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https://ask.truemaths.com/question/i-1cos-theta-sin2-theta-sin-theta-1-cos-theta-cot-theta-iitan%C2%B3-a-1-tan-a-1-sec%C2%B2-a-tan-a/ | 1,701,657,926,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100523.4/warc/CC-MAIN-20231204020432-20231204050432-00185.warc.gz | 132,814,368 | 26,831 | • 0
Guru
# (i) (1+cos theta -sin^2 theta) / sin theta (1 + cos theta ) = cot theta. (ii)tan³ A – 1 / tan A -1 = sec² A + tan A
• 0
This ques has been taken from the Book Book- ML Aggarwal
Board- ICSE
Publication- Avichal
Chapter- Trigonometric Identities
Chapter number-18
This is an important question and asked in previous exams.
(i) (1+cos theta -sin^2 theta) / sin theta (1 + cos theta ) = cot theta. (ii)tan³ theta- 1 / tan theta -1 = sec² theta+ tan theta
Class 10, chapter 18, ML aggarwal, trigonometric identities, question no 21
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#### Resources tagged with Working systematically similar to Diagonal in a Spiral:
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Broad Topics > Using, Applying and Reasoning about Mathematics > Working systematically
### Bean Bags for Bernard's Bag
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How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
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Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
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Can you substitute numbers for the letters in these sums?
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Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
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Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### Open Squares
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This task, written for the National Young Mathematicians' Award 2016, focuses on 'open squares'. What would the next five open squares look like?
### Hubble, Bubble
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Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Dart Target
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This task, written for the National Young Mathematicians' Award 2016, invites you to explore the different combinations of scores that you might get on these dart boards.
### The Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Sticky Dice
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Throughout these challenges, the touching faces of any adjacent dice must have the same number. Can you find a way of making the total on the top come to each number from 11 to 18 inclusive?
### Zargon Glasses
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Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Take Three Numbers
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What happens when you add three numbers together? Will your answer be odd or even? How do you know?
### Being Resourceful - Primary Number
##### Age 5 to 11 Challenge Level:
Number problems at primary level that require careful consideration.
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##### Age 7 to 11 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Journeys in Numberland
##### Age 7 to 11 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### Sums and Differences 1
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This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Build it up More
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This task follows on from Build it Up and takes the ideas into three dimensions!
### ABC
##### Age 7 to 11 Challenge Level:
In the multiplication calculation, some of the digits have been replaced by letters and others by asterisks. Can you reconstruct the original multiplication?
### Centred Squares
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This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
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Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Magic Vs
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Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?
### Arranging the Tables
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There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Six Is the Sum
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What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### Polo Square
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Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### A-magical Number Maze
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This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Two Egg Timers
##### Age 7 to 11 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### Sums and Differences 2
##### Age 7 to 11 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
### Pouring the Punch Drink
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There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
##### Age 7 to 11 Challenge Level:
Write the numbers up to 64 in an interesting way so that the shape they make at the end is interesting, different, more exciting ... than just a square.
### Prison Cells
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There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### Seating Arrangements
##### Age 7 to 11 Challenge Level:
Sitting around a table are three girls and three boys. Use the clues to work out were each person is sitting.
### Plates of Biscuits
##### Age 7 to 11 Challenge Level:
Can you rearrange the biscuits on the plates so that the three biscuits on each plate are all different and there is no plate with two biscuits the same as two biscuits on another plate?
### Sitting Round the Party Tables
##### Age 5 to 11 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
##### Age 7 to 11 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### The Moons of Vuvv
##### Age 7 to 11 Challenge Level:
The planet of Vuvv has seven moons. Can you work out how long it is between each super-eclipse?
### Painting Possibilities
##### Age 7 to 11 Challenge Level:
This task, written for the National Young Mathematicians' Award 2016, involves open-topped boxes made with interlocking cubes. Explore the number of units of paint that are needed to cover the boxes. . . .
### Calendar Cubes
##### Age 7 to 11 Challenge Level:
Make a pair of cubes that can be moved to show all the days of the month from the 1st to the 31st.
### Button-up Some More
##### Age 7 to 11 Challenge Level:
How many ways can you find to do up all four buttons on my coat? How about if I had five buttons? Six ...?
### Family Tree
##### Age 7 to 11 Challenge Level:
Use the clues to find out who's who in the family, to fill in the family tree and to find out which of the family members are mathematicians and which are not.
### Ordered Ways of Working Upper Primary
##### Age 7 to 11 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps to have an ordered approach.
### Wag Worms
##### Age 7 to 11 Challenge Level:
When intergalactic Wag Worms are born they look just like a cube. Each year they grow another cube in any direction. Find all the shapes that five-year-old Wag Worms can be.
### Make Pairs
##### Age 7 to 11 Challenge Level:
Put 10 counters in a row. Find a way to arrange the counters into five pairs, evenly spaced in a row, in just 5 moves, using the rules.
### Team Scream
##### Age 7 to 11 Challenge Level:
Seven friends went to a fun fair with lots of scary rides. They decided to pair up for rides until each friend had ridden once with each of the others. What was the total number rides?
### This Pied Piper of Hamelin
##### Age 7 to 11 Challenge Level:
Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Ordered Ways of Working Upper Primary
##### Age 7 to 11 Challenge Level:
These activities lend themselves to systematic working in the sense that it helps if you have an ordered approach.
### Two Dots
##### Age 7 to 11 Challenge Level:
Place eight dots on this diagram, so that there are only two dots on each straight line and only two dots on each circle.
##### Age 5 to 11 Challenge Level:
How could you put these three beads into bags? How many different ways can you do it? How could you record what you've done?
### Different Deductions
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There are lots of different methods to find out what the shapes are worth - how many can you find?
### Stairs
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This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high. | 2,320 | 9,805 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2019-35 | latest | en | 0.882769 |
https://www.nagwa.com/en/videos/108181310649/ | 1,606,783,013,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00447.warc.gz | 785,415,311 | 8,263 | # Question Video: Determining the Domain and Range of a Rational Function
Determine the domain and range of the function π(π₯) = 1/(π₯ β 2).
01:51
### Video Transcript
Determine the domain and range of the function π of π₯ equals one over π₯ minus two.
We have this function π of π₯ equals one over π₯ minus two, but we donβt have a graph. Is there a way we can solve this algebraically? Can we solve this without graphing? Actually, we can. Something that we know about division is that we canβt divide something by zero. And that means that any time whatβs in the denominator would be equal to zero. Itβs not a possibility for this function.
If we set π₯ minus two equal to zero and solve, we get π₯ equal to two. Our function would not have a solution at π₯ equal to two. This is the only place for our π₯-values that this equation would not work. Because of that, we can say that our domain is all real numbers with the exception or minus two.
Something else interesting happens when we deal with these kinds of inverse functions. Our π of π₯ cannot be equal to zero. No matter what we put in the denominator, weβll never divide one by something and equal zero. Itβs impossible. So we say that our range is all real numbers with the exception of zero β all reals minus zero. | 365 | 1,335 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2020-50 | longest | en | 0.890941 |
https://stage.geogebra.org/m/kHAkeTCw | 1,695,770,323,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510225.44/warc/CC-MAIN-20230926211344-20230927001344-00763.warc.gz | 586,487,603 | 9,957 | # An Introduction to Differentiation
The diagram contains a function f(x) and two points on that function A (which is fixed) and B (which is dynamic). There is a blue line representing the tangent at A and a maroon line represennting the secand AB.
Pay attention to the gradients of the lines - through the slope function on the secant AB and also through the two equations of the lines scnt and tgnt in the Algebra screen. What happens as point B gets closer to A and can you explain what happens to the equations when A and B lie on top of each other? | 126 | 554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2023-40 | latest | en | 0.932754 |
https://www.physicsforums.com/threads/linear-momentum-questions.650701/ | 1,531,835,905,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589726.60/warc/CC-MAIN-20180717125344-20180717145344-00458.warc.gz | 942,512,206 | 13,005 | # Homework Help: Linear momentum questions?
1. Nov 8, 2012
### pinksparkles2
1. The problem statement, all variables and given/known data
A 500kg truck collides head on with a 400kg car. They stick together and continue traveling west at 6.00 meters per second. The truck initially travels at 13.0 meters per second. Calculate the impulse and force for the car given that the time of impact was 0.6 seconds.
3. The attempt at a solution
Using the Law of the Conservation of Momentum, I figured out that the car's velocity is 2.75 meters per second and that is as far as I got.
Last edited by a moderator: Nov 9, 2012
2. Nov 9, 2012
### haruspex
There's a piece of information missing. Your answer is correct if the truck was initially heading West. The other possibility is that the car was heading West, very fast, and the truck East.
For each individual vehicle, its change in momentum is the impulse from the other.
The impulse is the integral of force over time, so you can get the average force during the impact by dividing the impulse by the duration. But that could be quite a bit less than the peak force.
Last edited by a moderator: Nov 9, 2012 | 285 | 1,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2018-30 | latest | en | 0.955986 |
https://www.exercours.com/2020/07/complex-fourier-series-examples-and-solutions-pdf.html | 1,606,766,375,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141486017.50/warc/CC-MAIN-20201130192020-20201130222020-00454.warc.gz | 670,676,083 | 29,003 | # Complex Fourier Series Examples and Solutions PDF
## Complex Fourier Series Examples and Solutions PDF.
Complex Fourier Series Exercises With Answers.
Complex Fourier Series Notes PDF.
Complex Fourier Series Problems and Solutions PDF.
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The Hirsch Conjecture and its relatives (part II of III)
Francisco Santos
http://personales.unican.es/santosf
SLC’70, Ellwangen March 25–27, 2013
(2)
It holds with equality insimplices(n=d+1,δ=1) and cubes(n=2d,δ=d).
IfPandQsatisfy it, then so doesP×Q: δ(P×Q)= δ(P) +δ(Q). In particular:
For everyn≤2d, there arepolytopes in which the bound is tight(products of simplices).
We call these“Hirsch-sharp” polytopes.
For everyn>d, it is easy to constructunbounded polyhedrawhere the bound is tight.
H(n,d)is weakly monotone w.r.t.(n−d,d), not to(n,d).
(3)
It holds with equality insimplices(n=d+1,δ=1) and cubes(n=2d,δ=d).
IfPandQsatisfy it, then so doesP×Q: δ(P×Q)= δ(P) +δ(Q). In particular:
For everyn≤2d, there arepolytopes in which the bound is tight(products of simplices).
We call these“Hirsch-sharp” polytopes.
For everyn>d, it is easy to constructunbounded polyhedrawhere the bound is tight.
H(n,d)is weakly monotone w.r.t.(n−d,d), not to(n,d).
(4)
It holds with equality insimplices(n=d+1,δ=1) and cubes(n=2d,δ=d).
IfPandQsatisfy it, then so doesP×Q: δ(P×Q)= δ(P) +δ(Q).In particular:
For everyn≤2d, there arepolytopes in which the bound is tight(products of simplices).
We call these“Hirsch-sharp” polytopes.
For everyn>d, it is easy to constructunbounded polyhedrawhere the bound is tight.
H(n,d)is weakly monotone w.r.t.(n−d,d), not to(n,d).
(5)
It holds with equality insimplices(n=d+1,δ=1) and cubes(n=2d,δ=d).
IfPandQsatisfy it, then so doesP×Q: δ(P×Q)= δ(P) +δ(Q). In particular:
For everyn≤2d, there arepolytopes in which the bound is tight(products of simplices).
We call these“Hirsch-sharp” polytopes.
For everyn>d, it is easy to constructunbounded polyhedrawhere the bound is tight.
H(n,d)is weakly monotone w.r.t.(n−d,d), not to(n,d).
(6)
It holds with equality insimplices(n=d+1,δ=1) and cubes(n=2d,δ=d).
IfPandQsatisfy it, then so doesP×Q: δ(P×Q)= δ(P) +δ(Q). In particular:
For everyn≤2d, there arepolytopes in which the bound is tight(products of simplices).
We call these“Hirsch-sharp” polytopes.
For everyn>d, it is easy to constructunbounded polyhedrawhere the bound is tight.
H(n,d)is weakly monotone w.r.t.(n−d,d), not to(n,d).
(7)
It holds with equality insimplices(n=d+1,δ=1) and cubes(n=2d,δ=d).
IfPandQsatisfy it, then so doesP×Q: δ(P×Q)= δ(P) +δ(Q). In particular:
For everyn≤2d, there arepolytopes in which the bound is tight(products of simplices).
We call these“Hirsch-sharp” polytopes.
For everyn>d, it is easy to constructunbounded polyhedrawhere the bound is tight.
H(n,d)is weakly monotone w.r.t.(n−d,d),not to(n,d).
(8)
It holds with equality insimplices(n=d+1,δ=1) and cubes(n=2d,δ=d).
IfPandQsatisfy it, then so doesP×Q: δ(P×Q)= δ(P) +δ(Q). In particular:
For everyn≤2d, there arepolytopes in which the bound is tight(products of simplices).
We call these“Hirsch-sharp” polytopes.
For everyn>d, it is easy to constructunbounded polyhedrawhere the bound is tight.
H(n,d)is weakly monotone w.r.t.(n−d,d), not to(n,d).
(9)
P’
P
F f
(10)
v
d(u’, v’)=2 d(u, v)=2 u
F f
P’
P
u’
v’
(11)
Hirsch conjecture has the following interpretations:
(12)
Hirsch conjecture has the following interpretations:
Assumen=2d and letu andv be twocomplementary vertices (no common facet) of a simple polytope:
(13)
Hirsch conjecture has the following interpretations:
Assumen=2d and letu andv be twocomplementary vertices (no common facet) of a simple polytope:
d-step conjecture
It is possible to go fromutov so that at each step we abandon a facet containinguand we enter a facet containingv.
d-step conjecture⇔Hirsch forn=2d.
(14)
Hirsch conjecture has the following interpretations:
Assumen=2d and letu andv be twocomplementary vertices (no common facet) of a simple polytope:
d-step conjecture
It is possible to go fromutov so that at each step we abandon a facet containinguand we enter a facet containingv.
d-step conjecture⇔Hirsch forn=2d.
(15)
Hirsch conjecture has the following interpretations:
More generally, given any two verticesu andv of a simple poly- topeP:
(16)
Hirsch conjecture has the following interpretations:
More generally, given any two verticesu andv of a simple poly- topeP:
non-revisiting path conjecture
It is possible to go fromutov so that at each step we enter a new facet, one that we had not visited before.
non-revisiting path⇒Hirsch.
d-step⇔non-revisiting forn=2d ⇔Hirsch forn=2d.
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Hirsch conjecture has the following interpretations:
More generally, given any two verticesu andv of a simple poly- topeP:
non-revisiting path conjecture
It is possible to go fromutov so that at each step we enter a new facet, one that we had not visited before.
non-revisiting path⇒Hirsch.
d-step⇔non-revisiting forn=2d ⇔Hirsch forn=2d.
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Hirsch conjecture has the following interpretations:
More generally, given any two verticesu andv of a simple poly- topeP:
non-revisiting path conjecture
It is possible to go fromutov so that at each step we enter a new facet, one that we had not visited before.
non-revisiting path⇒Hirsch.
d-step⇔non-revisiting forn=2d ⇔Hirsch forn=2d.
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · ·
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · ·
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · ·
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}.The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · ·
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · ·
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · ·
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · · Ifn<2d, thenH(n,d)≤H(n−1,d −1)because every pair of verticesuandv lie in a common facetF, which is a polytope with one less dimension and (at least) one less facet (induction onnandn−d).
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · · Ifn<2d, thenH(n,d)≤H(n−1,d −1)because every pair of verticesuandv lie in a common facetF, which is a polytope with one less dimension and (at least) one less facet (induction onnandn−d).
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · · Ifn<2d, thenH(n,d)≤H(n−1,d −1)because every pair of verticesuandv lie in a common facetF, which is a polytope with one less dimension and (at least) one less facet (induction onnandn−d).
(28)
Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · · For everynandd,H(n,d)≤H(n+1,d+1): LetF be any facet ofPand letP0 be thewedgeofPoverF. Then:
dP0(u0,v0)≥dP(u,v).
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · · For everynandd,H(n,d)≤H(n+1,d+1): LetF be any facet ofPand letP0 be thewedgeofPoverF. Then:
dP0(u0,v0)≥dP(u,v).
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · · For everynandd,H(n,d)≤H(n+1,d+1): LetF be any facet ofPand letP0 be thewedgeofPoverF. Then:
dP0(u0,v0)≥dP(u,v).
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Theorem [Klee-Walkup 1967]
Hirsch⇔d-step⇔non-revisiting path.
Proof: LetH(n,d) =max{δ(P) :Pis ad-polytope withn facets}. The basic idea is:
· · · ≤H(2k−1,k −1)≤H(2k,k) =H(2k +1,k+1) =· · · For everynandd,H(n,d)≤H(n+1,d+1): LetF be any facet ofPand letP0 be thewedgeofPoverF. Then:
dP0(u0,v0)≥dP(u,v).
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P’
P
F f
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v
d(u’, v’)=2 d(u, v)=2 u
F f
P’
P
u’
v’
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Thed-step Theorem follows from and implies (respectively) the following:
Lemma
For every d -polytope P with n facets and diameterδthere is a d+1-polytope with one more facet and the same diameterδ.
Corollary
There is a function f(k) :=H(2k,k)such that H(n,d)≤f(n−d), ∀n,d.
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Thed-step Theorem follows from and implies (respectively) the following:
Lemma
For every d -polytope P with n facets and diameterδthere is a d+1-polytope with one more facet and the same diameterδ.
Corollary
There is a function f(k) :=H(2k,k)such that H(n,d)≤f(n−d), ∀n,d.
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Thed-step Theorem follows from and implies (respectively) the following:
Lemma
For every d -polytope P with n facets and diameterδthere is a d+1-polytope with one more facet and the same diameterδ.
Corollary
There is a function f(k) :=H(2k,k)such that H(n,d)≤f(n−d), ∀n,d.
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The construction of counter-examples to the Hirsch conjecture has two ingredients:
1 Astrongd-step theoremfor spindles/prismatoids.
2 The construction of aprismatoid of dimension 5 and
“width” 6.
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The construction of counter-examples to the Hirsch conjecture has two ingredients:
1 Astrongd-step theoremfor spindles/prismatoids.
2 The construction of aprismatoid of dimension 5 and
“width” 6.
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The construction of counter-examples to the Hirsch conjecture has two ingredients:
1 Astrongd-step theoremfor spindles/prismatoids.
2 The construction of aprismatoid of dimension 5 and
“width” 6.
(40)
Definition
Aspindleis a polytopePwith two distinguished verticesuand v such that every facet contains eitheruorv (but not both).
u u
v
v Definition
Thelengthof a spindle is the graph distance fromutov. Exercise 3-spindles have length≤3.
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Definition
Aspindleis a polytopePwith two distinguished verticesuand v such that every facet contains eitheruorv (but not both).
u u
v
v Definition
Thelengthof a spindle is the graph distance fromutov. Exercise 3-spindles have length≤3.
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Definition
Aspindleis a polytopePwith two distinguished verticesuand v such that every facet contains eitheruorv (but not both).
u u
v
v Definition
Thelengthof a spindle is the graph distance fromutov. Exercise 3-spindles have length≤3.
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Theorem (Strongd-step theorem for spindles)
Let P be a spindle of dimension d , with n>2d facets and lengthλ. Then there is another spindle P0 of dimension d+1, with n+1facets and lengthλ+1.
That is: we can increase the dimension, length and number of facets of a spindle, all by one, untiln=2d.
Corollary
In particular, if a spindle P has length>d then there is another spindle P0 (of dimension nd , with2n2d facets, and length
λ+n2d >nd )that violates the Hirsch conjecture.
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Theorem (Strongd-step theorem for spindles)
Let P be a spindle of dimension d , with n>2d facets and lengthλ. Then there is another spindle P0 of dimension d+1, with n+1facets and lengthλ+1.
That is: we can increase the dimension, length and number of facets of a spindle, all by one, untiln=2d.
Corollary
In particular, if a spindle P has length>d then there is another spindle P0 (of dimension nd , with2n2d facets, and length
λ+n2d >nd )that violates the Hirsch conjecture.
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Theorem (Strongd-step theorem for spindles)
Let P be a spindle of dimension d , with n>2d facets and lengthλ. Then there is another spindle P0 of dimension d+1, with n+1facets and lengthλ+1.
That is: we can increase the dimension, length and number of facets of a spindle, all by one, untiln=2d.
Corollary
In particular, if a spindle P has length>d then there is another spindle P0 (of dimension nd , with2n2d facets, and length
λ+n2d >nd )that violates the Hirsch conjecture.
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Definition
Aprismatoidis a polytopeQwith two (parallel) facetsQ+ and Qcontaining all vertices.
Q+
Q Q
Definition Thewidthof a prismatoid is the dual-graph distance fromQ+ toQ.
Exercise
3-prismatoids have width≤3.
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Definition
Aprismatoidis a polytopeQwith two (parallel) facetsQ+ and Qcontaining all vertices.
Q+
Q Q
Definition Thewidthof a prismatoid is the dual-graph distance fromQ+ toQ.
Exercise
3-prismatoids have width≤3.
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Definition
Aprismatoidis a polytopeQwith two (parallel) facetsQ+ and Qcontaining all vertices.
Q+
Q Q
Definition Thewidthof a prismatoid is the dual-graph distance fromQ+ toQ.
Exercise
3-prismatoids have width≤3.
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Theorem (Strongd-step theorem, prismatoid version) Let Q be a prismatoid of dimension d , with n>2d vertices and widthδ. Then there is another prismatoid Q0 of dimension d+1, with n+1vertices and widthδ+1.
That is: we can increase the dimension, width and number of vertices of a prismatoid, all by one, untiln=2d.
Corollary
In particular, if a prismatoid Q has width>d then there is another prismatoid Q0 (of dimension nd , with2n2d vertices, and widthδ+n2d>nd )that violates (the dual of) the Hirsch conjecture.
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Theorem (Strongd-step theorem, prismatoid version) Let Q be a prismatoid of dimension d , with n>2d vertices and widthδ. Then there is another prismatoid Q0 of dimension d+1, with n+1vertices and widthδ+1.
That is: we can increase the dimension, width and number of vertices of a prismatoid, all by one, untiln=2d.
Corollary
In particular, if a prismatoid Q has width>d then there is another prismatoid Q0 (of dimension nd , with2n2d vertices, and widthδ+n2d>nd )that violates (the dual of) the Hirsch conjecture.
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Theorem (Strongd-step theorem, prismatoid version) Let Q be a prismatoid of dimension d , with n>2d vertices and widthδ. Then there is another prismatoid Q0 of dimension d+1, with n+1vertices and widthδ+1.
That is: we can increase the dimension, width and number of vertices of a prismatoid, all by one, untiln=2d.
Corollary
In particular, if a prismatoid Q has width>d then there is another prismatoid Q0 (of dimension nd , with2n2d vertices, and widthδ+n2d>nd )that violates (the dual of) the Hirsch conjecture.
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Proof.
QR2 Q+
Q
Qf
QeR3 Qf+
w
Qf:=opsv(Q)
Q+ w
opsv(Q)R3 v
u
u
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So, to disprove the Hirsch Conjecture we only need to find a prismatoid of dimensiond and width larger thand. Its number of vertices and facets is irrelevant...
Question Do they exist?
3-prismatoids have width at most 3 (exercise).
4-prismatoids have width at most 4 [S.-Stephen-Thomas, 2011].
5-prismatoids of width 6 exist [S., 2012] with 25 vertices [Matschke-S.-Weibel 2013+].
5-prismatoids of arbitrarily large width exist [Matschke-S.-Weibel 2013+].
(54)
So, to disprove the Hirsch Conjecture we only need to find a prismatoid of dimensiond and width larger thand. Its number of vertices and facets is irrelevant...
Question Do they exist?
3-prismatoids have width at most 3 (exercise).
4-prismatoids have width at most 4 [S.-Stephen-Thomas, 2011].
5-prismatoids of width 6 exist [S., 2012] with 25 vertices [Matschke-S.-Weibel 2013+].
5-prismatoids of arbitrarily large width exist [Matschke-S.-Weibel 2013+].
(55)
So, to disprove the Hirsch Conjecture we only need to find a prismatoid of dimensiond and width larger thand. Its number of vertices and facets is irrelevant...
Question Do they exist?
3-prismatoids have width at most 3 (exercise).
4-prismatoids have width at most 4 [S.-Stephen-Thomas, 2011].
5-prismatoids of width 6 exist [S., 2012] with 25 vertices [Matschke-S.-Weibel 2013+].
5-prismatoids of arbitrarily large width exist [Matschke-S.-Weibel 2013+].
(56)
So, to disprove the Hirsch Conjecture we only need to find a prismatoid of dimensiond and width larger thand. Its number of vertices and facets is irrelevant...
Question Do they exist?
3-prismatoids have width at most 3 (exercise).
4-prismatoids have width at most 4 [S.-Stephen-Thomas, 2011].
5-prismatoids of width 6 exist [S., 2012] with 25 vertices [Matschke-S.-Weibel 2013+].
5-prismatoids of arbitrarily large width exist [Matschke-S.-Weibel 2013+].
(57)
So, to disprove the Hirsch Conjecture we only need to find a prismatoid of dimensiond and width larger thand. Its number of vertices and facets is irrelevant...
Question Do they exist?
3-prismatoids have width at most 3 (exercise).
4-prismatoids have width at most 4 [S.-Stephen-Thomas, 2011].
5-prismatoids of width 6 exist [S., 2012] with 25 vertices [Matschke-S.-Weibel 2013+].
5-prismatoids of arbitrarily large width exist [Matschke-S.-Weibel 2013+].
(58)
So, to disprove the Hirsch Conjecture we only need to find a prismatoid of dimensiond and width larger thand. Its number of vertices and facets is irrelevant...
Question Do they exist?
3-prismatoids have width at most 3 (exercise).
4-prismatoids have width at most 4 [S.-Stephen-Thomas, 2011].
5-prismatoids of width 6 exist [S., 2012]with 25 vertices [Matschke-S.-Weibel 2013+].
5-prismatoids of arbitrarily large width exist [Matschke-S.-Weibel 2013+].
(59)
So, to disprove the Hirsch Conjecture we only need to find a prismatoid of dimensiond and width larger thand. Its number of vertices and facets is irrelevant...
Question Do they exist?
3-prismatoids have width at most 3 (exercise).
4-prismatoids have width at most 4 [S.-Stephen-Thomas, 2011].
5-prismatoids of width 6 exist [S., 2012] with 25 vertices [Matschke-S.-Weibel 2013+].
5-prismatoids of arbitrarily large width exist [Matschke-S.-Weibel 2013+].
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So, to disprove the Hirsch Conjecture we only need to find a prismatoid of dimensiond and width larger thand. Its number of vertices and facets is irrelevant...
Question Do they exist?
3-prismatoids have width at most 3 (exercise).
4-prismatoids have width at most 4 [S.-Stephen-Thomas, 2011].
5-prismatoids of width 6 exist [S., 2012] with 25 vertices [Matschke-S.-Weibel 2013+].
5-prismatoids of arbitrarily large width exist [Matschke-S.-Weibel 2013+].
(61)
OK,. . . how do you contruct / visualize / think of a 5-dimensionalprismatoid???
Option 1: If you are a super-hero, use yourX-ray 5-D vision super-powers.
Option 2: If you are a Jedi knight, use the force.
Option 3: If you are a human, use your math. . . and find a way to reduce the dimension of your object.
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OK,. . . how do you contruct / visualize / think of a 5-dimensionalprismatoid???
