical of Earth’s weather, we could look at a temperature range of ${0^{\circ}\mbox{ C}}$ to ${40^{\circ}\mbox{ C}}$. From Schroeder’s Figure 5.11, we can use the values at ${T=25^{\circ}\mbox{ C}}$ to estimate ${K}$. Here, ${L=43.99\times10^{3}\mbox{ J mol}^{-1}}$ and ${P_{v}=0.0317\times10^{5}\mbox{ Pa}}$. The gas constant is ${R=8.314}$ in SI units. This gives
$\displaystyle K$ $\displaystyle =$ $\displaystyle P_{v}e^{L/RT}\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.0317\times10^{5}e^{43.99\times10^{3}/\left(8.314\times298\right)}\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1.63\times10^{11}\mbox{ Pa} \ \ \ \ \ (4)$
This gives a graph as follows:
We can see from the graph that ${P_{v}}$ roughly doubles for each increase in temperature of 10 degrees.
The relative humidity is defined as the ratio of the actual vapour pressure to the equilibrium vapour pressure at a given temperature. A high humidity is what makes it feel hotter than the actual temperature on a warm summer day, because it decreases the evaporation rate of the sweat that is attempting to cool you down.
The dew point is the temperature at which the current vapour pressure would be the equilibrium vapour pressure. At the dew point, the air is said to be saturated, meaning that liquid water will not spontaneously evaporate. If the vapour pressure exceeds the equilibrium pressure, vapour will condense spontaneously to form dew, fog or rain.
As an example, suppose the air temperature is ${30^{\circ}\mbox{ C}=303\mbox{ K}}$. From the above, this gives an equilibrium vapour pressure of
$\displaystyle P_{v}\left(303\right)=4.249\times10^{3}\mbox{ Pa} \ \ \ \ \ (5)$
If the relative humidity is 90%, the actual vapour pressure is
$\displaystyle P=0.9P_{v}\left(303\right)=3.824\times10^{3}\mbox{ Pa} \ \ \ \ \ (6)$
The dew point at this humidity is found by solving 1 for ${T}$:
$\displaystyle T=\frac{L}{R\ln\left(K/P\right)} \ \ \ \ \ (7)$
Plugging in the numbers gives
$\displaystyle T=301.2\mbox{ K}=28.2^{\circ}\mbox{ C} \ \ \ \ \ (8)$
For a relative humidity of 40%, the vapour pressure is
$\displaystyle P=0.4P_{v}\left(303\right)=1.700\times10^{3}\mbox{ Pa} \ \ \ \ \ (9)$
The dew point is now
$\displaystyle T=287.9\mbox{ K}=14.9^{\circ}\mbox{ C} \ \ \ \ \ (10)$
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2017-02-26 12:25:51
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|
https://www.studysmarter.us/explanations/math/calculus/divergence-test/
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# Divergence Test
Sometimes proving that a series diverges can be quite a challenge! Using the Divergence Test, also called the $$n^{th}$$ Term Test for Divergence, is a simple test you can do to see right away if a series diverges. This can save you considerable time in the long run.
## Divergence Tests in Calculus
Many of the tests used for series will have a part that also talks about divergence.
For example, the Direct Comparison Test and the Limit Comparison Test both have a part that talks about convergence and another that talks about divergence. The Integral Test, Ratio Test, and Root Test do as well. Some series, such as the P-series, Geometric series, and Arithmetic series, have known conditions for when they diverge and converge. So when you are looking for divergence tests, be sure to look at Convergence tests as well.
## Series Divergence Tests
Here you will see a test that is only good to tell if a series diverges. Consider the series
$\sum_{n=1}^{\infty} a_n,$
and call the partial sums for this series $$s_n$$. Sometimes you can look at the limit of the sequence $${a_n}$$ to tell if the series diverges. This is called the $$n^{th}$$ term test for divergence.
$$n^{th}$$ term test for divergence.
If
$\lim\limits_{n\to \infty}a_n$
does not exist, or if it does exist but is not equal to zero, then the series
$\sum_{n=1}^{\infty}a_n$
diverges.
What is the wrong way to use the test?
The most common mistake people make is to say that if the limit of the sequence is zero, then the series converges. Let's take a look at an example to show why that isn't true.
Can you use the $$n^{th}$$ term test for divergence to say that if
$\lim\limits_{n\to \infty}a_n=0$
the series converges?
Solution
Let's take a look at two examples.
First look at the Harmonic series
$\sum_{n=1}^{\infty}\frac{1}{n}.$
For this series, we have
$\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\frac{1}{n}=0,$
but you know that the series diverges.
Next look at the P-series with $$p=2$$,
$\sum_{n=1}^{\infty}\frac{1}{n^2}.$
If you look at the limit of the sequence of terms for this series, you get,
\begin{align}\lim\limits_{n\to\infty}a_n&=\lim\limits_{n\to \infty}\frac{1}{n^2}\\ &=0\end{align}
as well, but this series converges.
So in fact, if the limit is zero, the series might converge and it might diverge, you just can't tell.
## Proof of the $$n^{th}$$ Term Test for Divergence
Let's take a look at the $$n^{th}$$ term test for divergence is true. Sometimes in math, you prove a statement like "if A is true then B is true", and sometimes it is easier to prove the contrapositive, which is "if B is false then A is false".
For the $$n^{th}$$ term test for divergence, it is easier to show the contrapositive.
So what is the contrapositive for the $$n^{th}$$ term test for divergence?
The statement B is "the series diverges", and saying that "B is false" is the same as saying "the series converges".
The statement A is "the limit of the sequence either doesn't exist or does exist and isn't zero", and "A is false" is the same as saying "the limit of the sequence is zero". That means we will look at the proof of:
If $$\sum\limits_{n=1}^{\infty}a_n$$ converges then
$\lim\limits_{n\to \infty}a_n=0.$
To do this, you will need to look at the partial sums for the series. The sequence of partial sums is defined by $s_n=\sum_{k=1}^{n}a_k.$
The previous term in the sequence of partial sum would be $s_{n-1}=\sum_{k=1}^{n-1}a_k$
Subtracting them gives,
\begin{align} s_{n}-s_{n-1}&=\sum_{k=1}^{n} a_{k}-\sum_{k=1}^{n-1}a_{k}\\&=(a_{1}+a_{2}+\cdots+a_{n-1}+a_{n})-(a_{1}+a_{2}+\cdots+a_{n-1})\\&=a_{n}.\end{align}
You already know that the series converges, which implies that the sequence of partial sums converges as well, or in other words
$\lim\limits_{n\to\infty}s_n=L$
for some real number $$L$$. Now taking the limit of the subtraction of partial sums,
\begin{align}\lim\limits_{n\to\infty}[s_n-s_{n-1}]&=\lim\limits_{n\to \infty}s_n-\lim\limits_{n\to\infty}s_{n-1}\\&=L-L\\&=0.\end{align}
But you also know that
$\lim\limits_{n\to\infty}[s_{n}-s_{n-1}]=\lim\limits_{n\to\infty} a_n,$
which means that
$\lim\limits_{n\to\infty}a_n=0.$
## Examples Using the $$n^{th}$$Term Test for Divergence
Let's look at some examples of how to properly use the $$n^{th}$$ term test for divergence.
What can you say about the convergence or divergence of the series
$\sum_{n=1}^{\infty}\frac{2n+3}{7n-1}$
using the $$n^{th}$$term test for divergence?
Solution
For this series, $a_{n}=\frac{2n+3}{7n-1},$
and
\begin{align}\lim\limits_{n\to\infty}a_n &=\lim\limits_{n\to\infty}\frac{2n+3}{7n-1} \\ &=\frac{2}{7}.\end{align}
So the sequence converges, but the limit isn't zero. Then by the $$n^{th}$$ term test for divergence, the series diverges.
Let's take a look at another example.
What can you say about the convergence or divergence of the series $\sum_{n=1}^{\infty}(-1)^n,$
using the $$n^{th}$$ term test for divergence?
Solution
For this series, $$a_{n}=(-1)^n$$, and the limit of this sequence doesn't exist. So by $$n^{th}$$ term test for divergence, the series diverges.
## Integral Divergence Test
As mentioned already, the Integral Test has a part that talks about divergence. So for more information on divergence when using the Integral Test, see Integral Test.
## Divergence Test - Key takeaways
• Many tests can be used to tell if a series converges or diverges, such as the Integral Test or the Limit Comparison Test.
• The $$n^{th}$$ term test for divergence is a good first test to use on a series because it is a relatively simple check to do, and if the series turns out to be divergent you are done testing.
• If $\sum\limits_{n=1}^{\infty}a_{n}$ converges then $\lim\limits_{n\to\infty}a_n=0.$
• $$n^{th}$$ term test for divergence: If $\lim\limits_{n\to\infty}a_{n}$
does not exist, or if it does exist but is not equal to zero, then the series $\sum_{n=1}^{\infty} a_n$ diverges.
It is a way to look at the limit of the terms of a series to tell if it diverges.
Look at the limit of the terms of the series. If that limit is not zero then the series diverges.
The divergence test is a good one to use on any series as a basic check. If the divergence test shows the series doesn't converge, then you are done testing.
By looking at the difference of the partial sums.
We say that a series is convergent if its partial sum converges to a finite limit. If the partial sum does not have a finite limit, or the limit does not exist, the series is divergent.
## Final Divergence Test Quiz
Question
What is a good first test to do to see if a series diverges or converges?
The nth term test for divergence. It is relatively simple to apply it to see if the series diverges. If it diverges, you are done testing.
Show question
Question
Is the nth term test for divergence the only divergence test?
No, there are lots of tests that also look at divergence in addition to convergence, e.g. the Integral Test, the Ratio Test, and the Root Test. Not to mention the Direct Comparison Test and the Limit Comparison Test.
Show question
Question
What is the wrong way to use the $$n^{th}$$ term divergence test?
The most common mistake people make is to say that if the limit of the sequence is zero, then the series converges.
Show question
Question
If
$\lim\limits_{n\to \infty}a_n$
does not exist what can you say about convergence of
$\sum_{n=1}^{\infty}a_n?$
By the $$n^{th}$$ term test for divergence, you can say the series diverges.
Show question
Question
If
$\lim\limits_{n\to \infty}a_n \ne 0$ what can you say about convergence of the series
$\sum_{n=1}^{\infty}a_n?$
By the $$n^{th}$$ term test for divergence, you can say the series diverges.
Show question
Question
If
$\lim\limits_{n\to \infty}a_n = 0$ what do you know about convergence of the series
$\sum_{n=1}^{\infty}a_n ?$
Nothing. That isn't enough information to tell if the series converges or diverges.
Show question
Question
Is it true that if $$\sum\limits_{n=1}^{\infty}a_n$$ converges then
$\lim\limits_{n\to \infty}a_n=0?$
Yes. This is the contrapositive of the $$n^{th}$$ term test for divergence.
Show question
Question
Is it true that if $$\sum\limits_{n=1}^{\infty}a_n$$ diverges then
$\lim\limits_{n\to \infty}a_n=0?$
No, just knowing the series diverges doesn't tell you anything about the limit of the sequence.
Show question
Question
Why do you use the $$n^{th}$$ term test on a new series as a first test?
Because it is relatively simple to use and can save you a lot of work checking the various convergence tests.
Show question
Question
Other than the $$n^{th}$$ term test, what is another test used to check for divergence?
The Integral Test.
Show question
Question
What is the contrapositive of the statement "if A is true then B is true"?
The contrapositive is "if B is false then A is false"
Show question
Question
Give an example of a series where
$\lim\limits_{n\to \infty}a_n=0$
but the corresponding series converges.
The P-series with $$p=2$$
Show question
Question
Give an example of a series where
$\lim\limits_{n\to \infty}a_n=0$
but the corresponding series diverges.
The Harmonic Series.
Show question
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2022-11-30 16:51:07
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https://pabloariasal.github.io/2018/06/26/std-variant/
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# std::variant Doesn't Let Me Sleep
Where I come from we have a saying that I like very much: de la risa al llanto, which translates to from laughter to tears. It’s used in situations that appear beneficial at first, but end up turning into a curse; when the line between the bad and the good dissipates, when the dream becomes the nightmare.
Last month I attended a talk by Juanpe Bolivar: the most valuable values. In his talk he made use of C++17’s new sum type proposal: std::variant. Sum types are objects that can vary their type dynamically. A std::variant is similar to a union: it allocates a fixed portion of memory and reuses it to hold a value of one of several predefined alternative types at a time:
#include <variant>
std::variant<int, float, char> a = 42; //a is an int
a = 3.141f; //a holds now a float
char my_char = std::get<char>(a); //error: a holds a float, not a char
a = 'c'; //all cool bro
std::variants are superior to unions because they are type-safe: the variant object knows which type is currently being held. Self harm becomes a little bit more difficult if you use variants over unions.
Often you want to do something different with a variant depending on the type currently being held. In such cases you need to visit the variant with std::visit. std::visit receives two parameters: the variant itself and a visitor, which is a function object that is callable for all types supported by the variant:
struct Visitor
{
void operator()(int a)
{
//Called if variant holds an int
}
void operator()(float a)
{
//Called if variant holds a float
}
void operator()(char a)
{
//Called if variant holds a char
}
};
std::visit(a, Visitor{});
In his talk Juanpe presented a convenience function for creating such visitors, make_visitor(), that constructs a visitor like the one above from a list of lambdas:
auto visitor = make_visitor(
[](int b)
{
//Called if variant holds an int
},
[](float b)
{
//Called if variant holds a float
},
[](char b)
{
//Called if variant holds a char
}
);
make_visitor() is not part of the standard and it instantly caught my eye. How would the implementation of such function look like? I closed my eyes and after some consideration I managed to scribble something like this in my head:
template<typename ...Ts>
struct Visitor : Ts...
{
Visitor(const Ts&... args) : Ts(args)...
{
}
}
template<typename ...Ts>
auto make_visitor(Ts... lamdbas)
{
return Visitor<Ts...>(lambdas...);
}
This isn’t precisely first semester programming class, but I assure you that it looks much more scary than it really is. For the above code to make any sense there are two language features that need to be understood: parameter packs and closure classes. Let’s revisit them very quickly.
# Parameter Packs
I usually tell people that my favorite C++11 feature are parameter packs, also known in the streets as variadic templates. A parameter pack is a template that can be instantiated with any number of types. For example, you can use parameter packs to define functions that take arbitrary arguments:
template<typename ...Ts>
void f(Ts... args)
{
}
f(), f(8, 9.0f), f("Hello") are all possible calls to f. For every call, the compiler will perform template type deduction and instantiate definitions of f with Ts = {}, Ts = {int, float} and Ts={const char*}, respectively.
You may have noticed the ellipsis operator ... in f(Ts... args). This is called a pack expansion and it lets you unroll a parameter pack. Unrolling basically means constructing a comma-separated list from a pattern containing a parameter pack.
For example, if Ts={int, char}, Ts... will expand the pack to int, char. If you give the pack a name, the name will be expanded as well: Ts... args becomes int args0, char args1. This also works on more complex patterns: const Ts&... args, will be const int& args0, const char& args1, Ts(args)... expands to int(args0), char(args1). I hope you get the idea.
The Visitor struct above inherits from all types in its parameter pack Ts. This is achieved by expanding Ts in the inheritance list: struct Visitor: Ts.... Note that pack expansion is also used to define a copy constructor that propagates the call to the base classes constructors.
This is how an instantiation of Visitor<A,B,C> would look like after expansion:
struct Visitor : A, B, C
{
Visitor(const A& arg0, const B& arg1, const C& arg2) : A(arg0), B(arg1), C(arg2)
{
}
}
Visitor<A,B,C> make_visitor(A lambda1, B lambda2, C lambda3)
{
return Visitor<A,B,C>(lambda1, lambda2, lambda3);
}
# Closure Classes
The implementation of make_visitor(Ts... args) above returns an object that inherits from all the types in Ts. If this function is called with a set of lambdas, which is what we want, this object will be a struct inheriting from all lambdas provided. Wait, what? Can you inherit from a lambda? What is actually the type of a lambda?
Yes, you can inherit from a lambda and the type of a lambda is, well, only your compiler knows. Despite their incredible impact after their introduction, lambdas are just syntactic sugar. Lambdas are just a shortened notation for constructing callable, a.k.a function objects.
Consider the following example:
const int n{42};
auto lambda = [n](int a){return n + a;};
const auto m = lambda(8); //m is now 50
When the compiler sees this it will translate lambda into a function object. This object will be instantiated from a compiler-generated class specifically for this lambda, called a closure class. This is the the generated closure class for lambda:
class ClosureClass
{
public:
ClosureClass(int i): n(i)
{
}
int operator()(int a)
{
return n + a;
}
private:
int n;
}
Note that the captured variables of the lambda are translated as member variables in the closure class.
The snippet above effectively becomes:
const int n{42};
auto lambda = ClosureClass(n);
const auto m = lambda(8); //m is now 50
Hopefully by now the implementation of make_visitor() should be clear: it returns an object that inherits from all closure classes deduced from the lambdas provided as argument. Each one of the closure classes defines an operator() for a different type in the variant, creating an overload set in the derived class, just like the very first visitor we built.
You might be amazed by my intelligence and creativity, but don’t get fooled. This doesn’t work.
# From Laughter to Tears
Some nights ago while browsing reddit I stumbled across a blog post by Matt Klein on, guess what, std::visit . In his post, Matt makes a reasonable critique on the growing complexity of C++ and how absurdly difficult it sometimes is to solve common tasks. To defend his thesis Matt presented the same problem as Juanpe: construct a visitor for a variant from a set of lambdas. This time, however, he provided an implementation:
template <class... Ts>
struct Visitor;
template <class T, class... Ts>
struct Visitor<T, Ts...> : T, Visitor<Ts...>
{
Visitor(T t, Ts... rest) : T(t), Visitor<Ts...>(rest...) {}
using T::operator();
using Visitor<Ts...>::operator();
};
template <class T>
struct Visitor<T> : T
{
Visitor(T t) : T(t) {}
using T::operator();
};
I studied Matt’s solution for some minutes but was unable to fully grasp his intentions. I closed my browser and went to sleep, hoping to forget about it in the morning. I didn’t. Nights went by and I became more anxious; I couldn’t sleep. Why does Matt need the recursive overload? Why are those using operator() all over the place? Matt saw something I didn’t and I wasn’t going to rest until I had seen it too.
But what is Matt doing anyway?
As we saw early, parameter packs let you to define templated classes and functions that can be instantiated with arbitrary types:
template<typename ...Ts>
void f(Ts... args)
{
}
But what if you want to do anything with the parameters of f() at all? Like, for example, write all arguments of f() to stdout? In that case you need to define a recursive overload set:
template<typename T>
void f(const T& t)
{
std::cout << t << std::endl;
}
template<typename T, typename ...Ts>
void f(const T& head, const Ts&... tail)
{
std::cout << head << ", ";
f(tail...);
}
Recursive overload sets are a common pattern when dealing with variadic templates. The key idea is simple: split the parameter pack into two parts: a head and a tail. The head is the first parameter in the pack, the tail are the rest.
The pattern is always the same. You provide two overloads for your function: a recursive case and a base case. The recursive variant does something with the head and recursively call itself with the tail. Note that the size of the pack decreases after each recursive call of f().
f() will keep invoking itself with a smaller list each time until the base case is reached: the parameter pack contains just the last item. Here the compiler will, during overload resolution, call the overload that receives a single argument: f(const T& t). This breaks the recursion and we are done.
Matt does uses the same strategy in his implementation: he defines an overloaded recursive struct. In the recursive case, his struct inherits from head and recursively from an instantiation of itself using tail. Yes, you heard that right, Matt’s visitor inherits from itself:
template <class T, class... Ts>
struct Visitor<T, Ts...> : T, Visitor<Ts...>
{
Visitor(T t, Ts... rest) : T(t), Visitor<Ts...>(rest...) {}
using T::operator();
using Visitor<Ts...>::operator();
};
The recursion continues to unroll constructing an inheritance hierarchy with all types in the parameter pack, until the base case is reached and the recursion stops.
Matt brings the operator() of the super classes (remember we are dealing with closure classes here) into the scope of the derived class. He does this recursively: operator()s from all classes in the hierarchy end in the scope of the very bottom class.
# Reaching Nirvana
Similarly to my original idea, Matt constructs an object that inherits from all the closure classes in the pack. But what kept me awake at night was the usage of the recursive overload and the pulling of the operator() into the derived classes scope. Luckily a miraculous search landed me on the C++ FAQ:
In C++, there is no overloading across scopes – derived class scopes are not an exception to this general rule.
Overload sets don’t work across multiple scopes! Off course, A::operator() and B::operator() inherited from two base classes don’t overload each other in the derived class. I wasn’t building an overload set in Visitor at all! And then it finally hit me: Matt needs the recursive overload because he needs to bring all operator() into the scope of the bottom class to form an overload set!
I don’t know is Matt’s solution is a strike of genious or the product of a very troubled mind.
This story still has a happy ending, though. My solution was not entirely wrong, it was, well, ahead of its time. The C++17 standard introduces pack expansion with the using keyword:
template<typename ...Ts>
struct Visitor : Ts...
{
Visitor(const Ts&... args) : Ts(args)...
{
}
using Ts::operator()...;
}
I sleep like a baby ever since.
|
2019-08-18 03:35:03
|
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|
http://mathoverflow.net/questions/107981/generating-function-for-random-walk-hitting-time-taking-the-wrong-root
|
# Generating function for Random Walk Hitting Time, taking the wrong root
In a calculation of the hitting time for a Bernoulli random walk we have to calculate the hitting time $\tau(1)=\inf\{n\ge 0:S_n=1\}$ to reach $+1$ and the generating function has the recursion relationship $E[s^\tau]=G_\tau(s)=ps+qsE(s^{\tau'+\tau''})$ where $\tau'$ and $\tau''$ have the same distribution as $\tau$. The generating function is then one of the two solutions to $G(s)=\frac{1\pm\sqrt{1-4pqs^2}}{2qs}$, and it is easily seen to be the negative $\frac{1-\sqrt{1-4pqs^2}}{2qs}$.
My question is: What is the other solution $\left(\frac{1+\sqrt{1-4pqs^2}}{2qs}\right)$ a solution to? Does it have any meaningful interpretation?
-
## 1 Answer
It's $\dfrac{1}{qs} - G(s)$, so you could call it $E\left[ \dfrac{1}{qs} - s^\tau \right]$. Of course it's not a probability generating function, because it has negative coefficients except for the $1/(qs)$ and is not $1$ at $s=1$.
-
Well yes, but I was looking for an answer more in the vein of the Dirac's positron solution... perhaps there is nothing to see here. – David Sep 25 '12 at 1:25
|
2015-07-30 02:23:06
|
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|
https://www.expii.com/t/what-is-degree-n-nth-order-taylor-approximation-327
|
Expii
# What Is Degree N (Nth Order) Taylor Approximation? - Expii
The nth Taylor approximation of f(x) at a point x=a is a degree n polynomial, namely P(x) = ∑^(n)_(k=0) (f^k a)/(k!) (x-a)^k = f(a) + f'(a)(x-a)^1 + 1/2 f''(a)(x-a)^2 + ... + 1/(n!) f^n (a)(x-a)^n. This make sense, at least, if f is n-times-differentiable at x=a. The intuition is that f(a)=P(a), f'(a)=P'(a), f''(a)=P''(a), etc., up to f^n (a)=P^n (a): the "zeroth", first, second, etc., up to nth, derivatives match. (More concisely, f^k (a)=P^k (a) for 0 <= k <= n: the kth derivatives match for 0 <= k <= n.).
|
2021-07-30 02:12:43
|
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|
https://www.quantconnect.com/docs/v2/lean-cli/live-trading/algorithm-control
|
## Algorithm Control
### Introduction
The algorithm control features let you adjust your algorithm while it executes live so that you can perform actions that are not written in the project files. The control features let you intervene in the execution of your algorithm and make adjustments. The control features that are available to you depend on if you deploy the algorithm on your local machine or on the QuantConnect cloud servers.
### Control Local Algorithms
While your local algorithms run, you can add security subscriptions, submit orders, adjust orders, and stop their execution.
You can manually create security subscriptions for your algorithm instead of calling the AddsecurityType methods in your code files. If you add security subscriptions to your algorithm, you can place manual trades without having to edit and redeploy the algorithm. To add security subscriptions, open a terminal in the workspace that contains the project and then run lean live add-security "My Project".
#### Update Orders
To update an existing order, open a terminal in the workspace that contains the project and then run lean live update-order "My Project".
#### Liquidate Positions
To liquidate a specific asset in your algorithm, open a terminal in the workspace that contains the project and then run lean live liquidate "My Project".
### Control Cloud Algorithms
While your cloud algorithms run, you can liquidate their positions and stop their exeuction.
#### Liquidate Positions
The lean cloud live liquidate command acts as a "kill switch" to sell all of your portfolio holdings. If your algorithm has a bug in it that caused it to purchase a lot of securities that you didn't want, this command let's you easily liquidate your portfolio instead of placing many manual trades. When you run the command, if the market is open for an asset you hold, the algorithm liquidates it with market orders. If the market is not open, the algorithm places market on open orders. After the algorithm submits the liquidation orders, it stops executing.
To stop an algorithm, open a terminal in the workspace that contains the project and then run lean cloud live liquidate "My Project".
|
2023-01-27 00:45:44
|
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|
https://math.stackexchange.com/questions/1641110/proof-that-the-rubik-s-cube-group-is-2-generated/1641564
|
# Proof that the Rubik’s Cube group is 2-generated
Singmaster (1981) writes, on page 32 of his Notes on Rubik’s Magic Cube:
Frank Barnes observes that the group of the cube is generated by two moves: \begin{align*} \alpha &= L^2 B R D^{-1} L^{-1} &=(RF,RU,RB,UB,LD,LB,LU,BD,DF,FL,RD)& \\ &&\cdot (FUR,UBR,LDB,LBU,DLF,BDR,DFR)\\ \beta &= UFRUR^{-1}U^{-1}F^{-1} &=(UF,UL)_+(UR)_+(UBR,UFL)_-(URF)_+ \end{align*} Observe that $\alpha^7$ is an $11$-cycle of edges and $\alpha^{11}$ is a $7$-cycle of corners, that $\beta$ affects the edge and corner left fixed by $\alpha$, and that $$\beta^2 = (UF)_+(UL)_+(UBR)_-(UFL)_-(UFR)_-$$ [...] The remaining details are left as an exercise.
I hadn't seen this notation before, so I'll explain it here.
Notation like $(LU, BD, DF)$ means an edge cycle, in which:
• The $L$-$U$ edge moves to the $B$-$D$ edge's place, with the $L$ half ending up on the $B$ face, and the $U$ half ending up on the $D$ face.
• Similarly $BD \to DF$ and $DF \to LU$.
The notation for corners is similar.
Notation like $(UF, UL)_+$ is a twisted cycle: again, $UF \to UL$, but now $UL \to FU$; the final edge gets flipped when cycling back to the first edge. For corners, the notation is similar, but corners rotate, they don't flip. A subscript $+$ means clockwise rotation, a subscript $-$ means counterclockwise rotation.
$(UR)_+$ means a single edge is flipped. $(UBR)_-$ means a single corner is rotated counterclockwise.
I would like to show that this is indeed true, by writing each element in $\{F,B,L,R,U,D\}$ as a product of elements in $\{\alpha, \beta, \alpha^{-1}, \beta^{-1}\}$ – preferably, having those product be as short as possible. How would I go about finding them? (I'm okay with using software like GAP – if it is at all computationally possible.)
To start, we will use the example from "Analyzing Rubik's Cube with GAP". It creates the group generated by the six generators, corresponding to the six faces of the cube:
+--------------+
| |
| 1 2 3 |
| |
| 4 up 5 |
| |
| 6 7 8 |
| |
+--------------+--------------+--------------+--------------+
| | | | |
| 9 10 11 | 17 18 19 | 25 26 27 | 33 34 35 |
| | | | |
| 12 left 13 | 20 front 21 | 28 right 29 | 36 back 37 |
| | | | |
| 14 15 16 | 22 23 24 | 30 31 32 | 38 39 40 |
| | | | |
+--------------+--------------+--------------+--------------+
| |
| 41 42 43 |
| |
| 44 down 45 |
| |
| 46 47 48 |
| |
+--------------+
It is easy to identify which of the six permutations corresponds to the rotation of the upper (U), left (L), front (F), right (R), back (B) and down (D) sides to use the same letters as in the question above. We will now create them in GAP:
gap> U:=( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19);;
gap> L:=( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35);;
gap> F:=(17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11);;
gap> R:=(25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24);;
gap> B:=(33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27);;
gap> D:= (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40);;
Next, we will construct a group generated by these permutations:
gap> G:=Group(F,B,L,R,U,D);
<permutation group with 6 generators>
gap> Size(G);
43252003274489856000
We may now try to use SmallGeneratingSet to find a generating set that has fewer elements. SmallGeneratingSet does not guarantee to return a non-redundant list of minimal possible length, but this time we're lucky:
gap> sgs:=SmallGeneratingSet(G);
[ (1,32,41,3,6,19,35,48,22,27,11,25,9,38,16,33,17,8)(2,15,37,7,5,28,45,23,10,
20)(4,13,34,44,12,18,26,21,31,42)(14,46,40)(29,36)(39,47),
(1,43,27,41,11,48)(2,29,23)(3,16,6,32,9,24)(4,20,5,18,21,37,45,15,39)(7,28,
12,31,44,47,10,13,26)(8,40)(14,25)(17,38,35,30,33,22)(19,46)(34,36,42) ]
gap> Length(sgs);
2
Thus, indeed the group is 2-generated. Now let's create permutations a and b corresponding to $\alpha$ and $\beta$ from the question:
gap> a:=L^2*B*R*D^-1*L^-1;
(1,22,32,30,25,27,40)(2,15,12,10,39,42,20,31,28,26,29)(3,14,9,41,38,43,19)(4,
47,23,13,45,21,5,36,34,44,37)(8,33,46,35,16,48,24)
gap> b:=U*F*R*U*R^-1*U^-1*F^-1;
(3,6,27,11,33,17)(4,18,10,7)(5,26)(8,25,19)
It is easy to check that they indeed generate the same group:
gap> H:=Group(a,b);
<permutation group with 2 generators>
gap> Size(G)=Size(H);
true
gap> G=H;
true
It remains to show how to factorise, for example, $U$ in terms of $\alpha$ and $\beta$ and their inverses. We will use the same approach that is used here to solve the puzzle.
gap> K:=FreeGroup("a","b");
<free group on the generators [ a, b ]>
gap> hom := GroupHomomorphismByImages( K, H, GeneratorsOfGroup(F), GeneratorsOfGroup(H) );
[ a, b ] -> [ (1,22,32,30,25,27,40)(2,15,12,10,39,42,20,31,28,26,29)(3,14,9,
41,38,43,19)(4,47,23,13,45,21,5,36,34,44,37)(8,33,46,35,16,48,24),
(3,6,27,11,33,17)(4,18,10,7)(5,26)(8,25,19) ]
gap> w:=PreImagesRepresentative( hom, U );
b*a^2*b*a^-5*b*a^-1*b^-1*(a*b*a)^2*b*a^-2*b*a^-1*b^-1*a*b^-1*a^-1*b^-1*a^3*b^-\
1*a^-2*(b^-1*(a*b)^2*a^4*b*a^-6*(b^-1*a^-1*b^-1*a)^2)^2*b^-1*a^3*b*(b*a)^4*a^2\
*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*a*(a*b)^\
4*a^3*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*(a*\
(a*b)^4*a^3*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)\
^2*a^2*b^-1*a^-2*b)^2*(a*b)^2*(b*a)^4*a^2*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a\
^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*a^2*b^-1*a^-2*(b*a)^2*a^2*b^3*a^2*b*a^-1*b^-1\
*a^22*b*a^-1*b^-1*a^-6*b^-2*a^2*b^-1*(b^-1*a^-1)^2*b^-6*a^4*b*(b*a^4*b*a^-7*b^\
-1*a^3)^2*b*((b*a)^4*a^2*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b\
*a^-1*b^-1*a)^2*a^2)^2*b^-1*a^-2*b*a^2*b^-1*a^-2*(b*a)^2*b*((b*a)^4*a^2*b^2*a^\
-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*a^2)^3*b^-1*a^-\
2*((b*a)^2*b)^2*a*b*a^3*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*\
a^-1*b^-1*a)^2*a^2*b^-5*a^-1*b^2*a^-1*b*(a*b*a^2)^2*a*b*a^-1*b*a^13*b*a^-1*b^-\
1*a^-11*b*a*b^-1*a^3*(b*a^-1)^2*b^-1*a*b^-1*a^-1*b^-1*a^3*b^-1*a^-2*b^-1*a*b^2\
*a*b*a^-1*((b*a)^2*a^3*b*a^-6*(b^-1*a^-1*b^-1*a)^2*b^-1*a)^2*(b*a)^2*b^2*a^4*b\
*a^-7*b^-1*(a*b)^3*a^4*b*a^-7*b^-1*a^3*b*(b*a)^4*a^2*b^2*a^-3*b*a^-1*b^-1*a^-2\
*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*(a*(a*b)^4*a^3*b^2*a^-3*b*a^-1*b^-\
1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*a^2*b^-1*a^-2*b)^2*(a*b)^2*(\
b*a)^4*a^2*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^\
2*a^2*b^-1*a^-2*b*a^4*b^3*a^2*b*a^-1*b^-1*a^22*b*a^-1*b^-1*a^-6*b^-2*a^2*b^-1*\
(b^-1*a^-1)^2*b^-6*a^4*b*(b*a^4*b*a^-7*b^-1*a^3)^2*b*((b*a)^4*a^2*b^2*a^-3*b*a\
^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*a^2)^2*b^-1*a^-2*b*a^\
2*b^-1*a^-2*(b*a)^2*b*((b*a)^4*a^2*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-\
1*a*b*(a*b*a^-1*b^-1*a)^2*a^2)^3*b^-1*a^-2*((b*a)^2*b)^2*a*b*a^3*b^2*a^-3*b*a^\
-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*a^2*b^-1*a^-2*(b*a)^2*\
a^3*b*a^-6*(b^-1*a^-1*b^-1*a)^2*b^-1*a*b^2*a*b*a^-1*(b*a)^2*a^3*b*a^-6*(b^-1*a\
^-1*b^-1*a)^2*b^-1*(a*b)^2*a^4*b*a^-6*(b^-1*a^-1*b^-1*a)^2*b^-1*a^-1*b^-6*a*b^\
6*a^2*b^5*a^-1*b*a*b^-5*a*(b*a^2*b)^2*a^-3*b^4*a^-5*b^-2*a^2*b^-1*(b^-1*a^-1)^\
2*b^-6*a^4*b*(b*a^4*b*a^-7*b^-1*a^3)^2*b^2*a^-1*b*a*b^-1*a^-1*b^-1*a*b^3*a^-1*\
b*(b*a^3)^2*b*a^-1*b*a^13*b*a^-1*b^-1*a^-11*b*a*b^-1*a^3*(b*a^-1)^2*b^-1*a*b^-\
1*a^-1*b^-1*a^3*b^-1*a^-2*b^-1*a*b^2*a*b*a^-1*((b*a)^2*a^3*b*a^-6*(b^-1*a^-1*b\
^-1*a)^2*b^-1*a)^2*(b*a)^2*b^2*a^4*b*a^-7*b^-1*(a*b)^3*a^4*b*a^-7*b^-1*a^3*b*(\
b*a)^4*a^2*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^\
2*(a*(a*b)^4*a^3*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^\
-1*a)^2*a^2*b^-1*a^-2*b)^2*(a*b)^2*(b*a)^4*a^2*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-\
5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*a^2*b^-1*a^-2*b*a^4*b^3*a^2*b*a^-1*b^-1*\
a^22*b*a^-1*b^-1*a^-6*b^-2*a^2*b^-1*(b^-1*a^-1)^2*b^-6*a^4*b*(b*a^4*b*a^-7*b^-\
1*a^3)^2*b*((b*a)^4*a^2*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*\
a^-1*b^-1*a)^2*a^2)^2*b^-1*a^-2*b*a^2*b^-1*a^-2*(b*a)^2*b*((b*a)^4*a^2*b^2*a^-\
3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a^-1*b^-1*a)^2*a^2)^3*b^-1*a^-2\
*((b*a)^2*b)^2*a*b*a^3*b^2*a^-3*b*a^-1*b^-1*a^-2*b*a^-5*b*a^-1*b^-1*a*b*(a*b*a\
^-1*b^-1*a)^2*a^2*b^-1*a^-2*(b*a)^2*a^3*b*a^-6*(b^-1*a^-1*b^-1*a)^2*b^-1*a*b^2\
*a*b*a^-1*(b*a)^2*a^3*b*a^-6*(b^-1*a^-1*b^-1*a)^2*b^-1*(a*b)^2*a^4*b*a^-6*(b^-\
1*a^-1*b^-1*a)^2*b^-1*a^-1*b^-6*a*b^6*a^2*b^5*a^-1*b*a*b^-5*a^-1*b^-1*(a^-1*b^\
-1*a*b^2)^2*b^4
It's not guaranteed that this factorisation is the shortest - finding that would be much more difficult computational task.
Now one could similarly calculate factorisation for other five permutations. They will be computed faster than in the first call to PreImagesRepresentative since some data needed for the algorithm are already computed and stored in the group.
• Also, using LaTeX(w) (note that this function is obsolete in GAP), one could export the answer to LaTeX. – Alexander Konovalov Feb 5 '16 at 10:22
• Amazing answer! That product is a lot longer than I expected, but it’s still nice to see it really work out. Thank you for showing me the GAP functions I need to hack at this myself. – Lynn Feb 5 '16 at 13:37
• @Lynn: thank you - it's a nice question, too! – Alexander Konovalov Feb 5 '16 at 15:20
• Or -- somewhat pathetic random search: Size(Group(B/R*U,R*L/D)); shows that these two short products generate the same group. – ahulpke Feb 5 '16 at 16:07
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2020-09-18 08:15:58
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https://tex.stackexchange.com/questions/249467/png-in-plain-tex
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# PNG in Plain TeX [duplicate]
This question already has an answer here:
I have been presented with graphic files in the format .png. I need to include them in a Plain TeX document. How is that possible? Is there a way to convert them to .eps?
## marked as duplicate by Svend Tveskæg, Heiko Oberdiek, user13907, Andrew Swann, barbara beetonJun 12 '15 at 19:21
• are you using classic dvi tex or pdftex? – David Carlisle Jun 9 '15 at 21:10
• Thanks for the 3 replies. The day after I asked the question, I realized that my issue is not plain vs. LaTeX. but dvi vs. pdf[la]tex. I have both .eps and .png figures, so I followed the advice of the 3rd respondent and used "convert" to get .eps files that work in dvips (which I'm used to). A junior colleague got slightly better results with gimp instead of convert. @lhf – S. A. Fulling Jun 11 '15 at 19:25
Use pdftex and the graphicx package from LaTeX with eplain
\input eplain
\beginpackages
\usepackage{graphicx}
\endpackages
Load your graphics with \includegraphics{yourpng}, and typeset your document with pdftex (not pdflatex).
If you find loading eplain too bloated, you may try graphics.tex, bundled in the graphics-pln bundle.
\input graphicx
\includegraphics{yourpng}
Beware you may need to set the driver to match your engine. In the graphics.tex file change \def\Gin@driver{dvips.def} for \def\Gin@driver{pdftex.def}.
To convert an image to eps, you can use convert, which you probably already have.
Just do convert foo.png foo.eps.
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2019-07-22 07:33:16
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https://physics.stackexchange.com/questions/123783/uncertainty-principle-with-two-photons
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# Uncertainty principle with two photons
Imagine an experimental setup in which you have to measure the momentum and location of a particle. To measure it we know we will have to affect it, and the uncertainty principle would come into the picture, but I have a different setup. The classical setup is that you fire a photon to measure the location of the particle, but the particle will change its momentum due to the collision with the photon.
I decided to take two photons. I will shoot one photon from either side of the particle, so the effects of the two photons cancel each other, giving an accurate measurement. To understand this, see the picture below.
1. The classic experiment
2. My thought experiment
In the second experiment, we shoot a photon of the same energy as the first one and counteract the effect of the first photon, so the electron would continue on its original path. Please tell me where I am wrong.
EDIT
We will have to take multiple photons but equal from both sides and in opposite directions.
• How do you know how do time the photons? You are assuming you already know position and velocity beforehand. – pfnuesel Jul 9 '14 at 10:55
• Well in that case we could use multiple photons but from opposite directions. Nice point though. – rahulgarg12342 Jul 9 '14 at 10:56
• So this is a three body collision, I don't think the electron is constrained to "continue on its original path" – Antonio Ragagnin Jul 9 '14 at 10:58
• Bigger problem at a fundamental level. For the two photons to have the exact same effect, they need to collide precisely at the same spot. You will thus not measure the location of the electron, since this location is a constraint of your setup... I did not do the math, but it may be that your experiment have results (from the scattered photons) independent from the momentum of the electron... To be checked! In that case you have not measured momentum either. – Martigan Jul 9 '14 at 11:09
First of all, the uncertainty principle is more than just disturbance of observation.
From the Wikipedia article "Uncertainty principle":
Historically, the uncertainty principle has been confused with a somewhat similar effect in physics, called the observer effect, which notes that measurements of certain systems cannot be made without affecting the systems. Heisenberg offered such an observer effect at the quantum level (see below) as a physical "explanation" of quantum uncertainty.
It has since become clear, however, that the uncertainty principle is inherent in the properties of all wave-like systems, and that it arises in quantum mechanics simply due to the matter wave nature of all quantum objects. Thus, the uncertainty principle actually states a fundamental property of quantum systems, and is not a statement about the observational success of current technology. It must be emphasized that measurement does not mean only a process in which a physicist-observer takes part, but rather any interaction between classical and quantum objects regardless of any observer.
Now, you've drawn 'the' path of the electron as if the electron has a definite trajectory and that two photons of equal and opposite momentum interact with the electron at a definite location.
However, the state of definite position has maximum 'uncertainty' in momentum! Not only can there not be a definite trajectory but, if the electron is localized by an interaction, one cannot escape the inherent uncertainty of that localized state.
The problem with your set up is that you have ignored the quantum mechanical nature of the photons. The photons are subject to the uncertainty principle as well as the electron, and so there is no way to send in 2 photons with precisely the same momentum at precisely the same time, and we can't guarantee they will scatter of in precisely the same way. The setup you have described is, therefore, impossible.
What makes you think an electron reflects photons as you have drawn?
• Electrons scatter photons in any direction, although not uniformly. (Examples: Thomson scattering, Rayleigh scattering, X-ray crystallography)
• The electron may absorb the photon for an arbitrary period of time, changing momentum and thus position, then release a photon of a different wavelength. The electron may absorb one photon and release many photons. (Examples: Compton scattering, inelastic scattering)
• The photons could interact with each other and ignore the electron. (Examples: Delbrück scattering, additional two-photon physics)
• Even if you are lucky enough to have an interaction in which the electron and photons play along with your plan... Your experiment cannot distinguish which photon is which at the exit. They could have been "scattered horizontally" or "scattered vertically", leading to a great deal of position and momentum uncertainty in your system.
Your classic experiment 1) is quantum mechanical. Electrons and photons are elementary particles and interact individually as quantum mechanical entities.
The plot shows the elastic scattering of photon on electron, a computable process quantum mechanically thus the angular distribution is known. As we are talking quantum mechanics there is a probability for each angle where the electron may scatter that is given by the Klein-Nishina formula. So the individual photon-electron pair does not have a fixed angle which you could use in your thought experiment to counteract the effect on the electron's direction. ( always presuming you could time the photons to hit at the same delta(t) of Heisenberg's uncertainty principle) .
The thought set up 2) is impossible.
It has been said that your second experiment is impossible. For me it is possible but can't be used to predict the position of the electron and its momentum.
• First consider that your "classic experiment" is true. In this experiment, the important fact is that one can't obtain the momentum of the electron when the photon angle is measured.
• Now, consider your "thought experiment" with this modification: we send two photons as you said, but we are not able to let them hit the electron at the sime time. Instead they interact with the electron in two different times. And this difference in time is $\Delta t$ and can be arbitrarly small.
• Now, let's (reasonabely) suppose that if I compute the state of the final system for my "thought experiment", then it will depend by $\Delta t$. So, if I send $\Delta t\rightarrow 0$ then I obtain a prediction for your "thought experiment."
• In my experiment one photon hit the electron and then it works like the "classic experiment": the first photon scatter randomly and you can't infer the new momentum of the electron. The second photon will hit the scattered electron and will do the same thing as the first one: you can't obtain the new momentum of the elctron from this new photon either.
• This result does not depend on $\Delta t,$ so it should hold also in the limit $\Delta t=0.$ (that is your experiment)
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2019-08-21 20:20:51
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https://mathhelpboards.com/threads/evaluating-arithmetic-expression.4802/
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# Evaluating Arithmetic Expression
#### Britt
##### New member
I need help figuring out what the little 2 next to an equation means this is what my problem looks like I have the answer but I can't figure out how it was found out. (3)(-4)2 - (3)(-5) so the 2 next to the 4 in parentheses is little.
#### Ackbach
##### Indicium Physicus
Staff member
I need help figuring out what the little 2 next to an equation means this is what my problem looks like I have the answer but I can't figure out how it was found out. (3)(-4)2 - (3)(-5) so the 2 next to the 4 in parentheses is little.
Do you mean $(3)(-4)^{2}-(3)(-5)?$ If so, the small raised '2' is an exponent. That is a shorthand notation for multiplying something by itself a certain number of times.
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2020-09-22 00:54:37
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https://www.tutordale.com/how-do-you-find-time-in-physics/
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Monday, May 16, 2022
# How Do You Find Time In Physics
## How Do You Find The Period In Physics Circular Motion
Different way to determine time in air | Two-dimensional motion | Physics | Khan Academy
4.4/5circlecircleperiodspeedcircular motionmore on it
A period T is the time required for one complete cycle of vibration to pass a given point. As the frequency of a wave increases, the period of the wave decreases. Frequency and Period are in reciprocal relationships and can be expressed mathematically as: Period equals the Total time divided by the Number of cycles.
Beside above, what is the unit for period? second
Simply so, what is a period in physics circular motion?
Uniform circular motion. When an object is experiencing uniform circular motion, it is traveling in a circular path at a constant speed. If r is the radius of the path, and we define the period, T, as the time it takes to make a complete circle, then the speed is given by the circumference over the period.
What is the SI unit of centripetal force?
Centripetal force is measured in Newtons and is calculated as the mass , multiplied by tangential velocity squared, divided by the radius . This means that if tangential velocity doubles, the force will quadruple.
## Newton’s Physics: Linear Time
In or around 1665, when Isaac Newton derived the motion of objects falling under gravity, the first clear formulation for mathematical physics of a treatment of time began: linear time, conceived as a universal clock.
Absolute, true, and mathematical time, of itself, and from its own nature flows equably without regard to anything external, and by another name is called duration: relative, apparent, and common time, is some sensible and external measure of duration by the means of motion, which is commonly used instead of true time such as an hour, a day, a month, a year.
The water clock mechanism described by Galileo was engineered to provide laminar flow of the water during the experiments, thus providing a constant flow of water for the durations of the experiments, and embodying what Newton called duration.
In this section, the relationships listed below treat time as a parameter which serves as an index to the behavior of the physical system under consideration. Because Newton’s fluents treat a linear flow of time , time could be considered to be a linearly varying parameter, an abstraction of the march of the hours on the face of a clock. Calendars and ship’s logs could then be mapped to the march of the hours, days, months, years and centuries.
Prerequisites
## Galileo: The Flow Of Time
In 1583, Galileo Galilei discovered that a pendulum’s harmonic motion has a constant period, which he learned by timing the motion of a swaying lamp in harmonic motion at mass at the cathedral of Pisa, with his pulse.
In his Two New Sciences , Galileo used a water clock to measure the time taken for a bronze ball to roll a known distance down an inclined plane this clock was
“a large vessel of water placed in an elevated position to the bottom of this vessel was soldered a pipe of small diameter giving a thin jet of water, which we collected in a small glass during the time of each descent, whether for the whole length of the channel or for a part of its length the water thus collected was weighed, after each descent, on a very accurate balance the differences and ratios of these weights gave us the differences and ratios of the times, and this with such accuracy that although the operation was repeated many, many times, there was no appreciable discrepancy in the results.”
Galileo’s experimental setup to measure the literal flow of time, in order to describe the motion of a ball, preceded Isaac Newton‘s statement in his Principia:
I do not define time, space, place and motion, as being well known to all.
The Galilean transformations assume that time is the same for all reference frames.
## How Do You Calculate Echo In Physics
4.9/5echoHearing echoes
• d = the distance the sound wave traveled back and forth,
• v = velocity of sound, and.
• t = the time it takes the sound to go back and forth.
• Similarly one may ask, what do you mean by echo in physics?
Echo is a distinct, reflected sound wave from a surface. A reflected sound can be heard separately from the original sound if the sound source is closer to the receiver while the reflecting hard surface is sufficiently far from receiver. Such reflected sound is called an echo.
One may also ask, how does echo measure distance? If you produce an impulsive sound and measure the time until you hear the echo from this object, just multiply the number of seconds by 170 and get the distance to the object. Multiplication by 170 because the sound has travelled the distance twice.
Similarly, you may ask, what is Echo give formula for time of Echo?
The distance traveled by sound to create an echo is two times the distance between the source and obstacle. So, the distance traveled by sound is 2d. Let velocity is the speed of sound in air which is equal to 343 m/s and time is 0.1 s as this the time required by human to hear two sounds clearly.
What is echo method?
the echo method is used to measure large distance Explanation: The reception of wave after reflection is called ‘echo‘. When echo method is used to measure large distances, we send the wave towards the object and the reflected wave is again received.
## The Average Velocity Formula And Velocity Units
The average velocity formula describes the relationship between the length of your route and the time it takes to travel. For example, if you drive a car for a distance of 70 miles in one hour, your average velocity equals 70 mph. In the previous section, we have introduced the basic velocity equation, but as you probably have already realized, there are more equations in the velocity calculator. Let’s list and organize them below:
• Simple velocity equation:
• Velocity after a certain time of acceleration:
• final velocity = initial velocity + acceleration * time
• Average velocity formula – weighted average of velocities:
• average velocity = velocityâ * timeâ + velocityâ * timeâ + …
You should use the average velocity formula if you can divide your route into few segments. For example, you drive a car with a speed of 25 mph for 1 h in the city and then reach 70 mph for 3 h on the highway. What is your average velocity? With the velocity calculator, you can find that it will be about 59 mph.
From the above equations, you can also imagine what are velocity units. British imperial units are feet per second ft/s and miles per hour mph. In the metric SI system the units are meters per second m/s and kilometers per hour km/h. Remember you can always easily switch between all of them in our tool!
Also Check: What Does Cyte Mean
## What Is The Formula For Velocity
The velocity is said to as the time rate of change of displacement. So, the velocity formula physics is:
$$v = \frac$$
Velocity After a Certain Time of Acceleration:
$$Final velocity = Initial Velocity + Acceleration \times Time$$
Our total velocity calculator also uses the same velocity calculating formula to calculating velocity accurately.
## What Is The Formula For Work
Definition: In our daily life work implies an activity resulting in muscular or mental exertion. However, in physics the term work is used in a specific sense involves the displacement of a particle or body under the action of a force. work is said to be done when the point of application of a force moves. Work done in moving a body is equal to the product of force exerted on the body and the distance moved by the body in the direction of force.Work = Force × Distance moved in the direction of force.
The work done by a force on a body depends on two factors Magnitude of the force, and Distance through which the body moves
Unit of Work When a force of 1 newton moves a body through a distance of 1 metre in its own direction, then the work done is known as 1 joule. Work = Force × Displacement 1 joule = 1 N × 1 m or 1 J = 1 Nm
You May Like: Altogether Math Definition
## What Is Velocity In Physics
A velocity definition of physics is referred to as a vector measurement of the rate and direction of motion. In other words, velocity is a measure of how quickly an object moves. When it comes to the velocity equation, it is stated as the change in the position of an object, divided by the time. You get more clearance with the formula for velocity.
## Causality Correlations And Quantum Mechanics
Solving for time | One-dimensional motion | Physics | Khan Academy
This finding did not go undisputed. Many authors attempted to locate the flaw in Einstein’s and Bell’s argument, but logically it seemed to be impeccable. Bell assumed that determinism means that one can build a model, any model, in which classical equations control the behavior of dynamical variables, and where, at the tiniest scales where these variables describe the data, the evolution laws do not leave the slightest ambiguity there are no wave functions, no statistical considerations, as everything that happens is controlled by certainties. Moreover, there is some sense of locality: the laws control all processes using only the data that are situated at given localities, while action at a distance, or backwards in time, are forbidden. The classical degrees of freedom that really exist were called beables.
Here, the first topic for discussion arises: what does action backwards in time mean? In La Nouvelle Cuisine, Bell formulated as precisely as he could what causality forward in time means:
A theory is said to be locally causal if the probabilities attached to values of local beables in a space-time region 1 are unaltered by specification of values of local beables in a space-like separated region 2, when what happens in the backward light cone of 1 is already sufficiently specified
Read Also: Algebra With Pizzazz Answer Key Page 34 Books Never Written
## How To Use Time Formula
Time formula can be used to find the time taken by an object, given the distance and unit speed. Let’s take a quick look at an example showing how to use the formula for time.
Example: What will be the total time to cover 3600 m at 2 meter per seconds?
Solution: Using Formula for time,
Time = Distance ÷ SpeedTime = 3600 ÷ = 1800 seconds.
Answer: Total time taken to cover the distance of 3600 m is 1800 seconds.
## The Bell And Chsh Inequalities
Bell’s Gedanken experiment is in essence much the same as the Einstein Rosen Podolsky set-up. A local device is constructed that can emit two entangled particles, and , which leave the machine in opposite directions. Alice and Bob , both choose whether to measure property X or property Y of the particles they can see. Alice chooses setting a to measure and Bob setting b to measure .
The correlations needed to explain the quantum mechanical result require that the settings a and b chosen by Alice and bob, must be correlated with one another as well as the spins of the two entangled particles. The author calculated the minimal amount of correlation that is needed to produce the quantum result. We found the following distribution :
where a is the angle chosen by Alice for her measurement, b is Bob’s angle, and a parameter describing the polarization of the entangled photons produced by the sourceand detected by Alice and Bob. W is the probability distribution, and C is a normalization constant. It features a 3 body correlation: whenever we integrate over all values of a, or all values of b, or all values of , we get a flat distribution.
So let us emphasize and summarize this essential point:
## Technology For Timekeeping Standards
The primary time standard in the U.S. is currently NIST-F1, a laser-cooled Cs fountain, the latest in a series of time and frequency standards, from the ammonia-based atomic clock to the caesium-based NBS-1 to NIST-7 . The respective clock uncertainty declined from 10,000 nanoseconds per day to 0.5 nanoseconds per day in 5 decades. In 2001 the clock uncertainty for NIST-F1 was 0.1 nanoseconds/day. Development of increasingly accurate frequency standards is underway.
In this time and frequency standard, a population of caesium atoms is laser-cooled to temperatures of one microkelvin. The atoms collect in a ball shaped by six lasers, two for each spatial dimension, vertical , horizontal , and back/forth. The vertical lasers push the caesium ball through a microwave cavity. As the ball is cooled, the caesium population cools to its ground state and emits light at its natural frequency, stated in the definition of second above. Eleven physical effects are accounted for in the emissions from the caesium population, which are then controlled for in the NIST-F1 clock. These results are reported to BIPM.