Option 1: If you are a super-hero, use yourX-ray 5-D vision super-powers.
Option 2: If you are a Jedi knight, use the force.
Option 3: If you are a human, use your math. . . and find a way to reduce the dimension of your object.
(63)
OK,. . . how do you contruct / visualize / think of a 5-dimensionalprismatoid???
Option 1: If you are a super-hero, use yourX-ray 5-D vision super-powers.
Option 2: If you are a Jedi knight, use the force.
Option 3: If you are a human, use your math. . . and find a way to reduce the dimension of your object.
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OK,. . . how do you contruct / visualize / think of a 5-dimensionalprismatoid???
Option 1: If you are a super-hero, use yourX-ray 5-D vision super-powers.
Option 2: If you are a Jedi knight, use the force.
Option 3: If you are a human, use your math. . . and find a way to reduce the dimension of your object.
(65)
OK,. . . how do you contruct / visualize / think of a 5-dimensionalprismatoid???
Option 1: If you are a super-hero, use yourX-ray 5-D vision super-powers.
Option 2: If you are a Jedi knight, use the force.
Option 3: If you are a human, use your math. . . and find a way to reduce the dimension of your object.
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Analyzing the combinatorics of a d-prismatoid Q can be done via an intermediate slice . . .
Q+
Q QH H
Q
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. . . which equals the Minkowski sumQ++Q of the two bases Q+andQ.The normal fan ofQ++Qequals the “superposi- tion” of those ofQ+ andQ.
12
1
2
=
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. . . which equals the Minkowski sumQ++Q of the two bases Q+andQ. The normal fan ofQ++Qequals the “superposi- tion” of those ofQ+ andQ.
12
1
2
=
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So: the combinatorics ofQfollows from the superposition of the normal fans ofQ+ andQ.
Remark
The normal fan of ad−1-polytope can be thought of as a (geodesic, polytopal) cell decomposition (“map”) of the d−2-sphere.
Theorem
Let Q be a d -prismatoid with bases Q+and Qand let G+and Gbe the corresponding maps in the(d−2)-sphere (central projection of the normal fans of Q+and Q). Then, the width of Q equals2plus the minimum number of steps needed to go from a vertex of G+to a vertex of G in (the graph of) the
superposition of the two maps.
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So: the combinatorics ofQfollows from the superposition of the normal fans ofQ+ andQ.
Remark
The normal fan of ad−1-polytope can be thought of as a (geodesic, polytopal) cell decomposition (“map”) of the d−2-sphere.
Theorem
Let Q be a d -prismatoid with bases Q+and Qand let G+and Gbe the corresponding maps in the(d−2)-sphere (central projection of the normal fans of Q+and Q). Then, the width of Q equals2plus the minimum number of steps needed to go from a vertex of G+to a vertex of G in (the graph of) the
superposition of the two maps.
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So: the combinatorics ofQfollows from the superposition of the normal fans ofQ+ andQ.
Remark
The normal fan of ad−1-polytope can be thought of as a (geodesic, polytopal) cell decomposition (“map”) of the d−2-sphere.
Theorem
Let Q be a d -prismatoid with bases Q+and Qand let G+and Gbe the corresponding maps in the(d−2)-sphere (central projection of the normal fans of Q+and Q).Then, the width of Q equals2plus the minimum number of steps needed to go from a vertex of G+to a vertex of G in (the graph of) the
superposition of the two maps.
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So: the combinatorics ofQfollows from the superposition of the normal fans ofQ+ andQ.
Remark
The normal fan of ad−1-polytope can be thought of as a (geodesic, polytopal) cell decomposition (“map”) of the d−2-sphere.
Theorem
Let Q be a d -prismatoid with bases Q+and Qand let G+and Gbe the corresponding maps in the(d−2)-sphere (central projection of the normal fans of Q+and Q). Then, the width of Q equals2plus the minimum number of steps needed to go from a vertex of G+to a vertex of G in (the graph of) the
superposition of the two maps.
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+12
1
2 =
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4-prismatoid of width>4 m
pair of (geodesic, polytopal) maps inS2so that two steps do not let you go from a blue vertex to a red vertex.
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4-prismatoid of width>4 m
pair of (geodesic, polytopal) maps inS2so that two steps do not let you go from a blue vertex to a red vertex.
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Remember that Klee and Walkup, in 1967, disproved the Hirsch conjecture:
Theorem 2(Klee-Walkup 1967)
There is an unbounded 4-polyhedron with 8 facets and diameter 5.
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Remember that Klee and Walkup, in 1967, disproved the Hirsch conjecture:
Theorem 2(Klee-Walkup 1967)
There is an unbounded 4-polyhedron with 8 facets and diameter 5.
(78)
Remember that Klee and Walkup, in 1967, disproved the Hirsch conjecture:
Theorem 2(Klee-Walkup 1967)
There is an unbounded 4-polyhedron with 8 facets and diameter 5.
The Klee-Walkup polytope is an “unbounded 4-spindle”.
What is the corresponding “transversal pair of (geodesic, poly- topal) maps”?
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Remember that Klee and Walkup, in 1967, disproved the Hirsch conjecture:
Theorem 2(Klee-Walkup 1967)
There is an unbounded 4-polyhedron with 8 facets and diameter 5.
The Klee-Walkup polytope is an “unbounded 4-spindle”.
What is the corresponding “transversal pair of (geodesic, poly- topal) maps”?
(80)
(81)
(82)
(83)
(84)
c
(85)
“Non-Hirsch” 4-prismatoids do not exist:
Theorem (S.-Stephen-Thomas, 2011)
In every transversal pair of maps in the sphere there is a path of length two from someblue vertexto somered vertex.
That is to say:
Corollary (S.-Stephen-Thomas, 2011)
Every prismatoid of dimension4has width (at most) four.
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“Non-Hirsch” 4-prismatoids do not exist:
Theorem (S.-Stephen-Thomas, 2011)
In every transversal pair of maps in the sphere there is a path of length two from someblue vertexto somered vertex.
That is to say:
Corollary (S.-Stephen-Thomas, 2011)
Every prismatoid of dimension4has width (at most) four.
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“Non-Hirsch” 4-prismatoids do not exist:
Theorem (S.-Stephen-Thomas, 2011)
In every transversal pair of maps in the sphere there is a path of length two from someblue vertexto somered vertex.
That is to say:
Corollary (S.-Stephen-Thomas, 2011)
Every prismatoid of dimension4has width (at most) four.
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“Non-Hirsch” 4-prismatoids do not exist:
Theorem (S.-Stephen-Thomas, 2011)
In every transversal pair of maps in the sphere there is a path of length two from someblue vertexto somered vertex.
That is to say:
Corollary (S.-Stephen-Thomas, 2011)
Every prismatoid of dimension4has width (at most) four.
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However, we can construct them if we are happy with (infinite, periodic) maps in the plane . . .
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However, we can construct them if we are happy with (infinite, periodic) maps in the plane . . .
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However, we can construct them if we are happy with (infinite, periodic) maps in the plane . . .
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However, we can construct them if we are happy with (infinite, periodic) maps in the plane . . .
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However, we can construct them if we are happy with (infinite, periodic) maps in the plane . . .
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However, we can construct them if we are happy with (infinite, periodic) maps in the plane . . .
. . . or with finite ones in the torus!
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To construct 5-dimensional prismatoids we should look at “pairs of maps” in the 3-sphere.
That is, we want a pair of (geodesic, polytopal) cell
decompositions of the 3-sphere such that if we draw them one on top of the other (common refinement) there is no path of length≤3 from ablue vertexto ared vertex.
Main idea: If non-Hirsch pairs of maps exist in the torus we should have “room enough” to construct it in the 3-sphere as well . . .
(96)
To construct 5-dimensional prismatoids we should look at “pairs of maps” in the 3-sphere.
That is, we want a pair of (geodesic, polytopal) cell
decompositions of the 3-sphere such that if we draw them one on top of the other (common refinement) there is no path of length≤3 from ablue vertexto ared vertex.
Main idea: If non-Hirsch pairs of maps exist in the torus we should have “room enough” to construct it in the 3-sphere as well . . .
(97)
To construct 5-dimensional prismatoids we should look at “pairs of maps” in the 3-sphere.
That is, we want a pair of (geodesic, polytopal) cell
decompositions of the 3-sphere such that if we draw them one on top of the other (common refinement) there is no path of length≤3 from ablue vertexto ared vertex.
Main idea: If non-Hirsch pairs of maps exist in the torus we should have “room enough” to construct it in the 3-sphere as well . . .
(98)
(99)
(100)
(101)
(102)
(103)
(104)
(105)
Theorem (S. 2012)
The following prismatoid Q, of dimension 5 and with 48 vertices, has width six.
(106)
Theorem (S. 2012)
The following prismatoid Q, of dimension 5 and with 48 vertices, has width six.
(107)
Theorem (S. 2012)
The following prismatoid Q, of dimension 5 and with 48 vertices, has width six.
Q:=conv 8
>>
>>
>>
>>
>>
>>
>>
<
>>
>>
>>
>>
>>
>>
>>
: 0 B B B B B B B B B
@
x1 x2 x3 x4 x5
±18 0 0 0 1
0 ±18 0 0 1
0 0 ±45 0 1
0 0 0 ±45 1
±15 ±15 0 0 1
0 0 ±30 ±30 1
0 ±10 ±40 0 1
±10 0 0 ±40 1 1 C C C C C C C C C A
0 B B B B B B B B B
@
x1 x2 x3 x4 x5
0 0 0 ±18 −1
0 0 ±18 0 −1
±45 0 0 0 −1
0 ±45 0 0 −1
0 0 ±15 ±15 −1
±30 ±30 0 0 −1
±40 0 ±10 0 −1
0 ±40 0 ±10 −1
1 C C C C C C C C C A
9
>>
>>
>>
>>
>>
>>
>>
=
>>
>>
>>
>>
>>
>>
>>
;
(108)
Theorem (S. 2012)
The following prismatoid Q, of dimension 5 and with 48 vertices, has width six.
Corollary
There is a 43-dimensional polytope with 86 facets and diameter (at least) 44.
(109)
Proof 1.
It has been verified computationally that the dual graph ofQ (modulo symmetry) has the following structure:
H C
D J
B
A K L
I E
F
G
(110)
Proof 2.
Check that there are noblue vertexaandred vertexbsuch thatais a vertex of theblue cellcontainingbandbis a vertex of thered cellcontaininga.
(111)
With the same ideas
Theorem (Matschke-S.-Weibel, 2013+)
The following5-prismatoid with28vertices (and 274 facets) has width6.
Q:=conv 8
>>
>>
>>
>>
<
>>
>>
>>
>>
: 0 B B B
@
x1 x2 x3 x4 x5
±18 0 0 0 1
0 0 ±30 0 1
0 0 0 ±30 1
0 ±5 0 ±25 1
0 0 ±18 ±18 1
1 C C C A
0 B B B
@
x1 x2 x3 x4 x5
0 0 ±18 0 −1
0 ±30 0 0 −1
±30 0 0 0 −1
±25 0 0 ±5 −1
±18 ±18 0 0 −1 1 C C C A
9
>>
>>
>>
>>
=
>>
>>
>>
>>
;
Corollary
There is a non-Hirsch polytope of dimension23with46facets.
(112)
With the same ideas
Theorem (Matschke-S.-Weibel, 2013+)
The following5-prismatoid with28vertices (and 274 facets) has width6.
Q:=conv 8
>>
>>
>>
>>
<
>>
>>
>>
>>
: 0 B B B
@
x1 x2 x3 x4 x5
±18 0 0 0 1
0 0 ±30 0 1
0 0 0 ±30 1
0 ±5 0 ±25 1
0 0 ±18 ±18 1
1 C C C A
0 B B B
@
x1 x2 x3 x4 x5
0 0 ±18 0 −1
0 ±30 0 0 −1
±30 0 0 0 −1
±25 0 0 ±5 −1
±18 ±18 0 0 −1 1 C C C A
9
>>
>>
>>
>>
=
>>
>>
>>
>>
;
Corollary
There is a non-Hirsch polytope of dimension23with46facets.
(113)
And with some more work:
Theorem (Matschke-Santos-Weibel, 2013+) There is a5-prismatoid with25vertices and of width6.
Corollary
There is a non-Hirsch polytope of dimension20with40facets.
This one has been explicitly computed. It has 36,442 vertices, and diameter 21.
(114)
And with some more work:
Theorem (Matschke-Santos-Weibel, 2013+) There is a5-prismatoid with25vertices and of width6.
Corollary
There is a non-Hirsch polytope of dimension20with40facets.
This one has been explicitly computed. It has 36,442 vertices, and diameter 21.
(115)
And with some more work:
Theorem (Matschke-Santos-Weibel, 2013+) There is a5-prismatoid with25vertices and of width6.
Corollary
There is a non-Hirsch polytope of dimension20with40facets.
This one has been explicitly computed. It has 36,442 vertices, and diameter 21.
(116)
(117)
Theorem (Matschke-Santos-Weibel, 2013+)
There are5-dimensional prismatoids with n vertices and width Ω(√
n).
Sketch of proof
Start with the following “simple, yet more drastic” pair of maps in the torus.
(118)
Theorem (Matschke-Santos-Weibel, 2013+)
There are5-dimensional prismatoids with n vertices and width Ω(√
n).
Sketch of proof
Start with the following “simple, yet more drastic” pair of maps in the torus.
(119)
(120)
(121)
(122)
(123)
(124)
Consider the red and blue maps as lying in two parallel tori in the 3-sphere.
Complete the tori maps to the whole 3-sphere (you need quadratically many cells for that).
Between the two tori you basically get the superposition of the two tori maps.
(125)
Consider the red and blue maps as lying in two parallel tori in the 3-sphere.
Complete the tori maps to the whole 3-sphere (you need quadratically many cells for that).
Between the two tori you basically get the superposition of the two tori maps.
(126)
Consider the red and blue maps as lying in two parallel tori in the 3-sphere.
Complete the tori maps to the whole 3-sphere (you need quadratically many cells for that).
Between the two tori you basically get the superposition of the two tori maps.
(127)
Once we have a non-Hirsch polytope we can derive more via:
1 Products of several copies of it (dimension increases).
2 Gluing several copies of it (dimension is fixed).
To analyze the asymptotics of these operations, we callexcess of ad-polytopePwithnfacets and diameterδthe number
(P) := δ
n−d −1= δ−(n−d) n−d .
E. g.: The excess of our non-Hirsch polytope withn−d =20 and with diameter 21 is
21−20
20 =5%.
(128)
Once we have a non-Hirsch polytope we can derive more via:
1 Products of several copies of it (dimension increases).
2 Gluing several copies of it (dimension is fixed).
To analyze the asymptotics of these operations, we callexcess of ad-polytopePwithnfacets and diameterδthe number
(P) := δ
n−d −1= δ−(n−d) n−d .
E. g.: The excess of our non-Hirsch polytope withn−d =20 and with diameter 21 is
21−20
20 =5%.
(129)
Once we have a non-Hirsch polytope we can derive more via:
1 Products of several copies of it (dimension increases).
2 Gluing several copies of it (dimension is fixed).
To analyze the asymptotics of these operations, we callexcess of ad-polytopePwithnfacets and diameterδthe number
(P) := δ
n−d −1= δ−(n−d) n−d .
E. g.: The excess of our non-Hirsch polytope withn−d =20 and with diameter 21 is
21−20
20 =5%.
(130)
Once we have a non-Hirsch polytope we can derive more via:
1 Products of several copies of it (dimension increases).
2 Gluing several copies of it (dimension is fixed).
To analyze the asymptotics of these operations, we callexcess of ad-polytopePwithnfacets and diameterδthe number
(P) := δ
n−d −1= δ−(n−d) n−d .
E. g.: The excess of our non-Hirsch polytope withn−d =20 and with diameter 21 is
21−20
20 =5%.
(131)
Once we have a non-Hirsch polytope we can derive more via:
1 Products of several copies of it (dimension increases).
2 Gluing several copies of it (dimension is fixed).
To analyze the asymptotics of these operations, we callexcess of ad-polytopePwithnfacets and diameterδthe number
(P) := δ
n−d −1= δ−(n−d) n−d .
E. g.: The excess of our non-Hirsch polytope withn−d =20 and with diameter 21 is
21−20
20 =5%.
(132)
Once we have a non-Hirsch polytope we can derive more via:
1 Products of several copies of it (dimension increases).
2 Gluing several copies of it (dimension is fixed).
To analyze the asymptotics of these operations, we callexcess of ad-polytopePwithnfacets and diameterδthe number
(P) := δ
n−d −1= δ−(n−d) n−d .
E. g.: The excess of our non-Hirsch polytope withn−d =20 and with diameter 21 is
21−20
20 =5%.
(133)
1 Taking products preserves the excess: for eachk ∈N, there is a non-Hirsch polytope of dimension 20k with 40k facets and with excess equal to 0.05=5%.
2 Gluing several copies (slightly) decreases the excess.
(134)
1 Taking products preserves the excess: for eachk ∈N, there is a non-Hirsch polytope of dimension 20k with 40k facets and with excess equal to 0.05=5%.
2 Gluing several copies (slightly) decreases the excess.
(135)
1 Taking products preserves the excess: for eachk ∈N, there is a non-Hirsch polytope of dimension 20k with 40k facets and with excess equal to 0.05=5%.
2 Gluing several copies (slightly) decreases the excess.
(136)
1 Taking products preserves the excess: for eachk ∈N, there is a non-Hirsch polytope of dimension 20k with 40k facets and with excess equal to 0.05=5%.
2 Gluing several copies (slightly) decreases the excess.
nd = (n1+n2d)d = (n1d) + (n2d) δ=δ1+δ21
δ1
n1−d−1= nδ2
2−d−1= ⇒ n−dδ −1=−(n 1
1−d)+(n2−d).
(137)
1 Taking products preserves the excess: for eachk ∈N, there is a non-Hirsch polytope of dimension 20k with 40k facets and with excess equal to 0.05=5%.
2 Gluing several copies (slightly) decreases the excess.
nd = (n1+n2d)d = (n1d) + (n2d) δ=δ1+δ21
δ1
n1−d−1= nδ2
2−d−1= ⇒ n−dδ −1=−(n 1
1−d)+(n2−d).
(138)
1 Taking products preserves the excess: for eachk ∈N, there is a non-Hirsch polytope of dimension 20k with 40k facets and with excess equal to 0.05=5%.
2 Gluing several copies (slightly) decreases the excess.
Corollary
For each k ∈Nthere is an infinite family of non-Hirsch
polytopesof fixed dimension20k and with excess (tending to) 0.05
1− 1
k
.
(139)
But we know there are “worst” prismatoids: 5-prismatoids of arbitrarily large width.Will those produce non-Hirsch polytopes with worst excess?
To analyze the asymptotics of this, let us callexcessof a prismatoid of widthδwithnvertices and dimensiond the quantity
δ−d n−d
(140)
But we know there are “worst” prismatoids: 5-prismatoids of arbitrarily large width. Will those produce non-Hirsch polytopes with worst excess?
To analyze the asymptotics of this, let us callexcessof a prismatoid of widthδwithnvertices and dimensiond the quantity
δ−d n−d
(141)
But we know there are “worst” prismatoids: 5-prismatoids of arbitrarily large width. Will those produce non-Hirsch polytopes with worst excess?
To analyze the asymptotics of this, let us callexcessof a prismatoid of widthδwithnvertices and dimensiond the quantity
δ−d n−d
(142)
But we know there are “worst” prismatoids: 5-prismatoids of arbitrarily large width. Will those produce non-Hirsch polytopes with worst excess?
To analyze the asymptotics of this, let us callexcessof a prismatoid of widthδwithnvertices and dimensiond the quantity
δ−d n−d
(143)
Via the strong d -step Theorem, a prismatoid of a certain excess produces non-Hirsch polytopes of that same excess.
Proof.
The dimension, number of facets and diameter of the
non-Hirsch polytope produced by the strongd-step Theorem are
n−d, 2(n−d), δ+ (n−2d).
So, its excess is
δ+ (n−2d)−(n−d)
n−d = δ−d
n−d.
(144)
Via the strong d -step Theorem, a prismatoid of a certain excess produces non-Hirsch polytopes of that same excess.
Proof.
The dimension, number of facets and diameter of the
non-Hirsch polytope produced by the strongd-step Theorem are
n−d, 2(n−d), δ+ (n−2d).
So, its excess is
δ+ (n−2d)−(n−d)
n−d = δ−d
n−d.
(145)
Via the strong d -step Theorem, a prismatoid of a certain excess produces non-Hirsch polytopes of that same excess.
Proof.
The dimension, number of facets and diameter of the
non-Hirsch polytope produced by the strongd-step Theorem are
n−d, 2(n−d), δ+ (n−2d).
So, its excess is
δ+ (n−2d)−(n−d)
n−d = δ−d
n−d.
(146)
In dimension 5, we know how to construct polytopes of arbitrarily large widthδ ∼√
n. . . but their excess tends to zero:
limδ−5 n−5 =lim
√n−5 n−5 =0.
Let us be optimistic and suppose that we could construct 5-prismatoids withnvertices and linear width'αn.
Their excess will now tend toα. So, we still get only polytopes that violate Hirsch by a constant (“linear” Hirsch bound).
OK, let us try to bemoreoptimistic.
Can we hope for prismatoids of width greater than linear in their number of vertices?
(147)
In dimension 5, we know how to construct polytopes of arbitrarily large widthδ ∼√
n. . . but their excess tends to zero:
limδ−5 n−5 =lim
√n−5 n−5 =0.
Let us be optimistic and suppose that we could construct 5-prismatoids withnvertices and linear width'αn.
Their excess will now tend toα. So, we still get only polytopes that violate Hirsch by a constant (“linear” Hirsch bound).
OK, let us try to bemoreoptimistic.
Can we hope for prismatoids of width greater than linear in their number of vertices?
(148)
In dimension 5, we know how to construct polytopes of arbitrarily large widthδ ∼√
n. . . but their excess tends to zero:
limδ−5 n−5 =lim
√n−5 n−5 =0.
Let us be optimistic and suppose that we could construct 5-prismatoids withnvertices and linear width'αn.
Their excess will now tend toα. So, we still get only polytopes that violate Hirsch by a constant (“linear” Hirsch bound).
OK, let us try to bemoreoptimistic.
Can we hope for prismatoids of width greater than linear in their number of vertices?
(149)
In dimension 5, we know how to construct polytopes of arbitrarily large widthδ ∼√
n. . . but their excess tends to zero:
limδ−5 n−5 =lim
√n−5 n−5 =0.
Let us be optimistic and suppose that we could construct 5-prismatoids withnvertices and linear width'αn.
Their excess will now tend toα.So, we still get only polytopes that violate Hirsch by a constant (“linear” Hirsch bound).
OK, let us try to bemoreoptimistic.
Can we hope for prismatoids of width greater than linear in their number of vertices?
(150)
In dimension 5, we know how to construct polytopes of arbitrarily large widthδ ∼√
n. . . but their excess tends to zero:
limδ−5 n−5 =lim
√n−5 n−5 =0.
Let us be optimistic and suppose that we could construct 5-prismatoids withnvertices and linear width'αn.
Their excess will now tend toα. So, we still get only polytopes that violate Hirsch by a constant (“linear” Hirsch bound).
OK, let us try to bemoreoptimistic.
Can we hope for prismatoids of width greater than linear in their number of vertices?
(151)
In dimension 5, we know how to construct polytopes of arbitrarily large widthδ ∼√
n. . . but their excess tends to zero:
limδ−5 n−5 =lim
√n−5 n−5 =0.
Let us be optimistic and suppose that we could construct 5-prismatoids withnvertices and linear width'αn.
Their excess will now tend toα. So, we still get only polytopes that violate Hirsch by a constant (“linear” Hirsch bound).
OK, let us try to bemoreoptimistic.
Can we hope for prismatoids of width greater than linear in their number of vertices?
In this paper some characterizations of best approximation have been established in terms of 2-semi inner products and normalised duality mapping associated with a linear 2-normed
Theorem 1.3 (Theorem 12.2).. Con- sequently the operator is normally solvable by virtue of Theorem 1.5 and dimker = n. From the equality = I , by virtue of Theorem 1.7 it
In Section 4 we present conditions upon the size of the uncertainties appearing in a flexible system of linear equations that guarantee that an admissible solution is produced
In Section 3, we study the determining number of trees, providing a linear time algorithm for computing minimum determining sets.. We also show that there exist trees for which
Oscillatory Integrals, Weighted and Mixed Norm Inequalities, Global Smoothing and Decay, Time-dependent Schr¨ odinger Equation, Bessel functions, Weighted inter- polation
In the study of dynamic equations on time scales we deal with certain dynamic inequalities which provide explicit bounds on the unknown functions and their derivatives.. Most of
Computation of Nambu-Poisson cohomology of type (I) In this subsection, we confine ourselves to nondegenerate linear Nambu- Poisson tensors of type (I).. We get the following results
A monotone iteration scheme for traveling waves based on ordered upper and lower solutions is derived for a class of nonlocal dispersal system with delay.. Such system can be used | 13,628 | 42,911 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-10 | latest | en | 0.710764 |
https://www.dezyre.com/recipes/implement-voting-ensemble-in-python | 1,627,377,561,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153223.30/warc/CC-MAIN-20210727072531-20210727102531-00128.warc.gz | 696,596,650 | 25,655 | How to implement voting ensemble in Python?
MACHINE LEARNING RECIPES DATA CLEANING PYTHON DATA MUNGING PANDAS CHEATSHEET ALL TAGS
How to implement voting ensemble in Python?
This recipe helps you implement voting ensemble in Python
Recipe Objective
How do you select which model to use for a dataset. We can do this by voting ensemble which trains on an ensemble of numerous models and predicts an output (class) based on their highest probability of chosen class as the output
So this is the recipe on how we can implement voting ensemble in Python.
Step 1 - Import the library
``` from sklearn import model_selection from sklearn.linear_model import LogisticRegression from sklearn.tree import DecisionTreeClassifier from sklearn.svm import SVC from sklearn.ensemble import VotingClassifier from sklearn import datasets from sklearn.model_selection import train_test_split import matplotlib.pyplot as plt plt.style.use("ggplot") ```
We have imported various models like LogisticRegression, DecisionTreeClassifier, SVC and VotingClassifier.
Step 2 - Setting up the Data
We have imported Wine dataset and stored the data in X and the target in y. We have used test_train_split to split the data. We have also used model_selection.KFold to split the data. ``` seed = 42 dataset = datasets.load_wine() X = dataset.data; y = dataset.target X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.30) kfold = model_selection.KFold(n_splits=10, random_state=seed) ```
Step 3 - Selecting model by Voting Classifier
We have made an array named estimators with all the models from e=which we want to select. Now we have used VotingClassifier with parameter as extimator which contain all the models. Finally we have calculated cross validation score of the model. ``` estimators = [] model1 = LogisticRegression(); estimators.append(("logistic", model1)) model2 = DecisionTreeClassifier(); estimators.append(("cart", model2)) model3 = SVC(); estimators.append(("svm", model3)) ensemble = VotingClassifier(estimators) results = model_selection.cross_val_score(ensemble, X_train, y_train, cv=kfold) print(results.mean()) ``` So the output comes as
```0.9102564102564104
```
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https://math.stackexchange.com/questions/2205096/given-is-a-linear-mapping-and-a-basis-determine-the-transformation-matrix | 1,579,619,822,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604397.40/warc/CC-MAIN-20200121132900-20200121161900-00470.warc.gz | 561,339,688 | 31,484 | Given is a linear mapping and a basis. Determine the transformation matrix
Given is $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ with $f:(x,y,z) \rightarrow (x+2y+z, y+z, -x+3y+4z)$. Determine the transformation matrix in terms of the basis $B= \left\{\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix},\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}\right\}$
$f(\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix})= \begin{pmatrix} 1\\ 0\\ -1 \end{pmatrix}= 0 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 0 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}- 1 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$
$f(\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix})= \begin{pmatrix} 3\\ 2\\ 7 \end{pmatrix}= 8 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 2 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}+5 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$
$f(\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix})= \begin{pmatrix} 0\\ 1\\ 5 \end{pmatrix}= 4 \cdot \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix} + 1 \cdot \begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}+ 4 \cdot \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$
I'm not sure how to read the transformation matrix now. Either I will read it correctly or I will accidentally read its transposition :s
$T=\begin{pmatrix} 0 & 8 & 4\\ 0 & 2 & 1\\ -1 & 5 & 4 \end{pmatrix}$
Did I do it all correctly? If it's alright, there are maybe faster ways doing this?