Additionally, a reference hydrogen maser is also reported to BIPM as a frequency standard for TAI .
## How Do You Find Vf In Physics
Final Velocityfinal velocity
. Just so, what does VF mean in physics?
The first equation describes your free fall. Your final velocity is equal to your initial velocity plus the acceleration you are undergoing times the amount of time you are falling .
Additionally, what does VF stand for? ventricular fibrillation
Beside this, how do you find W in physics?
The work is calculated by multiplying the force by the amount of movement of an object . A force of 10 newtons, that moves an object 3 meters, does 30 n-m of work. A newton-meter is the same thing as a joule, so the units for work are the same as those for energy joules.
What does M stand for in physics?
meter. atomic mass number. acceleration. meters per second squared magnetic flux density.
Also Check: 2.2 Sneak Peek
## Work Done By A Force Example Problems With Solutions
Example 1: How much work is done by a force of 10N in moving an object through a distance of 1 m in the direction of the force ?Solution: The work done is calculated by using the formula: W = F × S Here, Force F = 10 N And, Distance, S = 1 m So, Work done, W = 10 × 1 J = 10 J Thus, the work done is 10 joules
Example 2: Find the work done by a force of 10 N in moving an object through a distance of 2 m.Solution: Work done = Force × Distance moved Here, Force = 10 N Work done, W = 10 N × 2 m = 20 Joule = 20 J
Example 3: Calculate the work done in pushing a cart, through a distance of 100 m against the force of friction equal to 120 N.Solution: Force, F = 120 N Distance, s = 100 m Using the formula, we have W = Fs = 120 N × 100 m = 12,000 J
Example 4: A body of mass 5 kg is displaced through a distance of 4m under an acceleration of 3 m/s2. Calculate the work done.Sol. Given: mass, m = 5 kg acceleration, a = 3 m/s2 Force acting on the body is given by F = ma = 5 × 3 = 15 N Now, work done is given by W = Fs = 15 N × 4 m = 60 J
Example 5: Calculate the work done in raising a bucket full of water and weighing 200 kg through a height of 5 m. .Solution: Force of gravity mg = 200 × 9.8 = 1960.0 N h = 5 m Work done, W = mgh or W = 1960 × 5 = 9800 J
Example 7: An engine does 64,000 J of work by exerting a force of 8,000 N. Calculate the displacement in the direction of force.Solution: Given W = 64,000 J F = 8,000 N Work done is given by W = Fs or 64000 = 8000 × s or s = 8 m
## What Causes A Change In Velocity
Experts depicts that forces are something that affect how objects move they may cause motion, also, they may stop, slow or even change the direction of motion of an object . As force cause changes in the speed or direction of an object, it is said to be that forces causes changes in velocity. Remember that acceleration is said to be change in velocity.
You May Like: Holt Mcdougal Geometry Practice Workbook Answer Key
## The Unit Of Measurement Of Time: The Second
In the International System of Units , the unit of time is the second . It is a SI base unit, and has been defined since 1967 as “the duration of 9,192,631,770 of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom”. This definition is based on the operation of a caesium atomic clock. These clocks became practical for use as primary reference standards after about 1955, and have been in use ever since.
## Terminal Velocity Escape Velocity And Relativistic Velocity
Velocity Time Graphs, Acceleration & Position Time Graphs – Physics
Velocity is present in many aspects of physics, and we have created many calculators about it! The first velocity is the so-called terminal velocity, which is the highest velocity attainable by a free falling object. Terminal velocity occurs in fluids and depends on the fluid’s density. You can read more about it in our free fall with air resistance calculator.
Knowing how to calculate velocity is of particular importance in astrophysics since results have to be very accurate. If you’re interested in the world of massive celestial objects like suns or planets, visit our escape velocity calculator or orbital velocity calculator. You can learn a lot there!
In the high energy region, there is another important velocity – relativistic velocity. It results from the fact that no object with a non-zero mass can reach the speed of light. Why? When it approaches light speed, it’s kinetic energy becomes unattainable, very large or even infinite. You can check it with the relativistic kinetic energy calculator by filling the velocity field with the speed of light 299,792,458 m/s or 2.998e8 m/s in scientific notation. Moreover, this is a cause of other phenomena like relativistic velocity addition, time dilation, and length contraction. Also the Albert Einstein’s famous E = mc2 formula bases on the relativistic velocity concept.
Don’t Miss: What Happened To Beth The Child Of Rage
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2022-05-18 03:04:42
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https://examproblems4lossmodels.wordpress.com/tag/bayesian-probability/page/2/
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## Exam C Practice Problem 9 – Examples of Claim Frequency Models
Problem 9-A
A portfolio consists of independent risks divided into two classes. Eighty percent of the risks are in Class 1 and twenty percent are in Class 2.
• For each risk in Class 1, the number of claims in a year has a Poisson distribution with mean $\theta$ such that $\theta$ follows a Gamma distribution with mean 1.6 and variance 1.28.
• For each risk in Class 2, the number of claims in a year has a Poisson distribution with mean $\delta$ such that $\delta$ follows a Gamma distribution with mean 2.5 and variance 3.125.
An actuary is hired to examine the claim experience of the risks in this portfolio. What proportion of the risks can be expected to incur exactly 1 claim in one year?
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.24$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.25$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.26$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.27$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.28$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
Problem 9-B
A portfolio consists of independent risks divided into two classes. Sixty percent of the risks are in Class 1 and fourty percent are in Class 2.
• For each risk in Class 1, the number of claims in a year has a Poisson distribution with mean $\theta$ such that $\theta$ follows a Gamma distribution with mean 2.4 and variance $\displaystyle \frac{48}{25}$.
• For each risk in Class 2, the number of claims in a year has a Poisson distribution with mean $\delta$ such that $\delta$ follows a Gamma distribution with mean 3.75 and variance $\displaystyle \frac{75}{16}$.
An actuary is hired to examine the claim experience of the risks in this portfolio. Of the risks that incur exactly 2 claims in a year, what proportion of the risks can be expected to come from Class 2?
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.34$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.35$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.36$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.37$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.38$
___________________________________________________________________________________
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$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
___________________________________________________________________________________
___________________________________________________________________________________
$\copyright \ 2013 \ \ \text{Dan Ma}$
## Exam C Practice Problem 8 – Bayesian Estimates of Claim Frequency
Problem 8-A
A portfolio consists of ten independent risks divided into two classes. Class 1 contains 6 risks and Class 2 contains 4 risks.
The risks in each class are assumed to follow identical annual claim frequency distribution. The following table shows the distributions of the number of claims in a calendar year.
$\displaystyle \begin{bmatrix} \text{ }&\text{ }&\text{Class 1} &\text{ }&\text{Class 2} \\X=x&\text{ }&P(X=x) &\text{ }&P(X=x) \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ 0&\text{ }&0.50 &\text{ }&0.20 \\ 1&\text{ }&0.25&\text{ }&0.25 \\ 2&\text{ }&0.12&\text{ }&0.30 \\ 3&\text{ }&0.08&\text{ }&0.15 \\ 4&\text{ }&0.05&\text{ }&0.10 \end{bmatrix}$
The risks in the portfolio are observed for one calendar year. The following table shows the observed results.
$\displaystyle \begin{bmatrix} \text{Number of Claims}&\text{ }&\text{Number of Risks} \\\text{ }&\text{ }&\text{ } &\text{ } \\ 0&\text{ }&4 \\ 1&\text{ }&2 \\ 2&\text{ }&2 \\ 3&\text{ }&1 \\ 4&\text{ }&1 \end{bmatrix}$
Determine the Bayesian expected number of claims per risk in the next year.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.10$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.15$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.22$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.29$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.315$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
Problem 8-B
A portfolio consists of ten independent risks divided into two classes. Both classes contain the same number of risks.
The risks in each class are assumed to follow identical annual claim frequency distribution. The following table shows the distributions of the number of claims in a calendar year.
$\displaystyle \begin{bmatrix} \text{ }&\text{ }&\text{Class 1} &\text{ }&\text{Class 2} \\X=x&\text{ }&P(X=x) &\text{ }&P(X=x) \\\text{ }&\text{ }&\text{ } &\text{ }&\text{ } \\ 0&\text{ }&0.5 &\text{ }&0.1 \\ 1&\text{ }&0.3&\text{ }&0.4 \\ 2&\text{ }&0.1&\text{ }&0.3 \\ 3&\text{ }&0.05&\text{ }&0.1 \\ 4&\text{ }&0.05&\text{ }&0.1 \end{bmatrix}$
The risks in the portfolio are observed for one calendar year. The following table shows the observed results.
$\displaystyle \begin{bmatrix} \text{Number of Claims}&\text{ }&\text{Number of Risks} \\\text{ }&\text{ }&\text{ } &\text{ } \\ 0&\text{ }&2 \\ 1&\text{ }&2 \\ 2&\text{ }&3 \\ 3&\text{ }&1 \\ 4&\text{ }&2 \end{bmatrix}$
What is the probability that a randomly selected risk in this portfolio will incur exactly 2 claims in the next year?
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.26$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.27$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.28$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.29$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.30$
___________________________________________________________________________________
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
___________________________________________________________________________________
___________________________________________________________________________________
$\copyright \ 2013 \ \ \text{Dan Ma}$
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2019-09-17 09:13:05
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https://www.hackmath.net/en/math-problem/315?tag_id=13_80
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# Trigonometric functions
In right triangle is:
$\text{ tg} \ \alpha = \dfrac{ 2} 1$
Determine the value of s and c:
$\text{ sin } \alpha = \dfrac{ s} { \sqrt{ 5 }}$
$\text{ cos } \alpha = \dfrac{ c} { \sqrt{ 5 }}$
Correct result:
s = 2
c = 1
#### Solution:
$\text{ tg} \ \alpha = \dfrac{ 2} { 1} = \dfrac{ a }{ b } \ \\ \text{ sin} \ \alpha = \dfrac{ a} {c} = \dfrac{ a} { \sqrt{ a^2+b^2}} = \dfrac{ 2 } { \sqrt{ 5 }}$
$\text{ cos} \ \alpha = \dfrac{ b} {c} = \dfrac{ b} { \sqrt{ a^2+b^2}} = \dfrac{ 1 } { \sqrt{ 5 }}$
We would be very happy if you find an error in the example, spelling mistakes, or inaccuracies, and please send it to us. We thank you!
Tips to related online calculators
Pythagorean theorem is the base for the right triangle calculator.
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
## Next similar math problems:
• Traffic sign
There is a traffic sign for climbing on the road with an angle of 7%. Calculate at what angle the road rises (falls).
• Powerplant chimney
From the window of the building at a height of 7.5 m, the top of the factory chimney can be seen at an altitude angle of 76° 30 ′. The base of the chimney can be seen from the same place at a depth angle of 5° 50 ′. How tall is the chimney?
• The tower
The observer sees the base of the tower 96 meters high at a depth of 30 degrees and 10 minutes and the top of the tower at a depth of 20 degrees and 50 minutes. How high is the observer above the horizontal plane on which the tower stands?
The height of a regular quadrilateral prism is v = 10 cm, the deviation of the body diagonal from the base is 60°. Determine the length of the base edges, the surface, and the volume of the prism.
• Sailboat
The 20 m long sailboat has an 8 m high mast in the middle of the deck. The top of the mast is fixed to the bow and stern with a steel cable. Determine how much cable is needed to secure the mast and what angle the cable will make with the ship's deck.
• Space diagonal angles
Calculate the angle between the body diagonal and the side edge c of the block with dimensions: a = 28cm, b = 45cm and c = 73cm. Then, find the angle between the body diagonal and the plane of the base ABCD.
• Distance of points
A regular quadrilateral pyramid ABCDV is given, in which edge AB = a = 4 cm and height v = 8 cm. Let S be the center of the CV. Find the distance of points A and S.
In a regular quadrilateral pyramid, the height is 6.5 cm and the angle between the base and the side wall is 42°. Calculate the surface area and volume of the body. Round calculations to 1 decimal place.
• An observer
An observer standing west of the tower sees its top at an altitude angle of 45 degrees. After moving 50 meters to the south, he sees its top at an altitude angle of 30 degrees. How tall is the tower?
• The staircase
The staircase has a total height of 3.6 m and forms an angle of 26° with the horizontal. Calculate the length of the whole staircase.
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2020-09-25 03:56:00
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http://www.mathnet.ru/php/archive.phtml?jrnid=mzm&wshow=issue&year=2007&volume=81&volume_alt=&issue=4&issue_alt=&option_lang=eng
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Matematicheskie Zametki
RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
General information Latest issue Forthcoming papers Archive Impact factor Subscription Guidelines for authors License agreement Submit a manuscript Search papers Search references RSS Latest issue Current issues Archive issues What is RSS
Mat. Zametki: Year: Volume: Issue: Page: Find
On the Normability of Marcinkiewicz ClassesS. V. Astashkin 483 On Subspaces of $C[0,1]$ Consisting of Nonsmooth FunctionsE. I. Berezhnoi 490 Non-Self-Adjoint Sturm–Liouville Operators with Matrix PotentialsO. A. Veliev 496 The Buffer Phenomenon in One-Dimensional Piecewise Linear Mapping in RadiophysicsS. D. Glyzin, A. Yu. Kolesov, N. Kh. Rozov 507 Pseudofinite Homogeneity, Isolation, and ReducibilityS. M. Dudakov 515 On Some Questions Related to the Krichever CorrespondenceA. B. Zheglov, D. V. Osipov 528 On a Method for Studying the Norm and the Stability of SolutionsYu. A. Konyaev 540 Approximation of Functions on the Real Axis by Féjer-Type Operators in the Generalized Hölder MetricR. A. Lasuriya 547 Minimization of the Uncertainty Constant of the Family of Meyer WaveletsE. A. Lebedeva 553 On Complex Submanifolds Whose Grassmann Image Has Maximal Holomorphic CurvatureO. V. Leibina 561 Estimates of the Solutions of Difference-Differential Equations of Neutral TypeA. A. Lesnykh 569 Uniqueness of Multiple Walsh Series for the Convergence on Binary CubesN. S. Moreva 586 General Linear Problem of the Isomonodromic Deformation of Fuchsian SystemsV. A. Poberezhnyi 599 A Theorem on Martingale Selection for Relatively Open Convex Set-Valued Random SequencesD. B. Rokhlin 614 Brief Communications On Morphisms onto QuadricsE. Yu. Amerik 621 Description of Spectral Functions of Differential Operators with Arbitrary Deficiency IndicesV. I. Mogilevskii 625 Elliptic Equation with Nonsymmetric Matrix: Averaging of the “Variational Solutions”S. E. Pastukhova, S. V. Tikhomirova 631 Regular and Completely Regular Differential OperatorsE. A. Shiryaev, A. A. Shkalikov 636
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2021-06-20 08:52:04
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https://physexams.com/blog/kinemtic_c2
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# Flash Cards
kinematic
## Acceleration on Position-Time Graph
How to find acceleration from a position vs. time graph? Answer: Acceleration on a position vs. time graph can be obtained, numerically by having the initial position and velocity of a moving object Or graphically, by observing the curvature of the $x-t$ graph. A graph, looking like an upside-down bowl, represents a negative acceleration and vice versa. In this long article, we want to show you how to find constant acceleration from a position-time graph with some solved problems. You can skip this introduction and refer to the worked examples. Types of Motion: An object can move at a constant speed or have a changing velocity. Suppose you a
kinematic
## Displacement and distance problems with solutions
Problems and Solutions about displacement and distance are presented which is helpful for high school and college students. In the following, displacement is computed for simple cases. Displacement Problems: Problem (1): An object moves from point A to B, C, and D finally, along a rectangle. (a) Find the magnitude and direction of the displacement vector of the object? (b) Find the distance traveled by that object? (c) Suppose the object returns to point A, its initial position. Now, Find the displacement and distance. Solution: (a) By definition of displacement, connect the initial (A) and final (D) points together. As shown, displace
The projectile motion formulas along with numerous solved examples for a better understanding of their application are presented. Definition of projectile motion: Any object that is thrown into the air with an angle $\theta$ is projectile and its motion is called projectile motion. In other words, any motion in two dimensions and only under the effect of gravitational force is called projectile motion. Projectile Motion Formulas: The following are all projectile motion equations in vertical and horizontal directions. In horizontal direction: \[\begin{aligned} \text{Displacement}&:\,\Delta x=\underbrace{\left(v_0 \cos \theta\right)}_{v_{0x}}t\\ \text{Velocity
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2023-03-27 16:01:09
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https://www-math.umd.edu/newsletter-single/919-happy-birthday-coach.html
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Celebrating 80 years with Distinguished University Professor Emeritus Jim Yorke, the “Father of Chaos.”
James “Jim” Yorke has held many titles during his career as a mathematician at the University of Maryland—from Ph.D. student to Distinguished University Professor—but he prefers just one: coach.
“A coach is supposed to help you excel, but I don't know what a professor is supposed to do, profess?” said Yorke (Ph.D. ’66, mathematics). “If I want facts, I can get them from books.”
Yorke wasn’t being snarky. He was posing a challenge to see things differently, more deeply—the way he always has throughout his nearly 60-year career.
“I still think of him as Coach,” said Laura Tedeschini-Lalli (Ph.D. ’86, applied mathematics), a professor of mathematics at Roma Tre University in Rome, Italy, who was one of Yorke’s students. “He used to tell us, ‘A teacher can teach you part of what he knows and that’s about it, but a coach helps you excel at what you do well.’ And really, he goes beyond teaching about math. He gets to know people, and through his humanity he helps people excel as mathematicians, but also as better human beings.”
That sentiment was echoed over and over as more than 100 of Yorke’s colleagues, friends and former students celebrated his 80th birthday on September 18, 2021. The celebration, held virtually on Zoom, included six scientific talks and two hours of toasts.
During his toast, S. James Gates Jr.—a College Park Professor of Physics, National Medal of Science awardee and the current president of the American Physical Society—called Yorke the closest thing to a Renaissance man he had ever met and said he appreciated “opportunities to follow such a broadly intellectual and forceful personality.”
DJ Patil (M.A. ’99, Ph.D. ’01, applied mathematics), one of Yorke’s graduate students who went on to serve as the first U.S. chief data scientist, said Yorke helped set the tenor for the open, creative culture that taught him what a collaborative environment should be.
Equally impressive was the number of younger scientists who said Yorke made them feel valued and at home, including high school student Ishan Saha, who reached out to Yorke last year with questions about chaos. Saha chimed in to raise a water-filled glass to the famed mathematician who generously met with him weekly to talk about math and science.
Wayne Hayes, a former postdoctoral associate who is now an associate professor in the Donald Bren School of Information and Computer Sciences at UC Irvine, joined the celebration while on a motorcycle trip. He tipped his helmet to Yorke and reminisced about his first few days at UMD when Yorke lent him cash to pay rent because he was unable to access his Canadian bank account.
Yorke’s humanity and generosity made an impact on students and colleagues far and wide, and although he has published more than 300 papers, his priority has always been connecting with the people around him.
“I like to work with people,” he said. “I haven’t had a single-author paper since 1970-something. You bring in more people and you end up with better papers.”
Nor has he stayed in his lane as a mathematician. Over his career, Yorke’s co-authors have hailed from fields as diverse as epidemiology, meteorology, ecology and artificial intelligence. The six invited speakers at his celebration presented recent research they collaborated on with Yorke that spanned topics ranging from laying the groundwork for ensemble weather prediction models to simplifying methods for reconstructing the largest genome ever assembled (eight times larger than the human genome) to understanding species interactions and research on artificial intelligence.
Yorke’s ability to contribute to such a broad range of disciplines comes from his unique ability to cut to the core of what matters.
### Seeing Through the Chaos
Yorke came to UMD in 1963 as a graduate student after earning his bachelor’s degree in mathematics from Columbia University. He was attracted to the interdisciplinary work at the Institute for Fluid Dynamics and Applied Mathematics (IFDAM), which is now known as the Institute for Physical Science and Technology (IPST). He earned his Ph.D. in mathematics in 1966 and joined IFDAM as a faculty member.
Yorke held a joint appointment in the Department of Mathematics beginning in 1976 and the Department of Physics since 2000. He also served as IPST director from 1981 to 2005 and as chair of the Department of Mathematics from 2007 to 2013.
When Yorke began his career, mathematicians and scientists tended to think that all physical phenomena were predictable and cases that defied prediction were merely examples of flawed experiments. The idea that dynamical systems can be both guided by precise rules and yet remain unpredictable in the long run (such as weather, which is governed by the laws of physics), seemed to stand only in isolated situations.
In the early 1970s, a colleague gave Yorke a copy of a 1963 paper from the Journal of the Atmospheric Sciences that would lay the foundation for the rest of his career. The paper, by American mathematician and meteorologist Edward Norton Lorenz, focused on mathematical methods used in numerical weather forecast models. It derived three differential equations from a complicated atmospheric circulation system and showed how small changes to data fed into the equations could result in considerably different outcomes. It was a very narrowly focused paper that didn’t initially rattle scientific cages, but Yorke recognized that Lorenz’s equations described a phenomenon that was applicable to all complex, dynamical systems.
“Lorenz’s 1963 paper was very insightful, but he hadn’t related it to the broader world,” Yorke said. “It didn't lead anybody to say, ‘Oh, maybe that's what's happening here.’ What I did was explain and generalize the phenomena that Lorenz’s paper described.”
Yorke also gave the phenomenon a name: chaos. In 1975, Yorke and his graduate student Tien-Yien Li published the explanation in an article titled “Period Three Implies Chaos,” in the journal American Mathematical Monthly.
Suddenly, there was a scientific principle to describe how a dynamic system could be so sensitive to small, random perturbations as to be unpredictable, and it opened the gates for scientists and mathematicians to investigate randomness and instability in everything from weather and fluid dynamics to financial markets and the spread of disease.
“The last people to know about chaos were the scientists,” Yorke said. “Because we all know how small things can have a big effect on our lives.”
Chaos became a new branch of science well suited to exploiting the emergence of personal computers and might have evolved in a completely different direction or never come about at all if Yorke’s colleague had never handed him that paper by Lorenz. Or if Yorke had not seen the bigger picture in Lorenz’s work.
“I am pretty good as a mathematical problem-solver, but there are much better problem-solvers,” Yorke explained. “I see connections between things which don't look connected to other people until after the fact, and then, they seem obvious.”
Yorke has been recognized for his work at the highest levels. In 1995, he was named Distinguished University Professor, the highest academic honor bestowed by UMD. In 2003, he was awarded the Japan Prize for finding universality in complex systems such as chaos and fractals. The Japan Prize, administered by the Emperor of Japan, carries nearly the prestige of the Fields Medal and the Nobel Prize. And in 2013, he was awarded the UMD President’s Medal.
That same year, the rotunda of the Mathematics building (now called William E. Kirwan Hall) was named for Yorke in commemoration of his 50 years of scholarship and service to the university. The plaque memorializing the dedication includes a few quotes that summarize Yorke’s philosophy on life and mathematics, including one that was referenced more than once during his 80th birthday celebration: “A degree in mathematics is a license to explore the universe.”
### Curiosity as a Driving Force
It was an exploration of the universe that first drew Yorke to science. In the fourth grade, on a trip to the Hayden Planetarium in New York City, he fell in love with the solar system. It was the imagery of the planets that captivated him at first, perhaps spurred by his budding interest in photography. (Yorke’s father had a darkroom at that time, and Yorke’s passion for photography has carried on, with photographs he took hanging in the lower level of Kirwan Hall.)
The beauty of those solar system images at the planetarium also inspired Yorke to read about astronomy, which led to an interest in physics and mathematics. In college, Yorke discovered he wasn’t excited by lab experiments or finding solutions to problems. He wanted to look at problems in new ways, to take an alternative approach that made the solutions seem more obvious.
“Doing physics experiments was like something equivalent to doing my tax returns,” he said. “I’m not really interested in solving problems, I’m interested in taking a problem and a method that don’t work together, and then you’ve got to change the problem and change the method until they work together.”
Perhaps just as central to Yorke’s success is his drive to root mathematics and physics in the tangible world and to apply his unique vision in areas that resonate with people outside of math and physics.
“My goal is not to make discoveries that only the five smartest people in the world will understand,” Yorke said. “My goal is to tell physicists how to think about mathematics. I’m interested in examples that people can understand, and I’m always asking, ‘Can you come up with research that will surprise people?’”
Call it curiosity or wonder, but even now the element of surprise seems to light a visible spark in Yorke’s eyes when he speaks. It’s as if he’s perpetually challenging the world to throw something new at him.
As he said during his 80th birthday celebration, “The question is always, ‘What do we do next?’”
Written by Kimbra Cutlip
• #### Happy Birthday, Coach!
Celebrating 80 years with Distinguished University Professor Emeritus Jim Yorke, the “Father of Chaos.” James “Jim” Yorke has held many Read More
• #### Technique, Hard Work and Fearlessness
Brin Postdoctoral Fellow Agnieszka Zelerowicz took a chance by leaving Poland, and it set her on a challenging new path. Read More
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2022-01-20 05:00:08
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https://guitarknights.com/guitar-tuner-microphone-guitar-prices.html
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In major-thirds tuning, the interval between open strings is always a major third. Consequently, four frets suffice to play the chromatic scale. Chord inversion is especially simple in major-thirds tuning. Chords are inverted simply by raising one or two notes by three strings. The raised notes are played with the same finger as the original notes.[22][23] In contrast, in standard tuning, the shape of inversions depends on the involvement of the irregular major-third.[24]
Piezoelectric, or piezo, pickups represent another class of pickup. These employ piezoelectricity to generate the musical signal and are popular in hybrid electro-acoustic guitars. A crystal is located under each string, usually in the saddle. When the string vibrates, the shape of the crystal is distorted, and the stresses associated with this change produce tiny voltages across the crystal that can be amplified and manipulated.
While the free guitar lessons here will help you get started, we always recommend committed students to invest in their guitar skills by starting a Guitareo membership. That’s where you’ll get a more comprehensive library of step-by-step video lessons so you always know exactly what to learn next, play-along songs so you can apply your skills to real music, and community support so you’ll get all of your questions answered. Click here to learn more about Guitareo.
Vesala - Muitaki Ihmisii The Ridleys - Maybe The Payolas - Christmas Is Coming The Skywalkers - My Sense Of Fear Tony Cetinski - Sto Si Ti Meni Hotel Books - I Knew Better But Did Nothing The Grays - Everybodys World The Tide - Hey Everybody AVVAH - Kaleidoskop Hearts & Colors - Can't Help Falling In Love Cashew Chemists - Common Equation Sakari Kuosmanen - Laulajan Helmi Remo Drive - Strawberita Jordan Fisher - Mess Kaseva - Silloin Kun
I was lucky enough to meet Justin at the Guitar Institute during a summer school in 2004, and to have some private lessons with him afterwards. He was the teacher who kickstarted my guitar career and persuaded me that I was ready to join a band. That was 14 years ago and many dozens of gigs later. I’m now just finishing a degree in Popular Music Performance. Justin's online lessons are easy to follow and he has a manner about him which makes you believe that you can achieve. Where he demonstrates songs, I have found his versions to be consistently more accurate and easy to follow than those of any other online teacher. On this website you really will find all the skills and information you need to become an excellent musician. Many thanks. Ian.
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###### The BMus in performance with a concentration in guitar is a program that focuses on the study of classical guitar literature and techniques. Goals include enabling students to express themselves musically while emphasizing the skills necessary to pursue careers as professional musicians. The course of study includes extensive performance experiences.
In major-thirds tuning, the interval between open strings is always a major third. Consequently, four frets suffice to play the chromatic scale. Chord inversion is especially simple in major-thirds tuning. Chords are inverted simply by raising one or two notes by three strings. The raised notes are played with the same finger as the original notes.[22][23] In contrast, in standard tuning, the shape of inversions depends on the involvement of the irregular major-third.[24]
Instructor ProfileArlen RothThe King of All Guitar TeachersMusic lesson pioneer Arlen Roth is the quintessential guitarist. An accomplished and brilliant musician — and one of the very few who can honestly say he’s done it all — Roth has, over the course of his celebrated 35-year career, played on the world’s grandest stages, accompanied many of the greatest figures in modern music and revolutionized the concept of teaching guitar. Read More...Lessons Wes Montgomery-style Octaves
When playing seventh chords, guitarists often play only subset of notes from the chord. The fifth is often omitted. When a guitar is accompanied by a bass, the guitarist may omit the bass note from a chord. As discussed earlier, the third of a triad is doubled to emphasize its major or minor quality; similarly, the third of a seventh is doubled to emphasize its major or minor quality. The most frequent seventh is the dominant seventh; the minor, half-diminished, and major sevenths are also popular.[79]
The intervals between the notes of a chromatic scale are listed in a table, in which only the emboldened intervals are discussed in this article's section on fundamental chords; those intervals and other seventh-intervals are discussed in the section on intermediate chords. The unison and octave intervals have perfect consonance. Octave intervals were popularized by the jazz playing of Wes Montgomery. The perfect-fifth interval is highly consonant, which means that the successive playing of the two notes from the perfect fifth sounds harmonious.
When you’re learning a new chord, make the shape and leave it on the guitar for about thirty seconds. Then remove your hand, shake it out, and make the chord shape again. It may take some time for you to make the chord shape again, but that’s okay because you’re working on your muscle memory. Repeating this process a few times is a great way of memorizing your chords.
School of Rock has a finely-tuned preschoolers program called Little Wing that offers all the benefits of our beginner lessons, but is tailored to capture the attention of these young students and set them on a path towards music proficiency. Through playful exploration of rhythm, song structure, and melody kids are introduced to the guitar and other instruments.
Ernie Ball is the world's leading manufacturer of premium electric, acoustic, and classical guitar strings, bass strings, mandolin, banjo, pedal steel strings and guitar accessories. Our strings have been played on many of the best-selling albums of all time and are used by some of history’s greatest musicians including Paul McCartney, Eric Clapton, Jimmy Page, Slash, The Rolling Stones, Angus Young, Eagles, Jeff Beck, Pete Townshend, Aerosmith, Metallica, and more.
Guitar Chords are a group of at least 3 notes played together, this means three different notes, i.e. notes with 3 different pitches. If, for example, you select an E major chord on the guitar chord generator on this page, you can see the 3 notes E, B and G# (Ab) make up this chord. Some notes can be expressed as either sharp or flat (enharmonic spelling), the notes sound just the same but the naming of them is decided by which key the song is in. You can find out more about this in our music theory section.
I design my guitar lessons around each individual student and strive to keep the lesson light and enjoyable while also aiming for a high standard of education. Whether you are looking to play for pleasure at home,around the campfire,or want to learn practical music theory for composition and/or improvisation. I have instruction programs for all levels and styles from beginner to advanced and from ages 7 years on up.
I use this book in my Music Therapy sessions with Geriatric clients--and they love it! The songs are well-known and fun--mostly folk tunes like I've Been Working on the Railroad and My Bonnie Lies Over the Ocean. Chords are SO simple--no bar chords (so you may need a capo to adjust the key). If you're looking for the most basic way to play favorite songs...this is the way to go!
### In music, a guitar chord is a set of notes played on a guitar. A chord's notes are often played simultaneously, but they can be played sequentially in an arpeggio. The implementation of guitar chords depends on the guitar tuning. Most guitars used in popular music have six strings with the "standard" tuning of the Spanish classical-guitar, namely E-A-D-G-B-E' (from the lowest pitched string to the highest); in standard tuning, the intervals present among adjacent strings are perfect fourths except for the major third (G,B). Standard tuning requires four chord-shapes for the major triads.
Learning to play guitar is loads of fun, though playing chords may seem a little intimidating at first. Fear not, it is not much different than playing single notes: you're just playing them all at once! This article will walk you through the process of working out the fingering, and show you how to play some common chords. Pull out your axe, and rock on!
The neck joint or heel is the point at which the neck is either bolted or glued to the body of the guitar. Almost all acoustic steel-string guitars, with the primary exception of Taylors, have glued (otherwise known as set) necks, while electric guitars are constructed using both types. Most classical guitars have a neck and headblock carved from one piece of wood, known as a "Spanish heel." Commonly used set neck joints include mortise and tenon joints (such as those used by C. F. Martin & Co.), dovetail joints (also used by C. F. Martin on the D-28 and similar models) and Spanish heel neck joints, which are named after the shoe they resemble and commonly found in classical guitars. All three types offer stability.
I think for all parties, if we had the ability to send a "two day backstage pass" (you can use that), where we can send a friend a link and they could try the sight for free but couldn't submit videos. I have a few friends that I keep trying to sell them on but they just are not totally sold and I think it's because the beauty of this site is the personalized VE's are. The interaction between student and teacher is what really makes this magical and it's really hard to describe. The search feature could be a little better, more precise and sometimes it finds no VE hits on simple searches like street. Otherwise I am one of your biggest fans.
Are you stuck in a musical rut? New tunings and tricks can help you keep learning guitar in fresh, fun ways. Try one of these great tips from guitar teacher Samuel B. to breathe new life into your guitar playing... One of the first things I tell any new student is that I don't specialize in a formal discipline. If jazz or classical training is your objective, then I'm not your guy. Instead, I specialize primarily in American roots music (that which we tend to casually lump together as "folk"
The ratio of the spacing of two consecutive frets is {\displaystyle {\sqrt[{12}]{2}}} (twelfth root of two). In practice, luthiers determine fret positions using the constant 17.817—an approximation to 1/(1-1/ {\displaystyle {\sqrt[{12}]{2}}} ). If the nth fret is a distance x from the bridge, then the distance from the (n+1)th fret to the bridge is x-(x/17.817).[15] Frets are available in several different gauges and can be fitted according to player preference. Among these are "jumbo" frets, which have much thicker gauge, allowing for use of a slight vibrato technique from pushing the string down harder and softer. "Scalloped" fretboards, where the wood of the fretboard itself is "scooped out" between the frets, allow a dramatic vibrato effect. Fine frets, much flatter, allow a very low string-action, but require that other conditions, such as curvature of the neck, be well-maintained to prevent buzz.
### With that in mind, the inverse to this rule isn’t always true. During the Folk Boom of the 1950s and 60s, there were actually quite a few musicians who put nylon strings on steel string acoustics. This gave the guitar a very warm and relaxed tone, though should you choose to do this be aware that you’re going to get a lot less volume and a reduced response across the entire frequency range.
Lucky you, guitar players from all over the world, to take advantage of the tutorials presented in Justin's comprehensive website! Whatever the style you fancy, Justin is there for you with generous and precise guidance to help you enhance your playing and by doing so, introduce you to so many ways of approaching the guitar and discover new artists along the way. I wish there would have been such a medium and dedicated host around when I started to learn how to play. Bravo Justin, and my gratitude for bringing music to the heart and ears of many!
A string’s gauge is how thick it is. As a general rule, the thicker a string is the warmer its response will be and the more volume it will produce. However, thicker strings are also stiffer. This makes it harder to fret the string and makes it more difficult to execute heavy string bends. Thinner strings are generally brighter and easier to play, but on some instruments they can sound thin and tinny.
The pickguard, also known as the scratchplate, is usually a piece of laminated plastic or other material that protects the finish of the top of the guitar from damage due to the use of a plectrum ("pick") or fingernails. Electric guitars sometimes mount pickups and electronics on the pickguard. It is a common feature on steel-string acoustic guitars. Some performance styles that use the guitar as a percussion instrument (tapping the top or sides between notes, etc.), such as flamenco, require that a scratchplate or pickguard be fitted to nylon-string instruments.
You need to place one finger on whatever fret you want to bar and hold it there over all of the strings on that fret. The rest of your fingers will act as the next finger down the line (second finger barring, so third finger will be your main finger, and so on). You can also buy a capo, so that you don't have to deal with the pain of the guitar's strings going against your fingers. The capo bars the frets for you. This also works with a ukulele.
Flat-top or steel-string guitars are similar to the classical guitar, however, within the varied sizes of the steel-stringed guitar the body size is usually significantly larger than a classical guitar, and has a narrower, reinforced neck and stronger structural design. The robust X-bracing typical of the steel-string was developed in the 1840s by German-American luthiers, of whom Christian Friedrich "C. F." Martin is the best known. Originally used on gut-strung instruments, the strength of the system allowed the guitar to withstand the additional tension of steel strings when this fortunate combination arose in the early 20th century. The steel strings produce a brighter tone, and according to many players, a louder sound. The acoustic guitar is used in many kinds of music including folk, country, bluegrass, pop, jazz, and blues. Many variations are possible from the roughly classical-sized OO and Parlour to the large Dreadnought (the most commonly available type) and Jumbo. Ovation makes a modern variation, with a rounded back/side assembly molded from artificial materials.
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###### Paul Gilbert has made Guitar World's "50 Fastest Guitarists of All Time", and his guitar shredding is legendary. Paul picked up his first guitar when he was 5 years old. By the time he was 15, he was featured in Guitar Player magazine. After graduating from the Guitar Institute of Technology (GIT) in Los Angeles, Paul immediately joined the faculty as a guitar teacher. While teaching, he formed the band Racer X and Grammy-nominated group Mr. Big. Paul breaks down his iconic playing techniques and practice drills. He's known for his enthusiastic and approachable guitar lessons, which are founded on decades of experience.
The Spanish vihuela, called in Italian the "viola da mano", a guitar-like instrument of the 15th and 16th centuries, is widely considered to have been the single most important influence in the development of the baroque guitar. It had six courses (usually), lute-like tuning in fourths and a guitar-like body, although early representations reveal an instrument with a sharply cut waist. It was also larger than the contemporary four-course guitars. By the 16th century, the vihuela's construction had more in common with the modern guitar, with its curved one-piece ribs, than with the viols, and more like a larger version of the contemporary four-course guitars. The vihuela enjoyed only a relatively short period of popularity in Spain and Italy during an era dominated elsewhere in Europe by the lute; the last surviving published music for the instrument appeared in 1576.[9]
Solo, lead, and rhythm guitarists everywhere can now access the best selection of instantly downloadable digital sheet music and guitar tab on the internet. Put down the pick for just a moment and put your fingers to work browsing through Musicnotes.com's vast archives of guitar tabs ready to be enjoyed by musicians of all ages. Our collection features a weekly updated catalogue of some of guitars greatest compilations.
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2019-06-25 02:13:23
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https://www.projecteuclid.org/euclid.ecp/1463434786
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Electronic Communications in Probability
A Non-Ballistic Law of Large Numbers for Random Walks in I.I.D. Random Environment
Martin Zerner
Abstract
We prove that random walks in i.i.d. random environments which oscillate in a given direction have velocity zero with respect to that direction. This complements existing results thus giving a general law of large numbers under the only assumption of a certain zero-one law, which is known to hold if the dimension is two.
Article information
Source
Electron. Commun. Probab., Volume 7 (2002), paper no. 19, 191-197.
Dates
Accepted: 30 September 2002
First available in Project Euclid: 16 May 2016
https://projecteuclid.org/euclid.ecp/1463434786
Digital Object Identifier
doi:10.1214/ECP.v7-1060
Mathematical Reviews number (MathSciNet)
MR1937904
Zentralblatt MATH identifier
1008.60107
Subjects
Primary: 60K37: Processes in random environments
Secondary: 60F15: Strong theorems
Rights
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2019-12-08 08:48:18
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http://clanzd.com/mathjax-basic-tutorial-crossing-things-out.html/
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# LaTeX & MathJax basic tutorial – Crossing things out
## Crossing things out
Use \require{cancel} in the first formula in your post that requires cancelling; you need it only once per page. Then use:
$$\require{cancel}\begin{array}{rl} \verb|y+\cancel{x}| & y+\cancel{x}\\ \verb|\cancel{y+x}| & \cancel{y+x}\\ \verb|y+\bcancel{x}| & y+\bcancel{x}\\ \verb|y+\xcancel{x}| & y+\xcancel{x}\\ \verb|y+\cancelto{0}{x}| & y+\cancelto{0}{x}\\ \verb+\frac{1\cancel9}{\cancel95} = \frac15+& \frac{1\cancel9}{\cancel95} = \frac15 \\ \end{array}$$
Use \require{enclose} for the following:
$$\require{enclose}\begin{array}{rl} \verb|\enclose{horizontalstrike}{x+y}| & \enclose{horizontalstrike}{x+y}\\ \verb|\enclose{verticalstrike}{\frac xy}| & \enclose{verticalstrike}{\frac xy}\\ \verb|\enclose{updiagonalstrike}{x+y}| & \enclose{updiagonalstrike}{x+y}\\ \verb|\enclose{downdiagonalstrike}{x+y}| & \enclose{downdiagonalstrike}{x+y}\\ \verb|\enclose{horizontalstrike,updiagonalstrike}{x+y}| & \enclose{horizontalstrike,updiagonalstrike}{x+y}\\ \end{array}$$
\enclose can also produce enclosing boxes, circles, and other notations; see MathML menclose documentation for a complete list.
## Continued fractions
To make a continued fraction, use \cfrac, which works just like \frac but typesets the results differently:
$$x = a_0 + \cfrac{1^2}{a_1 + \cfrac{2^2}{a_2 + \cfrac{3^2}{a_3 + \cfrac{4^4}{a_4 + \cdots}}}}$$
Don’t use regular \frac or \over, or it will look awful:
$$x = a_0 + \frac{1^2}{a_1 + \frac{2^2}{a_2 + \frac{3^2}{a_3 + \frac{4^4}{a_4 + \cdots}}}}$$
You can of course use \frac for the compact notation:
$$x = a_0 + \frac{1^2}{a_1+} \frac{2^2}{a_2+} \frac{3^2}{a_3 +} \frac{4^4}{a_4 +} \cdots$$
Continued fractions are too big to put inline. Display them with $$$$ or use a notation like $[a_0; a_1, a_2, a_3, \ldots]$.
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2018-05-26 19:20:06
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https://byjus.com/maths/vertical-angles/
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# Vertical Angles (Vertically Opposite Angles)
When two lines intersect each other, then the opposite angles, formed due to intersection are called vertical angles or vertically opposite angles. A pair of vertically opposite angles are always equal to each other. Also, a vertical angle and its adjacent angle are supplementary angles, i.e., they add up to 180 degrees. For example, if two lines intersect and make an angle, say X=45°, then its opposite angle is also equal to 45°. And the angle adjacent to angle X will be equal to 180 – 45 = 135°.
When two lines meet at a point in a plane, they are known as intersecting lines. When the lines do not meet at any point in a plane, they are called parallel lines. Learn about Intersecting Lines And Non-intersecting Lines here.
## Definition
As we have discussed already in the introduction, the vertical angles are formed when two lines intersect each other at a point. After the intersection of two lines, there are a pair of two vertical angles, which are opposite to each other.
The given figure shows intersecting lines and parallel lines.
In the figure given above, the line segment $\overline{AB}$ and $\overline{CD}$ meet at the point $O$ and these represent two intersecting lines. The line segment $\overline{PQ}$ and $\overline{RS}$ represent two parallel lines as they have no common intersection point in the given plane.
In a pair of intersecting lines, the angles which are opposite to each other form a pair of vertically opposite angles. In the figure given above, ∠AOD and ∠COB form a pair of vertically opposite angle and similarly ∠AOC and ∠BOD form such a pair. Therefore,
∠AOD = ∠COB
∠AOC = ∠BOD
For a pair of opposite angles the following theorem, known as vertical angle theorem holds true.
Note: A vertical angle and its adjacent angle is supplementary to each other. It means they add up to 180 degrees
## Vertical Angles: Theorem and Proof
Theorem: In a pair of intersecting lines the vertically opposite angles are equal.
Proof: Consider two lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{CD}$ which intersect each other at $O$. The two pairs of vertical angles are:
i) ∠AOD and ∠COB
ii) ∠AOC and ∠BOD
It can be seen that ray $\overline{OA}$ stands on the line $\overleftrightarrow{CD}$ and according to Linear Pair Axiom, if a ray stands on a line, then the adjacent angles form a linear pair of angles.
Therefore, ∠AOD + ∠AOC = 180° —(1) (Linear pair of angles)
Similarly, $\overline{OC}$ stands on the line $\overleftrightarrow{AB}$.
Therefore, ∠AOC + ∠BOC = 180° —(2) (Linear pair of angles)
From (1) and (2),
∠AOD + ∠AOC = ∠AOC + ∠BOC
⇒ ∠AOD = ∠BOC —(3)
Also, $\overline{OD}$ stands on the line $\overleftrightarrow{AB}$.
Therefore, ∠AOD + ∠BOD = 180° —(4) (Linear pair of angles)
From (1) and (4),
∠AOD + ∠AOC = ∠AOD + ∠BOD
⇒ ∠AOC = ∠BOD —(5)
Thus, the pair of opposite angles are equal.
Hence, proved.
### Solved Example
Consider the figure given below to understand this concept.
In the given figure ∠AOC = ∠BOD and ∠COB = ∠AOD(Vertical Angles)
⇒ ∠BOD = 105° and ∠AOD = 75°
## Frequently Asked Questions – FAQs
### What is vertical angles?
When two lines intersect each other, then the angles opposite to each other are called vertical angles.
### How to measure vertical angles?
If the angle next to the vertical angle is given to us, then we can subtract it from 180 degrees to get the measure of vertical angle, because vertical angle and its adjacent angle are supplementary to each other.
### If x=30 degrees is a vertical angle, when two lines intersect, then find all the angles?
Given, vertical angle, x = 30
Let y is the angle vertically opposite to x, then y = 30 degrees
Now, as we know, vertical angle and its adjacent angle add up to 180 degrees, therefore,
The other two angles are: 180 – 30 = 150 degrees
### What are complementary angles with example?
The angles which are adjacent to each other and their sum is equal to 90 degrees, are called complementary angles. For example, x = 45 degrees, then its complement angle is: 90 – 45 = 45 degrees
#### 1 Comment
1. Pralipta Das
Thank you sir or mam this is helpful in my examination also .a lots of thank you sir or mam
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2020-10-25 20:09:49
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http://cms.math.ca/cjm/msc/46J40
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location: Publications → journals
Search results
Search: MSC category 46J40 ( Structure, classification of commutative topological algebras )
Expand all Collapse all Results 1 - 1 of 1
1. CJM 2007 (vol 59 pp. 966)
Forrest, Brian E.; Runde, Volker; Spronk, Nico
Operator Amenability of the Fourier Algebra in the \$\cb\$-Multiplier Norm Let \$G\$ be a locally compact group, and let \$A_{\cb}(G)\$ denote the closure of \$A(G)\$, the Fourier algebra of \$G\$, in the space of completely bounded multipliers of \$A(G)\$. If \$G\$ is a weakly amenable, discrete group such that \$\cstar(G)\$ is residually finite-dimensional, we show that \$A_{\cb}(G)\$ is operator amenable. In particular, \$A_{\cb}(\free_2)\$ is operator amenable even though \$\free_2\$, the free group in two generators, is not an amenable group. Moreover, we show that if \$G\$ is a discrete group such that \$A_{\cb}(G)\$ is operator amenable, a closed ideal of \$A(G)\$ is weakly completely complemented in \$A(G)\$ if and only if it has an approximate identity bounded in the \$\cb\$-multiplier norm. Keywords:\$\cb\$-multiplier norm, Fourier algebra, operator amenability, weak amenabilityCategories:43A22, 43A30, 46H25, 46J10, 46J40, 46L07, 47L25
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2014-10-23 01:35:15
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https://wiki.iota.org/shimmer/learn/maintenance/
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# Maintaining Shimmer
At its core, Shimmer needs few resources to run. To make a transaction, you need to cryptographically sign a short data sequence and send it to a node to turn it into a block and propagate the block through the network. When another node receives that block, it runs it through a series of simple checks, like verifying the signatures and checking what blocks it refers to and adds a new entry to its database of blocks. An average personal computer is enough to run a Shimmer node.
## Incentives
Fundamentally, Shimmer regulates only the communication between its nodes, not the nodes and the end users, and a random node has no incentives to take your transaction into work. Public nodes process strangers' transactions out of good spirit, but they can always go offline or put your transaction to the very end of the queue. If you need a reliable connection to the Tangle, the only way to get it is to run your own Shimmer node.
### Proof of Work
A node could start issuing so many blocks that it would clog the network. To prevent this, the Shimmer network requires the issuing node to perform some empty work. It affects only a single node and not the rest of the network, and it distributes the throughput of the network fairly between all its nodes.
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2023-03-29 10:38:27
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https://datascience.stackexchange.com/questions/37941/what-is-the-difference-between-sgd-classifier-and-the-logisitc-regression
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# What is the difference between SGD classifier and the Logisitc regression?
To my understanding, the SGD classifier, and Logistic regression seems similar. An SGD classifier with loss = 'log' implements Logistic regression and loss = 'hinge' implements Linear SVM. I also understand that logistic regression uses gradient descent as the optimization function and SGD uses Stochastic gradient descent which converges much faster. But which of the two algorithms to use in which scenarios? Also, how are SGD and Logistic regression similar and how are they different?
Welcome to SE:Data Science.
SGD is a optimization method, while Logistic Regression (LR) is a machine learning algorithm/model. You can think of that a machine learning model defines a loss function, and the optimization method minimizes/maximizes it.
Some machine learning libraries could make users confused about the two concepts. For instance, in scikit-learn there is a model called SGDClassifier which might mislead some user to think that SGD is a classifier. But no, that's a linear classifier optimized by the SGD.
In general, SGD can be used for a wide range of machine learning algorithms, not only LR or linear models. And LR can use other optimizers like L-BFGS, conjugate gradient or Newton-like methods.
• So, if I write clf = SGDClassifier( class_weight='balanced', alpha=i, penalty='l2', loss='hinge', random_state=42) it is an implementation of Linear SVM and if I write clf = SGDClassifier( class_weight='balanced', alpha=i, penalty='l2', loss='log', random_state=42) . It is an implementation of Logisitic regression. Am I right ? – Akash Dubey Sep 7 '18 at 18:37
• @AkashDubey Yes – user12075 Sep 7 '18 at 21:18
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2019-12-09 08:40:16
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http://physics.stackexchange.com/tags/temperature/hot
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# Tag Info
22
I think you are right. A perhaps more precise relation between temperature and velocity is the Maxwell–Boltzmann distribution: \begin{equation*} P(\textbf{v}) = \left( \frac{m}{2\pi k_B T} \right)^{3/2} \text{exp} \left[-\frac{m ( \textbf{v} - \textbf{v}_0)^2}{2 k_B T} \right]. \end{equation*} where you see that the mean velocity $\textbf{v}_0$ and the ...
12
I think your view is correct, and you can think about the following real word example. In labs here on earth, we can use laser cooling techniques to cool atoms to $\mu$K scales in the lab frame. But the lab is on earth, and the earth is moving very fast around the sun, and the sun is moving very fast around the galactic center and so on. We don't take ...
6
You're right that the classical idea of radiation emission from an accelerated charge cannot be applied to electrons in orbit around nuclei, and thus they do not emit radiation (unless they're in an exited state and decay to a lower state). The same thing does not apply to the nuclei. As you suspect, they will, over time, lose energy and vibrate less and ...
5
When two quantities of water ($m_1$ and $m_2$) at different temperatures (resp. $T_1$ and $T_2$) are mixed in adiabatic conditions (no heat loss and no external heating during mixing) the temperature $T$ of the resulting mixture can be calculated from the heat balance (no heat is lost or added so the heat contained in both masses is found again in the ...
4
Why does heat added to a system cause an increase in entropy that is independent of the amount of particles in the system? Short answer: it doesn't The systems won't end up with the same entropy. Your intuition is correct that the change in entropy depends on the number of particles. The reason why you can't just reason directly from $dS = \delta Q/T$ ...
4
A single particle can be a 'system' within itself having modes depending on the particle ' s structure. These modes may be in 'tune' with the incident radiation and thus capture the energy which can increase the particle ' s momentum and therefore its velocity. We never say the particle's temperature has increased but rather it's momentum. When a system of ...
3
The temperature is really only negative in the sense of the classical definition of temperature. What is actually happening in a population inversion is the particles aren't following Boltzmann distribution of energies anymore. Comparing the temperature of a Boltzmann distributed system to a non-Boltzmann system might not be meaningful at all. People say ...
2
It's simple. You can think of temperature as being the standard deviation of KE among all components (atoms) of a mass. This is significant because KE is a relative quantity, but temperature is absolute, and this relationship makes that possible. If all atoms are moving uniformly in the same direction, then the temperature would be 0.
2
There seems to be some confusion here - the perception of "heat" from an IR lamp relates to the amount of energy absorbed by the body. This depends on the reflectivity of the skin at that particular wavelength; also, you would have to normalize this in some way. Now we know from Planck's Law that there is a distribution of wavelengths from a black body: the ...
2
Therefore, when we say, for example, that the energy of the ideal gas at temperature T is E=32NkBT, we should really be saying "the energy of the ideal gas immersed in a heat bath at temperature T"? Is this reasoning valid? This is true. What is also true is that you can also say that the temperature of the (completely isolated) gas of energy E is T=2/3 ...
2
Since the question is rather vague, I will just give you some key points: Debye's model treats oscillation modes of a solid as sound waves (phonons) with frequency $\omega(\mathbf{k})=v|\mathbf{k}|$ ($v$ the sound velocity). As a result, with this model, Debye shows how the heat capacity is directly related to the rate of change of the energy expectation ...
2
As march pointed out, your reasoning is incorrect for finite $Q$ and $\Delta S$. However, the equation is true for differentials, so we have to address that. Possible 'conceptual' answers: Stat mech: the increase in $S$ is related to how much extra disorder is produced. When there are less particles, they each get a bigger share of the $dQ$, so they each ...
2
To add a little detail about radiative thermal equilibrium: As atoms at nonzero temperature collide with each other, they do emit electromagnetic radiation, and if they were in an empty universe, they would approach zero temperature. However, since the universe isn't empty, they also absorb electromagnetic radiation coming from all the other atoms around. ...
1
Cubero et al. 2007: Thermal equilibrium and statistical thermometers in special relativity (http://arxiv.org/abs/0705.3328) came to the conclusion that 'temperature' can be statistically defined and measured in an observer frame independent way. With fully relativistic 1D molecular dynamics simulations they verified that the temperature definition ...
1
By the Ideal gas law, $PV=nRT$, or "pressure times volume equals the number of molecules times a constant times temperature". So, all else being the same, as the temperature goes up, the pressure goes up in an exact ratio. However, all else does not have to be the same. So, for instance, if you reduce the number of molecules in a container ($n$), the ...
1
Of course, they are relate to each other but that doesn't mean they are the same things. Temperature is the average kinetic energy of the molecules while pressure is the force they exert perpendicularly on any surface. Of course, more the temperature, more would be the pressure. While the former is related to the energy, the later is related to the ...
1
Because conventionally we assume constant temperature, and length and density are also assumed to be constants for a given resistor. Of course, this is not true. In some circuit designs I have to pay very careful attention to resistance changes with temperature, and indeed this is sometimes used to provide temperature measurements in the form of RTD ...
1
The electric fan increases the velocity and hence the kinetic energy of the molecules in the air. this would mean that the temperature has increased. I think that there is a bit of a problem here. Kinetic energy is not quite the same as thermal energy and temperature. Thermal energy and temperature is a measure of random, thermal motion of atoms or ...
1
Temperature is connected with the kinetic degrees of freedom of the atoms or molecules . In a gas , temperature is given by directly proportional to the root mean square of velocity, an average kinetic energy. Raising the temperature raises the probability of atoms/molecules scattering against each other. Attraction with neutral atoms and molecules ...