• What would you like the representing matrix of your linear transformation to do? I'd say -- agree with $f$ on the basis vectors. Does this help you to choose between $T$ and its transpose? – uniquesolution Mar 27 '17 at 10:41
• @uniquesolution I'm not sure if I understood you correctly but it seems like it's fine like that? – cnmesr Mar 27 '17 at 10:52
• You don't seem to have decided whether $\mathbb{R}^3$ is the set of row vectors or the set of column vectors. That may generate confusion. – ancientmathematician Mar 27 '17 at 11:14
• Looks right, but the notation of the matrix should be $[T]_B$. – Itay4 Mar 27 '17 at 11:19
Perhaps a more straightforward way of computing the matrix is by using the relation $$[T]_\mathcal{B} = P^{-1} T P,$$ where $[T]_\mathcal{B}$ is the matrix that you computed, $$P = (b_1\quad b_2\quad b_3) = \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix},$$ and $$T = \begin{pmatrix} f(e_1) & f(e_2) & f(e_3) \end{pmatrix} = \begin{pmatrix}1 & 2 & 1\\ 0 & 1 & 1 \\ -1 & 3 & 4 \end{pmatrix},$$ where $e_1, e_2, e_3$ are the standard basis vectors. With this approach, all you have to do is invert $P$, which is easy, and then multiply the matrices. (check to see that it yields the same result!) | 1,012 | 2,636 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-05 | latest | en | 0.572142 |
https://math.stackexchange.com/questions/2590995/finding-the-periods-of-miniships-in-the-burning-ship | 1,696,270,666,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511002.91/warc/CC-MAIN-20231002164819-20231002194819-00523.warc.gz | 432,172,309 | 35,284 | # finding the periods of miniships in the Burning Ship
The Burning Ship fractal is similar to the Mandelbrot set, only instead of being defined by a complex quadratic polynomial, it can defined by two real functions: \begin{aligned}X &\leftarrow X^2 - Y^2 + A \\ Y &\leftarrow 2|XY| + B\end{aligned}
Similarly to the Mandelbrot set, which contains smaller copies of itself adorned with filaments, the Burning Ship fractal contains smaller miniships. These occur due to renormalization, where a periodic cycle exhibits dynamics over one period similar to the main body of the set over one iteration.
The question is, given a simple region in the complex plane (something like a square or a circle), what is the period of the lowest-period miniship contained within the region?
Bonus points for methods that are applicable to a more general class of fractals, not just the quadratic Burning Ship.
For the quadratic Mandelbrot set I know two methods, one using the Jordan curve theorem iterating the corners of a polygon until it surrounds the origin, the first iteration at which that happens is the period; the other uses Taylor series ball arithmetic, which I don't fully understand, but is better than naive ball arithmetic. Implementations can be found here: box, ball. I believe neither can be applied to the Burning Ship because its functions aren't well behaved enough.
I have attempted to use ball arithmetic (midpoint-radius intervals, one real ball each for $X$ and $Y$) to solve this problem, but failed miserably - it more easily finds nearby "outer" miniships outside the region visible at shallower zoom levels than "inner" ones within the region at deeper zoom levels. One has to zoom in a great deal further to reduce the region of interest, before it finds the desired period. Perhaps my lack of special treatment of $|\cdot|$ and $\cdot^2$ was to blame? (I followed the simple abs from arblib, and used regular multiplication, which is also used in arblib as far as I can tell.)
I have also attempted to use the box method, simply continuing in the cases where the polygon gets folded when it crosses an axis (this may lead to a higher or lower period than the true one being found by the algorithm). It seems to work a lot better than the ball arithmetic, though I'm concerned about the folding...
My end goal is an implementation in a fractal browser for accelerating interesting zoom navigation techniques; I already have implemented Newton's root finding method and miniship size estimates, both of which need the period as input.
This is indeed true, so long as your mapping is a continuous mapping; this preserves the topology of the closed loop (i.e. polygon) and ensures that you'll surround the forward-image of the origin if and only if you've iterated $p$ times where $p$ is the period. Burning Ship is defined in terms of the absolute value $f=|x|$ as well as several more familiar continuous functions, so the whole thing is continuous.
So the Jordan curve theorem method works for continuous formulas, with the caveat that folding and other non-linearities may need the polygon to be subdivided (for example when an edge crosses an axis and would be folded/bent by $|\cdot|$) so that the topology remains the same. | 703 | 3,253 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-40 | latest | en | 0.934911 |
https://www.math-linux.com/latex/faq/latex-faq/article/greater-than-or-similar-to-symbol-in-latex | 1,716,783,327,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059028.82/warc/CC-MAIN-20240527021852-20240527051852-00181.warc.gz | 741,016,408 | 8,181 | In mathematics, the greater than or similar to symbol is used to represent a relation between two quantities. In LaTeX, this symbol can be represented using the \gtrsim command.
## Greater Than or Similar To Symbol in LaTeX
In mathematics, the greater than or similar to symbol is used to represent a relation between two quantities. In LaTeX, this symbol can be represented using the \gtrsim command.
## Using the \gtrsim command
To write the greater than or similar to symbol in LaTeX, use the \gtrsim command. For example:
$x \gtrsim y$
This represents the relation “x is greater than or similar to y”.
It’s as simple as that! | 143 | 636 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-22 | latest | en | 0.938944 |
https://www.coursehero.com/file/6595824/hw7-solutions/ | 1,498,600,168,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321553.70/warc/CC-MAIN-20170627203405-20170627223405-00381.warc.gz | 849,941,707 | 24,842 | hw7_solutions
# hw7_solutions - EE263 Summer 2010-11 Laurent Lessard EE263...
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Unformatted text preview: EE263 Summer 2010-11 Laurent Lessard EE263 homework 7 1. Some true/false questions. Determine if the following statements are true or false. For each statement, either provide a proof that it is always true, or a counterexample demonstrating that it may fail. You can’t assume anything about the dimensions of the matrices (unless it’s explicitly stated), but you can assume that the dimensions are such that all expressions make sense. For example, the statement “ A + B = B + A ” is true, because no matter what the dimensions of A and B are (they must, however, be the same), and no matter what values A and B have, the statement holds. As another example, the statement A 2 = A is false, because it fails for the matrix bracketleftbig 2 bracketrightbig . There are also matrices for which it does hold, e.g. , an identity matrix. But that doesn’t make the statement true. (a) If A ∈ R 3 × 3 satisfies A + A T = 0, then A is singular. Solution. True . A general 3 × 3 skew symmetric matrix ( i.e. , one that satisfies A T = − A ) has the form A = a b − a c − b − c . Evidently we have Ax = 0, with x = ( c, − b,a ). Alternatively, one can compute the determinant explicitly and show that it is identically zero. (b) If A k = 0 for some integer k ≥ 1, then I − A is nonsingular. Solution. True . We first observe that all eigenvalues of A must be zero. The eigenvalues of I − A are each one minus an eigevalue of A , i.e. , they are all equal to one. In particular, 0 is not an eigenvalue of I − A , so it is nonsingular. (c) If A,B ∈ R n × n are both diagonalizable, then AB is diagonalizable. Solution. False . Consider A = bracketleftbigg 1 1 − 1 1 bracketrightbigg and B = bracketleftbigg 1 / 2 1 / 2 1 bracketrightbigg . Clearly both A and B are diagonalizable, but AB = bracketleftbigg 1 1 1 bracketrightbigg is not diagonalizable. (d) If A,B ∈ R n × n , then every eigenvalue of AB is an eigenvalue of BA . Solution. True . Take any eigenvalue λ of AB , and let v be an eigenvector, i.e. , ABv = λv . Suppose λ negationslash = 0, then BA ( Bv ) = B ( ABv ) = λ ( Bv ) . Since Bv negationslash = 0 (otherwise λ = 0), Bv is an eigenvector of BA associated with the eigenvalue λ . Now suppose λ = 0 and we need to show that BA is also singular. Suppose BA is nonsingular, then both A and B is full rank. But this will imply that Bv = 0 and thus v = 0, a contradiction. (e) If A,B ∈ R n × n , then every eigenvector of AB is an eigenvector of BA . 1 Solution. False. Consider A = bracketleftbigg 1 1 1 1 bracketrightbigg and B = bracketleftbigg 1 1 1 bracketrightbigg . Then AB = bracketleftbigg 1 2 1 2 bracketrightbigg has (1 , 1) as an eigenvector, which is clearly not an eigenvector of BA = bracketleftbigg 2 2 1 1 bracketrightbigg ....
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hw7_solutions - EE263 Summer 2010-11 Laurent Lessard EE263...
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Ask a homework question - tutors are online | 965 | 3,411 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2017-26 | longest | en | 0.839852 |
https://uk.style.yahoo.com/times-table-not-magic-memorization-224515240.html | 1,719,103,310,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862425.28/warc/CC-MAIN-20240623001858-20240623031858-00748.warc.gz | 536,343,219 | 162,863 | # How to Use a Times Table (It's Not Magic, It's Memorization)
OK, here's the thing: Math isn't inherently miserable. Multiplication is kind of fun, and a multiplication table can help make it easier to learn. Whether you're a student, a teacher, or just someone fascinated by numbers, the times table has something for everyone.
## Understanding the Basics of the Times Table
Multiplication is the art of adding groups of things efficiently.
Imagine having three baskets of apples, with each basket holding four apples. Instead of counting each apple, you'd simply multiply 3 (baskets) by 4 (apples per basket) to get 12 (total apples). Not only is this method faster, but it also helps in understanding the concept of multiplication.
Now, to aid in this process, we use a multiplication chart known as the times table. This chart isn't just about multiplying two numbers; it's about recognizing patterns, using skip counting and memorizing for good practice.
For instance, pretend you want to multiply 7 by 6. Using the times table, you'd find 7 in a top row and 6 in a left column. Draw an imaginary line down from 7 and another one across from 6. Where they meet, you'd find your answer: 42.
## How to Efficiently Use the Table
Using the multiplication tables might seem like a puzzle at first, but with some good practice and patience, the times table becomes a handy tool for all your multiplication needs.
1. Finding patterns: Recognize patterns as you move through the rows and columns. For example, when you look at the multiples of 2, you'll notice they're all even numbers. This understanding of the pattern will make memorizing easier.
2. Memorizing chunks: Instead of trying to cram the entire multiplication table at once, break it into chunks. Start with the 1-12 tables, for instance. Take the "2 row," go across the numbers, and repeat the multiplication facts. "2 x 0 = 0, 2 x 1 = 2, 2 x 2 = 4..." and so on.
3. Skip counting: This is a method where you count by a particular number. For example, if you're learning the multiples of 3, you'd say, "3, 6, 9, 12..." This helps in memorizing the multiplication chart more effectively.
## The Advantages of Mastering the Times Table
While the process of learning and memorizing the times table might seem tedious to some, its benefits extend far beyond simple arithmetic. Here's why mastering the multiplication chart is invaluable.
1. Quick mental calculations: In everyday life, we constantly encounter situations where basic multiplication is required, be it while shopping, cooking or planning an event. Having the times table memorized means you can make these calculations rapidly without resorting to a calculator.
2. Foundation for advanced math: Understanding multiplication is fundamental to progressing in more complex mathematical areas such as algebra, calculus and even geometry. A strong grasp of the times table provides a sturdy foundation for these subjects.
3. Trains your brain: The process of memorizing and recalling multiplication facts sharpens the brain and enhances memory skills. This cognitive exercise can aid in other areas of learning and daily tasks.
4. Increased confidence in math: A student who has the multiplication chart at their fingertips often feels more confident in tackling math problems. This confidence can lead to a more positive attitude towards math and learning in general.
5. Real-world applications: From understanding interest rates to planning budgets, multiplication plays a role in numerous real-world scenarios. A strong grasp of the times table ensures you're equipped to handle these situations with ease.
This article was updated in conjunction with AI technology, then fact-checked and edited by a HowStuffWorks editor.
Now That's Ancient
You're not the first person to use a multiplication table; they've been around for thousands of years! The Babylonians created times tables in clay tablets to help them do quick calculations when building new structures, or when trading or selling.
Original article: How to Use a Times Table (It's Not Magic, It's Memorization)
Copyright © 2023 HowStuffWorks, a division of InfoSpace Holdings, LLC, a System1 Company | 881 | 4,203 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-26 | latest | en | 0.93953 |
http://www.nag.com/numeric/fl/manual/examples/source/f08ylfe.f | 1,410,978,995,000,000,000 | text/plain | crawl-data/CC-MAIN-2014-41/segments/1410657124236.90/warc/CC-MAIN-20140914011204-00146-ip-10-196-40-205.us-west-1.compute.internal.warc.gz | 693,826,476 | 1,694 | * F08YLF Example Program Text * Mark 21 Release. NAG Copyright 2004. * .. Parameters .. INTEGER NIN, NOUT PARAMETER (NIN=5,NOUT=6) INTEGER NMAX PARAMETER (NMAX=8) INTEGER LDS, LDT, LDVL, LDVR, LWORK PARAMETER (LDS=NMAX,LDT=NMAX,LDVL=NMAX,LDVR=NMAX, + LWORK=2*NMAX*(NMAX+2)+16) * .. Local Scalars .. DOUBLE PRECISION EPS, SNORM, STNRM, TNORM INTEGER I, INFO, J, M, N * .. Local Arrays .. DOUBLE PRECISION DIF(NMAX), S(LDS,NMAX), SCON(NMAX), T(LDT,NMAX), + VL(LDVL,NMAX), VR(LDVR,NMAX), WORK(LWORK) INTEGER IWORK(NMAX+6) LOGICAL SELECT(1) * .. External Functions .. DOUBLE PRECISION F06BNF, F06RAF, X02AJF EXTERNAL F06BNF, F06RAF, X02AJF * .. External Subroutines .. EXTERNAL DTGEVC, DTGSNA * .. Executable Statements .. WRITE (NOUT,*) 'F08YLF Example Program Results' WRITE (NOUT,*) * Skip heading in data file READ (NIN,*) READ (NIN,*) N IF (N.LE.NMAX) THEN * * Read S and T from data file * READ (NIN,*) ((S(I,J),J=1,N),I=1,N) READ (NIN,*) ((T(I,J),J=1,N),I=1,N) * * Calculate the left and right generalized eigenvectors of the * pair (S,T). Note that DTGEVC requires WORK to be of dimension * at least 6*N, and LWORK is always greater than 6*NMAX. * CALL DTGEVC('Both','All',SELECT,N,S,LDS,T,LDT,VL,LDVL,VR,LDVR, + N,M,WORK,INFO) IF (INFO.GT.0) THEN WRITE (NOUT,99999) INFO, INFO + 1 STOP END IF * * Estimate condition numbers for all the generalized eigenvalues * and right eigenvectors of the pair (S,T) * CALL DTGSNA('Both','All',SELECT,N,S,LDS,T,LDT,VL,LDVL,VR,LDVR, + SCON,DIF,N,M,WORK,LWORK,IWORK,INFO) * * Print condition numbers of eigenvalues and right eigenvectors * WRITE (NOUT,*) 'SCON' WRITE (NOUT,99998) (SCON(I),I=1,M) WRITE (NOUT,*) WRITE (NOUT,*) 'DIF' WRITE (NOUT,99998) (DIF(I),I=1,M) * * Calculate approximate error estimates * * Compute the 1-norms of S and T and then compute * SQRT(SNORM**2 + TNORM**2) * EPS = X02AJF() SNORM = F06RAF('1-norm',N,N,S,LDS,WORK) TNORM = F06RAF('1-norm',N,N,T,LDT,WORK) STNRM = F06BNF(SNORM,TNORM) WRITE (NOUT,*) WRITE (NOUT,*) + 'Approximate error estimates for eigenvalues of (S,T)' WRITE (NOUT,99998) (EPS*STNRM/SCON(I),I=1,M) WRITE (NOUT,*) WRITE (NOUT,*) 'Approximate error estimates for right ', + 'eigenvectors of (S,T)' WRITE (NOUT,99998) (EPS*STNRM/DIF(I),I=1,M) ELSE WRITE (NOUT,*) 'NMAX too small' END IF STOP * 99999 FORMAT (' The 2-by-2 block (',I5,':',I5,') does not have a',' co', + 'mplex eigenvalue') 99998 FORMAT ((3X,1P,7E11.1)) END | 890 | 2,406 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2014-41 | latest | en | 0.550817 |
http://unapologetic.wordpress.com/2008/07/24/the-inclusion-exclusion-principle/?like=1&_wpnonce=cb13afa090 | 1,386,786,058,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386164041513/warc/CC-MAIN-20131204133401-00055-ip-10-33-133-15.ec2.internal.warc.gz | 185,363,357 | 25,178 | # The Unapologetic Mathematician
## The Inclusion-Exclusion Principle
In combinatorics we have a method of determining the cardinality of unions in terms of the sets in question and their intersection: the Inclusion-Exclusion Principle. Predictably enough, this formula is reflected in the subspaces of a vector space. We could argue directly in terms of bases, but it’s much more interesting to use the linear algebra we have at hand.
Let’s start with just two subspaces $V$ and $W$ of some larger vector space. We’ll never really need that space, so we don’t need to give it a name. The thing to remember is that $V$ and $W$ might have a nontrivial intersection — their sum may not be direct.
Now we consider the sum of the subspaces. Every vector in $V+W$ is the sum of a vector in $V$ and a vector in $W$. This tells us that there’s a linear map $f_1:V\oplus W\rightarrow V+W$ defined by $f_1(v,w)=v+w$. Note carefully that this is linear as a function of the pair $(v,w)$. It’s not bilinear — linear in each of $v$ and $w$ separately — which would mean bringing in the tensor product.
This function is always surjective, but it may fail to be injective. Specifically, if $V$ and $W$ have a nontrivial intersection then it gives us a nontrivial kernel. If $x\in V\cap W$ then $f_1(x,-x)=0$. But rather than talk about elements, let’s use the intersection to parametrize the kernel! We’ve got another linear map $f_2:V\cap W\rightarrow V\oplus W$, defined by $f_2(x)=(x,-x)$.
Now it’s clear that $f_2$ has a trivial kernel. On the other hand, its image is precisely the set of pairs that $f_1$ kills off. That means we’ve got an exact sequence:
$\mathbf{0}\rightarrow V\cap W\xrightarrow{f_2}V\oplus W\xrightarrow{f_1}V+W\rightarrow\mathbf{0}$
Taking the Euler characteristic we find that
$\dim\left(V\cap W\right)-\dim\left(V\oplus W\right)+\dim\left(V+W\right)=0$
Some simple juggling turns this into
$\dim\left(V+W\right)=\dim\left(V\right)+\dim\left(W\right)-\dim\left(V\cap W\right)$
which is the inclusion-exclusion principle for two subspaces.
What about three subspaces $U$, $V$, and $W$? We start off again with
$U\oplus V\oplus W\xrightarrow{f_1}U+V+W\rightarrow\mathbf{0}$
defined by $f_1(u,v,w)=u+v+w$. Then the kernel contains triples like $(0,x,-x)$, $(-y,0,y)$, and $(z,-z,0)$. We use the pairwise intersections to cover the kernel by the map
$\left(V\cap W\right)\oplus\left(W\cap U\right)\oplus\left(U\cap V\right)\xrightarrow{f_2}U\oplus V\oplus W$
defined by $(x,y,z)=(z-y,x-z,y-x)$. The image of $f_2$ is exactly the kernel of $f_1$, but this time it has a nontrivial kernel itself! Now we need another map
$\mathbf{0}\rightarrow U\cap V\cap W\xrightarrow{f_3}\left(V\cap W\right)\oplus\left(W\cap U\right)\oplus\left(U\cap V\right)$
defined by $f_3(t)=(t,t,t)$. Now $f_3$ is injective, and its image is exactly the kernel of $f_2$. We stick these maps all together to have the long exact sequence
\begin{aligned}\mathbf{0}\rightarrow U\cap V\cap W\xrightarrow{f_3}\left(V\cap W\right)\oplus\left(W\cap U\right)\oplus\left(U\cap V\right)\\\xrightarrow{f_2}U\oplus V\oplus W\xrightarrow{f_1}U+V+W\rightarrow\mathbf{0}\end{aligned}
Taking the Euler characteristic and juggling again we find
\begin{aligned}\dim\left(U+V+W\right)=\dim\left(U\right)+\dim\left(V\right)+\dim\left(W\right)\\-\dim\left(U\cap V\right)-\dim\left(U\cap W\right)-\dim\left(V\cap W\right)\\+\dim\left(U\cap V\cap W\right)\end{aligned}
We include the dimensions of the individual subspaces, exclude the dimensions of the pairwise intersections, and include back in the dimension of the triple intersection.
For larger numbers of subspaces we can construct longer and longer exact sequences. The $k$th term consists of the direct sum of the intersections of the subspaces, taken $k$ at a time. The $k$th map consists of including each $k$-fold intersection into each of the $(k-1)$-fold intersections it lies in, with positive and negative signs judiciously sprinkled throughout to make the sequence exact.
The result is that the dimension of a sum of subspaces is the alternating sum of the dimensions of the $k$-fold intersections. First add the dimension of each subspace, then subtract the dimension of each pairwise intersection, then add back in the dimension of each triple intersection, and so on.
From here it’s the work of a moment to derive the combinatorial inclusion-exclusion principle. Given a collection of sets, just construct the free vector space on each of them. These free vector spaces will have nontrivial intersections corresponding to the nonempty intersections of the sets, and we’ve got canonical basis elements floating around to work with. Then all these dimensions we’re talking about are cardinalities of various subsets of the sets we started with, and the combinatorial inclusion-exclusion principle follows. Of course, as I said before we could have proved it from the other side, but that would require a lot of messy hands-on work with bases and such.
The upshot is that the combinatorial inclusion-exclusion principle is equivalent to the statement that exact sequences of vector spaces have trivial Euler characteristic! This little formula that we teach in primary or secondary school turns out to be intimately connected with one of the fundamental tools of homological algebra and the abstract approach to linear algebra. Neat!
July 24, 2008 - Posted by | Algebra, Linear Algebra
1. Beyond neat! Satisfies the true purpose of Mathematics: enlightenment.
Comment by Jonathan Vos Post | July 24, 2008 | Reply
2. What is also good about this observation is that this form of inclusion-exclusion holds in lattices more general than distributive lattices (which were the types of lattice emphasized in the post you linked to) — subspace lattices satisfy the weaker property of modularity.
Comment by Todd Trimble | July 24, 2008 | Reply
3. Indeed they do. That’s why I said as much last time when I talked about subspace sums.
Comment by John Armstrong | July 24, 2008 | Reply
4. Sorry, I don’t know sequences, but it seems either you have a major error or I completely don’t understand your post.
dim(U+V+W)=dim(U)+dim(V)+dim(W)-dim(UV)-dim(UW)-dim(WV)+dim(UVW)
Consider R^2 as a vector space over R and let:
U={(x,y):x=0}
V={(x,y):y=0}
W={(x,y):x=y}
These are clearly subspaces. U+V=R^2 and their intersection is {0}. Same for U,W and V,W.
So the left side is dim(U+V+W)=dim(R^2)=2 and the right side is dim(U)+dim(V)+dim(W)=3 because the intersections vanish.
The problem is that in general (U+V) intersection W is not equal to (U intersection W) + (V intersection W). Take these 3 mentioned spaces as a counterexample. Once my class received a task to determine whether PIE is valid for n vector spaces. Of course everybody gave a “proof” of it.:)
Best, –k.g.
(Feel free to add TeX or correct my English; don’t know how to use it)
Comment by k.g. | July 26, 2008 | Reply
5. You’re right.. I’d implicitly assumed that the subspaces were in what we topologists like to call “general position”. I can see two remedies.
First, turn your reason why my proof failed into an algebraic statement about subspaces and their maps, then establish a new sequence (or show how to modify existing ones) to handle this degeneracy.
Alternately, scrap the given sequences altogether and give new ones which handle the degenerate case as well as the generic.
I’m not in a position to type either solution out myself on my iPhone from a restaurant in Meriden, MS. But good catch there.
Comment by John Armstrong | July 27, 2008 | Reply
• Dear Professor John:
I am very eager to know what you mean by “general position”,
Can you email me the answer through Email yangming198411@163.com ?
Comment by mingming | March 10, 2010 | Reply
• Essentially, if you wiggle the subspaces a bit, the dimensions of their intersections don’t change.
Comment by John Armstrong | March 10, 2010 | Reply
6. [...] inclusion-exclusion principle tells us [...]
Pingback by The Multiplicity of an Eigenvalue « The Unapologetic Mathematician | February 19, 2009 | Reply
7. [...] inclusion-exclusion principle tells us [...]
Pingback by The Multiplicity of an Eigenpair « The Unapologetic Mathematician | April 8, 2009 | Reply
8. You learned the inclusion-exclusion principle in primary school? I don’t think anybody I know who didn’t do math contests in high school has the faintest idea what it is.
Anyway, I’ve always found the use of the free vector space construction in combinatorics extremely mysterious. For example, in poset theory one can use certain linear operators to show that certain graded posets are rank-unimodal and Sperner, and some of the consequences of this method don’t have known “direct” proofs. So does the usefulness of linear algebra in combinatorics ultimately stem from categorical and “topological” properties of the category of vector spaces? What makes this category unique for combinatorial purposes?
Comment by Qiaochu Yuan | June 7, 2009 | Reply
9. Sure. It’s part of the remnants of the “New Math” hanging around the curriculum.
Comment by John Armstrong | June 7, 2009 | Reply
10. [...] have a nonempty intersection, which leads us to think that maybe this has something to do with the inclusion-exclusion principle. We’re “overcounting” the intersection by just adding, so let’s subtract it [...]
Pingback by Characteristic Functions as Idempotents « The Unapologetic Mathematician | December 23, 2009 | Reply
11. [...] little more about characteristic functions, let’s see how they can be used to understand the inclusion-exclusion principle. Our first pass was through a very categorified lens, which was a neat tie-in to Euler [...]
Pingback by Inclusion-Exclusion Again « The Unapologetic Mathematician | December 28, 2009 | Reply | 2,565 | 9,827 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 47, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2013-48 | longest | en | 0.831786 |
https://stats.stackexchange.com/questions/196601/what-is-the-intuition-behind-defining-completeness-in-a-statistic-as-being-impos?noredirect=1 | 1,660,943,590,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573760.75/warc/CC-MAIN-20220819191655-20220819221655-00723.warc.gz | 471,950,132 | 71,613 | # What is the intuition behind defining completeness in a statistic as being impossible to form an unbiased estimator of $0$ from it?
In classical statistics, there is a definition that a statistic $$T$$ of a set of data $$y_1, \ldots, y_n$$ is defined to be complete for a parameter $$\theta$$ it is impossible to form an unbiased estimator of $$0$$ from it nontrivially. That is, the only way to have $$E h(T (y )) = 0$$ for all $$\theta$$ is to have $$h$$ be $$0$$ almost surely.
Is there a intuition behind this? Is seems like a rather mechanical way of defining this, I am aware this has been asked before, but was wondering if there was a very easy to understand intuition that would make introductory students have an easier time digesting the material.
• That is a very good question, I had to dig into it myself. It turns out that the reason it is such a mechanical definition and does not appear intuitively meaningful to a standard practicioner like me is that it is primarily used for proving fundamental contributions in mathematical statistics. In particular, my short search revealed that the Lehmann-Scheffé theorem and Basu's theorem require completeness of a statistic in order to hold. These are contributions of the mid 1950s. I cannot offer you an intuitive explanation - but if you really want to build one, maybe the proofs associ Feb 15, 2016 at 21:43
I will try to add to the other answer. First, completeness is a technical condition which is justified mainly by the theorems that use it. So let us start with some related concepts and theorems where they occur.
Let $$X=(X_1,X_2,\dotsc,X_n)$$ represent a vector of iid data, which we model as having a distribution $$f(x;\theta), \theta \in \Theta$$ where the parameter $$\theta$$ governing the data is unknown. $$T=T(X)$$ is sufficient if the conditional distribution of $$X \mid T$$ does not depend on the parameter $$\theta$$. $$V=V(X)$$ is ancillary if the distribution of $$V$$ does not depend on $$\theta$$ (within the family $$f(x;\theta)$$). $$U=U(X)$$ is an unbiased estimator of zero if its expectation is zero, irrespective of $$\theta$$. $$S=S(X)$$ is a complete statistic if any unbiased estimator of zero based on $$S$$ is identically zero, that is, if $$\DeclareMathOperator{\E}{\mathbb{E}} \E g(S)=0 (\text{for all \theta})$$ then $$g(S)=0$$ a.e. (for all $$\theta$$).
Now, suppose you have two different unbiased estimators of $$\theta$$ based on the sufficient statistic $$T$$, $$g_1(T), g_2(T)$$. That is, in symbols $$\E g_1(T)=\theta ,\\ \E g_2(T)=\theta$$ and $$\DeclareMathOperator{\P}{\mathbb{P}} \P(g_1(T) \not= g_2(T) ) > 0$$ (for all $$\theta$$). Then $$g_1(T)-g_2(T)$$ is an unbiased estimator of zero, which is not identically zero, proving that $$T$$ is not complete. So, completeness of an sufficient statistic $$T$$ gives us that there exists only one unique unbiased estimator of $$\theta$$ based on $$T$$. That is already very close to the Lehmann–Scheffé theorem.
Let us look at some examples. Suppose $$X_1, \dotsc, X_n$$ now are iid uniform on the interval $$(\theta, \theta+1)$$. We can show that ($$X_{(1)} < X_{(2)} < \dotsm < X_{(n)}$$ are the order statistics) the pair $$(X_{(1)}, X_{(n)})$$ is sufficient, but it is not complete, because the difference $$X_{(n)}-X_{(1)}$$ is ancillary; we can compute its expectation, let it be $$c$$ (which is a function of $$n$$ only), and then $$X_{(n)}-X_{(1)} -c$$ will be an unbiased estimator of zero which is not identically zero. So our sufficient statistic, in this case, is not complete and sufficient. And we can see what that means: there exist functions of the sufficient statistic which are not informative about $$\theta$$ (in the context of the model). This cannot happen with a complete sufficient statistic; it is in a sense maximally informative, in that no functions of it are uninformative. On the other hand, if there is some function of the minimally sufficient statistic that has expectation zero, that could be seen as a noise term; disturbance/noise terms in models have expectation zero. So we could say that non-complete sufficient statistics do contain some noise.
Look again at the range $$R=X_{(n)}-X_{(1)}$$ in this example. Since its distribution does not depend on $$\theta$$, it doesn't by itself alone contain any information about $$\theta$$. But, together with the sufficient statistic, it does! How? Look at the case where $$R=1$$ is observed.Then, in the context of our (known to be true) model, we have perfect knowledge of $$\theta$$! Namely, we can say with certainty that $$\theta = X_{(1)}$$. You can check that any other value for $$\theta$$ then leads to either $$X_{(1)}$$ or $$X_{(n)}$$ being an impossible observation, under the assumed model. On the other hand, if we observe $$R=0.1$$, then the range of possible values for $$\theta$$ is rather large (exercise ...).
In this sense, the ancillary statistic $$R$$ does contain some information about the precision with which we can estimate $$\theta$$ based on this data and model. In this example, and others, the ancillary statistic $$R$$ "takes over the role of the sample size". Usually, confidence intervals and such need the sample size $$n$$, but in this example, we can make a conditional confidence interval this is computed using only $$R$$, not $$n$$ (exercise.) This was an idea of Fisher, that inference should be conditional on some ancillary statistic.
Now, Basu's theorem: If $$T$$ is complete sufficient, then it is independent of any ancillary statistic. That is, inference based on a complete sufficient statistic is simpler, in that we do not need to consider conditional inference. Conditioning on a statistic which is independent of $$T$$ does not change anything, of course.
Then, a last example to give some more intuition. Change our uniform distribution example to a uniform distribution on the interval $$(\theta_1, \theta_2)$$ (with $$\theta_1<\theta_2$$). In this case the statistic $$(X_{(1)}, X_{(n)})$$ is complete and sufficient. What changed? We can see that completeness is really a property of the model. In the former case, we had a restricted parameter space. This restriction destroyed completeness by introducing relationships on the order statistics. By removing this restriction we got completeness! So, in a sense, lack of completeness means that the parameter space is not big enough, and by enlarging it we can hope to restore completeness (and thus, easier inference).
Some other examples where lack of completeness is caused by restrictions on the parameter space,
• see my answer to: What kind of information is Fisher information?
• Let $$X_1, \dotsc, X_n$$ be iid $$\mathcal{Cauchy}(\theta,\sigma)$$ (a location-scale model). Then the order statistics are sufficient but not complete. But now enlarge this model to a fully nonparametric model, still iid but from some completely unspecified distribution $$F$$. Then the order statistics are sufficient and complete.
• For exponential families with canonical parameter space (that is, as large as possible) the minimal sufficient statistic is also complete. But in many cases, introducing restrictions on the parameter space, as with curved exponential families, destroys completeness.
Some intuition may be available from the theory of best (minimum variance) unbiased estimators (Casella and Berger's Statistical Inference (2002), Theorem 7.3.20).
If $$E_\theta W=\tau(\theta)$$ then $$W$$ is a best unbiased estimator of $$\tau(\theta)$$ iff $$W$$ is uncorrelated with all other unbiased estimators of zero.
Proof: Let $$W$$ be an unbiased estimator uncorrelated with all unbiased estimators of zero. Let $$W'$$ be another estimator such that $$E_\theta W'=E_\theta W=\tau(\theta)$$. Write $$W'=W+(W'-W)$$. By assumption, $$Var_\theta W'=Var_\theta W+Var_\theta (W'-W)$$. Hence, for any $$W'$$, $$Var_\theta W'\geq Var_\theta W$$.
Now assume that $$W$$ is a best unbiased estimator. Let there be some other estimator $$U$$ with $$E_\theta U=0$$. $$\phi_a:=W+aU$$ is also unbiased for $$\tau(\theta)$$. We have $$Var_\theta \phi_a:=Var_\theta W+2aCov_\theta(W,U)+a^2Var_\theta U.$$ If there were a $$\theta_0\in\Theta$$ such that $$Cov_{\theta_0}(W,U)<0$$, we would obtain $$Var_\theta \phi_a for $$a\in(0,-2Cov_{\theta_0}(W,U)/Var_{\theta_0} U)$$. $$W$$ could then not be the best unbiased estimator. QED
Intuitively, the result says that if an estimator is optimal, it must not be possible to improve it by just adding some noise to it, in the sense of combining it with an estimator that is just zero on average (being an unbiased estimator of zero).
Unfortunately, it is difficult to characterize all unbiased estimators of zero. The situation becomes much simpler if zero itself is the only unbiased estimator of zero, as any statistic $$W$$ satisfies $$Cov_\theta(W,0)=0$$. Completeness describes such a situation.
• This excerpt is taken almost verbatim and without attribution from Casella and Berger's Statistical Inference (2002) . May 26 at 2:20
• (Theorem 7.3.20) May 26 at 2:32
• You are right, thanks for the pointer. I had taken it from some unpublished lecture notes which I felt weren't worth citing. May 27 at 5:58 | 2,326 | 9,211 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 94, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2022-33 | latest | en | 0.939943 |
http://www.free-ebooks.net/ebook/Amusements-in-Mathematics/html/130 | 1,485,006,660,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560281084.84/warc/CC-MAIN-20170116095121-00523-ip-10-171-10-70.ec2.internal.warc.gz | 482,652,844 | 12,695 | # Amusements in Mathematics
straight lines in his path, but after many attempts he improved upon this. Can you get
more than fifty-five? You may end your path in any cell you like. Try the puzzle with a
pencil on chessboard diagrams, or you may regard them as rooks' moves on a board.
324.—THE LION AND THE MAN.
In a public place in Rome there once stood a prison divided into sixty-four cells, all open
to the sky and all communicating with one another, as shown in the illustration. The
sports that here took place were watched from a high tower. The favourite game was to
place a Christian in one corner cell and a lion in the diagonally opposite corner and then
leave them with all the inner doors open. The consequent effect was sometimes most
laughable. On one occasion the man was given a sword. He was no coward, and was as
anxious to find the lion as the lion undoubtedly was to find him.
The man visited every cell once and only once in the fewest possible straight lines until
he reached the lion's cell. The lion, curiously enough, also visited every cell once and
only once in the fewest possible straight lines until he finally reached the man's cell. They
started together and went at the same speed; yet, although they occasionally got glimpses
of one another, they never once met. The puzzle is to show the route that each happened
to take.
325.—AN EPISCOPAL VISITATION.
The white squares on the chessboard represent the parishes of a diocese. Place the bishop
on any square you like, and so contrive that (using the ordinary bishop's move of chess)
he shall visit every one of his parishes in the fewest possible moves. Of course, all the
parishes passed through on any move are regarded as "visited." You can visit any squares
more than once, but you are not allowed to move twice between the same two adjoining
squares. What are the fewest possible moves? The bishop need not end his visitation at
the parish from which he first set out.
326.—A NEW COUNTER PUZZLE.
Here is a new puzzle with moving counters, or coins, that at first glance looks as if it
must be absurdly simple. But it will be found quite a little perplexity. I give it in this
place for a reason that I will explain when we come to the next puzzle. Copy the simple
diagram, enlarged, on a sheet of paper; then place two white counters on the points 1 and
2, and two red counters on 9 and 10, The puzzle is to make the red and white change
places. You may move the counters one at a time in any order you like, along the lines
from point to point, with the only restriction that a red and a white counter may never
stand at once on the same straight line. Thus the first move can only be from 1 or 2 to 3,
or from 9 or 10 to 7.
327.—A NEW BISHOP'S PUZZLE.
This is quite a fascinating little puzzle. Place eight bishops (four black and four white) on
the reduced chessboard, as shown in the illustration. The problem is to make the black
bishops change places with the white ones, no bishop ever attacking another of the | 715 | 3,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2017-04 | latest | en | 0.979401 |
https://mydowntownsmyrna.org/and-pdf/285-class-8-maths-direct-and-inverse-proportions-pdf-974-864.php | 1,642,492,389,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320300805.79/warc/CC-MAIN-20220118062411-20220118092411-00467.warc.gz | 480,527,584 | 6,819 | and pdfWednesday, June 2, 2021 8:37:35 AM2
# Class 8 Maths Direct And Inverse Proportions Pdf
File Name: class 8 maths direct and inverse proportions .zip
Size: 11950Kb
Published: 02.06.2021
The book can be downloaded in pdf for Class 8 Direct and Inverse Proportions. You can download the entire textbook or each chapter in pdf, NCERT Books are suggested by CBSE for Class 8 Direct and Inverse Proportions exams, as they have been prepared as per syllabus issued by CBSE, download and read latest edition books and these have very important questions and exemplar problems for which studiestoday. Click on the below links to access books.
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A mixture of paint is prepared by mixing 1 pan of red pigments with 8 parts of base. In the following table, find the parts of base that need to be added. It is given that parts of red pigment, say x and parts of base, say y are in direct proportion. Therefore, the ratio of the correspondng values of x and y remain constant. So, x and y are in direct variation with the constant of variation equal to.
The solutions are also printable so that you could even have a group study with the hard copy of the PDF. Exercise In the first section of Class 8 Maths Chapter Direct and Inverse Proportion, there are many examples given from daily life where the knowledge of proportions is applied. For example:. If two students arrange chairs for an assembly in 20 minutes, how much time will 5 students take? If you have to make tea for 1 person and you use 1 teaspoon of sugar then how many teaspoons would you need for making tea for 5 people? From these examples, it can be figured that there is a direct relation in a change of one quantity on another quantity.
## Direct And Inverse Proportions NCERT Solutions : Class 8 Maths
Thanks for visiting our website. Our aim is to help students learn subjects like physics, maths and science for students in school , college and those preparing for competitive exams. All right reserved. All material given in this website is a property of physicscatalyst. For example: i As the speed of a vehicle increases, the time taken to cover the same distance decreases ii More apples cost more money iii More interest earned for more money deposited. Direct Proportion Two quantities x and y are said to be in direct proportion if they increase decrease together in such a manner that the ratio of their corresponding values remains constant.
DIRECT AND INVERSE PROPORTIONS. MATHEMATICS. Similarly, we can find the cost of 5 kg or 8 kg of sugar. prepare halwa for your class. 3.
## Direct and Inverse Proportions Class 8 Extra Questions Maths Chapter 13
Practicing the textbook questions assist the students in analyzing their preparation level and knowledge of all the topics. The students get familiarized with the format of the Mathematics class 8 th question paper so as to obtain high ranks in the exam. Frequent practice of RS Aggarwal solutions class 8 Maths chapter 12 also increases the speed of writing the exam. These solutions assist the students in gaining knowledge and strong command over the mathematics subject. If the students want to score excellent marks, it is necessary to practice RS Aggarwal solutions class 8 chapter
Question 1. How far will it travel in 20 minutes? Solution: Let the distance travelled by train in 20 minutes be x km.
Two persons could fit new windows in a house in 3 days. How long would the job take now? If, he uses 4 persons instead of three, how long should they take to complete the job?
For any inconvenience regarding website, please call us, we will immediately solve the problem. Offline apps are more helpful in case of slow internet. The number of workers and the time to complete the job is in inverse proportion because less workers will take more time to complete a work and more workers will take less time to complete the same work. In Chapter 13 Direct and Inverse Proportions, we will study the applications in daily life.
Он, конечно, видел старинную мавританскую башню, но взбираться на нее не. - Алькасар. Беккер снова кивнул, вспомнив ночь, когда слушал гитару Пако де Лючии - фламенко под звездами в крепости XV века. Вот бы побывать здесь вместе со Сьюзан.
Вдруг это вирус. Ты раньше говорил что-то про вирус.
1. ## Megan B.
07.06.2021 at 20:38 | 997 | 4,338 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-05 | latest | en | 0.942629 |
https://marketreachcalculator.com/gfa-calculation-australia/ | 1,686,057,372,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224652569.73/warc/CC-MAIN-20230606114156-20230606144156-00497.warc.gz | 426,912,700 | 15,574 | # gfa calculation australia
Introduction:
When it comes to investing in real estate, one of the most important factors to consider is the Gross Floor Area (GFA) calculation. GFA is a measure of the total floor space of a building, and it plays a critical role in determining the value of a property.
In Australia, GFA calculations are a crucial part of the property market, and they have a direct impact on property prices, development applications, and building regulations. In this article, we will take a closer look at GFA calculations in Australia, how they work, and their importance in the real estate industry.
## What Is GFA Calculation?
Gross Floor Area (GFA) calculation is a method used to determine the total floor space of a building, including all enclosed spaces such as rooms, hallways, stairwells, and other areas. It is a standard measurement used in the construction industry to determine the amount of rentable or usable space in a building.
The GFA calculation is also used to calculate the building’s density ratio, which is the number of square meters of floor space divided by the land area of the property. This ratio is used to determine the maximum allowable floor space for a given property, based on the zoning and planning regulations in the area.
### How Is GFA Calculated in Australia?
GFA calculations in Australia vary depending on the state or territory where the property is located. However, the basic formula used to calculate GFA is generally the same across the country.
To calculate the GFA of a building, you need to measure the total floor area of each level of the building, including any mezzanine levels or basements. You then add up the total floor areas of all levels, including any common areas such as lobbies or corridors.
Once you have the total floor area, you can subtract any areas that are not considered part of the GFA, such as stairwells, elevator shafts, or mechanical rooms. This will give you the final GFA measurement for the building.
### Importance of GFA Calculation in Australia
The GFA calculation is an essential factor in determining the value and potential use of a property in Australia. Property developers and investors use GFA calculations to determine the maximum allowable floor space for a given property, based on the zoning and planning regulations in the area.
This calculation is also critical in development applications, where the local council or planning authority will review the proposed floor space against the maximum allowable floor space determined by the GFA calculation. If the proposed development exceeds the maximum allowable floor space, the application may be rejected or require modifications to comply with local regulations.
The GFA calculation is also used to determine the rentable or usable space in commercial properties, which directly affects the rental income for the property owner. The larger the GFA, the higher the potential rental income for the property owner.
### How do you calculate GFA?
Gross Floor Area (GFA) is calculated by measuring the total floor area of a building, including all enclosed spaces such as rooms, hallways, stairwells, and other areas. To calculate GFA, you need to measure the floor area of each level of the building, add them together, and then subtract any areas that are not considered part of the GFA, such as stairwells or mechanical rooms.
### What’s included in GFA?
GFA includes all enclosed spaces within a building, including rooms, hallways, stairwells, and other areas. However, it does not include areas that are not enclosed, such as balconies or patios.
### What is the AIQS definition of GFA?
The Australian Institute of Quantity Surveyors (AIQS) defines GFA as “the total area of all floors of the building or buildings measured to the external face of the external walls, or to the centre line of walls separating two buildings.”
### What is the difference between GFA and GBA?
Gross Building Area (GBA) is similar to GFA but includes additional areas such as covered walkways, mechanical rooms, and other non-enclosed areas. GFA only includes enclosed spaces within a building.
NSW gross floor area definition
In New South Wales (NSW), GFA is defined as “the total area of all floors of the building or buildings measured to the internal face of external walls, ignoring any articulation or architectural projections.” This definition is used by the NSW Department of Planning, Industry, and Environment.
Gross Floor Area Australia
GFA is an important factor in the Australian real estate industry, as it is used to determine the value and potential use of a property, comply with zoning and planning regulations, and calculate the rentable or usable space in commercial properties.
How to calculate floor space ratio
Floor Space Ratio (FSR) is calculated by dividing the GFA of a building by the total land area of the property. FSR is used to determine the maximum allowable floor space for a given property, based on the zoning and planning regulations in the area.
Gross building area definition Australia
GBA is defined as “the total area of a building, including all enclosed spaces, covered walkways, mechanical rooms, and other non-enclosed areas.” This definition is used in some parts of Australia, but it is not as commonly used as GFA.
Is garage included in floor area Australia
Whether a garage is included in the floor area calculation depends on whether it is enclosed or not. If the garage is enclosed, it is generally included in the GFA calculation. However, if it is not enclosed (such as a carport), it is not included in the GFA calculation.
GFA calculation Victoria
GFA calculations in Victoria are similar to those used in other parts of Australia. GFA is calculated by measuring the floor area of each level of the building, adding them together, and then subtracting any areas that are not considered part of the GFA.
Gross floor area vs gross building area
The main difference between GFA and GBA is that GFA only includes enclosed spaces within a building, while GBA includes additional areas such as covered walkways, mechanical rooms, and other non-enclosed areas.
FSR calculation city of Sydney
The City of Sydney uses a formula to calculate FSR, which is based on the maximum allowable floor space for a given property, the total land area of the property, and any applicable planning regulations. The formula is designed to ensure that developments are in line with the density and character of the surrounding area.
### What is Gross Floor Area (GFA)?
Gross Floor Area (GFA) is a measure of the total floor space of a building, including all enclosed spaces such as rooms, hallways, stairwells, and other areas.
### Why is GFA important in Australia?
GFA is an essential factor in determining the value and potential use of a property in Australia. It is used to comply with zoning and planning regulations, calculate the rentable or usable space in commercial properties, and determine the maximum allowable floor space for a given property.
### How is GFA calculated in Australia?
To calculate GFA in Australia, you need to measure the floor area of each level of the building, add them together, and then subtract any areas that are not considered part of the GFA, such as stairwells or mechanical rooms.
### What areas are included in GFA calculation?
GFA includes all enclosed spaces within a building, including rooms, hallways, stairwells, and other areas. However, it does not include areas that are not enclosed, such as balconies or patios.
### What is the difference between GFA and GBA?
The main difference between GFA and Gross Building Area (GBA) is that GFA only includes enclosed spaces within a building, while GBA includes additional areas such as covered walkways, mechanical rooms, and other non-enclosed areas.
### Is garage included in floor area calculation in Australia?
Whether a garage is included in the floor area calculation depends on whether it is enclosed or not. If the garage is enclosed, it is generally included in the GFA calculation. However, if it is not enclosed (such as a carport), it is not included in the GFA calculation.
### What is the AIQS definition of GFA?
The Australian Institute of Quantity Surveyors (AIQS) defines GFA as “the total area of all floors of the building or buildings measured to the external face of the external walls, or to the centre line of walls separating two buildings.”
### How does GFA calculation affect development applications in Australia?
GFA calculation is critical in development applications, where the local council or planning authority will review the proposed floor space against the maximum allowable floor space determined by the GFA calculation. If the proposed development exceeds the maximum allowable floor space, the application may be rejected or require modifications to comply with local regulations.
### How do you calculate Floor Space Ratio (FSR) in Australia?
FSR is calculated by dividing the GFA of a building by the total land area of the property. FSR is used to determine the maximum allowable floor space for a given property, based on the zoning and planning regulations in the area.
### Are there any standard GFA calculation formulas used in Australia?
While GFA calculations may vary slightly depending on the state or territory, the basic formula used to calculate GFA is generally the same across the country.
## Conclusion:
In conclusion, Gross Floor Area (GFA) calculation is a crucial factor in the Australian real estate industry. It plays a significant role in determining the value and potential use of a property, as well as complying with zoning and planning regulations in the area.
Investors and developers must understand the GFA calculation and its significance when investing in real estate in Australia. By doing so, they can make informed decisions that maximize the potential of their investments while complying with local regulations. | 2,015 | 10,023 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-23 | latest | en | 0.940474 |
https://www.jobduniya.com/Private-Companies/3i-Infotech/Fully-Solved-Papers/Part-8.html | 1,638,024,314,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358189.36/warc/CC-MAIN-20211127133237-20211127163237-00244.warc.gz | 924,739,010 | 6,789 | # 3i Infotech Placement: Sample Questions 17 - 18 of 1245
Glide to success with Doorsteptutor material for competitive exams : get questions, notes, tests, video lectures and more- for all subjects of your exam.
## Question 17
Edit
### Write in Short
A man ate 100 bananas in five days, each day eating 6 more than the previous day. How many bananas did he eat on the first day?
### Explanation
Man ate 100 bananas in five days.
Each day man is eating 6 more than the previous day.
let՚s say, man is eating x bananas on the first day.
So, on second day, he will eat bananas.
On third day, he will eat bananas.
On fourth day, he will eat bananas.
On fifth day, he will eat bananas.
So,
So, man ate 8 bananas on the first day.
## Question 18
### Question
MCQ▾
Without the use of Cartesian product, how many joining conditions are required to join tables?
Choice (4)
a.
1
b.
3
c.
4
d.
2
b.
### Explanation
• To join ‘n’ tables ‘n-1’ conditions should be satisfied so to join tables conditions should be satisfied.
• Join is used to collate data from two more tables- joined to appear as single table of data.
• Combines columns from two or more tables by using values common to both table.
• “Join” keyword is used in SQL queries for joining two or more tables.
• Minimum required condition for joining table is (n-1) where n, is number of tables.
• A table can also join to itself, which is known as self join.
Developed by: | 368 | 1,454 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-49 | latest | en | 0.934951 |
http://www.veritasprep.com/blog/2013/06/gmat-tip-of-the-week-data-sufficiency-the-smartphone-way/ | 1,448,482,819,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445679.7/warc/CC-MAIN-20151124205405-00064-ip-10-71-132-137.ec2.internal.warc.gz | 767,329,719 | 16,868 | Let’s say you were in the market for some new technology, and let’s say your friend introduced you to a guy who sold used, refurbished gadgets at a huge discount. And let’s say he gave you this choice – you could buy:
A) An iPhone 5 for \$50
or
B) A digital camera for \$40
or
C) Both an iPhone 5 and a digital camera for \$75
Now, you have a few tech goals in mind. You want to be able to send text messages, update Twitter, use Google maps on the go, and upload pictures to Facebook and Instagram. Which deal do you take?
You take the iPhone only deal for \$50, right? Why don’t you take the camera too? That’s right – because option A already contains a digital camera! You don’t need to pay for another one. And because you know that the first option already gives you everything you need, you don’t pay for both products together. Agreed?
Well, that’s the game on many Data Sufficiency questions. Often times in a Data Sufficiency question one of the statements will already (but subtly) include the information from the other one. Which means, like in the case above, before you pick option C (both together) you’d better make sure you can’t do it with option A or option B alone. Here’s what we mean in a couple examples:
EXAMPLE 1:
Is 0 < x < 1 ? (1) x^2 < x (2) x > 0
In this case, many people pick option C, both together. But wait – you don’t want to buy a digital camera (the fact that x is positive) if your iPhone already has one. So let’s spend a little time playing with the features of that iPhone, or statement 1. Would a negative number even be possible given statement 1, or do we know already that x is positive? Try it: if x were to be -2, x^2 is 4…in that case x is not greater than x^2, so -2 wouldn’t work. And if x were -1/2, then x^2 is POSITIVE 1/4. Again, x is less than (not greater than) x^2. So neither a negative integer nor a negative fraction will work. Statement 1 already tells us that x is positive, so we don’t need to “buy” statement 2. It pays to take the time to – to continue the analogy – play with the features of the iPhone (statement 1) to see whether it already gives us the camera features we might want in statement 2.
Now let’s see a more advanced example.
EXAMPLE 2:
Julie is selling lemonades in two sizes, small and large. Small lemonades cost \$0.52 and large lemonades cost \$0.58. How many small lemonades did Julie sell?
(1) Julie sold a total of 9 lemonades.
(2) Julie’s total revenue from the sale of lemonades was \$4.92
Now, by the time you take your GMAT you should be able to come up with formulas for these statements pretty quickly:
(1) S + L = 9
(2) .52S + .58L = 4.92
And you should really quickly recognize that with two variables and two equations, you’ll be able to solve for S with both statements together. But that’s a little too easy for a test like the GMAT – especially when there’s a hidden piece of information that you can add to those two statements. Julie can’t sell 2.75 small lemonades – the values of S and L must be integers. So that should give you pause – if you take both statements together you’re leaving important information on the table. So at this point it pays to see whether the iPhone (statement 2) already has a digital camera (statement 1) embedded in it. Remember – the GMAT is a business test…it will reward you for being efficient with resources and for maximizing your ROI. It’s foolish in business to pay \$75 for two products when one would accomplish the task for \$50; similarly, it’s foolish to blindly pick C in 30 seconds when there’s a decent chance that the answer could be B. And how can you tell? If it were, indeed, an iPhone, you could spend some time playing with it to see if it would do exactly what the camera would. And that’s the goal here.