1
Background Let us assume we have a function, $f_{s}(\mathbf{x},\mathbf{v},t)$, which defines the number of particles of species $s$ in the following way: $$dN = f_{s}\left( \mathbf{x}, \mathbf{v}, t \right) \ d^{3}x \ d^{3}v$$ which tells us that $f_{s}(\mathbf{x},\mathbf{v},t)$ is the particle distribution function of species $s$ that defines a ...
1
On the phase diagram of methane, you can see that at RT (20°C), methane can only be gas (or super-critical if pressure is enough). The pressure can be calculated with the ideal gas equation $\frac{pV}{T}=nR$ You need to calculate quantity of n (in moles, given the volume, density of liquid methane, and the weight of the molecule). R is a constant, V is ...
1
If we were to send a unmannedspaceship through the Sun. What material can survive? A wide variety of materials might survive passing through the outer layer of the sun, but only if the spaceship is big enough, fast enough and has a thick enough sacrificial/ablative shell. If the ship is slow, it doesn't really matter if the hull survives, most stuff ...
1
Writing $(Q_m)$ (and so on) for the numerical value of the constants in the units stated, \begin{align} \frac{Q_m}{\omega_b \rho_b c_b} &= \frac{(Q_m)\mathrm W/\mathrm m^3}{(\omega_b)\frac{\mathrm{ml}}{\mathrm{ml\: s}} \times (\rho_b) \frac{\mathrm {kg}}{\mathrm m^3} \times (c_b)\frac{\mathrm J}{\mathrm{kg}\:^{\circ}\mathrm C}} \\ & = ...
1
Can a single particle be “heated” by radiation? Not really, because heat is an emergent macroscopic property of an ensemble of particles. But since you put the word "heated" in quotes, we can allow a yes of sorts. Take a look at the Wikipedia temperature page, where we can see this picture: CCASA image by Greg L, see Wikipedia It's to do with the ...
1
The conventional method of nuclear waste disposal is just fine. Just bury it deep into some geologically inactive rock. Consider this: If you dig a deep hole into ground in some geologically inactive area, you will find rocks and minerals that have been there for millions of years. If you put your nuclear waste there, there is no reason why it wouldn't ...
1
The energy per unit area radiated by an object at a temperature $T$ is given by the Stefan-Boltzmann law: $$J = \varepsilon\sigma T^4 \tag{1}$$ where $\sigma$ is the Stefan-Boltzmann constant and $\varepsilon$ is the emissivity. A spaceship in a vacuum can only lose heat by radiation, so it will heat up until the energy loss given by equation (1) is ...
1
Planck temperature is the maximum temperature on the Planck scale. Some sources call it absolute hot. The Planck temperature is defined as $$T_P = \frac{m_P c^2}{k} = \sqrt{\frac{\hbar c^5}{G k^2}} = 1.416833(85)\times10^{32} \text{K}$$ where $m_P$ is the Planck mass, $c$ is the speed of light, $\hbar = h/2\pi$ is the reduced Planck constant, $k$ is ...
1
Ultimately, Newton's law of cooling is a simplification that can be obtained from the full heat equation, i.e. $$\rho c\frac{\partial T}{\partial t} = - \kappa \nabla \cdot T.$$ The heat equation itself can be derived from first principles, assuming Fourier's law for heat flow, namely that it is proportional microscopically to the difference in temperature ...
Only top voted, non community-wiki answers of a minimum length are eligible
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2015-12-02 07:03:20
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https://www.gradesaver.com/textbooks/science/chemistry/chemistry-a-molecular-approach-3rd-edition/chapter-3-sections-3-1-3-12-exercises-problems-by-topic-page-133/93a
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## Chemistry: A Molecular Approach (3rd Edition)
$C_{12}H_{14}N_{2}$
Calculate the molar mass of the empirical formula by adding the molar masses of its constituent elements together. $6(12.01)+7(1.01)+1(14.0)=93.13$ Divide the molar mass given by this number. $\frac{186.24}{93.13}=2$ Multiply the empirical formula by this number to get the molecular formula. $2(C_{6}H_{7}N)=C_{12}H_{14}N_{2}$
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2018-12-10 11:35:58
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https://www.physicsforums.com/threads/question-about-non-linear-second-order-differential-equation.407481/
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# Question About Non-Linear Second Order Differential Equation
1. Jun 2, 2010
### mmmboh
Hi, I just had my ordinary differential equation final exam today (summer class), the exam was fine. We had a bonus question though, it was a nonlinear second order differential equation, we didn't learn anything about those except for one existence and uniqueness theorem. I am curious if I got it right. The question was, solve:
x2y''-4(y')2=0
Clearly C (constant) is a solution, for the other solution I made an educated guess that y is x2, that didn't quite work, but I figured out it was x2/8...so I put
y=x2/8 + C
I know this solution works because I've plugged it in, the thing I am wondering is when I solve it on wolframalpha.com it gives
y[x] = -4 ln(C1x-4)/(C12)-x/C1+C2.
I have noticed that these two solutions (mine and wolframs) are linearly independent, but since the equation is nonlinear I don't think combining the two would also be a solution....so I am just wondering if I got the answer right? Also, why doesn't wolfram also give my answer. I mean is it not a completely right answer? I guess the same rules don't apply to nonlinear and linear equations, could there be other answers as well?
Thanks!
2. Jun 2, 2010
### Dickfore
The equation does not contain y explicitly, so you can reduce its order by introducing a new unknown:
$$\begin{array}{l} y' = z \\ y'' = z'$$
The first order equation that you will get is with separable variables and can be integrated.
3. Jun 2, 2010
### mmmboh
Hm yes that worked and I got the answer I got, but what about the answer wolframalpha gives?
4. Jun 2, 2010
### mmmboh
I'm assuming you're talking about the wolfram alpha solution....well, as C1 approaches zero the equation approaches infinity...I'm not sure..
5. Jun 2, 2010
### Dickfore
If you reduce your equation to the first order equation:
$$x^{2} z' = 4 z^{2}$$
and you integrate this equation by separating variables:
$$\frac{z'}{z^{2}} = \frac{4}{x^{2}}$$
$$\frac{dz}{z^{2}} = \frac{4 \, dx}{x^{2}}$$
$$-\frac{1}{z} = -\frac{4}{x} + C_{1}$$
$$z = \frac{x}{4 - C_{1} \, x}$$
you can see that your solution corresponds to the value $C_{1} = 0$. The reason why this is inconvenient in the general solution is because $C_{1}$ is in the denominator and taking this limit requires that we use the absolute value in the logarithm (remember $\int{dx/x} = \ln|x|$) and adjust the value of $C_{2}$ so that we have an indeterminate form $0/0$ and then take L'Hospital's Rule. The way I am suggesting is more straightforward. So, you get:
$$z = y' = \frac{x}{4}$$
which is precisely what you have found.
6. Jun 3, 2010
### mmmboh
It doesn't seem like that is really the real answer then? You wouldn't know that C1=0. Sorry I am not familiar with non-linear equations, maybe I am applying first order logic that doesn't apply here or something, but does the equation have two linearly independent solutions that you can't combine to make one total solution? If you plug in different values for x in my solutions and in wolframs, you don't get the same answer.
7. Jun 3, 2010
### Dickfore
The red highlighted text addresses two different things: the order of a DE and (non)linearity. Also, the thing you are talking about the two linearly-independent solutions and constructing a general solution by making a linear combination of them is true only for second-order linear DE. As your equation is not linear, this is not the case.
The blue highlighted text also concerns different concepts. When the integrating constants in the general solution attain a particular value, we obtain a particular solution. This is still a function of x. You are not required to take a particular value for the argument of the function x.
As a side note, one particular solution that is "hidden" from the form of the general solution is
$$x = 0$$
To see this, treat y as an argument and x as a dependent variable. The derivatives become:
$$y' = \frac{1}{x'}$$
$$y'' = \frac{dy'}{dy} \, \frac{dy}{dx} = \frac{1}{x'} \, \frac{d}{dy}(\frac{1}{x'}) = -\frac{x''}{(x')^{3}}$$
Substituting these, we get:
$$x^{2} \left(-\frac{x''}{(x')^{3}}\right) - 4 \left(\frac{1}{x'}\right)^{2} = 0$$
$$x^{2} \, x'' + 4 x' = 0$$
Then, $x = 0$ ($x' = x'' = 0$) is an obvious solution.
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2017-12-12 00:37:58
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http://www.markjoshi.com/books/phpBB3/viewtopic.php?f=11&t=765&sid=45b3bbabf4c45c6d85a94ce36dbc41bb&p=2341
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## Chapter 0: Branching Processes
This forum is to discuss details in the book "Probability with Martingales" by David Williams.
### Chapter 0: Branching Processes
Hi Mark,
For the branching processes 'Concrete Example', section 0.9 (and page 11 of the eleventh printing), where \mu > 1, we find:
L(\lambda) = \frac{p\lambda+q-p}{q\lambda+q-p} = \pi exp{-\lambda x 0} + \int ... dx
I am fine with the first step, the second I am not. How do I take this Laplace transform and convert it into such as integral?
Trent
trent4213
Posts: 1
Joined: Tue Nov 30, 2010 1:45 am
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2018-06-19 22:23:16
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https://homework.cpm.org/category/MN/textbook/cc2mn/chapter/cc25/lesson/cc25.1.1/problem/5-13
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### Home > CC2MN > Chapter cc25 > Lesson cc25.1.1 > Problem5-13
5-13.
Simplify the following expressions. Homework Help ✎
1. $\frac { 2 } { 3 } + \frac { 4 } { 5 }$
Change both fractions to equivalent fractions with a common denominator.
$\frac{10}{15} + \frac{12}{15}$
$\frac{22}{15}$
1. $( \frac { 4 } { 5 } ) ( \frac { 2 } { 3 } )$
Multiply the numerators and denominators of both fractions.
$\frac{8}{15}$
1. $\frac{4}{5}-(-\frac{2}{3})$
Remember when you subtract a negative number, this has the same result as if you added it.
$\frac{4}{5} + \frac{2}{3}$
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2019-10-20 00:59:45
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https://tug.org/pipermail/pdftex/2002-August/002954.html
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# [pdftex] Position of hyperref targets
Ross Moore ross at ics.mq.edu.au
Thu Aug 29 10:27:27 CEST 2002
>
> I have a pdf document with a lot of labeled sections referred to by nameref. For a special version of the document I switched off section numbering and noticed that the targets for the hyper links shifted down to below the section titles (I took me a while to figure out the connection to the section numbers :-)
>
> I prepared a small sample document illustrating the problem. Source and pdf are available at
>
> ftp://www.ragtime.de/Public/hyperref/
>
You use:
\section{Willkommen, Bienvenue, Welcome}\label{TopicWelcome}
so that the \label comes *after* the \section has done its stuff.
Hence you are really marking the point at the beginning of the 1st paragraph
> What can I do to convince hyperref that my gentle readers want to see the section title after clicking a link?
>
Thus you really need to raise the anchor (sic :-) that is placed
inside the PDF by the \label command.
To do this, you could try something like:
\newcommand{\raiselabel}[2][25pt]{% default of 25pt
\llap{\rlap{$\smash{\raise #1\hbox{\label{#2}}}}$}%
\ignorespaces
}
\section{Willkommen, Bienvenue, Welcome}%
\raiselabel{TopicWelcome}
or
\section{Willkommen, Bienvenue, Welcome}%
\raiselabel[30pt]{TopicWelcome}
There are more elegant ways to do this, using deeper internal TeX coding;
but this should do it using macros for which you can easily find and
understand the documentation.
If the Acrobat positioning is not quite right in *all*
of the Acrobat page-view modes, then you can adjust the combination
of \llap and \rlap or add a left-shift using \kern-10pt, say.
Hope this helps,
Ross Moore
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2020-09-25 00:04:06
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https://socratic.org/questions/zinc-reacts-with-water-to-produce-zinc-hydroxide-and-hydrogen-gas-what-is-the-eq
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# Zinc reacts with water to produce zinc hydroxide and hydrogen gas. What is the equation for this reaction?
Nov 19, 2017
$Z n + 2 {H}_{2} O \to Z n {\left(O H\right)}_{2} + {H}_{2}$
#### Explanation:
To get this chemical equation, all you have to do is reference the question because it already gives you all of the compounds you need.
To get the formula for zinc (II) hydroxide, you need to know that hydroxide has a 1- charge and you need 2 of them to cancel out zinc's 2+ charge.
Hydrogen gas is a diatomic molecule from the mnemonic Br I N Cl H O F.
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2019-08-20 18:14:45
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https://mathoverflow.net/questions/301454/stability-and-complete-types-in-model-theory/302015
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# Stability and complete types (in Model Theory)
I read the following statement in these slides of Saharon Shelah: "$K$ is stable iff for every $M \in K$ there are only "few" complete types over $M$." About the notation: here $K$ consists of all structures $N$ with the same theory of $M$, that is, the same set of first order sentences which are true in $M$.
I am trying to understand this statement in a context far away from Model Theory.
Suppose $M$ is, e.g., a projective space $\mathbb{P}^n(k)$ with $k$ some field; then what would this statement mean exactly ?
• What makes you think this statement makes sense in a context "far away from Model Theory"? Shelah is only making the definition for elementary classes. In fact, it is possible to make sense of the definition in more general contexts (e.g. metric structures or abstract elementary classes), but my view is that making sense of this definition in some context would actually be evidence that that context is actually not so far away from classical model theory. – Alex Kruckman May 29 '18 at 14:36
• @Alex Kruckman: you should read it as "far away (at first sight)." I know that stability of certain types of graphs has been studied, so I could have asked the question about the point graph or incidence graph of $\mathbb{P}^n(k)$ instead. My question remains the same: what would the statement mean for this type of graph ? For instance, what are the "complete types" over such a graph ? – THC May 29 '18 at 14:49
• Well, the right way to make the definition depends on what properties you are interested in. Also, there are a number of definitions of stability which are equivalent in the first-order setting, but not necessarily in other settings. So for example, when people study stable classes of finite graphs, they usually define stability by a bound on the size of instances of the order property for the edge relation, which has nothing to do (in this context) with counting complete first-order types. In the case of a projective space in the incidence language (points, lines, incidence relation), ... – Alex Kruckman May 29 '18 at 15:01
• ... this is a perfectly good first-order structure, and you use the standard definition of stability from first-order model theory. In fact, $\mathbb{P}^n(k)$ (in the incidence language) is bi-interpretable with $k$ (in the field language), so it will be stable in the first-order model theory sense if and only if $k$ is stable (e.g. if $k$ is finite, algebraically closed, or separably closed). – Alex Kruckman May 29 '18 at 15:02
• Sure, but an abstract projective space in the incidence language is still a first-order structure, and you can take the usual definition of stability. All projective spaces of dimension at least $3$ satisfy Desargues's theorem, so there is at least a division ring around. You can google for "stable division ring" to get a sense for what these can look like. In the case $n = 2$, there are stable non-desarguesian projective planes, constructed using Hrushovski's method. For an exposition, see Section 10.4 of Tent and Ziegler's book A Course in Model Theory. – Alex Kruckman May 29 '18 at 15:27
My previous answer addressed the question "How can I make sense of the counting types definition of stability outside of a first-order model theory context?"
After discussion in the comments, I realized that you really wanted to know what stability means in the context of an (abstract) projective plane. So I'll try to address that question in this answer.
First some background in model theory, just to make sure if we're on the same page. Let $M$ be an $L$-structure and $C$ a subset of $M$. We write $L(C)$ for the set of all first-order formulas with parameters from $C$. For example, if $L = \{P,L,I\}$ is the language of incidence structures (where $P$ and $L$ are unary relations picking out the points and lines and $I$ is the binary incidence relation), then $\varphi(x_1,x_2): \forall y\, ((L(y) \land I(x_1,y)\land I(x_2,y))\rightarrow \lnot I(c,y))$ is a formula with a single parameter $c\in C$ expressing that no line through $x_1$ and $x_2$ is incident with $c$.
Let $x = (x_1,\dots,x_n)$ be a tuple of variables. A complete type $p(x)$ over $C$ (in variable context $x$) is a maximal set of $L(C)$-formulas with free variables from $x$ which is consistent with $\mathrm{Th}_{L(C)}(M)$, the set of all $L(C)$-sentences true in $M$. Maximality amounts to saying that for every formula $\varphi(x)\in L(C)$, either $\varphi(x)\in p(x)$ or $\lnot \varphi(x)\in p(x)$. We write $S_x(C)$ for the set of all complete types over $C$ in context $x$.
Let $a = (a_1,\dots,a_n)$ be an element of $M$. Then the type of $a$ over $C$ is the set of all formulas with parameters from $C$ satisfied by $a$ in $M$: $$\mathrm{tp}(a/C) = \{\varphi(x)\in L(C)\mid M\models \varphi(a)\}.$$
We could equivalently define $S_x(C) = \{\mathrm{tp}(b/C)\mid b\in N, M\preceq N\}$. Note here that in general we have to pass to an elementary extension of $M$: it may be that not every complete type over $C$ is the type of an element in $M$, but every complete type is realized in some elementary extension.
Now a theory $T$ is stable if there is some infinite cardinal $\kappa$ such that for all models $M\models T$, all sets $C\subseteq M$ with $|C| = \kappa$, and all variable contexts $x$, we have $|S_x(C)| = \kappa$.
In Shelah's slides, he gives a slightly different definition, depending on the following facts:
1. It suffices to check stability when the context is a single variable $x$ instead of a tuple.
2. It suffices to check stability when $C$ is an (elementary) submodel.
3. If $T$ is stable, then actually we have that for all infinite $\kappa$, $|C| = \kappa$ implies $|S_x(C)| = \kappa^{|L|}$, where $|L|$ is the number of formulas in the language (with no parameters). The fact that there are cardinals $\kappa$ such that $\kappa^{|L|} = \kappa$ gives the converse. Usually we're most interested in the case $|L| = \aleph_0$.
In practice, if you want to prove that a theory is stable by counting types, then you need some way to understand all the possible complete types. This can be hard, because first-order formulas can be very complicated in general. There are essentially two ways of doing this:
1. Quantifier elimination. A theory $T$ has quantifier elimination if for every first-order formula $\varphi(x)$, there is a quantifier-free formula $\psi(x)$ such that $T\models \forall x\, \varphi(x)\leftrightarrow \psi(x)$. Since quantifier-free formulas are usually much easier to understand than general formulas (they are just Boolean combinations of the atomic formulas - basic relations and equalities between terms), quantifier elimination can be very helpful for counting types. Some theories admit weaker forms of quantifier elimination, where you can show that every formula is equivalent to a Boolean combination of formulas in some restricted class that are still easy to understand.
2. Automorphisms. If $a,a'\in M$ and $\sigma\colon M\to M$ is an automorphism with $\sigma(c) = c$ for all $c\in C$ and $\sigma(a) = a'$, then automatically $\mathrm{tp}(a/C) = \mathrm{tp}(a'/C)$. So upper bounds on the number of orbits of automorphism groups of models of $T$ gives upper bounds on the number of complete types.
OK, now let's look at the example of projective planes. The easiest example of a stable theory of projective planes is the complete theory $T$ of $\mathbb{P}^2(k)$, where $k$ is an algebraically closed field. One way to see that $T$ is stable is to show that it is bi-interpretable with the complete theory of $k$. Any theory of algebraically closed fields is stable, and stability is preserved under bi-interpretation.
$T$ does not have quantifier elimination, but if you add a binary function $f$ to the language such that $f(p_1,p_2)$ is the unique line through the points $p_1$ and $p_2$, and $f(l_1,l_2)$ is the unique point at the intersection of $l_1$ and $l_2$, then $T$ has quantifier elimination in the expanded language (this fact rests heavily on the fact that the theory of algebraically closed fields has quantifier elimination).
Let's think about the counting types criterion. Let $M\models T$ (then $M = \mathbb{P}^2(K)$ where $K$ is algebraically closed of the same characteristic as $k$). Suppose $|M| = \kappa$. How big is $S_x(M)$ (where $x$ is a single variable)?
A type over $M$ could be the type of a point or a line. It suffices to count the types of points, by duality. It turns out that if we coordinatize $M = \mathbb{P}^2(K)$, the type of a point over $M$ is completely determined by the algebraic relationships between the coordinates which are definable over $K$. In short, the types are exactly the scheme-theoretic points of $\mathbb{P}^2(K)$.
The points of dimension $0$ are the types of points in $M$ (there are $\kappa$-many of these). The points of dimension $1$ are the generic types of curves in the projective plane defined over $K$ (there are $\kappa$-many of these, since each can be associated with a homogeneous polynomial over $K$). And there is a unique point of dimension $2$, the generic type of the plane. Altogether, there are only $\kappa$-many types, so $T$ is stable (in fact, it is $\omega$-stable, meaning that $|M| = \kappa$ implies $|S_x(M)| = \kappa$ for all infinite cardinals $\kappa$).
When the field is not algebraically closed, we do not get such a nice description of types thanks to the loss of quantifier elimination in the field (though understanding types relative to the theory of $\mathbb{P}^2(k)$ can still be reduced to understanding types relative to the theory of $k$).
The situation is even worse for non-Desarguesian projective planes, which are not coordinatized by fields or even division rings. As I alluded to in the comments, however, Hrushovski constructions can be used to produce non-Desarguesian projective planes with stable theories. These structures are extremely homogeneous, which allows stability to be checked using the automorphism criterion I described above.
• @ Alex Kruckman: great answer ! Just to be sure I understand, I have two small questions. (1) Why do you look at scheme-theoretic points ?; as a point-line geometry, the points of $\mathbb{P}^2(K)$ usually are the $K$-rational points; and (2) Where do the automorphisms come in (cf. your paragraph "2. Automorphisms ...") ? And what is $C$ here ? – THC Jun 5 '18 at 14:30
• @THC Good questions. (1) First of all, we're counting with parameters from a model $M = \mathbb{P}^2(K)$ (so $M = C$ from the preceding paragraphs). The $K$-rational points correspond to complete types which are realized in $M$. Such a type includes the formula $(x = p)$, where $p\in M$. But there are lots of other complete types over $M$ of points which are not in $M$ (types which contain the formulas $(x\neq p)$ for all $p\in M$). For a simple example, consider a line $\ell\in M$, and consider a type over $M$ which contains the formula $(x I \ell)$ but also $(x\neq p)$ for all $p\in M$. – Alex Kruckman Jun 5 '18 at 14:58
• Let's say our type is $\mathrm{tp}(a/M)$ for some point $a$ in an elementary extension $N$ of $M$. So $aI\ell$ in $N$, but $a\notin M$. Then $N = \mathbb{P}^2(F)$, where $F$ is an algebraically closed field containing $K$. If we coordinatize $\ell$ in $\mathbb{P}^2(F)$, it becomes a copy of $\mathbb{P}^1(F)$. Its points in $M$ correspond to elements of $K$ (plus $\infty$), while it's points in $N\setminus M$ correspond to elements of $F\setminus K$ (which are transcendental over $K$, since $K$ is algeabraically closed). – Alex Kruckman Jun 5 '18 at 15:02
• Now it turns out that there is a unique type with parameters from $M$ of a transcendental point $\ell$. This can be argued by quantifier elimination, but it's maybe a bit easier to argue using automorphisms (so here I guess I'm answering your second question as well). If $b$ is another transcendental point on the coordinatlized line $\ell$, there is an automorphism over $F$ which fixes $K$ pointwise and moves $a$ to $b$. This automorphism extends to an automorphism $\sigma$ of $N = \mathbb{P}^2(F)$ which fixes $M = \mathbb{P}^2(K)$ pointwise. Thus $\sigma(\ell) = \ell$ but $\sigma(a) = b$. – Alex Kruckman Jun 5 '18 at 15:05
• It follows that $\mathrm{tp}(a/M) = \mathrm{tp}(b/M)$. Ok, so we have a unique type of a point on $\ell$ which is not $K$-rational. It's natural to call this the "generic type" of the line $\ell$ and identify it with the scheme-theoretic generic point of $\ell$. It takes a bit more work to show that scheme-theoretic points which are generic points of plane curves correspond to complete types over $M$, since lines are baked into our language, while curves are not. But thanks to coordinatization and the interpretation of $+$ and $\times$ in the field, we can express polynomial relationships... – Alex Kruckman Jun 5 '18 at 15:08
Here is one way to make sense of the notion of complete type, outside the context of first-order model theory. It is inspired by the notion of Galois type in abstract elementary classes.
Let's say $M$ is a mathematical object of some kind that you're interested in, and let's assume that it is a set with extra structure of some kind. Let $A\subseteq M$ and $b,c\in M$. Then we can make the following definition: $b$ and $c$ have the same complete type over $A$ if and only if there is an automorphism $\sigma$ of $M$ such that $\sigma(a) = a$ for all $a\in A$ and $\sigma(b) = c$. This captures the idea that $b$ and $c$ relate to $A$ in exactly the same way (inside $M$). According to this definition, the set of complete types over $A$ (in $M$) is the set of orbits of the group of automorphisms of $M$ fixing $A$.
Now in the first-order setting, this definition agrees with the usual definition if $M$ is saturated. If $M$ is just any old model, then 1. it might not contain elements realizing every complete type over $A$, and 2. $M$ might not be homogeneous enough to have automorphisms linking any two elements with the same complete type over $A$.
So in a more general setting, you might also only want to compute "complete types" in analogues of saturated structures, and there's a question of what this means (the usual definition of saturation references complete types, defined as sets of formulas!). A natural choice would be to replace "saturated" by "universal-homogeneous", which is really what you need to make complete types agree with automorphism orbits in the first-order setting.
$M$ is universal universal-homogeneous for a class $K$ if every small structure in $K$ embeds in $M$, and if $f$ and $g$ are two embeddings of the small structure $N$ in $M$, then there is an automorphism $\sigma$ of $M$ such that $\sigma\circ f = g$.
OK, to make sense of that definition, you have to make sense of what "small" means (the usual approach in model theory is to fix some big cardinal $\kappa$, and take small to mean size $<\kappa$). And also you have to decide what counts as an embedding - you might want something stronger than the usual notion of embedding, to preserve properties you care about (e.g. taking elementary embeddings in the first-order context). Thinking about reasonable properties to assume about strong embeddings leads to the definition of abstract elementary classes.
Unfortunately, universal-homogeneous structures only exist if your class $K$ has the amalgamation property for strong embeddings. In the absence of the amalgamation property, here's a fix for the definition of complete type over a structure in $K$:
Let's say $M$ embeds (strongly) in $N_1$ and $N_2$ (and let's identify $M$ with its image in each of these structures), and $a_1\in N_1$ and $a_2\in N_2$. We say that $a_1$ and $a_2$ have the same complete type (Galois type) over $M$ if there are embeddings $f_1\colon N_1\to N_3$ and $f_2\colon N_2\to N_3$ such that $f_1(m) = f_2(m)$ for all $m\in M$, and $f_1(a_1) = f_2(a_2)$.
In other words, a complete type over $M$ is a class of elements in structures extending $M$ which can be made equal in further extensions. In the case when $N_1 = N_2$, this weakens the assumption that $a_1$ and $a_2$ are conjugate by an automorphism.