You know that if you have the statement “S + L = 9” you can pair that with statement 2 to solve for S. So try to see whether statement 2 includes that information by “borrowing” it. Could S + L be anything but 9? See if it can be 10:
If she sells 10 (the hypothetical) but still only makes \$4.92 (what we know from statement 2) she’d have to sell cheaper lemonades to keep the revenue down, so let’s try all 10 at the cheapest price. That’s 10(.52) = 5.20, which is already too much. There’s no way she can sell 10 or more!!!
If she sells 8 (another hypothetical) but still brings home \$4.92 (what we know from statement 2) she’d have to sell more expensive lemonades to make up for the fact that she’s selling fewer items. So let’s try 8 of the most expensive. That’s 8(.58) = 4.64, which isn’t enough. She can’t sell 8 or less, so we’ve just used statement 2 to prove that it already tells us the information from statement 1. Statement 2 is sufficient alone.
Now, this step is a little tricky for many, but look at what we just did – we had a hunch that we didn’t need to “buy” both statements because one might already have all the features of the other, like the iPhone with the built-in camera. So with those features in mind (S + L = 9) we set out to prove that the second statement included them, by “borrowing” them from statement 1. On tricky Data Sufficiency problems in which a trap answer like C in this case seems far too easy, this is an effective strategy to make sure you only pay for what you need. So keep this iPhone analogy in mind to maximize your statement efficiency on the GMAT; after all, at least for now, the GMAT won’t let you ask Siri.
Plan on taking the GMAT soon? We run a free online GMAT prep seminar every couple of weeks. And, be sure to find us on Facebook and Google+, and follow us on Twitter! | 1,411 | 5,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2015-48 | longest | en | 0.931848 |
https://www.onlinemathlearning.com/equation-of-a-line.html | 1,718,948,316,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862036.35/warc/CC-MAIN-20240621031127-20240621061127-00680.warc.gz | 809,049,381 | 9,445 | Find the Equation Of A Line Given Its Slope And A Point On The Line
Related Topics:
More Lessons for Geometry
Math Worksheets
In this lesson, we will learn how to get the equation of a line given its slope and a point on the line by
• using the Point-Slope Form.
• substituting into the Slope-Intercept equation.
Using the Point-Slope Form to get the Equation
We will now look at how to use the point-slope form to get the equation of a line given its slope and a point on the line.
Example:
Find the equation of a line with slope –3 and passing through (–2, 1).
Solution:
Step 1: Write out the Point-slope Form
yy1 = m(xx1)
Step 2: Substitute the slope –3 and the coordinates of the point (–2, 1) into the point-slope form.
y − 1 = –3(x −(−2))
Step 3: Simplify the equation
y − 1 = –3(x −(−2))
y − 1 = –3(x + 2)
y − 1 = –3x − 6
y = –3x − 6 + 1
y = –3x − 5
The required equation is y = –3x − 5
This video looks at writing linear equations in point-slope form, given a point and a slope, or two points. It includes four examples.
Using the Slope-Intercept to get the Equation
Example:
Find the equation of a line with slope – and passing through (–3, 1).
Solution:
Step 1: Substitute m = – , x = –3 and y = 1 into the equation y = mx+ c to obtain the value of c.
Step 2: Write out the equation of the line
The required equation is or 2y = –3x – 7
This will show you how to write an equation of a line that has a given slope and passes through a given point.
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. | 468 | 1,678 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-26 | latest | en | 0.884535 |
https://www.thestudentroom.co.uk/showthread.php?page=39&t=4013573 | 1,513,266,156,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948544677.45/warc/CC-MAIN-20171214144324-20171214164324-00035.warc.gz | 850,142,501 | 42,974 | You are Here: Home >< Maths
# [Official Thread] OCR MEI AS Mathematics 2016 (C1, C2 & S1) Watch
1. so.... do we need a Hitler video for this one ?
2. (Original post by GCSEsThen)
I got the same thing using area of a triangle formula 1/2*b*h
so 1/2*25*25/8
Your method would take longer and isnt as efficient but is perfectly valid
I feel so stupid for not doing that xD But at least I still got it right lol
3. On question 11 simplifying part. Did it ask you to
simplify (f(x+h) - f(x))/h when f(x) = x^2 - 2x
and not mention when x = 5?
4. What was the answer to the final question in c2? I rounded up :/??
5. FULL UNOFFICIAL MARK SCHEME
This took me so long, hope its helpful
Tweet me for any errors (@MilindS99)
6. (Original post by MilindS99)
FULL UNOFFICIAL MARK SCHEME
This took me so long, hope its helpful
Tweet me for any errors (@MilindS99)
Do you have a question paper so I can see question 11?
7. (Original post by Mr Moon Man)
Why can't you round up?
I was debating whether to round up and I did, but do u lose the mark if you rounded up?
8. (Original post by MilindS99)
FULL UNOFFICIAL MARK SCHEME
This took me so long, hope its helpful
Tweet me for any errors (@MilindS99)
On question 11 there is an error. It was x=5 and 5.1, not 8. 8 was gradient.
9. (Original post by Shipreck)
On question 11 there is an error. It was x=5 and 5.1, not 8. 8 was gradient.
for part (i) the gradient was 8.1 I think
10. I don't understand why you can't round up on the last question? Where does that rule come from?
11. (Original post by ineedA)
Did anyone get 4. Something for the sum of the sequence when A is 5, I'm sure it's not 25
For that question I got:
i) a=5
Sum = 25
ii) a=6
Sum = 0
12. Edit: Sorry the images are quite big for some reason
Both Wolfram Alpha and Mathway say that it is -m^3 although the question itself was really confusing. OCR are missing out brackets when needed especially on the sine question too.
13. (Original post by Akhawais)
Both Wolfram Alpha and Mathway say that it is -m^3 although the question itself was really confusing. OCR are missing out brackets when needed especially on the sine question too.
That isn't what the question was. The question was -loga(a^m)^3 = -loga(a^3m) = -3m.
14. (Original post by JoshP97)
For that question I got:
i) a=5
Sum = 25
ii) a=6
Sum = 0
How did you get that?
15. Also is y = sin x - 3
y = (sin x) - 3 OR y = sin (x - 3)
I was told that there was an error and it was supposed to be the 2nd one, but I can't find any where that OCR said this (I was told the exam board tweeted about it??). I really hope that's not true as I did the first one. I think both should be credited as it was very unclear.
16. (Original post by Akhawais)
Edit: Sorry the images are quite big for some reason
Both Wolfram Alpha and Mathway say that it is -m^3 although the question itself was really confusing. OCR are missing out brackets when needed especially on the sine question too.
If you read the actual 'input' in wolfram alpha i think it would have been put like that if the log was cubed, but i agree they needed brackets to make this clear
17. I can't be the only who thought I'd seen the paper before. Section B felt so familiar
18. What are you serious with that log question? I put -3m! Wtf is wrong with OCR why do they hate brackets?!?!?? Screwing us over for no reason we basically have to guess what they mean which doesn't test what we know.
19. (Original post by Mr Moon Man)
How did you get that?
i) When a = 5 the sequence oscillates, so the remainder would be 2, x 5 = 10 for the second term, the 3rd term would be 1 since 10/3 has a remainder of one, x 5 = 5 for the third term, this goes back to 10 for the fourth term. 10+5+10 = 25.
ii) When a = 6 there is no remainder to multiply by 5, so the sequence would be 0,0,0,0. the sum of these would be 0.
20. (Original post by ♥Samantha♥)
Also is y = sin x - 3
y = (sin x) - 3 OR y = sin (x - 3)
I was told that there was an error and it was supposed to be the 2nd one, but I can't find any where that OCR said this (I was told the exam board tweeted about it??). I really hope that's not true as I did the first one. I think both should be credited as it was very unclear.
Ocr haven't tweeted anything about that so I assume that the first one is correct
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Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice. | 1,536 | 5,553 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2017-51 | latest | en | 0.947779 |
http://magnetic-resonance.org/ch/06-05.html | 1,709,600,816,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476592.66/warc/CC-MAIN-20240304232829-20240305022829-00220.warc.gz | 19,940,829 | 5,693 | ## 06-05 Two-Dimensional Imaging
n an imaging experiment, definition and selection of a virtual (two-dimensional) slice through the examined object or a patient are of great importance. They are determined by characteristics of the excitation pulse. One distinguishes between shaped and hard pulses (cf. Figure 01-08).
## 06-05-01 Slice Selection
We can express the gradient strength in either mT/m or in Hz/m. Since the pulse has a fixed bandwidth (provided that the pulse duration is held constant), raising the gradient strength increases the number of Hz/m; this results in a decrease in slice thickness (Figure 06-15).
Figure 06-15:
Slice thickness: moving the gradient in the direction of the arrow increases the number of Hz/m, and thus gradient strength. It decreases slice thickness.
For example, for a sinc pulse with a bandwidth of 2 kHz, increasing the slice gradient from 4 mT/m (1.7 kHz/cm) to 8 mT/m (3.4 kHz/cm) reduces the slice thickness from 11.8 mm to 5.9 mm.
Applying an RF pulse in the absence of any field gradients will excite the whole sample. If a field gradient is applied at the same time as the pulse, the magnetic field, and therefore the resonance frequency, will change with position within the sample. For an RF pulse at the resonance frequency excitation will occur at the magnet center where the gradient has no effect (cf. Figure 06-05).
Off-center, the nuclei cannot be excited by RF pulses at the Larmor frequency. The distance (or slice thickness) over which the nuclei in the center resonate is determined by the range of frequencies (bandwidth) contained in the excitation pulse and the strength of the field gradient. If the RF pulse contains only a well defined band of frequencies, then excitation will occur for a well defined range of positions. This excitation corresponds to the selection of a slice in the sample.
The length of the RF pulse, and thus also its bandwidth, is the second factor influencing the slice thickness. The longer the pulse, the thinner the slice will be (Figure 06-16).
Figure 06-16:
(a) Long sinc pulses lead to thin slices, whereas (b) short sinc pulses increase slice thickness
The trade-off for thinner slices is the prolongation of the echo time (TE). Because TE is measured from the center of the pulse, longer pulses for thinner slices mean a longer initial TE, which, in turn, influences imaging time, image artifacts, and contrast.
Changing the frequency of the RF pulse corresponds to moving the position of the nuclei on resonance from the center of the sample. In this way we can move the slice to any desired location along the axis (Figure 06-17). For a transverse slice, the slice gradient is applied along the z-axis; for a coronal slice, the slice gradient is applied along the y-axis; and for a sagittal slice, it is applied along the x-axis.
Figure 06-17:
Moving the slice position: at 1.0 T, the resonance frequency in the center of the sample corresponds to 42.57 MHz. Changing the pulse frequency by several kHz moves the slice off-center.
## 06-05-02 Slice Definition
In analytical NMR studies, the maximal RF power is applied for a time sufficiently long to give the desired pulse angle (hard pulse). In more complicated experiments, it is necessary to adjust the pulse amplitude with time so as to give a better defined frequency content (shaped pulse).
The pulse shape is used to give an approximately rectangular slice profile for the slices in the imaging experiments (Gaussian and sinc pulses; see Figure 01-09) and can heavily influence image contrast in magnetic resonance imaging.
The phase of the RF pulse is also determined at this stage, with many MR machines only allowing phases of 0°, 90°, 180° or 270° to be selected. The resulting excitation pulses can be as short as 10 ms for non-selective hard pulses, and typically a few milliseconds for the frequency-selective shaped pulses used in magnetic resonance imaging with peak-to-peak amplitudes of up to several hundred volts. | 1,041 | 4,084 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2024-10 | latest | en | 0.878774 |
https://www.coursehero.com/file/6146342/CW2/ | 1,481,381,101,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543315.68/warc/CC-MAIN-20161202170903-00479-ip-10-31-129-80.ec2.internal.warc.gz | 914,336,510 | 21,148 | # CW2 - boland (kbb544) – CW2 – Mackie – (10611) This...
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Unformatted text preview: boland (kbb544) – CW2 – Mackie – (10611) This print-out should have 2 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A Cessna aircraft has a lift-off speed of 125 km/h. What minimum constant acceleration does this require if the aircraft is to be airborne after a take-off run of 214 m? Correct answer: 2.8169 m/s2 . Explanation: Let : vi = 0 km/h , v = 125 km/h , x = 214 m . 1 and Under constant acceleration, 2 2 vf = vi + 2 a x = 2 a x a= 2 vf 2 2x (125 km/h)2 1000 m = 2(214 m) 1 km = 2.8169 m/s2 . 1h 3600 s 2 002 (part 2 of 2) 10.0 points How long does it take the aircraft to become airborne? Correct answer: 12.3264 s. Explanation: Under constant acceleration, vf = vi + a t = a t vf t= a 125 km/h 1000 m = 2.8169 m/s2 1 km = 12.3264 s . 1h 3600 s ...
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Ask a homework question - tutors are online | 383 | 1,141 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2016-50 | longest | en | 0.754215 |
https://www.ukessayswriters.com/statistics-homework/ | 1,548,270,539,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547584336901.97/warc/CC-MAIN-20190123172047-20190123194047-00164.warc.gz | 972,199,669 | 12,923 | QUESTIONS
• (a) What are the steps involved in hypothesis testing? (b) How do you decide what the null hypothesis is and when would you reject a null hypothesis? (c) if you fail to reject the null hypothesis, what would be its implications? (5 pts.)
• The average work week for Americans has been recorded as 43 hours nationally. A local leisure product marketer randomly selects 44 sales people and asks them to keep a log book of hours worked. Suppose the sample average turns out to be 40.5 hours with a standard deviation of 9 hours. Conduct a hypothesis testing for the claim that Americans are now working less (show all steps and use 95 % confidence). Discuss the potential business implications from your result (10 pts- 7 pts for hypothesis testing and 3 pts for business implications).
• In 2014, the average number of years passenger cars were being used was 6.5 years. In 2015, a sample of 100 passenger cars showed a sample mean of 7.8 years and a sample standard deviation of 2.2 years. Use a .05 level of significance (p< .05) and test the claim that people use their cars for longer period of time. What marketing suggestions do you have for vehicle manufacturers? (10 pts- 7 pts for hypothesis testing and 3 pts for business implications).
• Q-Mart is interested in comparing its male and female customers. Q-Mart would like to know if its female customers spend more money, on average, than its male customers. They have collected random samples of 25 female customers and 22 male customers. On average, women customers spend \$102.23 with a sample standard deviation of 93.4 and men customers spend \$86.46 with a sample standard deviation of 59.7. Using a 10% level of significance, is there sufficient evidence for Q-Mart to conclude that women customers on average spend more than men customers? Please advise Q-Mart about potential business implications (10 pts- 7 pts for hypothesis testing and 3 pts for business implications).
• Suppose a Lightco produces light bulbs and wants to know whether it can claim that its light bulbs typically last more than 1500 hours. Hoping to find support for their claim, the firm collects a random sample of 25 lightbulbs and records the lifetime (in hours) of each bulb. The sample mean was 1509.5 hours with a sample standard deviation of 4.84. If you were to use 1 % significance level in this case, would you conclude that the mean life of the light bulbs is typically more than 1500 hours ? Please advise (10 pts- 7 pts for hypothesis testing and 3 pts for business implications).
• Please open the Santa Fe Grill dataset on SPSS and answer the following questions.
6.a The owners of the Santa Fe Grill want to find out whether their female customers have different opinions on several factors than their male customers: Fresh food (X_15), Food Taste (X_18), Friendly Employees (X_12). They are also curious whether it is the same for Jose’s customers. Set the hypothesis, report the SPSS analysis findings and advise Santa Fe Grill owners about potential implications (10 pts – 6 points setting the hypothesis and running the proper analysis, 4 points for the business advise).
1. b The owners of the Santa-Fe Grill noticed that their customers may have a higher likelihood of returning to the restaurant (X_23) than recommending the restaurant to their family and friends (X_24). Since these two variables are likely to be related, they want to know if the ratings for likely to return are significantly higher (more favorable) than for recommending the restaurant. They also would like to know what Jose’s customers are inclined to do in the same case. Choose the appropriate statistical test that allows you to examine whether two means from these two different questions answered by the respondents are significantly different. Set the hypothesis, report the SPSS analysis findings and advise Santa Fe Grill owners about potential implications (15 pts – 10 points setting the hypothesis and running the proper analysis, 5 points for the business advise).
6.c Santa Fe Grill owners want to find out whether their advertisements make a difference in their customers’ satisfaction level (X22), likelihood to come back (X23), and likelihood to recommend (X24). They decided to ask their customers whether they remember the advertisements (X31_ad recall), they also ask them whether they are satisfied with the service (X22), likely to return (X23), and likely to recommend (X24). They want to see whether there are differences in the levels of satisfaction, likely to return, and likely to recommend between people who recall the ads and who do not recall the ads. Set the hypothesis, report the SPSS analysis findings and advise Santa Fe Grill owners about potential implications (15 pts – 10 points setting the hypothesis and running the proper analysis, 5 points for the business advise).
6.d. Santa Fe Grill owners want to find out whether their customers’ (a) likelihood to return (X23) and (b) likelihood to recommend (X24) the restaurant are different based on how far their customers travel to get to the restaurant. In other words, do statistically significant differences exist for X23 and X24 between all the distance driven customer groups, or only some? Set the hypothesis, report the SPSS analysis findings and advise Santa Fe Grill owners about potential implications (15 pts – 10 points setting the hypothesis and running the proper analysis,5 points for the business advise). | 1,156 | 5,475 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.109375 | 3 | CC-MAIN-2019-04 | latest | en | 0.957606 |
https://codereview.stackexchange.com/questions/25666/my-implementation-of-lexicographic-permutation-in-f-is-3x-slower-than-c/25690 | 1,701,462,208,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100304.52/warc/CC-MAIN-20231201183432-20231201213432-00319.warc.gz | 214,451,748 | 43,699 | My implementation of lexicographic permutation in F# is 3x slower than C#
Can someone help me with the following micro optimization for the F# code for lexicographic permutation?
I have code in C# which runs for 0.8s. As a learning practice, I translated it into F#. However, it becomes 2.9s. Just out of curiosity, I am wondering why my code in F# runs that slow? Are there any improvement can be made to my F# code without changing the algorithm?
C#
static bool ToNextLexicographic(int[] myArray)
{
int pivot = -1;
for (int i = myArray.Length - 1; i > 0; i--)
{
if (myArray[i] > myArray[i - 1])
{
pivot = i - 1;
break;
}
}
if (pivot == -1)
return false;
for (int j = myArray.Length - 1; j > pivot; j--)
{
if (myArray[j] > myArray[pivot])
{
// swap
var tmp = myArray[j];
myArray[j] = myArray[pivot];
myArray[pivot] = tmp;
// reverse
for (int i = pivot + 1, k = myArray.Length - 1; i < k;i++,k-- )
{
var tmp = myArray[i];
myArray[i] = myArray[k];
myArray[k] = tmp;
}
break;
}
}
return true;
}
static IEnumerable<int[]> GetPermutationsLexicographic(int[] myArray)
{
Array.Sort(myArray);
yield return myArray;
while (ToNextLexicographic(myArray))
{
//yield return myArray.ToArray();
yield return myArray;
}
}
let inline toNextLexicographic (myArray: _[]) =
let rec findPivot i =
if i = 0 then -1
else
if myArray.[i] > myArray.[i-1] then i - 1
else findPivot (i - 1)
let rec findTarget value i =
if (myArray.[i] > value) then i
else findTarget value (i - 1)
let inline swap i j =
let tmp = myArray.[i]
myArray.[i] <- myArray.[j]
myArray.[j] <- tmp
let inline reverse i =
let mutable a = i
let mutable b = myArray.Length - 1
while a < b do
swap a b
a <- a + 1
b <- b - 1
let pivot = findPivot (myArray.Length - 1)
if pivot = -1 then false
else
let target = findTarget myArray.[pivot] (myArray.Length - 1)
swap pivot target
reverse (pivot + 1)
true;
let inline getPermutationsLexicographic myArray = seq {
Array.sortInPlace myArray
yield myArray
while toNextLexicographic myArray do
yield myArray
}
// benchmark
let mutable d = 0
for x in getPermutationsLexicographic([|1..11|]) do
d <- d + 1
d
• Well, have you tried profiling the code to find out what slows it down? May 1, 2013 at 0:35
• @svick the profiling info shows difference between x64 and x86, even though I struggled to find anything useful May 8, 2013 at 15:58
You're not comparing apples to apples. The F# code is generic whereas the C# only works with int arrays. Remove every occurrence of inline and add a type annotation to toNextLexicographic:
let toNextLexicographic (myArray: int[])
In my tests, that takes the time from 2.152 to 0.146 seconds.
The reason for the poor performance is inline isn't working. You can see this by decompiling. Most of the time is spent in LanguagePrimitives.HashCompare.GenericGreaterThanIntrinsic as a result. This is much more expensive than op_GreaterThan.
I'm not sure under what conditions inline is ignored, but I can only guess, in this case, it's due to the complexity of your code and the closures.
• I did what you suggested, it still takes 2.2s, not improved at all. Also I updated the C# code in question so it becomes generic. (time spent stays the same) May 1, 2013 at 15:10
• Something's amiss. You added the type annotation and it stayed the same? I see a huge difference. After the change to the C#, you're still not comparing the same code. Change the F# to use CompareTo. GenericGreaterThanIntrinsic is slow. You want to eliminate those calls. Also, are you compiling in Release mode? May 1, 2013 at 15:25
• I updated the post so that you can see my changed F#, (it is nothing new but to add int[] and remove some inline) Since the C# version is 0.8s, I don't think you can get less than that in F#. Your result of 0.146s is suspicious. The reason why I use inline at the first place is that I know GenericGreaterThanIntrinsic (doesn't implement IComparable<T>) is slower than CompareTo<T>. May 1, 2013 at 15:42
• Can you update to include your test code? May 1, 2013 at 15:54
• Ok, done. Btw it doesn't seem that inline was broken. Without inline and without type annotation, the program takes 20s. May 1, 2013 at 16:02
I can now confirm the performance difference is due to Seq / IEnumerable
Say, if I remove Seq / IEnumerable, and meausre speed simply use while loop, e.g.
while toNextLexicographic myArray do
d <- d + 2
The speed of C# and F# are now the same, being 0.6s. So I conclude it is the IEnumerable that make a difference. Still I don't get why Seq is slower than IEnumerable here, as they should be identical. | 1,310 | 4,569 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2023-50 | longest | en | 0.54431 |
https://aspirantszone.com/mixed-reasoning-questions-for-upcoming-exams-set-125/ | 1,675,493,312,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500094.26/warc/CC-MAIN-20230204044030-20230204074030-00398.warc.gz | 127,983,715 | 52,654 | # Mixed Reasoning Questions for Upcoming Exams – Set 125
Directions (1-5) : Study the information and answer the questions :
There are six cartons having different weight. Each box contains different number of chocolates- 2, 3, 4, 5, 6 and 1 but not necessarily in the same order.S contains more chocolate than N and S’s weight more than P, which contains odd number of chocolate. Carton O contains 4 chocolate. Q weighs more than only P. N weighs more than R and O, but less than S. Carton R contains 2 chocolate and R’s weight is more than O. The difference between the number of chocolate of carton N and O is 1 and carton N contains more than 3 chocolate.
1. Which of the following box contains highest number of chocolates?
N
P
Q
S
Cannot be determined
Option D
S(6)>N(5)>R(2)>O(4)>Q(1/3)>P(1/3)
2. Which of the following carton is second heaviest?
Q
N
P
S
None of these
Option B
3. What is the difference between number of chocolates between carton S and R?
1
2
3
4
None of these
Option D
4. Which of the following carton is heavier than O but lighter than N?
P
R
S
Cannot be determined
None of these
Option B
5. How many cartons are lighter than R?
3
1
2
4
None
Option A
6. Directions (6-8) : Study the information and answer the questions :
There are some candidates who all are sitting in a row facing towards the north direction. Lokesh sits fifth to the right of Kavya. Only two persons sits between Banya and Kavya. Nitesh sits seventh to the left of Banya.Himani sits third to the right of Nitesh. Yash is the immediate neighbour of Banya. Less than 16 persons sits in the row. Danish sits fifth to the right of Himani.
7. If ‘Lokesh’ sits third from the right end then how many candidates sits in the row?
12
11
13
15
18
Option A
8. How many candidates sit between Yash and Nitesh?
1
3
4
5
None of these
Option D
9. If ‘Priyanka’ sits between Kavya and Yash then what is the position of ‘Priyanka’ with respect to Nitesh?
Immediate left
Immediate right
4th to the right
5th to the right
None of these
Option D
10. Directions (9-10) : In these questions, relationship between different elements is shown in the statements. These statements are followed by two conclusions.
STATEMENTS: Rz>Qz>Jz>Dz, Mz≤Cz≤Jz
11. CONCLUSIONS:
a) Dz > Mz
b) Mz ≥ Dz
Conclusion a follows
Conclusion b follows
Either conclusion a or b follows
Both conclusion a and b follows
Neither conclusion a nor b follows
Option C
12. STATEMENTS: Zz≥Vz>Uz, Rz=EzMz Ez
b) Uz < Oz
Conclusion a follows
Conclusion b follows
Either conclusion a or b follows
Both conclusion a and b follows
Neither conclusion a nor b follows
Option A | 739 | 2,623 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2023-06 | latest | en | 0.927699 |
http://www.chegg.com/homework-help/questions-and-answers/identical-conducting-spheres-having-radius-0498-cm-connected-light-211-m-long-conducting-w-q1993425 | 1,369,341,779,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703788336/warc/CC-MAIN-20130516112948-00039-ip-10-60-113-184.ec2.internal.warc.gz | 380,087,237 | 7,914 | ## Gauss law
Two identical conducting spheres each having a radius of 0.498 cm are connected by a light 2.11 m-long conducting wire. Determine the tension in the wire if 55.1 µC is placed on one of the conductors.
• tension will be the electrostatic force between sphere
charge will be distributed in half in each sphere = 55.1/2
distance between center of spheres =.498cm+2.11m almost 2.11m
force
k.q1.q2/r2= k.(55.1/2)(55.1/2)/(2.11)2= 153.433N | 139 | 448 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2013-20 | longest | en | 0.851294 |
https://forum.pjrc.com/threads/60891-Vector-Math?s=df883d210baa4ff28b382576509d9f13&p=239834 | 1,590,475,985,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347390448.11/warc/CC-MAIN-20200526050333-20200526080333-00228.warc.gz | 325,020,277 | 11,292 | Forum Rule: Always post complete source code & details to reproduce any issue!
1. ## Vector Math
In reference to;
https://arm-software.github.io/CMSIS...groupMath.html
I'm trying to understand where vector math has usage given that it expects constants for the A/B array? Consider the vector float multiply;
Code:
```void arm_mult_f32 ( const float32_t * pSrcA,
const float32_t * pSrcB,
float32_t * pDst,
uint32_t blockSize
)```
So pSrcB & pSrcA have to be array defined at compile with static numbers? In the most basic way, how could one take advantage of
vector math like this for processing audio? like theoretically...
2. Originally Posted by martianredskies
So pSrcB & pSrcA have to be array defined at compile with static numbers?
No, it just means the function does not change the src.
3. Originally Posted by Frank B
No, it just means the function does not change the src.
Ok, that makes more sense. So if you were building an 8 channel mono mixer that did 8 multiplications, this could be done using the vector multiply? pSrcA could be an array of 8 values representing one sample of incoming audio, pSrcB could be multiplication values for each sample & pDst would be the results of each multiplcation? Does it matter how many samples are each array? 2/4/8/16 float32_t?
4. IMO, you should always first write your code yourself without using the CMSIS libraries. Only when you get it working should you introduce the CMSIS functions so as to (maybe) accelerate things.
If you're inexperienced with the CMSIS style, you're always going to be questioning if you're using them right. If you have your own implementation first, even if it is an inefficient implementation, you'll have something to compare to, which let you confirm that you're using the CMSIS functions correctly.
I've been there...I know...the CMSIS docs are not for newbies. So just implement your mixer in your own code first. 4/6/8 channels isn't really that complicated. You'll find it much easier to do it yourself than working through their docs. Then, once it's working, you can focus on the narrower task of fitting their functions to your known-working code. That'll be much easier, much less frustrating, and much more effective for learning this new skill.
Signed: your friend, a recent CMSIS newb himself, Chip
5. This is good advice, though just given the ambitiousness of the audio project I'm undertaking, i think I'm going to have to use every trick in the book to get the most out of a T3.6... either that or start offloading some things onto external hardware chips, which is not a bad idea Vs complicated coding tricks.
I've already designed the hardware interface and now I'm building the engine, every refinement means i can do a little more with the same design. Are the vector functions even available in the version that is packaged with teensyduino?
Originally Posted by chipaudette
IMO, you should always first write your code yourself without using the CMSIS libraries. Only when you get it working should you introduce the CMSIS functions so as to (maybe) accelerate things.
If you're inexperienced with the CMSIS style, you're always going to be questioning if you're using them right. If you have your own implementation first, even if it is an inefficient implementation, you'll have something to compare to, which let you confirm that you're using the CMSIS functions correctly.
I've been there...I know...the CMSIS docs are not for newbies. So just implement your mixer in your own code first. 4/6/8 channels isn't really that complicated. You'll find it much easier to do it yourself than working through their docs. Then, once it's working, you can focus on the narrower task of fitting their functions to your known-working code. That'll be much easier, much less frustrating, and much more effective for learning this new skill.
Signed: your friend, a recent CMSIS newb himself, Chip
6. Yes they are part of teensyduino. Just add to the top of your code
#include <arm_math.h>
As for learning how to use them, you don't have to prototype your use of vector math operations in your full system. I fact, you should not do it in your full system. You should do your trials in a super quick and simple program.
Write a very simple dumb "hello world" program that runs on the Teensy. you just need the teensy, no other hardware. Create a couple arrays of numbers. Fill the arrays with whatever values you want. Then, write a super basic function that multiplies them together (or whatever). Be as inefficient as you want. Just be simple so that you have confidence that you get the answer that you're expecting. Print the results over the Serial Monitor. Confirm you're getting what you expect.
Only then do you substitute one of the CMSIS functions for your hombrewed function. Print the results to the serial monitor. Does it match what you got with your code?
If so, you win! Now you know how to use the CMSIS function and you can deploy if with confidence into your full system.
If it does not give the same answer, try using it in a diff way. Or, you're using the wrong CMSIS function for the operation that yoj think that you want. Or, post your simple code here and ask questions.
Always test unfamiliar things in a short dumb simple program! Save yourself headaches!
7. I will take that advice. In the meantime I'm converting all my single variables into array groups, that should make it easier to integrate any possible vector math down the road conceptually. Thanks for your input.
#### Posting Permissions
• You may not post new threads
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• | 1,268 | 5,676 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2020-24 | latest | en | 0.930261 |
https://studdy.ai/shared-solution/ade144c3-6ae1-4595-9023-a8591291ffa5 | 1,726,385,434,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651616.56/warc/CC-MAIN-20240915052902-20240915082902-00109.warc.gz | 489,300,219 | 10,186 | # Chemistry Snap
## A student performs an experiment to determine the density of a sugar solution. she obtains the following results: 1.71 g/mL, 1.73 g/mL, 1.67 g/mL, 1.69 g/mL. If the actual value for the density of the sugar solution is 1.40 g/mL, which statement below best describes her results?
#### STEP 1
Assumptions1. We assume that the student's measurements are accurate and precise. . We assume that the actual value for the density of the sugar solution is correct.
#### STEP 2
Calculate the average of the student's measurements. The formula for the average (mean) is$\text{Average} = \frac{\text{Sum of all measurements}}{\text{Number of measurements}}$
#### STEP 3
Substitute the student's measurements into the above equation to find the average$\text{Average} = \frac{1.71 \, \text{g/mL} +1.73 \, \text{g/mL} +1.67 \, \text{g/mL} +1.69 \, \text{g/mL}}{}$
#### STEP 4
After performing the calculation, we find that the average of the student's measurements is approximately1.70 g/mL.
#### STEP 5
Compare the average of the student's measurements to the actual value for the density of the sugar solution. The student's average measurement is significantly higher than the actual value.
##### SOLUTION
Based on the comparison in5, we can conclude that the student's measurements are not accurate, but they are precise. The measurements are close to each other (precise), but they are not close to the actual value (not accurate). | 361 | 1,457 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 2, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2024-38 | latest | en | 0.833625 |
https://www.jianshu.com/p/9a3b3104776e | 1,568,564,906,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514571651.9/warc/CC-MAIN-20190915155225-20190915181225-00358.warc.gz | 923,739,283 | 35,175 | # Python:数据抽样平衡方法重写
R：
``````#设定工作目录
setwd(path)
# 安装包
install.packages("ROSE")
library(ROSE)
#检查数据
data(hacide)
table(hacide.train\$cls)
0 1
980 20
``````
``````data_balanced_over <- ovun.sample(cls ~ ., data = hacide.train, method = "over",N = 1960)\$data
table(data_balanced_over\$cls)
0 1
980 980
``````
``````data_balanced_under <- ovun.sample(cls ~ ., data = hacide.train, method = "under", N = 40, seed = 1)\$data
table(data_balanced_under\$cls)
0 1
20 20
``````
``````data_balanced_both <- ovun.sample(cls ~ ., data = hacide.train, method = "both", p=0.5, N=1000, seed = 1)\$data
table(data_balanced_both\$cls)
0 1
520 480
``````
`method`的不同值代表着不同的采样方法，p这边是控制正类的占比，seed保证抽取样本的固定，也就是种子值。
``````# -*- coding:utf-8 -*-
import pandas as pd
import random as rd
import numpy as np
import math as ma
class sample_s(object):
def __init__(self):
''''this is my pleasure'''
def group_sample(self, data_set, label, percent=0.1):
# 分层抽样
# data_set:数据集
# label:分层变量
# percent:抽样占比
# q:每次抽取是否随机,null为随机
# 抽样根据目标列分层，自动将样本数较多的样本分层按percent抽样，得到目标列样本较多的特征欠抽样数据
x = data_set
y = label
z = percent
diff_case = pd.DataFrame(x[y]).drop_duplicates([y])
result = []
result = pd.DataFrame(result)
for i in range(len(diff_case)):
k = np.array(diff_case)[i]
data_set = x[x[y] == k[0]]
nrow_nb = data_set.iloc[:, 0].count()
data_set.index = range(nrow_nb)
index_id = rd.sample(range(nrow_nb), int(nrow_nb * z))
result = pd.concat([result, data_set.iloc[index_id, :]], axis=0)
new_data = pd.Series(result['label']).value_counts()
new_data = pd.DataFrame(new_data)
new_data.columns = ['cnt']
k1 = pd.DataFrame(new_data.index)
k2 = new_data['cnt']
new_data = pd.concat([k1, k2], axis=1)
new_data.columns = ['id', 'cnt']
max_cnt = max(new_data['cnt'])
k3 = new_data[new_data['cnt'] == max_cnt]['id']
result = result[result[y] == k3[0]]
return result
def under_sample(self, data_set, label, percent=0.1, q=1):
# 欠抽样
# data_set:数据集
# label:抽样标签
# percent:抽样占比
# q:每次抽取是否随机
# 抽样根据目标列分层，自动将样本数较多的样本按percent抽样，得到目标列样本较多特征的欠抽样数据
x = data_set
y = label
z = percent
diff_case = pd.DataFrame(pd.Series(x[y]).value_counts())
diff_case.columns = ['cnt']
k1 = pd.DataFrame(diff_case.index)
k2 = diff_case['cnt']
diff_case = pd.concat([k1, k2], axis=1)
diff_case.columns = ['id', 'cnt']
max_cnt = max(diff_case['cnt'])
k3 = diff_case[diff_case['cnt'] == max_cnt]['id']
new_data = x[x[y] == k3[0]].sample(frac=z, random_state=q, axis=0)
return new_data
def combine_sample(self, data_set, label, number, percent=0.35, q=1):
# 组合抽样
# data_set:数据集
# label:目标列
# number:计划抽取多类及少类样本和
# percent：少类样本占比
# q:每次抽取是否随机
# 设定总的期待样本数量，及少类样本占比，采取多类样本欠抽样，少类样本过抽样的组合形式
x = data_set
y = label
n = number
p = percent
diff_case = pd.DataFrame(pd.Series(x[y]).value_counts())
diff_case.columns = ['cnt']
k1 = pd.DataFrame(diff_case.index)
k2 = diff_case['cnt']
diff_case = pd.concat([k1, k2], axis=1)
diff_case.columns = ['id', 'cnt']
max_cnt = max(diff_case['cnt'])
k3 = diff_case[diff_case['cnt'] == max_cnt]['id']
k4 = diff_case[diff_case['cnt'] != max_cnt]['id']
n1 = p * n
n2 = n - n1
fre1 = n2 / float(x[x[y] == k3[0]]['label'].count())
fre2 = n1 / float(x[x[y] == k4[1]]['label'].count())
fre3 = ma.modf(fre2)
new_data1 = x[x[y] == k3[0]].sample(frac=fre1, random_state=q, axis=0)
new_data2 = x[x[y] == k4[1]].sample(frac=fre3[0], random_state=q, axis=0)
test_data = pd.DataFrame([])
if int(fre3[1]) > 0:
i = 0
while i < (int(fre3[1])):
data = x[x[y] == k4[1]]
test_data = pd.concat([test_data, data], axis=0)
i += 1
result = pd.concat([new_data1, new_data2, test_data], axis=0)
return result
``````
``````#加载函数
import sample_s as sa
#这边可以选择你需要的分层抽样、欠抽样、组合抽样的函数
sample = sa.group_sample()
#直接调用函数即可
new_data3 = sample.combine_sample(data_train, 'label', 60000, 0.4)
#将data_train里面的label保持正样本（少类样本）达到0.4的占比下，总数抽取到60000个样本
``````
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https://howmonk.com/how-long-does-it-take-to-fall-1-meter/ | 1,718,766,397,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861797.58/warc/CC-MAIN-20240619025415-20240619055415-00563.warc.gz | 264,046,358 | 26,715 | # How Long Does It Take to Fall 1 Meter?
When it comes to falling, we often wonder about the speed at which it happens. How long does it take to fall 1 meter? Let’s break it down and find out.
Have you ever thought about how quickly you would fall if you were to drop just 1 meter? The answer may surprise you.
## Understanding the Physics of Falling
When it comes to understanding how long it takes to fall a specific distance, we need to first grasp the basic principles of gravity and acceleration. Gravity is the force that pulls objects towards the center of the Earth, causing them to fall when not supported. The acceleration due to gravity on Earth is approximately 9.81 meters per second squared. This means that an object in free fall will increase its speed by 9.81 m/s every second until it reaches terminal velocity or hits the ground.
## Factors Affecting Falling Speed
Several factors can impact how quickly an object falls, influencing the time it takes to fall a certain distance. The weight of an object plays a significant role, as heavier objects will fall faster due to the gravitational force acting on them. Additionally, air resistance can slow down the falling speed, especially for objects with larger surface areas. The height from which an object falls can also affect the time it takes to reach the ground, as a greater height allows for more acceleration.
1. Weight: Heavier objects fall faster due to gravity.
2. Air Resistance: Objects with larger surface areas experience more air resistance.
3. Height: Falling from a greater height leads to increased acceleration and shorter falling times.
For more in-depth understanding of the factors affecting falling speed, you can check out this resource from NASA: click here.
## Calculating Time to Fall 1 Meter
Have you ever wondered how long it would take to fall 1 meter from a certain height? Well, let’s break it down for you. When an object falls freely (without any air resistance), it accelerates at a constant rate due to gravity. The acceleration due to gravity on Earth is approximately 9.81 meters per second squared (m/s^2). To calculate the time it takes to fall a certain distance, we can use the equation:
$t = \sqrt{\frac{2d}{g}}$
Where: – t is the time in seconds – d is the distance fallen in meters (in this case, 1 meter) – g is the acceleration due to gravity (9.81 m/s^2)
By plugging in the values, we can calculate the time it would take to fall 1 meter. Isn’t physics fascinating?
## Real-Life Applications
Now, let’s see how understanding the concept of falling speed and distance can be applied to various real-life scenarios. In sports like skydiving or bungee jumping, knowing the time it takes to fall a certain distance is crucial for safety and performance. Engineers use this knowledge when designing structures like elevators or amusement park rides to ensure they are safe and efficient. Safety measures in buildings also take into account falling speed to protect occupants in case of emergencies.
So, the next time you look up at a tall building or go skydiving, remember the physics behind falling distances and speeds. It’s all around us!
• In sports like skydiving or bungee jumping, understanding falling speed and distance is vital for safety.
• Engineers use this knowledge when designing structures like elevators or amusement park rides.
• Safety measures in buildings consider falling speed to protect occupants during emergencies.
Remember to stay curious and keep learning about the world around you!
## Mythbusting Falling Speed
Let’s clear up a common misconception about falling speed: regardless of weight, all objects fall at the same rate in a vacuum. Yes, that’s right! Whether it’s a feather or a brick, they will hit the ground at the same time if dropped from the same height. This phenomenon, famously demonstrated by Galileo, is due to the acceleration caused by gravity being constant for all objects on Earth.
## The Impact of Gravity on Falling Time
Gravity is not just a force that keeps us grounded; it also determines how quickly things fall. When it comes to falling speed, gravity is the key player. The acceleration due to gravity on Earth is about 9.81 meters per second squared. This means that for every second an object falls, its speed increases by 9.81 meters per second.
So, how long does it take to fall 1 meter? Well, using the formula for falling distance (d = \frac{1}{2}gt^2), where (d) is the distance fallen, (g) is the acceleration due to gravity, and (t) is the time taken, we can calculate that it takes approximately: – 0.45 seconds to fall 1 meter.
Remember, this calculation assumes no air resistance and that the object is dropped from rest. If you’re intrigued by the science behind falling speed, check out this In-depth Guide to Gravity and Falling Phenomena for a more detailed understanding.
## How long does it take to fall 1 meter?
Have you ever wondered how quickly you would fall just 1 meter if you took a leap? Well, here’s the answer: it would typically take you around 0.45 seconds to fall that distance. That might seem like a very short amount of time, but it’s enough for gravity to pull you down to the ground at a surprising speed.
### Safety Precautions for Falling
When it comes to heights and falling, safety should always be a top priority. If you find yourself in a situation where falling is a possibility, make sure to follow these important precautions:
1. Wear proper protective gear: Whether it’s a helmet, harness, or knee pads, having the right gear can significantly reduce the risk of injury in case of a fall.
2. Get proper training: If you work in an environment where falling is a hazard, make sure to receive proper training on how to safely navigate heights and prevent falls.
3. Use fall protection equipment: This includes items like safety nets, guardrails, and anchor points to secure yourself and prevent falls in the first place.
Remember, it’s always better to be safe than sorry when it comes to dealing with heights and falling. | 1,283 | 6,071 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-26 | longest | en | 0.9296 |
https://kodlogs.com/index.php?qa=34809&qa_1=function-multiply-a-b-a-b | 1,596,697,517,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736883.40/warc/CC-MAIN-20200806061804-20200806091804-00379.warc.gz | 379,367,917 | 11,193 | Function multiply(a, b){ a * b }
0 votes
115 views
Problem :
I have the very unique question from the programming background. They want me to write the function which should multiply (a) and (b) but not writing it like below one:
``````multiplyfunc(a,b){
return a*b;
}``````
They want me to do a math with below approach
``multiplyfunc(a)(b)``
Is it possible to do it?
0 votes
Solution :
I had also faced the similar issue in the recent past. I did lot of research on it and found the solution on it. This is the problem with the people getting started..
You should make the function that returns another function as shown below.
``````const multiplfunc = a => b => a * b;
console.log(multiplyfunc(4)(3));``````
OR
Try with the function that returns the function as shown below :
``````function multiplyfunc (a) {
return function (b) {
return a * b;
};
};
console.log(multiplyfunc(4)(3));``````
OR
Currying is the process used to reduce the functions of more than one argument to the functions of one argument with the help of the lambda calculus.
``````f(n, m) --> f'(n)(m)
Example:
multiplyfunc = (n, m) => (n * m)
multiplyfunc(3, 4) === 12 // true
curryedMultiplyfunc = (n) => ( (m) => multiplyfunc(n, m) )
triplefunc = curryedMultiplyfunc(3)
triplefunc(4) === 12 // true``````
I hope it helps you in solving your issue.
36.1k points
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Problem: I am new, I need help, can anyone help me by filling in the blanks? The ________ built-in function is used to read a number that has been typed on the keyboard? A. get() B. input() C. keyboard() D. read() | 726 | 2,935 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2020-34 | latest | en | 0.854368 |
http://www.jiskha.com/display.cgi?id=1321923997 | 1,498,368,952,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320438.53/warc/CC-MAIN-20170625050430-20170625070430-00150.warc.gz | 571,743,009 | 3,721 | # trig
posted by .
if tan theta=7/24 and sin theta is greater than 0 what is the exact value of sin theta
• trig -
draw a triangle. If tanθ > 0 and sinθ > 0 then we are in the first quadrant.
the two legs are 7 and 24, the hypotenuse is 25
sinθ = 7/25 | 83 | 257 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2017-26 | latest | en | 0.873398 |
https://www.bookfusion.com/books/929589-essential-mathematics-for-cambridge-secondary-1-stage-7 | 1,627,836,293,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154214.63/warc/CC-MAIN-20210801154943-20210801184943-00378.warc.gz | 671,643,778 | 12,927 | \$32.09
# Essential Mathematics for Cambridge Secondary 1: Stage 7
By Sue Pemberton, Patrick Kivlin, Paul Winters
US\$ 32.09
The publisher has enabled DRM protection, which means that you need to use the BookFusion iOS, Android or Web app to read this eBook. This eBook cannot be used outside of the BookFusion platform.
Book Description
The Essential Mathematics for Cambridge Secondary 1 series has been created for the international student. Written by an expert author team with an experienced examiner, it provides complete coverage of the previous Cambridge syllabus. Providing comprehensive, clear theory notes followed by worked examples, the text helps students apply the knowledge they have learnt. Learning Outcomes are provided at the start of each chapter to clearly map topics in the text to the
syllabus. Numerous exercises are included, with clear progression, and problem-solving and real life applications embedded throughout. End-of-topic and end-of-section questions are designed to assess understanding, not only of the current topic but also of the ways in which different topics might be
combined in examinations.
• Front Cover
• Title Page
• Contents
• Introduction
• 1 Integers, powers and roots
• 1.1 Multiples
• 1.2 Factors
• 1.3 Prime numbers
• 1.4 Squares and square roots
• 1.5 Negative numbers
• 1.6 More adding and subtracting
• 1.7 Further calculations
• 2 Expressions
• 2.1 Using letters for unknown numbers
• 2.2 Simplifying expressions
• 2.3 Expanding brackets
• 2.4 Substitution into an expression
• 3 Shapes and geometric reasoning 1
• 3.1 Angles
• 3.2 Calculating angles
• 3.3 Triangles
• 3.5 Symmetry
• 4 Fractions
• 4.1 Equal parts
• 4.2 Equivalent fractions
• 4.3 Mixed numbers
• 4.4 Comparing, adding and subtracting fractions
• 4.5 Finding a fraction of an amount
• 5 Decimals
• 5.1 Place value
• 5.2 Rounding
• 5.3 Multiplying and dividing by 10, 100 and 1000
• 5.4 Calculating with decimals
• 5.5 Changing between fractions and decimals
• 6 Processing, interpreting and discussing data
• 6.1 Mode and median
• 6.2 Mean and range
• 6.3 Using frequency tables
• 6.4 Comparing two distributions
• 7 Length, mass and capacity
• 7.1 Length
• 7.2 Mass
• 7.3 Capacity
• 8 Equations
• 8.1 Equations with one operation
• 8.2 Equations with two operations
• 8.3 Equations with brackets
• 8.4 Using equations to solve problems
• 9 Shapes and geometric reasoning 2
• 9.1 Lines and angles
• 9.2 Measurement and construction
• 9.3 Solids
• 10 Presenting, interpreting and discussing data
• 10.1 Pictograms and frequency diagrams
• 10.2 Pie charts
• 11 Area, perimeter and volume
• 11.1 Area and perimeter
• 11.2 Compound shapes
• 11.3 Cuboids
• 11.4 Surface area
• 12 Formulae
• 12.1 Deriving formulae
• 12.2 Substitution into formulae
• 12.3 Further substitution into formulae
• 13 Position and movement
• 13.1 Reflection
• 13.2 Rotation
• 13.3 Translation
• 13.4 Using coordinates
• 14 Sequences
• 14.1 Number sequences
• 14.2 Sequences from patterns of shapes
• 15 Probability
• 15.1 Probability words
• 15.2 Probability scale and calculating probabilities
• 15.3 Estimating probabilities
• 16 Functions and graphs
• 16.1 Functions
• 16.2 Horizontal and vertical lines
• 16.3 Other straight lines
• 17 Fractions, decimalsand percentages
• 17.1 Describing part of an amount
• 17.2 Finding and using percentages of an amount
• 18 Planning and collecting data
• 18.1 Data collection sheets
• 18.2 Questionnaire design
• 18.3 Frequency tables
• 19 Ratio and proportion
• 19.1 Ratio
• 19.2 Dividing in a given ratio
• 19.3 Proportion
• 20 Time and rates of change
• 20.1 Time
• 20.2 Using timetables
• 20.3 Graphs
• 21 Sets
• 21.1 Describing sets
• 21.2 Intersection and union of sets
• 21.3 Elements and subsets
• 22 Matrices
• 22.1 Route matrices
• 22.2 Adding and subtracting matrices
• 22.3 Multiplying a matrix by a number
• Glossary
• Index
• Back Cover
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Essential Mathematics for Cambridge Secondary 1: Stage 8
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Essential Mathematics for Cambridge Secondary 1: Stage 9
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Also Available On
Categories | 1,214 | 4,164 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-31 | latest | en | 0.820918 |
https://hirecalculusexam.com/how-to-evaluate-the-trustworthiness-and-reliability-of-the-person-or-service-taking-my-calculus-test-through-a-background-check-or-verification-process | 1,726,357,961,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651601.85/warc/CC-MAIN-20240914225323-20240915015323-00437.warc.gz | 273,124,467 | 21,272 | # How to evaluate the trustworthiness and reliability of the person or service taking my calculus test through a background check or verification process?
How to evaluate the trustworthiness and reliability of the person or service taking my check test through a background check or verification process? After watching the television, a psychologist will usually check the degree or the reliability of the person, in my site link If a person complies with the application step, the algorithm will show the reliability. We don’t know the full description here about the background check. So the i loved this key question is about whether the “trustworthiness/recability” test is good or good enough to perform the calibration for a calibration test. The following exercises show our situation in the first example “real case sample” (case 1). The sample type is from a customer’s family, and for convenience purposes we will focus on customer family. Case 1 ——- If the customer’s family members is a customer. She can provide information about her family members or if they are older or under-age when they are younger. Before the calibration test, let’s compare her family members’s test. Since the family group is a small sample of children as opposed to the population, we can take a good sample of the mother-ruler of their family, or if they are 65 or older, we take a sample from child-parenting households of other family. Let’s webpage the example “family/child(child)/min(mine)”, the family/child(child)/min(mine) method. When one family mother died, the other mother became the father-mother. Therefore she might not have the problem. The family/child(child)/min(mine) method still applies as I said earlier. The problem that was observed for the study in the second example is the lack of a suitable calibration for the calibration test. At the table “practical calibration process”, I’ll show a brief “base-6 algorithm” for solving this problem with “minimal care”, which is based on an algorithm that I don’t have any idea about, which can solve that particular problem in another way. Case 1 ——-How to evaluate the trustworthiness and reliability of the person or service taking my find out this here test through a background check or verification process? I was given this challenge to try this out to do some good work. I want to take a look at how my work is perceived by me so I am considering “going ahead.” So this exercise is to let me decide a “challenge to proceed” (the task/situation/person can also be: something that is my best bet). I don’t always think about it. | 544 | 2,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2024-38 | latest | en | 0.936543 |
http://list.seqfan.eu/pipermail/seqfan/2015-October/015426.html | 1,585,935,135,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370515113.54/warc/CC-MAIN-20200403154746-20200403184746-00126.warc.gz | 113,206,681 | 3,348 | # [seqfan] Re: Rounding in A075465
Harvey P. Dale hpd at hpdale.org
Sat Oct 3 15:10:15 CEST 2015
``` I have added Bob Wilson's function and code to the Mathematica field in A075465.
A quick on-line search for rules about rounding disclosed no consensus about whether the fraction 1/2 should be rounded up or down. The code Bob provided is consistent with the rule that the fraction 1/2 should be rounded down.
Best,
Harvey
-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Robert G. Wilson v
Sent: Saturday, October 3, 2015 7:44 AM
To: 'Sequence Fanatics Discussion list' <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Rounding in A075465
et al,
I ran the following two versions of Mathematica:
Round[ Accumulate[ Prime[ Range[1000]]]/Range[1000]]; rnd[n_] := Block[{ip = IntegerPart[n]}, If[n > ip + 1/2, ip + 1, ip]];; r = Range at 1000; rnd@# & /@ (Accumulate[Prime[r]]/r);
I then compared their results; they were identical. BUT if the range was increased to 10000, they differ.
In fact that differ at: 1810, 2458, 240926, 317602, ..., .
Therefore I suggest that the Mathematica coding in A075465 be changed.
Bob.
-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of israel at math.ubc.ca
Sent: Friday, October 02, 2015 6:27 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Rounding in A075465
Harvey:
Would you mind making that modification to your existing Mathematica code in A075465?
Cheers,
Robert
On Oct 2 2015, Harvey P. Dale wrote:
> It may be of some passing interest that Mathematica's "Round"
> function does not always round down in the case of a tie. Instead, the
> documentation says that "Round rounds numbers of the form x.5 toward
> the nearest even integer." That means (1) that Mathematica rounds up
> or down, in the case of a tie, to whichever is the nearest even
> integer, and (2) that if one wants to control the rounding in
> Mathematica to be consistent one has to write a separate function to
> do it correctly. Best, Harvey
>
>
> -----Original Message----- From: SeqFan
> [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Neil Sloane Sent:
> Friday, October 2, 2015 8:08 AM To: Sequence Fanatics Discussion list
> <seqfan at list.seqfan.eu> Subject: [seqfan] Re: Rounding in A075465
>
>Unless Zak responds otherwise, I suggest we add a comment saying:
>
>In case of a tie, round down.
>
>That is consistent with the data, and is the usual convention anyway.
>
>Best regards
>Neil
>
>Neil J. A. Sloane, President, OEIS Foundation.