• @ Alex Kruckman: can you illustrate the definition in your final paragraph ("In other words ...") of complete type over $M$, and Shelah's characterization of stability, on the example of projective planes $\mathbb{P}^2(k)$, with $k$ algebraically closed ? – THC May 30 '18 at 9:54
• @THC: What Alex is describing in his answer is a very general setting to make sense of stability, (Galois) types etc. In order to fit this general setting to your question, you have to decide what notion of strong embedding you want to use. There are many options. Once you have your strong embedding notion, you have to verify that it satisfies the requirements for an Abstract Elementary Class and that is satisfies amalgamation for strong embeddings. Then everything that Alex describes applies. – Ioannis Souldatos May 30 '18 at 11:26
• @THC: To see some examples of AECs and their corresponding strong embeddings see Grossberg's math.cmu.edu/~rami/Rami-NBilgi.pdf and Baldwin's homepages.math.uic.edu/~jbaldwin/pub/AEClec.pdf – Ioannis Souldatos May 30 '18 at 11:28
• @THC Maybe I misinterpreted your question. This answer addresses the question "how can I make sense of the counting types definition of stability outside of a first-order model theory context?" On the other hand, the best notion of stability for a projective space is almost certainly the usual first-order model theory notion of stability. – Alex Kruckman May 30 '18 at 14:39
• @THC Oh, I completely misunderstood where you were coming from, sorry. But now I'm not sure what your background is... e.g. Do you what first-order formulas in the incidence language for projective planes look like? Do you know what a type is? Do you know what quantifier elimination means? – Alex Kruckman May 30 '18 at 15:18
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2019-06-24 14:30:44
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https://mathoverflow.net/questions/279315/does-every-finite-bridgeless-cubic-planar-simple-undirected-graph-admit-a-2-fact
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# Does every finite bridgeless cubic planar simple undirected graph admit a 2-factorization with at most two components each of which has even order?
Consider simple bridgeless cubic planar graphs.
• Does each such graph admit a 2-factorization with $\leq 2$ components each of which has even order?
• If not, does anyone know of an counterexample?
• More generally: each simple bridgeless cubic planar graph is either Hamiltonian or admits a 2-factorization each of which component has even order. But is there an upper bound of the number of these components?
• What is your definition of 2-factorization? My understanding is that a 2-factorization is a partition of the edge set into 2-factors. This does not seem possible in a cubic graph, since once you find a 2-factor and remove the edges you are left with a 1-factor. – fidbc Aug 22 '17 at 23:55
• A 2-factor is a regular subgraph of degree 2 containing all vertices of the original graph. The complement is indeed a 1 factor. – Jimmy Dillies Aug 23 '17 at 1:44
• Dear @JimmyDillies: I made several edits to the OP. If you have reasons to object, please correct or roll back. Most seriously, the paragraph starting with "Stated otherwise" was confusing (to me), at least one the level of English-composition: it left it unclear whether you claimed that what comes next is equivalent to what comes before (which it not really is). – Peter Heinig Aug 23 '17 at 7:33
• If one takes the question strictly literally, the answer is no: the infinite hexagonal grid is cubic bridgeless planar, yet does not admit any 2-factorization with an even number of components, for the boring cardinality reason that each 'factor' is by definition finite, so there must be $\aleph_0$-many components in any factorization, and $\aleph_0$ is not usually considered 'even'. Of course, this is not what the OP intends. I will not make this an answer, rather edit the OP. Yet it shows that any proof must make use of finiteness here. – Peter Heinig Aug 23 '17 at 7:45
If one takes a planar cubic non-Hamiltonian graph, and takes the vertex connect sum of 3 copies, then it cannot have such a 2-factor. It's not hard to check that a vertex connect sum of such graphs is also non-Hamiltonian. If there were a 2-factor with only two components, it would have to traverse two of the 3 edges connecting one of the pairs of subgraphs. But then one of the two sides must be Hamiltonian, a contradiction.
Here's a picture of a vertex sum of planar cubic graphs:
• Technical comments for inexperienced readers: the use of "vertex connect sum" is correct. In view of the usual connected sum, which is similar yet not the same, the term "connect sum" may sound wrong, yet the term is used in this form in Knot Theory. And the modifier 'vertex' in 'vertex connect sum' is to distinguish it from the 'connect sum' which is used in Knot Theory, too, and in which it is an edge which gets deleted. Both operations are discussed in this paper of D. P. Thurston. – Peter Heinig Aug 25 '17 at 7:51
This is not an answer, yet a very relevant comment which to give with due precision is not conveniently possibly via the comment boxes.
Worth pointing out: if the claim
each simple bridgeless cubic planar graph is either Hamiltonian or admits a 2-factorization each of which component has even length.
in the original OP is true1, then the hypothesis 'planar' in the OP is indispensable, because for the (non-planar) Petersen graph, the conclusion is false. More precisely, the Petersen graph is a simple bridgeless cubic graph, which
• is widely-known to be non-Hamiltonian,
• has the property that each of the 2-factorizations (which must exist for general reasons, by Petersen's theorem) consists of the disjoint union of two 5-circuits. ${}\qquad\qquad$ (claim.0)
(Hence, in the OP's terminology, the 'lengths of the components' of the 2-factorization both have odd length.)
Proof of (claim). Each component of a 2-factorization is a graph-theoretic cycle(=:circuit). Any circuit has length at least three. The Petersen graph does not contain any 3-circuit. Hence the only relevant partitions of 10 are $10 = 4+6$ and $10=5+5$.
Moreover, while the Petersen graph does contain both 4-circuits an 6-circuits, it
• does not contain any 2-factorization of the type (4-circuit)$\sqcup$(6-circuit) ${}\qquad\qquad$ (claim.1)
A proof of (claim.1) follows e.g. from what is described in
https://math.stackexchange.com/users/91818/rebecca-j-stones, the number of copy of 6-cycles in petersen graph, URL (version: 2015-05-29): https://math.stackexchange.com/q/1303377
from which I take the illustration
Briefly, a proof of (claim.1) is: fix the 'inner' 5-set in the 'usual' drawing of the Petersen graph; any 6-circuit must intersect the cut defined by the fixe 5-set in an even number of edges; it cannot intersect it in 0 or 4 edges, the latter since all the cut-edges form a matching, so if the 6-circuit did intersect the cut in 4 edges, then the 6-circuit would have to have at least 8 edges, which is impossible. So the 6-circuit must intersect the cut in precisely two edges. The rotational symmetries of the Petersen-graph now suffice (it has more symmetries...) to see that the 6-circuit up to isomorphism must look like one of the 6-circuits in Stones' illustrations, and then the non-existence of the 4-circuit in the putative 2-factorization follows by inspection of Stones' illustrations.
This completes a proof of (claim.1), and hence also of (claim.0).
To reiterate: if the above-cited claim in the OP is true, then the assumption of planarity is essential for its truth.
1 Which I for one do not find evident. It seems not to follow from superficial calculations alone.
• The claim is true; I discuss it in arXiv:1505.04487 and indeed requires planarity. The idea goes as follows. Take a 4-coloring of the faces of your cubic planar graph. Pick any two colors and look at the faces colored by these 2 colors. The border of these faces is is 2 factor where each cycle has even length. – Jimmy Dillies Aug 23 '17 at 19:39
While the accepted answer appears correct, and has nice constructive and even functorial properties, it does not (yet) address the generalized question
But is there an upper bound of the number of these components?
in the OP, and it now occurred to me that one can disprove this easily (albeit non-constructively), using old research literature on the statistics of path- and circuit-lengths in planar graphs. So here is another answer.
Since I prefer using a standard technical term, let me first point out that
'2-factor each of whose components has even order'
(also called a 'weak Hamiltonian' in Definition 2.1 of the article arXiv:1505.04487 linked to by the OP)
has been called
Tait subgraph
by both Dénes Kőnig in (an English translation of) his book Theory of Finite and Infinite Graphs, and by W.T. Tutte in the following excerpt from his book 'Graph Theory', Cambridge University Press, 2001,
so I will use 'Tait subgraph'$=$'2-factor each of whose components has even order' in the following.
Proposition(number of connected components of Tait subgraphs is not absolutely bounded for the class of all polyhedral graphs) . There does not exist any absolute constant $b$ such that every finite vertex-3-connected cubic planar simple undirected graph contains a Tait subgraph with at most $b$ components.
(Note that 'vertex-3-connected' implies 'bridgeless', so this even answers a weakening of the question in the OP, wherein only 'bridgeless' was assumed.)
Proof (by contradiction). If there were an absolute, constant bound $b\in\mathbb{Z}$ such that every $n$-vertex bridgeless cubic planar simple undirected graph contained a Tutte subgraph with at most $b$ components, then evidently (by the crudes counting argument imaginable) this would mean that
(assumption) every $n$-vertex bridgeless cubic planar simple undirected graph contains a circuit of length at least $\frac{1}{b} n$, in other words, has circumference at least that large.
Yet it is amply documented in the research literature that this is impossible.
We define some terminology.
If $\mathbb{K}$ is a class of graphs, a number $\sigma$ is called a shortness exponent for $\mathbb{K}$ if and only if there exists a constant $c>0$ and a sequence $G\colon\omega\rightarrow\mathbb{K}$ such that
• all $G_i$ are pairwise non-isomorphic (which implies, since we are speaking about finite graphs only, that $\lvert G_i\rvert\xrightarrow[]{i\to\infty}\infty$)
• $\mathrm{cir}(G_i)\leq c\cdot \lvert G_i \rvert^\sigma$.
Let $\sigma(\mathbb{K}) = \inf\{\sigma>0\colon\text{$\sigma$is a shortness exponent for$\mathbb{K}$}\}$.
Improving on a well-known result of Grünbaum and Motzkin, it is proved in
that $\sigma(\{\text{all$3$-regular vertex-$3$-connected plane graphs in which each face is bounded by at most$15$edges}\}) \leq \frac{\log 22}{\log 23} < 0.98583$
Fully spelled out, and taking (assumption) into consideration, this implies that there exists an absolute constant $c>0$ and a sequence $(G_i)_{i\in\omega}$ consisting only of $3$-regular vertex-$3$-connected graphs each of whose faces is bounded by at most $15$ edges such that
$\lvert G_i\rvert\xrightarrow[]{i\to\infty}\infty$, (growth)
and
for all $i\in\omega$, $\frac{1}{b}\lvert G_i\rvert \leq \mathrm{cir}(G_i)\leq c \cdot \lvert G_i\rvert^{0.99}$,
where in the latter for the lower bound (assumption) was used.
This implies that
for all $i\in\omega$, $\lvert G_i\rvert^{0.01} \leq b\cdot c$,
and this is impossible, no matter how large the constants $b$ and $c$ may be, because of (growth).
This completes the proof of the proposition.
(Incidentally, Grünbaum and Motzkin proved a statement similar to the one used above for graph-theoretic paths instead of graph-theoretic circuits, which allows for even stronger disproofs of the conjecture implicit in the OP.)
Again, this is a non-constructive argument, using real analysis, and Ian Agol's answer appears to have the nice property of being what one might call a 'constructive disproof'. Yet whether it gives the no-bound-whatsoever statement, too, remains to be checked.
• Ian Agol's gives a constructive proof: – Jimmy Dillies Aug 27 '17 at 4:23
• Let n(A) be the minimal number of components of a graph's Tait subgraphs, and let's denote by A' the vertex sum of three copies of A. Agol's answer shows that n(A')=3n(A)-2. By taking the iterated sequence A, A', (A')', ... one gets a family of planar cubic graphs with a strictly increasing number of components for its minimal Tait subgraph. – Jimmy Dillies Aug 27 '17 at 4:32
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2019-02-19 01:40:13
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https://technikblick.de/convert-image-to-pixel-values.html
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CV_32F is float - the pixel can have any value between 0-1. The user may then move the cursor, via the mouse, to a selected location on either the map or. This example creates an image with a bunch of random binary values. A companion reference to the article, The Incredible Em and Elastic Layouts With CSS. Indeed the value for the spacing of the image could be coming from: Imager Pixel Spacing (0018,1164) Nominal Scanned Pixel Spacing (0018,2010). Firstly I will read the sample image and then do the conversion. Make this an Add-In with a button the user can click on in the Ribbon. 2017-3-23 · Android’s dp units are density independent pixels. Unlike many other tools, we made our tools free, without intrusive ads, and with the simplest possible user interface. It happens anytime you resize or remap (distort) your image from one pixel grid to another. The most used color space is the RGB color space. The meaning of the various hex values in an image file depends entirely on the format of the file. How to purchase more credits? Use the slider (above the price display) in the Pricing section to change the number of Credits or Plan Type. Pillow is the Python imaging library that supports a range of image file formats such as PNG, JPEG, PPM, GIF, TIFF, and BMP. It is a very good idea to make a backup copy of your image before doing any processing. 2014-5-19 · How to convert image pixels values to the temperature values?, in image I found pixel value, like 255, and I know that it match 37,4 C, how to export that value to the screen? 0 Comments. It uses green background and white color for binary bits. #aseprite free #aseprite download #asprite #aesprite #pixel art generator #image to pixel art #convert image to pixel art #jpg to pixel art #convert png to pixel art #piskel #sprite maker My images are of raster pixels so obviously a conversion needs to take place. The original image does not change in any way. You can upload an image in JPEG, PNG, GIF or BMP format. show() Grayscale image conversion (L mode) You can tell the result is much smoother with black, white and gray shades filling each pixel accurately. Then explore the API and objects involved with opening said image and look at how the pixel values are stored within the specific object pertaining to the API. Convert image to Base64 online and use the result string as data URI, img src, CSS background-url, and others. The following steps need to be taken to normalize image pixels: Scaling pixels in the range 0-1 can be done by setting the rescale argument by dividing pixel's max value by pixel's min value: 1/255 = 0. 2013-10-20 · The commands I provided above convert an RGB image to L*A*B* all at once, so every pixel is modified to the new colorspace. The Image class is the heart of PIL, and its properties help manipulate the pixels, format, and contrast of the image. size: tuple [int] ¶ Image size, in pixels. For example the following is a bitmap which has 397 pixels horizontally, 294 pixels vertically, and each pixel contains a grey value from a possible 256 different greys. · My image size is 2464*2056, i. Get started by uploading your image by clicking the Upload image button. Once you click on it locate your image file in …. Remember a color image is nothing more than a matrix composed of numbers. how to convert the pixel values of the image in Learn more about image processing, convert, image analysis Image Processing Toolbox. pixilart, free online drawing editor and social platform for all ages. Free online service to convert a PDF file to a set of optimized JPG images. Hello, I have an image as a QImage object in format ARGB32. CV_8U is unsigned 8bit/pixel - ie a pixel can have values 0-255, this is the normal range for most image and video formats. Now pass X, Y of the center of the bounding box to this function projectPixelToRay((X, Y)). We can remove errant pixels and even, with the help of neural networks, recognize different breeds of dog. One mobile app that recently got a lot of attention was Instagram, which sold to Facebook for the bat shit crazy price of one billion dollars. So why do you want to work with ascii characters? You can load an image with img = loadImage(“yourImage. Use the Bitmap24 class to manipulate image pixels in …. However, I can't get the code to work. 08012633) => (lat,long) bottom-left coordinate: (35. Remember! ARGB will have an integer value in the range 0 to 255. Each gray value is a number proportional to the intensity of the pixel, adjusted by the ITU-R Recommendation BT. 2021-8-21 · Greyscale image Using OpenCV; Conclusion. Plot negative of an image in Python. This dpi calculator enables to find a dot per inch (DPI) that is referred to as the number of printed dots contained within one inch of an image printed by a printer. In this tutorial, we shall learn how to convert an image from color to black and white. Here's a simple code demonstrating the basics: from PIL import Image. For example, if the pixel values are concentrated in the far-left portion of the histogram (this would correspond to a very dark image), we can improve the image by shifting the values toward the center of the available range of. gif -resize [email protected] pixel_terminal. On the standard image, click on “File” and then choose “Properties”. Only one back-step is possible. 2022-2-1 · Show activity on this post. Convert Image to text or hexadecimal in c#. Download button is enabled, after image is converted to HTML file. WHAT IS THE DESIGN FACTORS TO BE CONSIDERED (FORMULA) TO OBTAIN THE SAME. 95275593333 Pixel Pixel↔cm 1 cm = 37. At times, you may need to convert an image to a binary image. 2020-4-8 · The pixel value (s) is stored in a 32-bit memory space holding ARGB values (8 bits each) in the same order. The larger the pixel size, the more image pixels will be merged together to create a pixelation pixel. Auto: Sets a grayscale mix based on the color values of your image. The word steganography is derived from the Greek words steganos (meaning hidden or covered) and graphe (meaning writing). convert (image, mode) Convert image to the given mode. Extend the data storage type defined on this page to support grayscale images. 2020-2-11 · Select a test image to load and work with Pillow (PIL) library. HOW TO CONVERT PIXEL VALUES TO MICROMETER. To convert inches to pixels, you have to multiply inches by resolution. size method returns the width and height (column and row) of the image (pixelmap or matrix). Can I convert RGB pixel values to Letters in an Image?. 2020-10-19 · The best way to convert an image to an array of values is by using either the function IMAQ ImageToArray or IMAQ ColorImageToArray. Go to the image converter on your browser. How to convert a range of pixel values to Learn more about medical image processing, grayscaleimage Image Processing Toolbox I have a DICOM image in which pixel values ranges from -2000 to 2666. 2020-4-10 · A digital image is stored as a 2D array of pixels and a pixel is the smallest element of a digital image. —numbers indicating variations of red, green, and blue at a particular location on a grid of pixels. Regards, Pramod Do you mean to output the pixel values in hex? You could use my EasyBMP C++ BMP library. Convert your photo into pixelart. And that too with 100% accuracy. Unlike the Edit>Invert command, pixels values are not altered, only the way the image is displayed on the screen. I have a 2D PNG map which I know the corners of my image and I also have the image size all as below: image size: 600x600 pixel top-right coordinate: (35. png', ' r') myfile is the name of the image to be read and give the appropriate file format. 2022-1-30 · Camera sensors convert photos striking the sensor in to an electrical charge that accumulates in each pixel. Select Pixel Size If you'd like to use the generated image for commercial purposes, please purchase some image credits: # STEP. 1 C, and 255 pixels = 37,4 C, I need to get intermediate proportional values, and I need, that matlab it automatically calculate, and derived each pixel temperature value. By default, when viewing an image service in the ArcGIS Online or Portal for ArcGIS map viewer, the pop-up displays the RGB values of the raster's classified symbology instead of the pixel values. This tool converts your image to JPEG - Joint Photographic Experts Group JFIF format (62) About 120 input formats are supported, including: BMP to JPEG, BRAILLE to JPEG, CIN to JPEG, CIP to JPEG, CLIP to JPEG, CMYK to JPEG, DCM to JPEG, DNG to JPEG, EPT to JPEG, FAX to JPEG, FITS to JPEG, FTS to JPEG, GIF to JPEG, ICON to. 2020-5-5 · With inverted LUTs, pixels with a value of zero are white and pixels with a value 255 are black. Note that "P6" PPM files can only be used for single byte colours. The following examples are two of the more common approaches used to convert images to grayscale or sepia. If you write your own grey scale converter, you can operate on the original ARGB bits without creating a new QImage. It allows you to set the hex number color, change the background color, set the image width and height, apply bold and italics. By a variety of resizing and scaling options, this tool enables you to easily scale up, scale down, proportionately distort and transform the photo, picture or image in size. 2022-3-15 · convert any image to image with symbols as pixel values - GitHub - vikash423q/ascii_image: convert any image to image with symbols as pixel values. If this white level value is much higher than the highest intensity value in the image, then using this white level value for scaling results in an image that is …. We delete uploaded jpg files instantly and converted ico files after 24 hours. Top SEO sites provided "Convert image to pixel art" keyword. Each Color of Pixel is Modular Divided(%) by 256 since the RGB Color Model holds the highest Color Range from 0 t0 255 Hexadecimal Values. The basic idea is to find a point between the peak of the foreground pixel values and the peak of the background pixel values. 2019-7-5 · Images are comprised of matrices of pixel values. It’s super easy to create a new Paint by Number. Get the answers of below questions: What is pixel? What is resolution? How to read image pixel? How to get image pixel? How to print image pixel? How to change the pixel value of an image in python […]. Upload your files to convert and optionally apply effects. p stands for pixels and Mp stands for megapixelss. Click any of the images to see them reconstructed. Millimeter value is from 1 to 3520, such as 53, an integer number. 2021-8-14 · In Python OpenCV Tutorial, Explained How to Change Pixels Color in an image using NumPy Slicing and indexing. A stop motion effect is a special technique with tiny pauses after each frame. turn it into a csv file with the columns being x and y coordinates of the pixels and a numerical value for their color. You can also use the DPI to change the image size when it comes to printing. How to Convert images to NumPy array?. setColorTable(), Sets the color table used to translate color indexes. Apply is altogether more dangerous. Just drop your image in tool and click convert to html button to write image as html file. 2021-4-28 · Note: Here the half of the image has been converted to grayscale, but the full can be done using the same code just by changing (width//2) to (width). You can also use LA mode with transparency to achieve the same result with the liberty of alpha. The first bit is at the rightmost side and is numbered or indexed 0 and the last bit is …. 2010-7-8 · Convert pixels to inches (Output to Monitors/Printers)Formula: Pixels ÷ DPI = Inches. 2022-2-11 · Font size: Smallest Small Medium Large Largest. PPI or pixels per inch is a measurement used to tell the resolution or quality of image. However, if we want to get the pixel values, we have to handle a single channel separately. You will also learn: What pixels are; How the image . The origin of the coordinate system is at the lower left corner of the image. that is if its a jpeg image then give it as myfile. How to convert byte array of image pixels data to grayscale using vector SSE operation It requires System. 0254 pixel/inch [ppi] From: To: Kinematic Viscosity. Upload your image and select between various filters to alter your image and apply digital effects. Video Tutorial: Convert rem to px. How to convert a color image into grayscale image in C#. Convert your image to $\mathrm{L}^{*} \mathrm{a}^{*} \mathrm{~b}^{*}$ space, or keep the original RGB colors, and augment them with the pixel $(x, y)$ locations. 2017-1-18 · Convert RGB pixel value of an image to limiting magnitude. Another possibility is to decompose to Hue/Saturation/Value components and consider the Value image (the other two are not usually useful for this purpose). Easy conversion of color images to black/white with treshold level trackbar slider or Dithering. ICO Convert( formerly ConvertICO. This free online service allows to convert your images to separate PDF files or to merge them together in one PDF file. 2013-2-28 · Hi i have to convert the image pixel values in the range of (0-255) into (0-1) Is there any function or program available? 0 Comments. 2015-10-1 · @Kazem Rangzan Ya sir I need to know the steps either in ENVI 4. grayscale: Convert an RGB image to grayscale in imager. DPI relates to the resolution of your screen. We need to know that to compute the proper image size in pixels to resize the image properly according to the print size and whether the 200 is dpi or dpc. Import Image module from PILLOW library of Python as PIL. I'm working on the esp32 cam this is the setting of cam. 2008-8-8 · image is created using paint brush (like any letter A) convert it to the binary matrix If you use a BufferedImage object to store the image in java then there are methods associated with that class to convert the image to an array of RGB values. Accept Solution Reject Solution. 2019-3-17 · I want to convert pixel values that store in my outout text file into image(png or jpg). 2014-12-11 · When you convert RGB images to CMYK, you lose those out-of-gamut colors, and they won’t come return if you convert back to RGB. Digital images are made of pixels. 2022-1-30 · How to convert RGB to grayscale. Fast, free, and without intrusive ads. The module also provides a number of factory functions, including functions to load images from files, and to create new images. Convert an image to JPG, PNG, WEBP, TIFF or GIF format. A lens of 4 diopters has a optical power of. A simple browser-based utility that converts ASCII to an image. You can use it to create fonts, menus, intros etc. For example here we resize both our images to a rough 64x64 size, or 4096 pixels in size. 2022-3-25 · Output Pixel Vaule of cell 5 = 4*Input Pixel value of cell 5 - (Sum of pixel values of cells 2,4,6,8) How to perform this operation for all the pixels in the tiff image?. The equivalent statement to "get (x, y)" using pixels [] is "pixels [y*width+x]". Each of these pixels is denoted as the numerical value and these numbers are called Pixel Values. So, to convert the color pixel into grayscale pixel we have to first find the average of R, G and B. Both the RGB and L*A*B* images should range from 0-255. You can convert grayscale image datasets to RGB. Import a PNG – get a grayscale PNG. To get luminance of a color use the formula recommended …. The first thing you need to do is to run test_video_capture. Images are comprised of matrices of pixel values. Either upload the image in the working directory or give your desired path. bpp= ( int )extract (fp1, 24, 2 ); How to read each pixel depends on the number of bits per pixel. 3 Indicate the start and end input ranges in the Range of input values group. org ) is a free online portable icon conversion software and favicon generator, it allows to make a single or multiple-size icon from png, gif, jpg and bmp format, or extract all the frames inside an icon to separate images. 2 and has a linear section for small intensities). 2018-1-1 · if you want to know the pixel size and its value in meters then just convert the image into UTM projection and the use image info tool in any …. EndInit() '///// Convert the BitmapSource to a new format ///// ' Use the BitmapImage created above as the source for a new BitmapSource object ' which is set to a two color pixel format …. If the input raster is a floating-point raster, you must use the Map Algebra Expression parameter to convert it to an integer raster. If we are dealing with a grayscale image, we are using numeric values from 0 (black pixels) up to 255 (white pixels). Digital image resolution unit conversion between dot/inch and pixel/millimeter, pixel/millimeter to dot/inch conversion in batch, dpi pixel/mm conversion chart. remove pixels from the image interior-clip: clip along the first path from the 8BIM profile-clahe geometry: contrast limited adaptive histogram equalization-clamp: set each pixel whose value is below zero to zero and any the pixel whose value is above the quantum range to the quantum range (e. Once we have byte array of Image file, apply below method to convert byte array into Base64 string. PIL module provides ImageOps class, which provides various methods that can help us to modify the image. Points are traditionally used in print media (anything that is to be printed on paper, etc. open() does not have any output, it open the image pointed out by the address "file" in the python background. How to convert RGB to grayscale. The basic form of this function is −. -implode amount, implode image pixels about the center ; -insert index, insert last image into the image sequence ; -integral, Calculate the sum of values (pixel . The value of Photometric Interpretation (0028,0004) specifies the intended interpretation of the image pixel data. How do you choose a good, or even best, pixel scaling method for your image classification or computer vision …. 2022-2-11 · To convert your images to JPG, you can use Aiseesoft Free Image Converter to do the job. I have a text file which is a matrix 48*48 pixels (pixel's values are between 0-255). The program allows the user to select a series of images from a file folder. describe the format and attributes of an image. Now our image has been uploaded, we need to convert it into gray colors. By using the text photo maker, the text will show up crisply and with a high resolution in the output image. 2022-1-15 · Pixel value is from 1 to 9998, such as 200, an integer number. 255 Convert pixel to grayscale: set red, green, and blue to be the average Standard code line to compute average within loop. List of Functions for the 'image' package. It works entirely in your browser and uses a fast base64 encoding algorithm to do the encoding. 2017-12-7 · An image histogram can help us to quickly identify processing operations that are appropriate for a particular image. 2018-1-4 · Easily convert your text to grayscale images and vice versa. The problem is that convert simply overlaps the images using the first image's size as the size of the animated gif, and since they have different sizes the …. alpha_composite if you want to combine images with respect to their alpha channels. Can I convert RGB pixel values to Letters in an Learn more about segmentation, landcover classification, image processing MATLAB I have classified an image using imsegkmeans with 5 classes and displayed it in RGB. This code converts a block of text to an image so that each character in the block defines one pixel in the image and each line in the block (delimited by \n's) builds one whole row of pixels in the image. In a colorful image, each pixel holds the information of Red, Green and Blue intensity at that pixel and the number of channels. 2021-2-26 · from PIL import Image image = Image. I want to convert pixel values that store in my outout text file into image(png or jpg). The module also provides a number of factory …. Convert Images to Base64 cross-browser testing tools. 2022-3-29 · The current size, both in pixels (px) and in megabytes (M), is found at the top. Automatically convert any px and em number to its equal em and pixle px value used by web designers in css files ( cascading style sheets ) or in a web page html source code files. array() method and pass the image data to the np. open ('image path') print (image. 2008-5-5 · The command mat2gray is useful if you have a matrix representing an image but the values representing the gray scale range between, let's say, 0 and 1000. If i am not wrong, there are some java commands like readImage and readPixel and raster that are related to digital image. In this theory part of the Image Processing Project we will learn about pixels. For a grayscale digital image, this process is straightforward. It tessellates the image file pixel-by-pixel until generating a complete topographic model of the image brightness. Convert Binary to an Image. You need to defi ne and implement a function to do the job. My preference is to scale the width to 100 px. Free tool to convert your color photo to black and white image. 26 illustrates some ratio images prepared from the visible and reflected TM bands of the Thermopolis subscene (Figure 4. To get the pixel value at one particular (row, column) location, you can just specify the index: grayLevel = grayImage (row, column); or you can use impixel (): rgbColor = impixel (rgbImage, column, row); Ben11 on 21 Aug 2014. Change pixel values of the canvas. You need to loop through each pixel in the image. 1dp might render as 1 screen pixel, or 2, or 3, or 4, or some other value, depending on the device and Android settings. 2010-2-15 · The pixels are either 8 bit or 10 bit depending on the image type. 2021-12-3 · Explanation: By using rgb2gray() function, the 3-channel RGB image of shape (400, 600, 3) is converted to a single-channel monochromatic image of shape (400, 300). World's simplest online Portable Network Graphics (PNG) color changer. How to convert diopters to focal length? A lens of focal length f has the “optical power” =. A black pixel will have a height of 0mm and not be included in the final 3D generated …. And as 8 bits equals 1 byte, each RGB pixel therefore equals 3 bytes in file size. So, we divide the image_array by 255 for normalization. However, since the perceived brightness is often dominated by the green component, a. png -fill "rgb(ff,00,00)" -draw "color 0,0 point" white-background. resize ( (70,100)) This imports the image into Python and note that I’ve re-sized the image to a 70×100 pixel image for further calculations. You want to encode the path, or the actual image ? Encoding the path is easy enough, just write an algorithm or use an existing one ( but not a hash, they are not reversable ). above max z: Your initial bit position above max z. Instant free online tool for pixel (X) to meter conversion or vice versa. 100 dpi is recommended for a poster. 2021-12-13 · Is there a way I can convert image (pixel) Learn more about convertion, graph, raman, spectroscopy 登录 到您的 MathWorks 帐户 登录到您的 MathWorks 帐户 Access your MathWorks Account 我的帐户 我的社区资料 关联许可证 登出 产品 解决方案. A pixel (short for picture element) is a small block that represents the amount of gray intensity to be displayed for that particular portion of the image. png"); how to retrieve the byte array value of this png image. Once process done, tool will preview your black and white image along with download button. By operating ndarray, you can get and set (change) pixel values, If you convert the image to grayscale with convert('L') and then pass . On the menu bar, click on File, and then Place. Converting Tensor to Image Let us define a function tensor_to_image to convert the input tensor to an image format. A simple browser-based utility for pixelating Joint Photographic Experts Group pictures. With these information the tool can convert everything to box-shadow. For every pixel $(L, a, b, x, y)$, compute the weighted mean of its neighbors using either a unit ball (Epanechnikov kernel) or finite-radius Gaussian, or some other kernel of your. Convert Numbers to a Picture. 2021-5-15 · OpenCV provides cvtColor function that allows to convert an image from one color space to another. "P6" image files are obviously smaller than "P3" and much faster to read. Here’s a simple code demonstrating the basics: from PIL import Image. Imagine an image with only zero brightness pixels and a single pixel different from 0, then only two peaks would be seen in the histogram (when bin width is 1), one peak at zero and one corresponding to the …. scoreatpercentile (read the docstring!) to saturate 5% of the darkest pixels and 5% of the lightest. We then apply the colormap to the image_array and multiply it by 255 again. Video Games Consoles and Accessories. I have been used image j program to calculate pixel intensity values in different DICOM CT images. $\begingroup$ This is exactly what float_buffer parameter does to an image. Given a PIL-image img, you can convert it to the numpy array:. Import array module from NUMPY library of Python. This way, you can convert images to text in Microsoft Word and can use it for further purposes. Image interpolation occurs in all digital photos at some stage — whether this be in bayer demosaicing or in photo enlargement. Pick a color and create Monochrome / Monotone images …. Just import your PNG image in the editor on the left and you will instantly get a grayscale PNG on the right. Convert your images (jpeg, jpg or png) into scalable and clear vector art (svg) A Raster graphics image is a rectangular grid of pixels, in which each pixel (or point) has an associated color value. 2020-5-21 · A histogram illustrates how pixels in an image are distributed by graphing the number of pixels at each color intensity level. Namely, you can add hex prefix "0x" to all values, and you can put a "0" in front of hex values that use a single digit. Upload your image and pick a color to colorize it with. A printer produces dots and a monitor produces pixels, so this name attempts to provide context when an image is digital. 2021-6-16 · The superclass of Image, ImageBase, defines the geometry of the image in terms of where the image sits in physical space, how the image is oriented in physical space, the size of a pixel, and the extent of the image itself. cimg: Convert to cimg object as. Modules Needed: NumPy: By default in higher versions of Python like 3. Montage: juxtapose image thumbnails on an image canvas. For that, we need to create a pixel map by creating another variable ' . The second step converts the radiance data to ToA reflectance. array: Turn an numeric array into a cimg object as. Each pixel of the image is stored an integer number. Also, explore tools to convert centimeter or pixel (X) to other typography units or learn more about typography conversions. Images are converted into Numpy Array in Height, Width, Channel format. This will change the image (all layers), and confine the color model to use to the 1bit palette - modify it back to RGB color model (Image->Mode->RGB) if you plan to continue editing the image. 39999999813 mm mm↔ft 1 ft = 304. 2017-4-14 · Convert pixel values stored in text file to image. argv[1]) # convert to RGB image = cv2. My platform, Godot, accepts only TrueType and FNT file formats. When we talk about RGB in an image, we talk about Red, Green, and Blue intensity values at each and every pixel inside the image. This tool allows users to convert texts and symbols into an image easily. Using OpenCV, it is easy to read the value of pixels. However, most images either have a narrower range of values (because of poor contrast), or have most pixel values concentrated in a subrange of the accessible values. We can see that the pixel values are converted from unsigned integers to 32-bit floating point values, and in this case, converted to the array format [height, width, channels]. Image Converter - Convert photos and images to JPG, PNG, PDF, WebP, BMP, and more online. How to get pixel values of an image and how to set pixel value of an image in Java programming language. You can also customize the hex numbers how they appear in the output image. Created by computer nerds from team Browserling. stack_o October 20, 2021, 8:17pm #1. we use a function of Image module called getdata() to extract the pixel values. size can be used to get the size of the original image. If your input image is a truecolor image of data type single or double with pixel values outside this range, then you can use the rescale function to scale …. So the integer values of pixel should convert to 7 or 8 binary bits. Pixel values are often unsigned integers in the range between 0 and 255. Any number in between these two is a shade of gray. 2022-2-26 · Image quality option only works on a few image formats such as JPG, WEBP or AVIF etc. Answers (3) Azzi Abdelmalek on 28 Feb 2013. 2021-3-10 · Digital images are made of pixels. When it has set each pixel's value, the program sets the picture box's Image property to the bitmap. We currently support the following image to text conversions: JPG to text, PNG to text, TIFF to text, SVG to text, BMP to text, WEBP to text, and many more! Rate this tool 4. This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. On the other hand, DPI or dots per inch is just a measurement in output device (ex: printer). 2022-3-27 · So the pixel values would be about 113. Built-in Functions¶ These functions are applied to each individual pixel. 2021-11-23 · To convert images select Raw Data Converter from the File menu:. A histogram of the input values is generated, and a lower limit value for the useful portion of the input. The surface reflectance image can be …. This is a string specifying the pixel format used by the image. 2022-3-29 · In this session, we are going to learn how we can convert a given image into its negative form. A byte is a unit of measurement of the amount of computer data. 2022-3-25 · An intuitive way to convert a color image 3D array to a grayscale 2D array is, for each pixel, take the average of the red, green, and blue pixel values to get the grayscale value. Just import your PNG image in the editor on the left and you will instantly get a base64-encoded string on the right. At first glance it wouldn't seems to be a big gain when using generic operations. 2011-9-25 · You can then use the scale bar tool in ImageJ to convert pixels to micrometers. We do that as follows: Make the pixel values from [0 , 1] to [0, 255]. read to convert it to a BufferedImage. It will return the array consists of pixel values. The size is given as a 2-tuple (width, height). Before you convert bitmaps to vector images The bitmap images that you are vectorizing will very likely come from one of two sources: a file that is prepared in a drawing or bitmap-editing application and exported to one of the many available bitmap formats, or a file acquired via an image-capturing device such as a scanner or digital camera. HOW TO CONVERT PIXEL VALUES TO MICROMETER (FORMULA) HOW TO MAGNIFY AND CAPTURE A IMAGE TO MEASURE ITS VALUES IN MICROMETERS. 2022-1-31 · I realise that this question has been asked before as Extract pixel values by points and convert to a table in Google Earth Engine at Stack Overflow. It will take any image and convert it to ICO file, for web site favicon or Windows applications. Image file: Advanced options: Image width: (1-400) characters: Text colour: Background: Invert image: Extra contrast: For help on using the converter, see the help page. Converting to grayscale removes the hue and saturation information from the pixels and leaves just the brightness values. Instant free online tool for pixel (X) to centimeter conversion or vice versa. The best way to convert an image to an array of values is by using either the function IMAQ ImageToArray or IMAQ ColorImageToArray. Hence the term '24-bit colour'. We are listing the pixels in the image. The result is 'pixel perfect'/lossless. Here, Few approaches are discussed. Hello Developer, Hope you guys are doing great. 2020-7-2 · Use for the gt() value the same value as you use in the threshold argument for the CannyEdgeDetector: var canny = ee. import cv2 import numpy as np import matplotlib. 2019-10-11 · The task is to convert the string value containing ‘px’ to the Integer Value with the help of JavaScript. The input dynamic range is determined by acquisition circuitry (20), while the output range is determined by display or output devices or circuitry (36). This question shows research effort; it is useful and clear. 2016-2-19 · Pixel Ruler 3 is a no-frills application which allows users to quickly measure images/elements. If you know the pixel width and height of an image, this section will calculate the physical size (in inches) of the image when it is printed or displayed on various devices. Inkscape is powered by libdepixelize with the ability to automatically vectorize these "special" Pixel Art images. ImageBase provides the methods to convert between the index and physical space coordinate frames. And all utilities work exactly the same way — load image, get result. red/green/blue all set to 75 Now the pixel is gray, red/green/blue all equal. It shouldn’t take more than a second to convert an image to text. So with a default of 1280 by 720 pixels, which corresponds to 96 dpi (dots or pixels per inch) , you have a lower image resolution than HD …. When you set out to print an image on paper, GIMP needs to know how big each pixels is. 2022-1-12 · To convert an image to bitmap mode, you must first convert it to grayscale mode, simplifying the color information in the image and reducing its file size. So, if you need to convert 100 images, you should purchase 100 credits. After we convert the image to indexed color, to limit the number of colors, and we choose Web Colors, since the Arduino code wants HTML color codes. Supports jpg, png, webp, bmp and svg image formats. · Note, If you image is not a gray-scale one, each pixel will likely be presented as a triple (r, g, b), where r, g, b are integer values from 0 to 255 (or floats in [0,1]); So, you would be needed to store that triple for every pixel. Understanding Digital Image Interpolation. select('B5'), threshold: threshold }); // mask everything but the edges var cannyMasked = canny. Actually, it’s an image sequence that creates an action. Convert all the elements into strings using map and str; Join all the elements using spaces. Qt provides four classes for handling image data: QImage, QPixmap, QBitmap and QPicture. How to Vectorize an Image in Photoshop. An image with indexed pixel values or indexed color, will have each of its pixel value referring to an index of an array which is known as color palette. 2018-5-29 · The parts that needed changing already have the values I substituted in so I can process the image in question. Alpha take the leftmost 8 bits while Blue takes the rightmost 8 bits of the pixel. unfortunately Image j program (8 bit. opencv Tutorial => Setting and getting pixel values of a. Your resulting intensity is also a byte with range [0. The "To Black and White Picture Converter" converts colored pictures to black and white, or to grayscale respectively. Image histogram can be used to automatically determine the value of the threshold to be used for converting a grayscale image to a binary image. One aspect of preparing image data is scaling pixel values, such as normalizing the values to the range 0-1, centering, standardization, and more. 2022-3-29 · Extract text from images, photos, and other pictures. I would like to convert those 5 colors into. With this app you can convert single pages, multiple pages or all pages of a PDF document into images. To get started, we need to import cv2 module, which will make available the functionalities required to read an original image and to convert it to grayscale. - GitHub - janic0/r-place-image-generator: This python script allows you to convert your images into pixel sequences that you can use for the r/place subreddit. 2022-1-17 · For example, to convert PNG to JPG, you would save the image to PDF, and then to JPG. If the pixel type is demoted (lowered), the raster values outside the valid range for that pixel depth will be truncated and lost. 2017-4-12 · Btw are there actually values in your alpha channel to begin with? If not you will have to manually set all other pixels to 255 or start with a filled/initialized matrix. For instance, if you have an image of 20 x 20 dimensions, it would be represented by a matrix of 20x20 (a total of 400-pixel values). The main task of the byte is to store a character (it can be a number, a letter, etc. Also, it is much better than visiting multiple websites and being on the Internet to search for the best image converter. convert the image into an array object (for this case, a NumPy array) (optional, but recommended) resize the image so that it won't be too heavy in excel. In ArcMap, the Raster To Polygon tool cannot be used to convert each cell of a raster into a polygon if the raster does not have unique values. Convert an image to grayscale using custom weights. C # generate images with Byte array; Convert byte to int and int to byte in C#; C# convert byte[] into corresponding struct. 2022-3-31 · We will also learn how we can convert RGB to HSV. 96 pixels To convert inches to px, just follow this rule: 1 inch = PPI. 2022-3-29 · Online image converter. Consider a color pixel with the following values A = 255 R = 100 G = 150 B = 200 Where A, R, G and B represents the Alpha, Red, Green and Blue value of the pixel. first i need to store the pixels in 2d array. 2022-3-27 · With an RGB colour image each pixel is made up of three numbers to represent the Red, Green and Blue values. The mode includes 1-bit and 8-bits pixel black and white images, RGB images, HSV images, BGR images and, LAB images. You can use Pixel It to be your jump start to make some pixel art. 2021-4-27 · Convert your raster images to vector online for free! Vectorize JPG, PNG, BMP, TIF to SVG or DXF instantly. | Hello, if you are a fan of pixel art then you are in the right place, I will convert your photo into pixel art. To do that, first convert the multi-band Landsat image into an “Array Image”, where each pixel is an Array of band values. You can then construct a valid PIL image object using your extracted RGB values. Wait for a second, and the image will be converted to …. The following values are defined. you need to set a radius value of 3 pixels for the larger image and 1 pixel for the smaller image. 2016-1-26 · $\begingroup$ The reason that not all possible brightness values are existing is because the corresponding values (brightness) are not contained in the image. Traditionally, the images would have to be scaled prior to the development of the model and stored in memory or on disk in the scaled format. 5×, for hdpi displays (~240PPI). COLOR_BGR2RGB) # reshape the image to a 2D array of …. You can use the optional settings to apply even more options to improve your image while converting it like applying filters or changing the size. Then, choose the image format you want to turn your picture into. Optical character recognition or optical character reader (OCR) is the electronic or mechanical conversion of images of typed, handwritten or printed text into machine-encoded text, whether from a scanned document, a photo of a document, a scene-photo (for example the text on signs and billboards in a landscape photo) or from subtitle text superimposed on an image (for …. 2020-8-28 · Image data must be prepared before it can be used as the basis for modeling in image classification tasks. If it is, try to use another pixel that is near to that 0 pixels. we want to identify the geo location coordinate of image coordinate(in pixel) that user touched on the image. This online free image to pixel art converter allows you to quickly change any image to pixel art with three easy steps, ensuring the best quality, safety. If you want to create a favicon. Leave the values at default to include all pixels of the image. Scrolling is contained by the Surface clip area. Format :695 columns,316 rows,219620 pixel. Optimize the quality of an image. Convert JPG to Word Online for Free. Here’s my initial Java BufferedImage example code. 2013-9-5 · Like other programming languages (e. 2022-3-29 · The maximum size for the ICO format is 256 pixel. Leave the values at default to. These HU values have to be converted to pixel intensity f (x,y) using f (x,y) = ( (HU-P1)*pow (2,i-1))/W, where P1 is the window left border, i is the radiometric resolution and W is the window. I have taken photos at different exposures from different places of the Orion constellation. reshape((-1, 3)) pixel_values = np. The raster will be converted to a polygon feature class using the ArcGIS Raster to Polygon tool. image Image Processing Toolbox. A byte array is just an array of bytes. The image shown below represent a single pixel value consisting of 32 bits. ico, you only have to set the size to 16x16 pixel with this tool. Did you know that ketchup, paints, and a mixture of corn starch and water have something in common? If not, click or tap to find out! Resolution and Pixel Dimension. I am sure you can find code for grey scale conversion online or in the QImage source code. InputStream is = new ByteArrayInputStream(bytes); BufferedImage bi = ImageIO. Pixels to Megapixelss Conversion. Quick online tool to convert your image to HTML file. How to get pixels (RGB values) of an image using Java. Here we are loading an image in the matrix named 'cimage', and then it converts the image using 'cvtColor (cimage, img, COLOR_BGR2GRAY); ' and store it in the matrix. If necessary, you can adjust certain parameters such margins, page orientation, angle of rotation in order to create your ideal PDF document. In this case, we can select the crop button and un-crop the bottom two lines to ensure all the data is included when we convert it into the spreadsheet form. 1 day ago · Convert dot/meter [dot/m] to pixel/inch [ppi] 1 dot/meter [dot/m] = 0. 2012-12-5 · In my algorithm result, I am getting final circle diametr value in pixel I want to convert that into mm. 2011-3-8 · But setting the density AND unit can be a bit more problematic. You can resize the image to fit your requirement, if you only input "width" or "height", the software will calculate other side automatically, if you input …. 2014-8-27 · The code then loops through the image’s pixels using the object’s ImageBytes array to access the bytes in the image and inverts them. 2022-3-28 · CV_8U is unsigned 8bit/pixel - ie a pixel can have values 0-255, this is the normal range for most image and video formats. Each pixel is in 16 bits (and i have another file with integer uint16). 2015-2-12 · I have a DICOM image in which pixel values ranges from -2000 to 2666. gt(threshold) // edges will be 1, everything else 0. The top-left pixel of your image will be the background color of this element and all the other pixels will be added as the box-shadow. Re: Convert pixels [] value into x and y coordinates. read(is); The idea is puts the byte[] into an ByteArrayInputStream object, and we can use ImageIO. For only \$5, Susan_costa will make convert your image to pixel art. 2022-3-28 · And in a sense it does because the more pixels you have to spread out, the higher the pixel density will be. 2020-12-11 · When an image with RGB color space is tried to be converted to grayscale, the pixel values are calculated as the weighted sum of the red, green and blue pixels. 2020-6-9 · This commit does not belong to any branch on this repository, and may belong to a fork outside of the repository. DecodePixelWidth = 200 myBitmapImage. With inverted LUTs, pixels with a value of zero are white and pixels with a value 255 are black. It is also being used in web, digital and print media. This python script allows you to convert your images into pixel sequences that you can use for the r/place subreddit. This free OCR converter allows you to grab text from images and convert it to a plain text TXT file. DPI is used to measure the resolution of an image both on screen and in print. 2022-3-30 · The first step is to convert the DNs to radiance values using the bias and gain values specific to the individual scene you are working with. 2021-10-25 · The Scale Pixel Value option appears when a different pixel type is chosen, which can be used to scale your pixel type from one bit depth to another. World's simplest online image base64 encoder for web developers and programmers. Converting in Python is pretty straightforward, and the key part is using the "base64" module which provides standard data encoding an decoding. After making those changes, checking in the terminal with identify -list resource returns the new values. 2013-10-20 · Re: Convert pixels [] value into x and y coordinates. CannyEdgeDetector({ image: image. 2022-1-31 · Detailed Description. How to: Convert a BitmapSource to an Indexed Pixel …. How to convert an image into its negative image in Python. 2015-3-2 · Figure 3: Converting an image URL to OpenCV format with Python.
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2022-05-23 08:41:56
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http://docs.oracle.com/cd/E19620-01/805-4055/reference-34/index.html
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Common Desktop Environment: Help System Author's and Programmer's Guide
### Syntax
<emph>text<\emph>
Or:
!!text!!
The shorthand form for the <emph> element is a set of double exclamation marks (!!) before and after the text.
If you use the <emph> start tag, the <\emph> end tag is required.
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2013-12-06 15:28:36
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https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-12-sequences-and-series-12-1-geometric-sequences-12-1-exercises-page-612/8
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# Chapter 12 - Sequences and Series - 12.1 Geometric Sequences - 12.1 Exercises - Page 612: 8
$${a_n} = 8{\left( 4 \right)^{n - 1}}{\text{ and }}{a_5} = 2048$$
#### Work Step by Step
\eqalign{ & {a_1} = 8,\,\,\,\,r = 4 \cr & {\text{The general term of a geometric sequence is }}{a_n} = a{r^{n - 1}}.{\text{ with first term }}a{\text{ and }} \cr & {\text{common ratio }}r. \cr & {\text{substituting }}{a_1} = a = 8,\,\,\,\,r = 4{\text{ into }}{a_n} = a{r^{n - 1}}{\text{ to obtain}} \cr & {a_n} = {a_1}{r^{n - 1}} = 8{\left( 4 \right)^{n - 1}} \cr & \cr & {\text{find }}{a_5},{\text{ substitute }}n = 5{\text{ into the general term formula}} \cr & {a_5} = 8{\left( 4 \right)^{5 - 1}} \cr & {a_5} = 8{\left( 4 \right)^4} \cr & {a_5} = 2048 \cr & {\text{then}} \cr & {a_n} = 8{\left( 4 \right)^{n - 1}}{\text{ and }}{a_5} = 2048 \cr}
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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2021-05-16 05:40:47
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https://brilliant.org/problems/modulus-inequality/
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# Modulus inequality
Algebra Level 3
$\large \dfrac{|x - 2| - 2}{|x - 1| - 1} \leq 0$
The values of $x$ satisfying the inequality above range from $(a,b]$ where $a < b$.
Find $a+b$.
×
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2021-06-19 04:23:34
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https://socratic.org/questions/what-is-the-highest-height-a-hill-could-be-if-the-roller-coaster-is-to-go-over-t
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# What is the highest height a hill could be if the roller coaster is to go over the top of it?
## I am studying for my physics final and need help with one of the questions from our midterm: A roller coaster, which has a mass of 5,000 kg (people riding + train), is on a near frictionless track. It is 10.54 m above the ground and has a velocity (at this point) of 4.26 m/s. You are coming up on a hill that will go up and then brings you all the way down to ground level. What is the highest height the hill you are coming up on could be, if the coaster is to go over the top? I tried using the relationship of KE=PE, because I know energy is conserved, but that didn't work. What do I do?
Jul 30, 2018
That has to work if energy is conserved.
$\frac{1}{2} m {v}^{2} = m g \Delta h$, where $\Delta h$ is the height relative to train's current height
$\therefore \Delta h = \frac{1}{2} {v}^{2} / g$
The actual height $H$ is therefore:
• $H = 10.54 + \frac{1}{2} \cdot {4.26}^{2} / 9.81 \approx 11.5 \text{ m}$
And $\Delta h \approx 0.92 \text{ m}$
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2019-11-22 22:28:26
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http://sciforums.com/threads/the-us-national-debt.105754/page-2
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# The US national debt
Discussion in 'Business & Economics' started by Mrs.Lucysnow, Jan 3, 2011.
1. ### Michael歌舞伎Valued Senior Member
Messages:
20,285
I wish I understood the process better, its at times counter-intuitive as well we use terms without really knowing exactly what they are, take money for example.
3. ### Michael歌舞伎Valued Senior Member
Messages:
20,285
Ｉ｀ｍ ｎｏｔ ｓｕｒｅ ｗｈａｔ ｙｏｕ ｍｅａｎ ｂｙ ｅｃｏｎｏｍｉｃ Ｆｅｕｄａｌｉｓｍ？ Ｌｉｆｅ ｈｅｒｅ ｉｎ Ｊａｐａｎ ｉｓ ａｃｔｕａｌｌｙ ｐｒｅｔｔｙ ｄｅｃｅｎｔ ｃｏｍｐａｒｅｄ ｗｉｔｈ ｔｈｅ ＵＳＡ ａｎｄ Ｉ ｆｅｅｌ ｔｈｉｎｇｓ ａｒｅ ｂｅｔｔｅｒ ｎｏｗ ｔｈａｎ ｔｈｅｙ ｗｅｒｅ ｌａｓｔ ｔｉｍｅ Ｉ ｗａｓ ｈｅｒｅ． Ｔｈｅ ｃｉｔｉｅｓ ａｒｅ ｃｌｅａｎ， ｐｅｏｐｌｅ ａｒｅ ｗｅｌｌ ｋｅｐｔ， ａｎｄ ｉｎｆｒａｓｔｒｕｃｔｕｒｅ ｓｅｅｍｓ ｅｖｅｎ ｂｅｔｔｅｒ ｔｈａｎ ｂｅｆｏｒｅ． Ｔｈｅ ｐｒｏｂｌｅｍ ｉｓ， ｌｉｋｅ ｍｉ， ａｌｌ ｔｈｅ ｏｌｄ ｐｅｏｐｌｅ ｈａｖｅ ａｌｌ ｔｈｅ ｍｏｎｅｙ ａｎｄ ｔｈｅｒｅ｀ｓ ａ ｌｏｔ ｏｆ ｔｈｅｍ．Ｏｌｄ ｐｅｏｐｌｅ ｄｏｎ｀ｔ ｉｎｖｅｓｔ ｉｎ ｔｈｅ ｆｕｔｕｒｅ ｔｈｅｉｒ ｉｎｔｅｒｅｓｔｅｄ ｉｎ ｅｎｊｏｙｉｎｇ ｔｈｅ ｌａｓｔ ｏｆ ｔｈｅ ｌｉｆｅ ｅａｔｉｎｇ ｏｕｔ ｏｒ ｓｉｇｈｔ ｓｅｅｉｎｇ． Ｔｈｅｙ｀ｒｅ ｃｅｒｔａｉｎｌｙ ｎｏｔ ｉｎ ｔｈｅ ｍｏｏｄ ｔｏ ｉｎｖｅｓｔ ｉｎ ａ ｌｏｎｇ－ｔｅｒｍ ｖｅｎｔｕｒｅ ｏｎ ｂｉｏ ｆｕｅｌ．
Ｔｈａｔ ｓａｉｄ， ｗｅ ｃｏｕｌｄ ｎｅｖｅｒ ｂｅｅｎ ｌｉｋｅ Ｊａｐａｎｅｓｅ． Ｗｈｉｌｅ Ｉ ｔｈｉｎｋ ｍｕｌｔｉｃｕｌｔｕｒｅａｌｉｓｍ ｉｓ ｍｏｒｅ ｄｙｎａｍｉｃ， Ｉ ｓｅｅ ｈｏｗ ｈｏｍｏｇｅｎｉｔｙ ｗｏｕｌｄ ｐｒｏｂａｂｌｙ ｗｏｒｋ ｗｅｌｌ ｉｎ ｔｉｍｅｓ ｏｆ ｕｎｃｅｒｔａｎｔｙ ａｎｄ ｄｕｒｅｓｓ ａｓ ｐｅｏｐｌｅ ｃａｎ ｐｕｌｌ ｔｏｇｅｔｈｅｒ ａｓ ａ ｕｎｉｔｅｄ ｗｈｏｌｅ ｉｎ ａ ｗａｙ ｗｅ ｉｎ ｔｈｅ Ｗｅｓｔ ｃａｎ｀ｔ．
Ｉ ｗａｓ ｉｎ ｔｈｅ ｍａｌｌｓ ｉｎ ＣＡ ａｎｄ ＭＩ ｉｎ ｔｈｅ ｒｕｎ ｕｐ ｔｏ Ｘｍａｓ ａｓ ｗｅｌｌ ａｓ ａｆｔｅｒ Ｘｍａｓ ａｎｄ ｔｈｅｙ ｗｅｒｅ ｄｅａｄ． Ｉｎ Ｊａｐａｎ ｔｈｅ ｓｔｏｒｅｓ ｗｅｒｅ ｐａｃｋｅｄ （ｉｎ ＡＵ ｔｈｅ ｓｔｏｒｅｓ ａｒｅ ａｌｗａｙｓ ｐａｃｋｅｄ）．
Ｉ ｔｈｉｎｋ ｍｏｓｔ Ａｍｅｒｉｃａｎｓ ｗｉｓｈｅｄ ｔｈｅｙ ｈａｄｎ｀ｔ ｂａｉｌｅｄ ｏｕｔ ｔｈｅ ｂａｎｋｓ ｎｏｗ。 Just ｌook at Iceland。
5. ### Michael歌舞伎Valued Senior Member
Messages:
20,285
By the Gods this keyboard is killing me! I keep hitting Shift Alt and switching font
Anyway, as I was saying, things are pretty good in Japan and one wouldnt know they were in deflation other than 100 yen sushi per plate is still 100 yen.