>11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
>Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
>Phone: 732 828 6098; home page: http://NeilSloane.com
>Email: njasloane at gmail.com
>
>
>On Fri, Oct 2, 2015 at 2:55 AM, <israel at math.ubc.ca> wrote:
>
>> A075465 is "Rounded average of first n primes", but nowhere does it
>> say what type of rounding is used in case of a tie (i.e. when the
>> actual average is half an odd integer). Since the Data gives a(2) =
>> 2,
>> 5/2 is being rounded to 2 rather than 3. That could indicate that
>> ties are rounded down, or that ties are rounded to the nearest even
>> integer. The next places the issue will arise in this sequence are
>> a(1810) (which could be 7265 or
>> 7266) and and a(2458) (10285 or 10286).
>>
>> Cheers, Robert
>>
>> _______________________________________________
>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
>_______________________________________________
>
>Seqfan Mailing list - http://list.seqfan.eu/
>
>_______________________________________________
>
>Seqfan Mailing list - http://list.seqfan.eu/
>
>
_______________________________________________
Seqfan Mailing list - http://list.seqfan.eu/
_______________________________________________
Seqfan Mailing list - http://list.seqfan.eu/
``` | 1,090 | 3,909 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-16 | latest | en | 0.850954 |
https://www.jiskha.com/display.cgi?id=1212964957 | 1,516,731,499,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084892059.90/warc/CC-MAIN-20180123171440-20180123191440-00445.warc.gz | 939,976,458 | 4,000 | # math
posted by .
use a graphing calculator use the graph y=e^x to evaluate e^1.7 to four decimal places.
• math -
I don't see why you need a graphing calculator. Enter 1.7 and then hit the e^x button, if it has one.
The answer is 5.4739, rounded to four decimal places after the decimal point. (That's five significant figures)
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More Similar Questions | 654 | 2,311 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-05 | latest | en | 0.815205 |
https://www.plati.market/itm/solution-14-5-15-from-the-collection-kep-oe-1989/2070961?lang=en-US | 1,576,142,210,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540542644.69/warc/CC-MAIN-20191212074623-20191212102623-00065.warc.gz | 807,007,645 | 12,824 | # Solution 14.5.15 from the collection Kep OE 1989
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## Description
14.5.15. In the plane Oxy moving mass points M1 and M2, which masses m1 = m2 = 1 kg. To determine the kinetic moment of the mass points of the system relative to the point O in a position where the velocity v1 = 2v2 = 4 m / s, distance = OM1 2OM2 = 4m and the angles α1 = α2 = 30 °. | 206 | 616 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-51 | latest | en | 0.711644 |
https://www.calculatorway.com/area-of-a-square-calculator/ | 1,679,947,529,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00629.warc.gz | 763,806,417 | 67,809 | # Area of a Square Calculator
Want to know how to calculate the area of a square? If you need help, check out the area of a square calculator
Square side length:
Units
Area of Square side² 00
Similar Calculator
## Area of a square
The area of a square is an easy math problem that is solved by using the formula A=s². The square has four equal sides, and each side has a length of s. So the area of the square is s squared or s². A square is just a rectangle that is perfectly square. It can be a square of any size. A square is a square, even if it is a square millimetre, square meter, square kilometre, square mile, square inch, square foot, square yard, square mile, square centimeter and square kilometer.
## Area of a square formula:-
Area of a square = (side)² = a² meter²
## How to find the area of a square
Example.1:-Square side length = 20 cm, find the area of a square
Area of a square = (side)² = (20)² = 400 cm²
## How to use this area of a square calculator
Enter only square side length value in the proper input field and then click calculate button, you get area of a square in answer box. if you want another value then first clear input value then entered new value in proper input field | 301 | 1,221 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.90625 | 4 | CC-MAIN-2023-14 | longest | en | 0.845636 |
https://homework.cpm.org/cpm-homework/homework/category/CON_FOUND/textbook/CA/chapter/Ch8/lesson/8.1.3/problem/8-29 | 1,590,361,052,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347385193.5/warc/CC-MAIN-20200524210325-20200525000325-00469.warc.gz | 393,684,472 | 15,367 | ### Home > CA > Chapter Ch8 > Lesson 8.1.3 > Problem8-29
8-29.
Use a generic rectangle and a diamond to solve.
(2x + 7)(3x − 2)
Insert a 0m between the 9m2 and the − 1. | 67 | 172 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2020-24 | latest | en | 0.538075 |
https://www.scribd.com/document/59271266/Metal-Cutting-Theory | 1,498,319,939,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320264.42/warc/CC-MAIN-20170624152159-20170624172159-00597.warc.gz | 918,306,274 | 30,762 | # METAL CUTTING THEORY All metal-cutting operation can be likened to the process shown in fig 1, where the
tool is wedge-shaped, has a straight cutting edge, and is constrained to move relative to the work piece in such a way that a layer of metal is removed in the form of a chip. Fig. 1b depicts the general case of cutting known as oblique cutting. A special case of cutting, where the cutting edge of the tool is arranged to be perpendicular to the direction of relative work-tool motion(fig 1a), is known as orthogonal cutting. Since orthogonal cutting represents a two-dimensional rather than a three-dimensional problem, it lends itself to research investigations where it is desirable to eliminate as many of the independent variables as possible. The relatively simple arrangement of orthogonal cutting is therefore widely used in theoretical and experiment work.
Fig. 3.1 , (a) Orthogonal cutting; (b) Oblique cutting The wedge-shaped cutting tool basically consists of two surface intersecting to form the cutting edge (fig. 3.2). The surface along which the chip flows is known as the rake face, or more simply as the face, and that surface ground back to clear the new or machined work piece surface is known as flank. Thus, during cutting a wedge-shaped “clearance crevice” exists between the tool flank and the new work piece surface. The depth of the individual layer of material removed by the action of the tool is known as the unreformed chip thickness (fig 3.2), and although in practical cutting operations this dimension often varies as cutting proceeds, for simplicity in much of the research work it is arranged to be constant. One of the most important variables in metal cutting is the slope of the tool face, and this slope, or angle, is specified in orthogonal cutting by the angle between the tool face and a line perpendicular to the new surface (fig 3.2). This angle is known as the rake or, in accordance with ISO terminology, the working normal rake, and fig 3 illustrates how the sign of the angle is defined. The rake angle is measured from the normal to the cut surface, with the positive direction such that the wedge angle is decreased. Negative rake angles result in a stronger cutting edge and consequently are often preferred for rough machining
The flank is the tool surface or surface over which the surface produced on the work piece passes. planers. however. where the angle is the included angle between the face and the flank. shapers. Thus from fig 3.2 the sum of the rake. These are shown in the figure and defined as follows: 1. more precisely. boring mills. In addition.operations. A typical single-point tool is illustrated in fig. . The tool flake plays no part in the process of chip removal. and similar machine tools. turret lathes. 2. Fig. 3. clearance. They are commonly used in lathes.3. and wedge angles is equal to 90 degrees. the working normal clearance.3 The most important feature are the cutting edges and adjacent surfaces. 3. (b) negative rake. (a) Positive rake.2 Term used in metal cutting. negative rake angles usually result in more usable cutting edges for replaceable tool inserts. The face is the surface or surface over which the chip flows. the angle between the flank and the new work piece surface can significantly affect the rate at which the cutting tool wears and is defined as the clearance angle or.2 The layer of material removed by single point cutting tool :single point tools are cutting tools having one cutting part (or chip-producing element)and one shank.
4.3.3 Typical single-point tool. The tool minor cutting edge is the remainder of the cutting edge. Fig. The cutting edge is that edge of which is intended to perform cutting. The tool major cutting edge is that entire part of the cutting edge which is intended to be responsible for the transient surface on the work piece. or it may be the actual intersection of these cutting edges. it may be curved or straight. 3. . The corner is the relatively small portion of the cutting edge at the junction of the major and minor cutting edges.
can vary along the major cutting edge. The resulting of these two tool motions is called the resulting cutting motion and is defined as the motion resulting from simultaneous primary and feed motions. The motion resulting from the primary motion of the machine tool. the angle between the direction of primary motion and the resultant cutting direction is called the resultant cutting-speed angle n. the instantaneous velocity of the resultant cutting motion of the selected point on the cutting edge relative to the work piece. its motion relative to the work piece has two components: 1.Fig. the instantaneous velocity of the primary motion of the selected point on the cutting edge relative to the work piece. which can be called the primary motion of the tool. the cutting speed v. the instantaneous velocity of the fed motion of the selected point on the cutting edge relative to the work piece. The motion resulting from feed motion of the machine. The feed speed . is given by = (2) . 3. Further. 2. Finally. when a tool is applied to a work piece. example). the resultant cutting motion is identical to the primary motion. the resultant cutting speed . Where the feed motion is applied continuously. is constant. This angle is usually extremely small and for most practical purpose can be assumed to be zero. It should be noted that in machine tools where the feed is applied the tool is not engaged with the work piece (as in shaping or planning.4 Resultant cutting motion in cylindrical turning In general.
significantly affects the power required to perform the operation. can be measured normal to the direction of primary motion. From fig 3. thus in Fig 3. as described above. Strictly. 3. where will be measured this way. therefore. the instantaneous engagement of the tool cutting edge with the work piece measured in the direction of feed motion.5. The thickness of the layer of material being removed at the selected point on the cutting edge.5 single-point tool operation. since n is small. where = .5 called the major cutting edge . However. is given by is equal to the is the feed engagement.But since for most practical operations n is very small. sin . For single-point cutting operations feed f. for all practical purposes. and therefore = The cross-section area of the layer of material being removed (cross-sectional area of the uncut chip) is approximately given by =f …(5) Fig.5 and all subsequently figures. known as the unreformed chip thickness . . this dimension should be measured both normal to the cutting edge and normal to the resultant cutting direction. it can generally be assumed that = (3) One of the important tool angles when considering the geometry of a particular machine operation is the angle in Fig 3.
A particular difficulty in formulating an analysis of the metal-cutting process is the lack of constraint in the process that reduces the range of boundary conditions that can be applied. In general the back engagement determines the depth of material removed from the work piece in a single-point cutting operation. Fig. cutting ratio. depend on the conditions that exist when the tool first contacts the work piece. = cutting . Fig. the simplify analysis of metal cutting developed do predict many of the general trends observed in the process. = shear force on shear plane resultant tool force. and so on that occur in a particular case. It has been suggested that a unique solution does not exist for a particular set of cutting conditions and further that parameters such as the cutting forces.3 THE APPARENT MEAN SHEAR STRENGTH OF THE WORK MATERIAL Many analyses of the metal-cutting processes have been developed. The back engagement is the instantaneous engagement of the tool with the work piece. In general. 3. previously known as depth of cut. A further complication is that different forms of chip formation mechanism occur at various cutting conditions.6 shows the idealized model of continuous chip formation employed in much of the previous work on the mechanism of the cutting process. It has been suggested that a unique solution does not exist for a particular set of cutting conditions that reduces the range f boundary conditions that can be applied. assumptions have to be made which are often valid only for a restricted range of conditions. However.4). measured perpendicular to the plane containing the directions of primary and feed motion (fig 3. where = force. A further complications that exist when the tool first contacts the work piece.Where is the back engagement. 3.3.6 Shear-plane model of continuous chip formation.
the working normal rake and the unreformed chip thickness are known. all the shear plane. or primary deformation zone.6 this force may be expressed in terms of cutting ( ) and trust ( ) components of the resultant tool force: =( cos ) – ( sin ) The area of shear is given by = = = /sin …(6) …(7) calculated in this way.2. and the chip thickness can be measured either directly with a ball-ended micrometer or obtained from the weight of a known length of a chip as follows: = Where = mass of the chip specimen = length of the chip specimen =width of the chip = density of the work piece material If the resultant tool force is resolved in a direction parallel to the plane. that at small feed . could be reasonably represented by a plane.3.3 From fig. As shown in fig. The angle of inclination of the shear plane to the direction of cutting. the force required to shear the work material and form the chip is obtained.Two of the earliest workers to employs this model where Ernst and merchant. …(3) …(2) Experimental work has shown that over a wide variety of cutting condition. however. could be determine as shown in the equation 1 . the length of the shear plane is given by … (1) And after rearrangement tan = the ratio is known as the cutting ratio and is denoted by tan = Where = shear angle = cutting ratio (given by = unreformed chip thickness = chip thickness = working normal rake In experimental work. who suggest that the shear zone. called shear angle φ. 3. remains constant for a given work material increases …(5) …(4) thus. It has been observed.6.
the strain rates are on the order of 103 to 105 s-1. is constant and independent of the cutting speed and rake angle under the range of this condition normally encountered in metal cutting. = …(8) It has been shown that if the components of are used to calculate the apparent shear strength of the work material.with decrease in feed (or unreformed chip thickness). This shear strength characteristic. However. is obtained . this apparent shear strength remain constant with respect to changes in feed. and under this conditions the shear straight of the metal could be expected to be constant and independent of strain rate. strain . Thus. strain rate and temperature is important.temperature. explain why in metal cutting the value of . …(9) . the trend to words the use of higher cutting speed means that accurate modeling of the constitutive behavior materials taking in to account strain. = Where = cutting component of = thrust component of = constant property of the work material Steady of the deformation of metals at high strain rates have shown that a material deforms at a relatively constant stress when the strain rate is sufficiently large. This exception to the constancy of explained by the existent of a constant plowing force force and . In metal cutting. it is suggest. the mean shear strength of the work material. If can be is subtracted from the resultant cutting the force required to remove the chip and acting on the tool face. | 2,397 | 11,714 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.953125 | 3 | CC-MAIN-2017-26 | latest | en | 0.956545 |
https://www.mail-archive.com/pgsql-hackers@postgresql.org/msg316570.html | 1,539,917,364,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00397.warc.gz | 1,002,328,696 | 4,128 | # Re: [HACKERS] Make ANALYZE more selective about what is a "most common value"?
```Dean Rasheed <dean.a.rash...@gmail.com> writes:
> I think we should attempt to come up with a more principled approach
> to this, taking into account the table and sample sizes. Here's what I
> found, after a bit of research:```
```
Thanks for doing some legwork on this!
> A common initial rule of thumb is that the value should occur at least
> 10 times in the sample - see, for example [1], [2]. ...
> Note that this says nothing about the margin of error of p, just that
> it is reasonable to treat it as having a normal distribution, which
> then allows the margin of error to be analysed using standard
> techniques.
Got it. Still, insisting on >= 10 occurrences feels a lot better to me
than insisting on >= 2. It's interesting that that can be correlated
to whether the margin of error is easily analyzed.
> The standard way of doing this is to calculate the "standard error" of
> the sample proportion - see, for example [3], [4]:
> SE = sqrt(p*(1-p)/n)
> Note, however, that this formula assumes that the sample size n is
> small compared to the population size N, which is not necessarily the
> case. This can be taken into account by applying the "finite
> population correction" (see, for example [5]), which involves
> multiplying by an additional factor:
> SE = sqrt(p*(1-p)/n) * sqrt((N-n)/(N-1))
It's been a long time since college statistics, but that wikipedia article
reminds me that the binomial distribution isn't really the right thing for
our problem anyway. We're doing sampling without replacement, so that the
correct model is the hypergeometric distribution. The article points out
that the binomial distribution is a good approximation as long as n << N.
Can this FPC factor be justified as converting binomial estimates into
hypergeometric ones, or is it ad hoc?
> This gives rise to the possibility of writing another rule for candidate MCVs:
> If there are Nd distinct values in the table, so that the average
> frequency of occurrence of any particular value is 1/Nd, then the test
> pmin > 1/Nd
Interesting indeed. We'd have to be working with our estimated Nd, of
course, not the true Nd. We know empirically that the estimate usually
lowballs the true value unless our sample is a fairly large fraction of
the whole table. That would tend to make our cutoff pmin too high,
so that we'd be excluding some candidate MCVs even though a better sample
would show they almost certainly had a frequency more common than average.
That behavior seems fine to me; it'd tend to drop questionable MCVs in
small samples, which is exactly what I think we should be doing. But it
is probably worth doing some empirical experiments with this rule to see
what we get.
regards, tom lane
--
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``` | 717 | 2,972 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2018-43 | latest | en | 0.931912 |
https://razonarte.org/free-printable-math-addition-worksheets-for-kindergarten/math-addition-worksheets-grade-piqqus-com/ | 1,550,308,777,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247480240.25/warc/CC-MAIN-20190216085312-20190216111312-00293.warc.gz | 681,077,510 | 5,063 | # Free Printable Math Addition Worksheets For Kindergarten Subtraction
By Ursula Wirtz on October 09 2018 18:24:27
People have to start from the ground, then first step, second, third and so on to reach their destination floor. Exactly the same way students have to start from Kindergarten, then grade one, grade two and three and so on to reach their math destination. Also, if some of the steps are broken in the staircase, it is still hard to reach the desired floor using those steps. Same way, if you are missing some of the basic concepts from elementary grades, math for you is still hard.
Parents and teachers are aware of the importance of math as well as all of the benefits. Taken in the account how important math is, parents will do whatever it takes to help their struggling children to effectively manage math anxiety. By using worksheets, it can play a major role in helping your kids cope with these stressful. This is a good way to show our children that practicing their math skills will help them improve. Here are some of the advantages using math and worksheets.
The math worksheet is not only for the young children in kindergarten and early primary school; they are also used for tutoring high school and university students to keep the students` math skills sharp. The sites that offer these worksheets have helped a lot and this resource is now a common thing to use for all kinds and levels of educators. The formats for the worksheets differ according to the level and content of the worksheets. For the young kids it is preferable to have the worksheet in large print, while the older students commonly use the small print ones that are simple and uncluttered.
Ratios and proportions are likewise wonderful math lessons with plenty of interesting practical applications. If three pans of pizza, one kilo of spaghetti, two buckets of chicken can properly feed 20 hungry friends, then how much pizza, spaghetti and chicken does mom need to prepare for birthday party with 30 kids?
There are several standard exercises which train students to convert percentages, decimals and fractions. Converting percentage to decimals for example is actually as simple as moving the decimal point two places to the left and losing the percent sign "%." Thus 89% is equal to 0.89. Expressed in fraction, that would be 89/100. When you drill kids to do this often enough, they learn to do conversion almost instinctively.
Another advantage of these math worksheets is that kids and parents will be able to keep them to serve as their references for review. Since worksheets are easy to correct, students will be able to identify the items and areas that they had mistakes so that they will be able to correct those deficiencies. Keeping record is really a good thing; As a parent, you will be able to go back through them and assess their strong and weak areas. Keeping track you will be able to track your child`s progress as empirical evidence. | 595 | 2,964 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2019-09 | latest | en | 0.953184 |
http://gamedev.stackexchange.com/questions/43828/how-to-rotate-an-object-around-a-fixed-point | 1,469,716,414,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257828283.6/warc/CC-MAIN-20160723071028-00227-ip-10-185-27-174.ec2.internal.warc.gz | 100,348,988 | 17,554 | # How to rotate an object around a fixed point?
I have a walking stick:
Some information:
• The handle is fixed, it cannot move
• The bottom part can move
Using the mouse click position and mouse movement, how can I rotate this stick with OpenGL?
-
I removed irrelevant text from your question, but it remains vague. To improve it: 1) Add an image of your stick being rendered, 2) add more useful "information": is the stick being held by a character, do you want a physics simulation, or just run the stick through an animation loop, 3) try to explain more carefully what mouse clicks and moves should do to the stick (preferably with a drawing), and 4) explain what you tried to fix this problem. – Eric Nov 16 '12 at 9:38
@Eric thanks for – gveaf Nov 16 '12 at 9:50
Try to make question titles as generic as possible. It helps for people searching in the future. I doubt "walking stick" is common enough to be searched for. – Byte56 Nov 17 '12 at 3:32
Generally, to rotate an object you'll want to use a rotation matrix. However, this will rotate your object around its origin - you want to rotate it around the handle.
So, if the handle isn't at the origin, you will want to apply a transformation matrix that moves the stick so that the handle is at the origin, then apply your rotation around the handle, and then possibly apply another transformation to position the stick in the world again.
Now, how you want to rotate your stick is entirely up to you, you could just map x/y movement to the theta/phi angles of a spherical coordinate system.
-
(T) * (R) * (T^(-1)) – Gustavo Maciel Nov 16 '12 at 16:07
I suppose so :) – melak47 Nov 16 '12 at 16:58
glPushMatrix();
glTranslatef(250,250,0.0); // 3. Translate to the object's position.
glRotatef(angle,0.0,0.0,1.0); // 2. Rotate the object.
glTranslatef(-250,-250,0.0); // 1. Translate to the origin.
// Draw the object
glPopMatrix();
-
You should really explain why this code works and the general idea behind how it's working. – Byte56 May 6 '14 at 22:22 | 527 | 2,029 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-30 | latest | en | 0.919485 |
https://www.english-arabic.org/english-to-arabic-meaning-longitude | 1,638,867,564,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363337.27/warc/CC-MAIN-20211207075308-20211207105308-00586.warc.gz | 834,527,380 | 17,525 | # English to Arabic Meaning :: longitude
Longitude :
خط الطول
خط الطول, طول - خط الطولخط الطولالطول
#### Show English Meaning (+)
Noun(1) the angular distance between a point on any meridian and the prime meridian at Greenwich
#### Show Examples (+)
(1) Her certificate of discharge even recorded the longitude and latitude at which the company's contractual obligations ended.(2) Every four minutes of difference would indicate 1 degree of longitude to the east or the west.(3) Each standard atlas covers thirty minutes of latitude and longitude at a scale of four miles to the inch, and fills one page in the book.(4) Firstly, remember that your longitude is the angular distance West of the Greenwich meridian.(5) This displaces the track of totality about 60 degrees east in longitude .(6) You may notice that both these institutions of higher education are a mere stone's throw from my current longitude and latitude.(7) Other published co-ordinates agree with the latitude, but longitude can vary by as much as 0.03 minutes west.(8) at a longitude of 2Ôö¼Ôûæ W(9) lines of longitude(10) Down the left of the chart Galileo lists the longitude and latitude for each planet.(11) At the right longitude and latitude, the resort has plentiful snowfall.(12) Any object on the same hour circle will have the same right ascension, just as any place on earth on the same meridian of longitude has the same longitude .(13) The zero degree line of longitude slices down through Greenwich, dividing London into western and eastern hemispheres.(14) Not by accident, he used Harrison's chronometer and lunar distances to calculate longitudes accurately.(15) It discusses topics such as geometry, geography and algebra with applications to the longitudes of the planets.(16) In such circumstances it is not surprising that a wrong reading is made of the latitudes and longitudes and very soon the young one finds himself sinking deeper and deeper into this bog of multiple explanations.
Related Words
(1) longitude ::
خط الطول
Synonyms
M
1. longitude ::
خط الطول
Different Forms
longitude, longitudes
English to Arabic Dictionary: longitude
Meaning and definitions of longitude, translation in Arabic language for longitude with similar and opposite words. Also find spoken pronunciation of longitude in Arabic and in English language.
Tags for the entry 'longitude'
What longitude means in Arabic, longitude meaning in Arabic, longitude definition, examples and pronunciation of longitude in Arabic language.
## English-Arabic.Org | English to Arabic Dictionary
This is not just an ordinary English to Arabic dictionary & Arabic to English dictionary. This dictionary has the largest database for word meaning. It does not only give you English toArabic and Arabic to English word meaning, it provides English to English word meaning along with Antonyms, Synonyms, Examples, Related words and Examples from your favorite TV Shows. This dictionary helps you to search quickly for Arabic to English translation, English to Arabic translation. It has more than 500,000 word meaning and is still growing. This English to Arabic dictionary also provides you an Android application for your offline use. The dictionary has mainly three features : translate English words to Arabic translate Arabic words to English, copy & paste any paragraph in the Reat Text box then tap on any word to get instant word meaning. This website also provides you English Grammar, TOEFL and most common words. | 718 | 3,486 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2021-49 | longest | en | 0.744802 |
https://consumerist.com/2012/04/02/target-using-its-signature-funny-math-to-trumpet-questionable-value/ | 1,500,893,335,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549424846.81/warc/CC-MAIN-20170724102308-20170724122308-00258.warc.gz | 650,020,261 | 33,467 | # Target Using Its Signature Funny Math To Trumpet Questionable "Value"
We’ve seen a lot of odd examples of “great deals,” “saving” and “new low prices,” but when it comes down to it, Target really takes the cake with what they call math. Hey, at least they’re consistent, right? In this latest example, Target says a four-pack of Lean Cuisine pizzas is a “great value,” when really, it’s cheaper to buy four individual pizzas.
Making the rounds over at Reddit is a picture wherein a hawk-eyed customer (who is no doubt better at basic math than whoever labels these products) notes the funny price discrepancy with the caption, “Target and I disagree on the definition of ‘value.’ “
See, it costs \$10.89 for the so-called value pack of four pepperoni pizzas, which comes out to \$2.72 each. Or, you can plunk down \$1.99 for each individual pizza. Hmm, which would you pick?
It’s not just Target, Walmart, Petco, Papa John’s and Walgreens, among others, have all been guilty of fuzzy math in the past.
*Thanks for the tip, Kevin! | 244 | 1,036 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2017-30 | longest | en | 0.931162 |
https://nbviewer.ipython.org/github/JuliaOpt/juliaopt-notebooks/blob/master/notebooks/Shuvomoy%20-%20Column%20generation.ipynb | 1,660,900,230,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882573630.12/warc/CC-MAIN-20220819070211-20220819100211-00777.warc.gz | 370,830,893 | 11,013 | Description: This notebook describes how to implement column generation, which is a large scale optimization scheme, in JuMP. The cutting stock problem has been used as an illustrative example.
Author: Shuvomoy Das Gupta
# Using Julia+JuMP for optimization - column generation¶
Implementing large scale optimization techniques such as column generation is really easy using JuMP. To explain how to implement column generation in JuMP, we consider the famous cutting stock problem. For more details about the problem, see pages 234-236 of Introduction to Linear Optimization by Bertsimas and Tsitsiklis.
## Notation and notions:¶
• Width of a large roll is $W$, and it needs to be cut into smaller width papers according to customer demand
• The set of indices of all feasible patterns is, $\mathcal{J}=\{1,2,\ldots,n\}$, where $n$ is a very large number
• A strict subset of $\mathcal{J}$ that is considered in the master problem is $\mathcal{J}'$
• The dummy index for a pattern is $j$
• The index set of all possible paper-widths is, $\mathcal{M}=\{1,2,\ldots,m\}$
• The width of the paper with index $i$ is $w_i$
• The demand for the paper of width $w_i$ is $b_i$
• Number of smaller rolls of width $w_i$ produced by pattern $j$ is denoted by $a_{ij}$
• Number of large rolls cut according to pattern $j$ is denoted by $x_j$
## Original unabridged problem:¶
\begin{align} &\text{minimize} && \sum_{j \in \mathcal{J}}{x_j} \\ &\text{subject to} &&\\ & &&\forall i \in \mathcal{M} \quad \sum_{j \in \mathcal{J}}{a_{ij} x_j}=b_i \\ & && \forall j \in \mathcal{J} \quad x_j \geq 0 \\ \end{align}
Because the set $\mathcal{J}$ can be astronomically large, even storing the problem is a challenge. So, we start with a smaller version of the problem, called the master problem, by replacing $\mathcal{J}$ with a strict subset $\mathcal{J}'$, which is much smaller than the original one.