I was watching PulpFiction the other day and having a laugh at Vincent Vegas exclamation when Mia Wallace orders a $5 shake! Hes like: WTF?!? 5 dollars for a shake? Do they put burbone in it??? Please Register or Log in to view the hidden image! Im sure theres a poster waiting to be made: Inflation What disconnects you from watching a great movie from the 1990s Please Register or Log in to view the hidden image! 6. ### Google AdSenseGuest Advertisement to hide all adverts. 7. ### jmpetValued Senior Member Messages: 1,891 I think we all underestimate the magical quality of the printed, in hand US dollar. 8. ### joepistoleDeacon BluesValued Senior Member Messages: 22,910 Then please explain it. You are definately using a different lexicon then as are those who share your political ideology. The Fed does not need to print money to increase the money supply and there are multiple definitions of money supply. http://en.wikipedia.org/wiki/Expansionary_monetary_policy 9. ### SyzygysAs a mother, I am telling youValued Senior Member Messages: 12,671 Sure, but what if: 1. It didn't exist in the first place? 2. It existed but was wasted on unnecessery wars? You might as well pay it back instead of spending it on warring... 10. ### SyzygysAs a mother, I am telling youValued Senior Member Messages: 12,671 11. ### joepistoleDeacon BluesValued Senior Member Messages: 22,910 12. ### quadraphonicsBloodthirsty BarbarianValued Senior Member Messages: 9,391 True enough, but not (particularly) because of the relatively high level of domestic debt ownership. It was more the industrial policy that they used all that savings to finance - or, rather, the attempt to unwind such when it became too unwieldy. Which is to say that there are upsides to owing debt to foreigners instead. You can rely on depreciation much more heavily, since the affected parties don't get to vote in your country. 13. ### Michael歌舞伎Valued Senior Member Messages: 20,285 (1) The USA is probably the wealthiest nation that has ever existed - although that seems (to me) that after transferring most of it to the rich Banking oligarchy what's left is rapidly coming to an end. (2) Unnecessary.... hmmmmm I often wonder if the USA doesn't ultimately drop the charaide of "Nation Building" and takes military control of Iraqi oil as a hedge on China's growth? Maybe not, but it may be the ONLY way to control China? (It's either that or invade Canada) As I understand it, in our system we don't "print" money, as in *poof* it just appears, we could do that, but we don't, we borrow it into existence, can you borrow yourself out of debt? Secondly, as the borrowed money at low interest enters the money supply all the other money is worth less, which means wealth is destroyed, a hidden tax as they say. 14. ### CarcanoValued Senior Member Messages: 6,865 My explanation isnt any different from what you read here: http://en.wikipedia.org/wiki/Money_multiplier#Mechanism But I know you wont bother reading it, because your EXTREME partisanism wont allow you to think independently of your favourite US vs THEM conflict. For example, an economic policy you might claim to HATE under Bush will sound to your ears like a revelation from heaven under Obama. You simply dont care about true vs false, you just want to wave a party flag...so whats the point in explaining anything to you? Really...whats my political ideology??? I've lived in Canada (dual citizenship) most of my life and have never associated myself with any ideology. I usually vote for the Green Party. I want congress to reinstate Glass-Steagle...what ideology does that fall under? I want a universal standard of value based on the gold gram...what ideology does that fall under? I want to establish international trade eqilibrium on manufactered goods...what ideology does that fall under? I want the US to copy Canada's wonderful health care system...what ideology does that fall under? I want the US to stop kissing Israel's ass and shut down the global military empire. I simply dont care what any of these ideas are labeled as, because unlike you Joe, I realize that ideological affiliations are usually an excuse to STOP THINKING! 15. ### joepistoleDeacon BluesValued Senior Member Messages: 22,910 Good now you defined what you meant. Now how the heck does that have anything thing to do with the national debt of the United States? http://www.sciforums.com/showpost.php?p=2670237&postcount=11 "Money is not issued multiple times but it is COUNTED multiple times under the fractional reserve multiplier effect. When banks stop lending this process is unwound. Money is created AS debt ONLY when the Federal Reserve creates it to buy US treasury bonds. And even then, the interest earned on that debt is almost entirely REMITTED to the Treasury" - Carcano The fractional banking system is not a Ponzi scheme. The fractional bankins system works in a well regulated environment. What does not work is deregulation of the banking and financial systems. And I don't need to read your reference, because I studied the banking system decades ago in business school. This has nothing to do with partisanship on my part friend. Additionally, you might be interested to know that the U S Government does not pay interest to itself. That would be silly. Additionally, money is not created by adding debt. If your claim were true, then you should be able to point out some examples. But you cannot because you accusation is not based in fact. Last edited: Jan 5, 2011 16. ### SyzygysAs a mother, I am telling youValued Senior Member Messages: 12,671 True, but 2 points: 1. Every and any wealth can be wasted, just give people enough time. Please Register or Log in to view the hidden image! 2. Germany was also a very wealthy country/empire and still ended up with a completely worthless currency in the 1920s. So a nation wealth doesn't always guarantee the stability of its currency. I guess we could agree that they (at least Iraq) were necessery for protecting the petrodollar's standing. I think eventually it is still just printing it, just by the Fed not by the government. The government borrowing it from the Fed is just one extra step in the process. No, we can not borrow ourselves out of debt, but we can print the situation until the debt's value goes to eventually zero. Remember, inflation is always good for the debtor, and bad for the creditor. Sure. But as long as the ruling class has its wealth in real estate, indusrty, gold,etc. they don't have to worry about the middle class' problem... 17. ### joepistoleDeacon BluesValued Senior Member Messages: 22,910 A few minor details, the story of post WWI German and the US is not comparable. Germany did not have a comperable banking/finance system nor did it have the natural wealth possessed by the U.S. Ultimately the wealth of a country is linked to its political stability. With the rise of the radical right; the US is certianly less stable that it used to be. You can make a case for the war in Afganistan. But you cannot make a case for screwing it up as George II did. And you cannot make are good case for the Iraq war or screwing it up as George II did. Not even the Arabs wanted the US to invade Iraq and install a Shite led government - just look at the Wikileaks material. First, the Fed is the government. And second we can "monetize" the debt. We have been doing it for a long time. And three the Fed can shrink the money supply by selling US debt. The process goes both ways. The Fed can expand the money supply and it can contract the money supply at will. Additionally, there is no evidence that the Fed is going on a printing binge. The money supply has remained stable and inflation is almost non existent. Last edited: Jan 5, 2011 18. ### cosmictravelerBe kind to yourself always.Valued Senior Member Messages: 33,264 I wonder why China never has to borrow any money to get out of a jam? :shrug: 19. ### joepistoleDeacon BluesValued Senior Member Messages: 22,910 Because it is a dictatorship and it is doing the same thing people accuse the Fed of doing - printing money. China has a wee bit of an inflation problem. I guess no one noticed. 20. ### SyzygysAs a mother, I am telling youValued Senior Member Messages: 12,671 I don't want to dwell on this for too long, but Germany was wealthy and it had a central bank. That's all we need to make the comparison valid. Also, it started eventually very costly wars... Sure I can. The US had no choice, because of peak oil, petrodollar dependancy (Saddam was selling in Euros). Now how much the plan was reality based is a different question. The original idea was a quick government takeover, the troops celebrated as giving liberty to the country, the oilrevenue paying for the costs and everything is fine. On paper it sounded so pretty... Or not. It is a quasi government entity, owned by the 12 regional banks. The President can not force policy on the Fed, unlike any other government agency... Except the government not publishing M3 anymore?? The reason inflation is low because there are several deflatory forces working at the same time, evening out the balance.... 21. ### joepistoleDeacon BluesValued Senior Member Messages: 22,910 Minor details friend, but you are wrong on all counts. http://en.wikipedia.org/wiki/Deutsche_Bundesbank The first independent German central bank was created in 1957. And Germany did not start the WWI. It started WWII. WWI began with Austria and an assination of a member of the royal family. http://en.wikipedia.org/wiki/World_War_I And finally, you ignored the most important point - the political instability which resulted from loosing WW I and the devistating terms of the Treaty of Versailles. LOL, yeah I suppose you can. But it will not be a rational reason. Peak oil had nothing to do with Gulf War II. Even if Iraq was selling oil on the side in Euros, it did not matter. For the record, the UN supervised Iraq oil sales prior to Gulf War II. It was not something controlled by Iraq. That might have been the George II plan, but it didn't work. And to anyone with half a brain, it would have been obvious the George II plan would not work. That is why the Arab states were against Gulf War II and why they opposed the US ouster of Saddam in Gulf War I. WRONG AGAIN, the Fed is not owned by the 12 regional Federal Reserve Banks. Those 12 regional Federal Reserve Banks are part of the Federal Reserve. The Federal Reserve was created by an act of Congress and is an independent agency of the Federal government just like many other agencies (e.g. Security and Exchange Commission). http://www.federalreserveeducation.org/about-the-fed/history/ http://www.federalreserve.gov/aboutthefed/fract.htm http://www.federalreserve.gov/aboutthefed/bios/board/default.htm Yes the government is no longer publishing M3 because it is essentially redundant - not needed. I suppose you can create some baseless fantasy as to why the Fed no longer reports M3. But it is only that, fantasy and not based in fact. The Fed stopped tracking M3 back in 2005. http://www.federalreserve.gov/Releases/h6/discm3.htm Here is where you are correct. There are deflationary forces at work in the economy. And that is exactly why the Federal Reserve has acted with quantative easing. The Fed is doing exactly what it should be doing, despite all of the clamour. The bottom line here is that there is essentially no inflation and the money supply has remained very stable. So for all of the screaming about the sky is falling, it has not fallen. Nor is it showing any signs of falling. Last edited: Jan 5, 2011 22. ### SyzygysAs a mother, I am telling youValued Senior Member Messages: 12,671 Man, this is going to be fun! Please Register or Log in to view the hidden image! I am seldom wrong and even more seldom on all accounts. Please Register or Log in to view the hidden image! Who cares? If your point is that there was no central bank in the Weimar Republic, YOU are wrong: http://en.wikipedia.org/wiki/Reichsbank "The Reichsbank was the central bank of Germany from 1876 until 1945. It was founded on 1 January 1876 (shortly after the establishment of the German Empire in 1871)." See? I can quote Wiki too! Again, who cares? The point was that they entered on their own, not because they were attacked, thus making it a costly and unnecessary war. So the analogy stands, thank you. yeah, as a result of an unnecessary war. We are in agreement here. If you think that 2 far away foreign wars can't break the US; back, just wait.$400 gas per gallon in Afghanistan can go a long way...
Sure, after all running out of the most important commodity has nothing to do with obtaining such reserves by military force.
Not on the side but OPENLY, and by this creating a very dangerous precedent. read up on petrodollars. The only thing what makes other countries not to switch to other currency is the US military power.
Completely irrelevant since Saddam did it openly. The UN couldn't force him to sell in dollars.
Tell it to Dick Cheney and the Reps. Sometimes governments (power) are put into a position where they don't really have a smart or an easy way out. the smart way is never easy, I should say...
I already mentioned that I am seldom wrong. Read The Creature from Jekyll Island. I can send it to you in pdf.
But if you can tell me who owns the Fed, I am curious to hear, because that is not the US taxpayer for sure.
Here it is, according to them:
"Who owns the Federal Reserve?
The Federal Reserve System is not "owned" by anyone and is not a private, profit-making institution. Instead, it is an independent entity within the government, having both public purposes and private aspects."
"The twelve regional Federal Reserve Banks, ... are organized much like private corporations. For example, the Reserve Banks issue shares of stock to member banks. However, owning Reserve Bank stock is quite different from owning stock in a private company."
http://www.federalreserve.gov/generalinfo/faq/faqfrs.htm#5
Even they acknowledge that the US taxpayer doesn't own them.
There are so many things not needed, whom publishing little M3 was hurting? It gave a picture of the money supply, and since they knew they are going to print, the just stopped publishing it.
The stockmarket must be going up then because of the strong economy.
Last edited: Jan 6, 2011
23. ### joepistoleDeacon BluesValued Senior Member
Messages:
22,910
LOL, you may not think so in a few minutes.
Well I will take your word for it. But on this topic, you have been all wrong as previously proven. And I think you should ask yourself, is it that you are not wrong or that you rarely admit that you are wrong?
The point was you were wrong. I said Germany did not have a modern central banking system. And as previously posted, Germany did not have a modern central banking system until 1957 - as previously proven.
The point is that you were wrong again. Germany did not start WW I as you previously claimed.
As for wither WW I was necessary, that is a topic for another thread. And frankly wither the war was or was not necessary has nothing to do with comparing post WW I Germany with the current state of the US.
The unpleasant fact remains, that comparing the two is like comparing apples to cats. The US has not been defeated in War. The US economic base is still intact. And the US is not constrained by a Treaty forced upon it by winners of WWI or any other war for that matter. And the US is not grossly expanding its money supply.
The US fought two world wars around the globe in WW II that make the Iraq and Afganistan wars look like childs play by any measure. And WWII did not break the back of the US. Your reasoning here is based on fantasy and extreme partisan politics...not reality.
We have not run out of "the most important commodity". You can go down to the local gas station and buy all of it you can afford. New oil is coming on line all the time. And it is not like there are not alternatives out there like natural gas.
You have made a claim, but you have not supported it with fact. I will agree that the decision to go to war was likely done in order to satisfy certian political interests here in the US.
But it was a stupid decision that was not supported by Arab states for reasons previously stated and which should be very obvious by now.
The only dangerous precident was in your mind. There was and is no danger that oil exporters will change the currency used to transact oil prices because a rennagade outcast like Saddam wanted to sell a little oil to support his life style in Euros.
I don't care if Iraq sold all of its oil in Euro's it would not make a difference. And certianly was no reason to go to war. It sounds like you are hanging around too many conspiracy whackos.
Extraordinary claims require some kind of evidence.
How is that relevant? The fact is the George II administration was incredibly incompetent.
The government owns the Federal Reserve System. Now who owns the government? Can you say the American people?
The point is you were wrong yet again. You claimed that banks owned the Federal Reserve....not the government. And you were wrong as I previously proved by pointing you to the Federal Reserve's web site. The site that you are now using to prove my point.
Do you remember your post #37?
"Or not. It is a quasi government entity, owned by the 12 regional banks. The President can not force policy on the Fed, unlike any other government agency... - Syzygys post 37
I repeat and you now seem to acknowlege that the Federal Reserve is not owned by the 12 regional Fed banks as you initially claimed. So you were wrong. And as previously pointed out, the Federal Reserve is an independent agency similar to the many other independent goverment agencies (e.g. Security Exchange Commission).
If you have to tell people that you are seldom wrong, perhaps you are wrong more than you would like to admit.
It not only stopped publishing it, it stopped tracking it per my previous post and Federal Reserve Press Release.
The Federal Reserve, the agency charged with managing the nation's money supply, felt that M3 no longer provided any meaningful information. So the Federal Reserve not only stopped publishing it, they stopped tracking it in 2005.
Indeed it is. If you have not noticed, sales and profits are up virtually across the board. The nation is now adding jobs versus loosing almost a million jobs a month as it was when Obama was sworn into office. Gross national product has been consistently growing for more than a year now.
And just today, the ADP employment numbers were unexpectedly good, forecasting an increase in jobs last month that was three times higher than expected.
http://www.bloomberg.com/markets/economic-calendar/
Not to mention the ISM manufacturing index has been on a tear and continues to rise.
http://www.bloomberg.com/markets/economic-calendar/
Last edited: Jan 6, 2011
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## Tuesday, February 20, 2018
### How to compute probabilities from list of boosted trees in xgboost
If number of trees == num of boosting rounds
score = sum over predicted leaf values over all trees + 0.5
prob = exp(score)/(1+exp(score)) or 1/(1+exp(-score))
If number of trees == num of boosting rounds * number of class
score_i for all class i = sum over predicted leaf values over all trees + 0.5 , for all class i
prob_i for i = exp(score_i)/sum_i_(exp(score_i))
References:
## Monday, July 6, 2015
### Extensive evaluation of different classifiers
TLDR: Random Forests is best thing common in both references.
From abstract of first reference "Do we Need Hundreds of Classifiers to Solve Real World Classification Problems?"
"We evaluate 179 classifiers arising from 17 families (discriminant analysis, Bayesian, neural networks, support vector machines, decision trees, rule-based classifiers, boosting, bagging, stacking, random forests and other ensembles, generalized linear models, nearest-neighbors, partial least squares and principal component regression, logistic and multinomial regression, multiple adaptive regression splines and other methods). We use 121 data sets, which represent the whole UCI database (excluding the large-scale problems) and other own real problems, in order to achieve significant conclusions about the classifier behavior, not dependent on the data set collection. The classifiers most likely to be the bests are the random forest(RF) versions."
From second reference:
"With excellent performance on all eight metrics, calibrated boosted trees were the best learning algorithm overall. Random forests are close second, followed by uncalibrated bagged trees, calibrated SVMs, and uncalibrated neural nets. The models that performed poorest were naive bayes, logistic regression, decision trees, and boosted stumps. Although some methods clearly perform better or worse than other methods on average, there is significant variability across the problems and metrics. Even the best models sometimes perform poorly, and models with poor average performance occasionally perform exceptionally well."
The two references perform an extensive evaluation of different classifiers across datasets and across performance metrics.
Do we Need Hundreds of Classifiers to Solve Real World Classifi cation Problems?
An Empirical Comparison of Supervised Learning Algorithms
## Wednesday, June 24, 2015
### Idea of modeling non-linearity as learning distrtibution over function spaces in feedforward neural network
I was thinking about relationship in graphical context of graphical models and feedforward neural network. On one hand, Feedforward Neural Network is a graph of deterministic functions and on the other hand, Graphical models are graph of dependence of random variable which are uncertain. Then I thought what if deterministic non-linearities can be replace with random process which generates functions and shared non-linearity can be inferred.
An interesting idea would be to learn a distribution over function space(which will be used as non-linearity in Feedforward Neural Networks) jointly with backpropagation in an EM like fashion.
To summarize, We want to replace the deterministic function with a learned function by modeling the distribution over function space and inferring the shared non-linearity in neural network.
## Sunday, June 7, 2015
### Squeezing space with LaTeX
I was trying to find ways to correct large vertical spaces between paragraphs. After serching a bit on internet, I got the following command along with other options. So I wanted to share it here and for my own future reference.
Remove the spacing between paragraphs and have a small paragraph indentation
\setlength{\parskip}{0cm}
\setlength{\parindent}{1em}
Source:
http://robjhyndman.com/hyndsight/squeezing-space-with-latex/
http://www.terminally-incoherent.com/blog/2007/09/19/latex-squeezing-the-vertical-white-space/
http://www-h.eng.cam.ac.uk/help/tpl/textprocessing/squeeze.html
https://ravirao.wordpress.com/2005/11/19/latex-tips-to-meet-publication-page-limits/
Make your text block as big as possible. The simplest way to do that is using the geometry package:
\usepackage[text={16cm,24cm}]{geometry}
Use a compact font such as Times Roman:
\usepackage{mathptmx}
\usepackage[compact]{titlesec} \titlespacing{\section}{0pt}{2ex}{1ex} \titlespacing{\subsection}{0pt}{1ex}{0ex} \titlespacing{\subsubsection}{0pt}{0.5ex}{0ex}
Beware of enumerated and itemized lists. Instead, replace them with compact lists.
\usepackage{paralist}
\begin{compactitem} \item ... \end{compactitem} \begin{compactenum} \item ... \end{compactenum}
If you are allowed, switching to double column can save heaps of space.
\usepackage{multicols}
\begin{multicols}{2} ... \end{multicols}
If the rules say 12pt, you can usually get away with 11.5pt without anyone noticing:
\begin{document}\fontsize{11.5}{14}\rm
When you get desperate, you can squeeze the inter-line spacing using
There is also a savetrees package which does a lot of squeezing, but the results don’t always look nice, so it is better to try one or more of the above tricks instead.
## Tuesday, May 26, 2015
### DL reading list for new students in LISA LAB
http://www.iro.umontreal.ca/~lisa/twiki/bin/view.cgi/Public/WebHome
## Tuesday, March 31, 2015
### Setting up Theano on Ubuntu 14.04
After 2 days of non-stop struggle, I was able to make theano work with my GPU NVIDIA GeForce GTX 860M.
Settings which worked are as follows.
DONOT install bumblebee
Run Following command to see if your NVIDIA GPU device is being detected. If not, this BLOG wont help. Sorry.
lspci | grep -i NVIDIA
Install the following:
Driver: NVIDIA 340.76
Cuda : 6.5 toolkit
• Switch to NVIDIA card (If you dont have this command try to get it, without changing the graphics drivers, conflicting drivers can be blacklisted, see next point).
prime-switch nvidia
• Blacklist other driver which can create conflicts:
create /etc/modprobe.d/blacklist-file-drivers.conf File with blacklisted drivers. Use command ubuntu-drivers devices to get a list of nvidia drivers:
blacklist nvidia-349
blacklist nvidia-346
blacklist xserver-xorg-video-nouveau
To list all installed Graphics Drivers (Useful while blacklisting drivers)
ubuntu-drivers devices
Note: DONOT blacklist nvidia-340
• Make sure the following command works without error.
nvidia-modprobe
nvidia-settings
nvidia-smi
• Find /usr/local/cuda-6.5/samples/1_Utilities/deviceQuery folder for your system.
use make command to create executable.
cd /usr/local/cuda-6.5/samples/1_Utilities/deviceQuery/
sudo make
run /usr/local/cuda-6.5/samples/1_Utilities/deviceQuery/deviceQuery
You should get the following results:
deviceQuery, CUDA Driver = CUDART, CUDA Driver Version = 6.5, CUDA Runtime Version = 6.0, NumDevs = 1, Device0 = GeForce GTX 860M
Result = PASS
• Install theano from GIT (First install dependencies from theano website):
sudo apt-get install python-numpy python-scipy python-dev python-pip python-nose g++ libopenblas-dev git
sudo pip uninstall theano
sudo pip install git+git://github.com/Theano/Theano.git
In /usr/local/cuda-6.5/targets/x86_64-linux/include/host_config.h, cuda-6.5 supports gcc-4.8 g++-4.8 so one needs to install these and make links to gcc and g++ respectively.
example: sudo ln -s /usr/bin/gcc-4.8 /usr/local/cuda/bin/gcc
• Create ~/.theanorc File with following content:
[global]
floatX = float32
device = gpu
[nvcc]
fastmath = True
[cuda]
root=/usr/local/cuda-6.5/
• Make a python file to test gpu say test.py
from theano import function, config, shared, sandbox
import theano.tensor as T
import numpy
import time
vlen = 10 * 30 * 768 # 10 x #cores x # threads per core
iters = 1000
rng = numpy.random.RandomState(22)
x = shared(numpy.asarray(rng.rand(vlen), config.floatX))
f = function([], T.exp(x))
print f.maker.fgraph.toposort()
t0 = time.time()
for i in xrange(iters):
r = f()
t1 = time.time()
print 'Looping %d times took' % iters, t1 - t0, 'seconds'
print 'Result is', r
if numpy.any([isinstance(x.op, T.Elemwise) for x in f.maker.fgraph.toposort()]):
print 'Used the cpu'
else:
print 'Used the gpu'
Once you run the above python file:
sudo python test.py
You might get an error:
Failed to compile cuda_ndarray.cu: libcublas.so.6.5: cannot open shared object file: No such file or directory
So you need to locate and configure that path:
locate libcublas.so.6.5
Add the following library path to ~/.bashrc (Second path has libcublas.so.6.5)
sudo ldconfig /usr/local/cuda-6.5/lib64/
sudo ldconfig /usr/local/cuda-6.5/targets/x86_64-linux/lib/
export LD_LIBRARY_PATH=/usr/local/cuda-6.5/targets/x86_64-linux/lib:$\$$LD_LIBRARY_PATH export PATH=/usr/local/cuda-6.5/bin:\$$PATH export PATH=/usr/local/cuda-6.5/targets/x86_64-linux/lib:$\PATH
Now run the gpu code.
sudo python test.py
I get roughly 10x speedup with GPU.
I screwed up many times with different versions of drivers, if you do the same. You might not get a login screen (black screen). The use ctrl + Alt + F1 to goto command mode.
Remove xorg.conf
sudo rm /etc/X11/xorg.conf
sudo service lightdm stop
sudo service lightdm start
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2018-11-18 08:52:54
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https://www.emathhelp.net/fr/calculators/differential-equations/half-life-calculator/
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# Calculateur de demi-vie
Cette calculatrice calculera la demi-vie, la quantité initiale, la quantité restante et le temps, avec les étapes indiquées.
There are units of mass of a substance with a half-life of units of time. In units of time, there will remain units of mass of the substance.
Enter any three values.
If the calculator did not compute something or you have identified an error, or you have a suggestion/feedback, please write it in the comments below.
## Solution
Your input: find $N(t)$ in $N(t)=N_0e^{-kt}$ given $N_0=250$, $t_h=15$, $t=100$.
$N(t)$ is the amount after the time $t$, $N_0$ is the initial amount, $t_h$ is the half-life.
First, find the constant $k$.
We know that after half-life there will be twice less the initial quantity: $N\left(t_h\right)=\frac{N_0}{2}=N_0e^{-k t_h}$.
Simplifying gives $\frac{1}{2}=e^{-k t_h}$ or $k=-\frac{\ln\left(\frac{1}{2}\right)}{t_h}$.
Plugging this into the initial equation, we obtain that $N(t)=N_0e^{\frac{\ln\left(\frac{1}{2}\right)}{t_h}t}$ or $N(t)=N_0\left(\frac{1}{2}\right)^{\frac{t}{t_h}}$.
Finally, just plug in the given values and find the unknown one.
From $N(t)=250\left(\frac{1}{2}\right)^{\frac{100}{15}}$, we have that $N(t)=\frac{125 \sqrt[3]{2}}{64}$.
Answer: $N(t)=\frac{125 \sqrt[3]{2}}{64}\approx 2.46078330057592$.
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2022-05-27 13:26:54
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https://brilliant.org/problems/is-this-polynomial-familiar-ii/
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# Is this polynomial familiar? - II
Algebra Level 2
$$f(x)$$ is a polynomial with integer coefficients. We have,
$$f(1)=1$$
$$f(2)=4$$
$$f(3)=9$$
$$f(4)=16$$
$$f(5)=25$$
$$f(6)=36$$
$$f(7)=49$$
$$f(8)=64$$
$$f(9)=81$$
$$f(10)=100$$
How many functions $$f(x)$$ satisfies the above conditions?
Details and assumptions:
• If your answer is infinity, enter $$0$$ as the answer.
More polynomials, anyone?
×
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2017-10-23 20:57:41
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