## Structure of the decomposition¶
Master Problem: \begin{align} &\text{minimize} && \sum_{j \in \mathcal{J}'}{x_j} \\ &\text{subject to} &&\\ & &&\forall i \in \mathcal{M} \quad \sum_{j \in \mathcal{J}'}{a_{ij} x_j}=b_i \\ & && \forall j \in \mathcal{J}' \quad x_j \geq 0 \\ \end{align}
After solving the master problem, we want to check the optimality status. Structure of the cutting stock problem allows us to construct a subproblem which can do this very easily.
Subproblem: \begin{align} &\text{minimize} && 1 - \sum_{i \in \mathcal{M}} \quad {p_i a_{i {j^*}}} \\ &\text{subject to} &&\\ & && \forall i \in \mathcal{M} \quad a_{i {j^*}} \geq 0, \quad a_{ij^*} \; \text{integer} \\ & && \sum_{i \in \mathcal{M}}{w_i a_{i{j^*}}} \leq W\\ \end{align}
The objective of the subproblem is the minimum of the reduced cost vector of the original problem. If the objective value of the subproblem is greater than or equal to $0$, then the current solution of the master problem is optimal for the original unabridged problem. Otherwise, add the resultant cost reducing column $(a_{i {j^*}})_{i \in \mathcal{M}}=A_{j*}$ and a corresponding new variable $x_{j*}$ is added to the master problem. The modified master problem is as follows:
Modified Master Problem \begin{align} &\text{minimize} && \sum_{j \in \mathcal{J}'}{x_j} + x_{j^*} \\ &\text{subject to} &&\\ & &&\forall i \in \mathcal{M} \quad \sum_{j \in \mathcal{J}'}{a_{ij} x_j}+a_{i j^*} x_{j^*}=b_i \\ & && \forall j \in \mathcal{J}' \quad x_j \geq 0, x_j^* \geq 0 \\ \end{align}
The pseudocode for the cutting stock problem is given below.
## Pseduocode¶
• Input preliminary data for starting the problem
• Solve the master problem with the initial data
\begin{align} &\text{minimize} && \sum_{j \in \mathcal{J}'}{x_j} \\ &\text{subject to} &&\\ & &&\forall i \in \mathcal{M} \quad \sum_{j \in \mathcal{J}'}{a_{ij} x_j}=b_i \\ & && \forall j \in \mathcal{J}' \quad x_j \geq 0 \\ \end{align}
• Collect the dual variables for the equality constraints and store them in an array $(p_i)_{i \in \mathcal{M}}$
• Solve the sub problem
\begin{align} &\text{minimize} && 1 - \sum_{i \in \mathcal{M}} \quad {p_i a_{i {j^*}}} \\ &\text{subject to} &&\\ & && \forall i \in \mathcal{M} \quad a_{i {j^*}} \geq 0, \quad a_{ij^*} \; \text{integer} \\ & && \sum_{i \in \mathcal{M}}{w_i a_{i{j^*}}} \leq W\\ \end{align}
• Flow control:
while ( $\text{optimal value of the subproblem} < 0$)
• Add the column $(a_{i {j^*}})_{i \in \mathcal{M}}=A_{j*}$ to $A$
• Add a corresponding new variable $x_{j*}$ to the list of variables
• Solve the modified master problem
\begin{align} &\text{minimize} && \sum_{j \in \mathcal{J}'}{x_j} + x_{j^*} \\ &\text{subject to} &&\\ & &&\forall i \in \mathcal{M} \quad \sum_{j \in \mathcal{J}'}{a_{ij} x_j}+a_{i j^*} x_{j^*}=b_i \\ & && \forall j \in \mathcal{J}' \quad x_j \geq 0 \\ & && \qquad \qquad \; \; x_{j^*} \geq 0 \end{align}
• Collect the dual variables for the equality constraints and store them in an array $(p_i)_{i \in \mathcal{M}}$
• Solve the sub problem as before
• Set $\mathcal{J}':=\mathcal{J}'\cup \{j^*\}$
end while
• Display the results
## Master Problem Modification in JuMP¶
The problem modification can be done by using the already mentioned @variable macro:
$$\texttt{@variable}(m, l \leq x_\text{new} \leq u, \texttt{Int}, \texttt{objective} = c_\text{new}, \texttt{inconstraints} = \text{arrayConstrrefs}, \texttt{coefficients} = \text{arrayCoefficients})$$
Here:
• The name of the original model is $m$.
• The new variable to be added is $x_\text{new}$ with lower bound $l$ and upper bound $u$.
• The type of the variable can be Int, Bin. For real variable the third argument is left vacant.
• The original objective, say $f_o(x)$ will become $f_o(x) + c_\text{new} x_\text{new}$ after modification
• The array $\texttt{arrayConstrrefs}$ contain references to those constraints that need to be modified by inclusion of $x_\text{new}$
• The array $\texttt{arrayCoefficients}$ contain the coefficients that have to multiplied with $x_\text{new}$ and then added to the constraints referenced by $\texttt{arrayConstrrefs}$ in an orderly manner. For example, if the $i$th element of $\texttt{arrayConstrrefs}$ refers to a constraint $a_i^T x \lesseqgtr b_i$, then after invoking the command, the constraint is modified as: $a_i^T x +\texttt{arrayCoefficients}[i] x_\text{new} \lesseqgtr b_i$
## Implementing one iteration of the column generation algorithm¶
To understand how the column generation is working in Julia, we implement one iteration of the column generation algorithm manually. The entire code is presented in the next section.
In [1]:
# Uploading the packages:
# -----------------------
using JuMP
# We will use default solvers
In [2]:
# Input preliminary data for starting the problem
# -----------------------------------------------
W=100
cardinalityM=5
M=collect(1:cardinalityM)
A=eye(cardinalityM)
p=zeros(5)
b=[45; 38; 25; 11; 12]
w=[22; 42; 52; 53; 78]
Out[2]:
5-element Array{Int64,1}:
22
42
52
53
78
In [3]:
# Description of the master problem with the initial data
#----------------------
cutstockMain = Model() # Model for the master problem
Jprime=collect(1:size(A,2)) # Initial number of variables
@variable(cutstockMain, 0 <= x[Jprime] <= 1000000) # Defining the variables
@objective(cutstockMain, Min, sum(1*x[j] for j in Jprime)) # Setting the objective
@constraint(cutstockMain, consRef[i=1:cardinalityM], sum(A[i,j]*x[j] for j in Jprime)==b[i])
# Here the second argument consRef[i=1:cardinalityM] means that the i-th constraint aᵢᵀx = bᵢ has the corresponding constraint reference
# consRef[i]
print(cutstockMain)
Min x[1] + x[2] + x[3] + x[4] + x[5]
Subject to
x[1] == 45
x[2] == 38
x[3] == 25
x[4] == 11
x[5] == 12
0 <= x[i] <= 1.0e6 for all i in {1,2,..,4,5}
In [4]:
# Solving the master problem with the initial data
# ------------------------------------------------
solve(cutstockMain)
println("Current solution of the master problem is ", getvalue(x))
println("Current objective value of the master problem is ", getobjectivevalue(cutstockMain))
Current solution of the master problem is x: 1 dimensions:
[1] = 45.0
[2] = 38.0
[3] = 25.0
[4] = 11.0
[5] = 12.0
Current objective value of the master problem is 131.0
In [5]:
#Collect the dual variables for the equality constraints and store them in an array p
for i in M
p[i] = getdual(consRef[i]) # These p[i] are the input data for the subproblem
end
println("The array storing the dual variables is ", p)
The array storing the dual variables is [1.0,1.0,1.0,1.0,1.0]
In [6]:
# Describe the sub problem
# ------------------------
cutstockSub=Model() # Model for the subproblem
@variable(cutstockSub, 0 <= Ajstar[M] <= 1000000, Int )
@objective(cutstockSub, Min, 1-sum(p[i]*Ajstar[i] for i in M))
@constraint(cutstockSub, sum(w[i]*Ajstar[i] for i in M) <= W)
print(cutstockSub)
Min -Ajstar[1] - Ajstar[2] - Ajstar[3] - Ajstar[4] - Ajstar[5] + 1
Subject to
22 Ajstar[1] + 42 Ajstar[2] + 52 Ajstar[3] + 53 Ajstar[4] + 78 Ajstar[5] <= 100
0 <= Ajstar[i] <= 1.0e6, integer, for all i in {1,2,..,4,5}
In [7]:
# Solve the sub problem
# ---------------------
solve(cutstockSub)
minreducedCost=getobjectivevalue(cutstockSub)
println("The minimum component of the reduced cost vector is ", minreducedCost)
The minimum component of the reduced cost vector is -3.0
The minimum component of the reduced cost vector is negative, so we have a suboptimal solution.
In [8]:
if minreducedCost >= 0
println("We are done, current solution of the master problem is optimal")
else
println("We have a cost reducing column ", getvalue(Ajstar))
end
We have a cost reducing column Ajstar: 1 dimensions:
[1] = 4.0
[2] = 0.0
[3] = 0.0
[4] = 0.0
[5] = 0.0
In [9]:
typeof(Ajstar)
Out[9]:
JuMP.JuMPArray{JuMP.Variable,1,Tuple{Array{Int64,1}}}
Now Ajstar is of type JuMPDict. To use it in the modified master problem, we have to store values from Ajstar in a column vector.
In [10]:
Anew=Float64[] # This Anew correspond to the newly added column to the A matrix
for i in 1:cardinalityM
push!(Anew, getvalue(Ajstar)[i])
end
When we add the cost reducing column Anew to the original matrix A, it also gives rise to a new variable xNew corresponding to Anew. Now we want to keep track of the new variables that are added by the subproblem. We do this by declaring an array of Variables named xNewArray, which will contain all such newly added variables in the process of column generation.
In [11]:
xNewArray=Variable[] # The newly added variables by flow control will be
# pushed to the new array of variables xNewArray
Out[11]:
$$Empty Array{Variable} (no indices)$$
Here we just illustrate one iteration of the while loop manually, because, for now, we are interested to understand how JuMP is managing the flow control and modifying the master problem and the sub problem.
Let's modify the master problem by adding the new column Anew to the old A matrix. Note that we do not have to rewrite the entire model.
In [12]:
# Modify the master problem by adding the new column Anew to the old A matrix
@variable(
cutstockMain, # Model to be modified
0 <= xNew <= 1000000, # New variable to be added
objective=1, # cost coefficient of new variable in the objective
inconstraints=consRef, # constraints to be modified
coefficients=Anew # the coefficients of the variable in those constraints
)
# The line above adds the column (aᵢⱼ*)ᵢ=Aⱼ* to A <br>
# and add a corresponding new variable xⱼ* to the list of variable
push!(xNewArray, xNew) # Pushing the new variable in the array of new variables
print(cutstockMain)
Min x[1] + x[2] + x[3] + x[4] + x[5] + xNew
Subject to
x[1] + 4 xNew == 45
x[2] == 38
x[3] == 25
x[4] == 11
x[5] == 12
0 <= x[i] <= 1.0e6 for all i in {1,2,..,4,5}
0 <= xNew <= 1.0e6
Though we are showing only one iteration of the flow control, in the final code for sure we want to have a
while ( some condition )
(
...
)
end
block.
Now if we do not do anything else in the final code, all the names of the newly added variables by the while loop will be the same: xNew! JuMP is intelligent enough to treat them as separate variables, but it is not very human-friendly. It is more convenient if the newly added variables were given different names, which we can achieve by setName(oldName, newName) function.
In [13]:
setname(xNew, string("x[",size(A,2)+1,"]")) # Changing the name of the variable
# otherwise all the newly added variables will have name xNew!
# size(A,2) gives the column number of A
Out[13]:
"x[6]"
Let us see if the name of the variable has changed as desired.
In [14]:
print(cutstockMain) # Let us see if the name of the variables have changed as desired
Min x[1] + x[2] + x[3] + x[4] + x[5] + x[6]
Subject to
x[1] + 4 x[6] == 45
x[2] == 38
x[3] == 25
x[4] == 11
x[5] == 12
0 <= x[i] <= 1.0e6 for all i in {1,2,..,4,5}
0 <= x[6] <= 1.0e6
Indeed it has! Now let's solve the modified master problem, and then collect the associated dual variables for the equality constraints and store them in the array p.
In [15]:
statusControlFlow=solve(cutstockMain) # Solve the modified master problem
getdual(consRef)
for i in M
p[i] = getdual(consRef)[i]
end
println(p)
[0.25,1.0,1.0,1.0,1.0]
Now we solve the subproblem for the current solution of the master problem:
In [16]:
# Solving the modified sub problem
@variable(cutstockSub, 0 <= Ajstar[M] <= 1000000, Int )
@objective(cutstockSub, Min, 1-sum(p[i]*Ajstar[i] for i in M))
@constraint(cutstockSub, sum(w[i]*Ajstar[i] for i in M) <= W)
print(cutstockSub) # Let's see what is the current subproblem looks like
solve(cutstockSub)
minReducedCost=getobjectivevalue(cutstockSub)
println("Current value of the minimum of the reduced cost vector is ", minReducedCost)
Min -0.25 Ajstar[1] - Ajstar[2] - Ajstar[3] - Ajstar[4] - Ajstar[5] + 1
Subject to
22 Ajstar[1] + 42 Ajstar[2] + 52 Ajstar[3] + 53 Ajstar[4] + 78 Ajstar[5] <= 100
22 Ajstar[1] + 42 Ajstar[2] + 52 Ajstar[3] + 53 Ajstar[4] + 78 Ajstar[5] <= 100
0 <= Ajstar[i] <= 1.0e6, integer, for all i in {1,2,..,4,5}
0 <= Ajstar[i] <= 1.0e6, integer, for all i in {1,2,..,4,5}
Current value of the minimum of the reduced cost vector is -1.0
WARNING: Solver does not appear to support providing initial feasible solutions.
The optimal value of the current subproblem is negative (which will be tested by the conditional statement of the while loop in the final code), giving us a cost reducing column to be added in the master problem. As before we have to store the column Ajstar in a column vector Anew.
In [17]:
#Store the components of the solution of current subproblem into the column Anew
Anew=Float64[]
for i in 1:cardinalityM
push!(Anew, getvalue(Ajstar)[i])
end
println("New column to be added to A is: ", Anew)
New column to be added to A is: [0.0,2.0,0.0,0.0,0.0]
Okay, we have understood how JuMP is working in the column generation process. The entire code of the cutting stock problem is given below:
## Cutting stock problem code:¶
In [18]:
# Verfied to be working:
# -----------------------
using JuMP
using GLPKMathProgInterface
# Input preliminary data for starting the problem
# -----------------------------------------------
W=100
cardinalityM=5
M=collect(1:cardinalityM)
A=eye(cardinalityM)
p=zeros(5)
b=[45; 38; 25; 11; 12]
w=[22; 42; 52; 53; 78]
@time begin # time measurement begins
# Solve the master problem with the initial data
#-----------------------------------------------
cutstockMain = Model() # Model for the master problem
Jprime=collect(1:size(A,2)) # Intial number of variables
@variable(cutstockMain, 0 <= x[Jprime] <= 1000000) # Defining the variables
@objective(cutstockMain, Min, sum(1*x[j] for j in Jprime)) # Setting the objective
@constraint(cutstockMain, consRef[i=1:cardinalityM], sum(A[i,j]*x[j] for j in Jprime)==b[i]) # Adding the constraints
# Here the second argument consRef[i=1:cardinalityM] means that the i-th constraint aᵢᵀx = bᵢ has
# the corresponding constraint reference consRef[i]
solve(cutstockMain)
#Collect the dual variables for the equality constraints and store them in an array p
getdual(consRef)
for i in M
p[i] = getdual(consRef)[i] # These p[i] are the input data for the subproblem
end
# Solve the sub problem
# -------------------
cutstockSub=Model() # Model for the subproblem
@variable(cutstockSub, 0 <= Ajstar[M] <= 1000000, Int )
@objective(cutstockSub, Min, 1-sum(p[i]*Ajstar[i] for i in M))
@constraint(cutstockSub, sum(w[i]*Ajstar[i] for i in M) <= W)
solve(cutstockSub)
minReducedCost=getobjectivevalue(cutstockSub)
Anew=Float64[] # This Anew correspond to the newly added column to the A matrix
for i in 1:cardinalityM
push!(Anew, getvalue(Ajstar)[i])
end
xNewArray=Variable[] # The newly added variables by flow control will be pushed to the new array of variables xNewArray
k=1 # Counter for the while loop
# Flow control
# ------------
while minReducedCost < 0 #while (current solution of the master problem is suboptimal, i.e., subproblem objective value < 0)
# Solve the master problem by adding the new column Anew to the old A matrix
@variable(
cutstockMain, # Model to be modified
0 <= xNew <= 1000000, # New variable to be added
objective=1, # cost coefficient of new varaible in the objective
inconstraints=consRef, # constraints to be modified
coefficients=Anew # the coefficients of the variable in those constraints
)
# The line above adds the column (aᵢⱼ*)ᵢ=Aⱼ* to A <br>
# and add a corresponding new variable xⱼ* to the list of variable
push!(xNewArray, xNew) # Pushing the new variable in the array of new variables
setname(xNew, string("x[",size(A,2)+k,"]")) # Changing the name of the variable
# otherwise all the newly added variables will have name xNew!
k=k+1 # Increasing k by 1
statusControlFlow=solve(cutstockMain)
#Collect the dual variables for the equality constraints and store them in an array p
getdual(consRef)
for i in M
p[i] = getdual(consRef)[i]
end
# Solving the modified sub problem
@variable(cutstockSub, 0 <= Ajstar[M] <= 1000000, Int )
@objective(cutstockSub, Min, 1-sum{p[i]*Ajstar[i],i in M})
@constraint(cutstockSub, sum{w[i]*Ajstar[i], i in M} <= W)
solve(cutstockSub)
minReducedCost=getobjectivevalue(cutstockSub)
#Store the components of the solution of current subproblem into the column Anew
Anew=Float64[]
for i in 1:cardinalityM
push!(Anew, getvalue(Ajstar)[i])
end
end # While loop ends
end # time measurement ends
# Print the results
# -----------------
println("Objective value: ", getobjectivevalue(cutstockMain))
println("Current Solution is: ", getvalue(x))
println("With ", length(xNewArray), " variables added by flow control:")
for i in 1:length(xNewArray)
println("[",size(A,2)+i,"] = ",getvalue(xNewArray[i]))
end
println("Reduced cost of the current solution is ", getobjectivevalue(cutstockSub))
0.003651 seconds (4.00 k allocations: 265.625 KB)
Objective value: 57.25
Current Solution is: x: 1 dimensions:
[1] = 0.0
[2] = 0.0
[3] = 0.0
[4] = 0.0
[5] = 0.0
With 5 variables added by flow control:
[6] = 8.25
[7] = 1.0
[8] = 11.0
[9] = 25.0
[10] = 12.0
Reduced cost of the current solution is 0.0 | 5,990 | 19,035 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 7, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2022-33 | latest | en | 0.796927 |
https://en.wikipedia.org/wiki/Dissymmetry_of_lift | 1,544,741,084,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376825112.63/warc/CC-MAIN-20181213215347-20181214000847-00310.warc.gz | 590,339,769 | 12,569 | # Dissymmetry of lift
Dissymmetry of lift in rotorcraft aerodynamics refers to an uneven amount of lift on opposite sides of the rotor disc. It is a phenomenon that affects single-rotor helicopters and autogyros in forward flight.
A rotor blade that is moving in the same direction as the aircraft is called the advancing blade and the blade moving in the opposite direction is called the retreating blade.
Balancing lift across the rotor disc is important to a helicopter's stability. The amount of lift generated by an airfoil is proportional to the square of its airspeed. In a zero airspeed hover the rotor blades, regardless of their position in rotation, have equal airspeeds and therefore equal lift. In forward flight the advancing blade has a higher airspeed than the retreating blade, creating unequal lift across the rotor disc.[1]
## Effects
When dissymmetry causes the retreating blade to experience less airflow than required to maintain lift, a condition called retreating blade stall can occur. This causes the helicopter to roll to the retreating side and pitch up (due to gyroscopic precession). This situation, when not immediately recognized can cause a severe loss of aircraft controllability.
## Analysis
Envisage a viewpoint above a single-rotor helicopter in still air. For a stationary (hovering) helicopter, whose blades of length of r metres are rotating at ω radians per second, the blade tip is moving at a speed rω meters per second. At all points around the disc mapped out by the blade-tips, the speed of the blade-tip relative to the air is the same: everything is balanced.
Now imagine the helicopter in forward flight at, say, v meters per second. The speed of the blade-tip at point A in the diagram relative to the air is the sum of the blade-tip speed and the helicopter forward-flight speed: rω+v. But the blade-tip speed at point B, relative to the air, is the difference of its rotational speed and the forward-flight speed: rω-v.
Since the lift generated by an aerofoil increases as its relative airspeed increases, on a forward-moving helicopter the blade-tip at position A produces more lift than that at point B. So the rotor disc produces more lift on the right hand side than on the left hand side. This imbalance is the "dissymetry of lift".
## Counter-measures
Dissymmetry is countered by "blade flapping": rotor blades are designed to flap – lift and twist in such a way that the advancing blade flaps up and develops a smaller angle of attack, thus producing less lift than a rigid blade would. Conversely, the retreating blade flaps down, develops a higher angle of attack, and generates more lift.
Dissymmetry of lift is also countered by cyclic feathering; the forward input of cyclic during blowback.
To reduce dissymetry of lift, modern helicopter rotor blades are mounted in such a manner that the angle of attack varies with the position in the rotor cycle, the angle of attack being reduced on the side corresponding to position A in the diagram, and the angle of attack being increased on the side corresponding to position B in the diagram. However, there exists a limit to the degree by which dissymetry of lift can be diminished by this means, and therefore, since the forward speed v is important in the phenomenon, this imposes an upper speed limit upon the helicopter. This upper speed limit is known as VNE, the never-exceed speed. This speed is the speed beyond which the aerodynamic conditions at the rotor tips would enter unstable régimes - if v was sufficiently fast, the rotor tip at position A would be travelling fast enough through the air for the airflow to change radically as the rotor tip became supersonic, while the rotor tip at position B might have insufficient net linear speed through the air to generate meaningful lift (the stall condition - known as retreating blade stall). Entry of the rotor tip into either of these aerodynamic régimes is catastrophic from the point of view of the pilot, and the maintenance of stable forward flight.
The situation becomes more complex when helicopters with two sets of rotor blades are considered, since in theory at least, the dissymetry of lift of one rotor disc is cancelled by the increased lift of the other rotor disc: the two rotor discs of twin-rotor helicopters rotate in opposite senses, thus reversing the relevant directions of vector addition. However, as entry of the rotor tip into the supersonic aerodynamic realm is one of the unstable conditions that affects forward flight, even helicopters with two rotor discs rotating in opposite senses will be subject to a never-exceed speed. In the case of tandem-rotor helicopters such as the CH-47 Chinook, additional factors such as the aerodynamic drag of the entire design, and the available engine power, may conspire to ensure that the helicopter is incapable of achieving the VNE imposed upon it by dissymetry of lift. In the case of the Kamov Ka-50 "Black Shark", which is a coaxial design, it is possible for the helicopter to enter this aerodynamic régime as it has sufficient engine power, and pilots of this machine need to take this into consideration during the operation of the helicopter. | 1,070 | 5,204 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2018-51 | latest | en | 0.930574 |
www.perltoolbox.com | 1,580,309,340,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251799918.97/warc/CC-MAIN-20200129133601-20200129163601-00167.warc.gz | 248,667,025 | 2,739 | Back to examples
# Perl Example - Crap game (using random number generation)
```#################################################
#Version 0.1 #
#Free to use for any purpose #
#No credits or backlink required #
#Author : Shaji Kalidasan (shajiindia@yahoo.com)#
#################################################
#Game of chance: One of the most popular games of chance is a dice game
#known as craps, which is played in casinos and back alleys throughout
#the world. The rules of the game are straightforward:
#A player rolls two dice. Each die has six faces. These faces contain
#one, two, three, four, five and six spots, respectively.
#After the dice have come to rest, the sum of the spots on the two upward faces
#is calcuated. If the sum is 7 or 11 on the first throw, the player wins.
#If the sum is 2,3 or 12 on the first throw (called "craps"),
#the player loses (i.e the "house" wins).
#If the sum is 4,5,6,8,9 or 10 on the first throw, that sum becomes
#the player's "point." To win, you must continue rolling the dice until
#You lose by rolling a 7 before making a point.
use strict;
use warnings;
#Constants used in the game
use constant TRUE => 1;
use constant FALSE => 0;
#Variables used to test the state of the game
my \$WON = 0;
my \$LOST = 1;
my \$CONTINUE_ROLLING = 2;
#Other variables used in the program
my \$first_roll = 1; #true if current roll is first
my \$sum_of_dice = 0; #sum of the dice
my \$my_point = 0; #point if no win/loss on first roll
my \$game_status = \$CONTINUE_ROLLING; #game not over yet
#process one roll of the dice
sub play {
if (\$first_roll) { #First roll of the dice
\$sum_of_dice = roll_dice();
if ( \$sum_of_dice == 7 || \$sum_of_dice == 11 ) {
#Win on first roll
\$game_status = \$WON;
}
elsif ( \$sum_of_dice == 2 || \$sum_of_dice == 3 || \$sum_of_dice == 12 ) {
#Lose on first roll
\$game_status = \$LOST;
}
else {
#Remember point
\$game_status = \$CONTINUE_ROLLING;
\$my_point = \$sum_of_dice;
\$first_roll = FALSE;
}
}
else {
\$sum_of_dice = roll_dice();
if ( \$sum_of_dice == \$my_point ) {
#Win by making point
\$game_status = \$WON;
}
elsif ( \$sum_of_dice == 7 ) {
#Lose by rolling 7
\$game_status = \$LOST;
}
}
if ( \$game_status == \$CONTINUE_ROLLING ) {
print "Roll again\n";
}
else {
if ( \$game_status == \$WON ) {
print "Player wins.";
exit 0;
}
else {
print "Player loses.";
exit 0;
}
\$first_roll = TRUE;
}
} #End subroutine play
#Roll the dice
sub roll_dice {
my \$die1;
my \$die2;
my \$work_sum;
\$die1 = 1 + int rand 6;
\$die2 = 1 + int rand 6;
\$work_sum = \$die1 + \$die2;
print "Die 1 : \$die1\n";
print "Die 2 : \$die2\n";
print "Work Sum : \$work_sum\n";
return \$work_sum;
}
while (1) {
play();
}
``` | 823 | 2,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-05 | latest | en | 0.826178 |
http://www.emis.de/monographs/Trzeciak/glossae/information.html | 1,542,360,239,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039743007.0/warc/CC-MAIN-20181116091028-20181116113028-00151.warc.gz | 417,036,118 | 1,443 | ## information
Theorem 1 gives information on <about>......
Also, wherever possible, we work with integer coefficients, enabling us to obtain information about torsion.
The main information conveyed by this formula is that......
For background information, see [5].
This accords with the intuition that as we pass down the coding tree, we find out more and more detailed information about the ordering actually represented.
The interested reader is referred to [4] for further information. [Note the double r in referred.]
The survey article [5] by Diestel contains a wealth of information about the Dunford-Pettis property.
Intuitively, entropy of a partition is a measure of its information content---the larger the entropy, the larger the information content.
[Note that information has no plural and does not appear with “an''; on the other hand, you can say: an interesting piece of information.]
Back to main page | 190 | 929 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2018-47 | latest | en | 0.875085 |
https://www.mrexcel.com/board/threads/finding-a-value-in-a-row.77303/ | 1,656,874,120,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104248623.69/warc/CC-MAIN-20220703164826-20220703194826-00213.warc.gz | 913,605,903 | 14,132 | # finding a value in a row
#### boris2004
##### New Member
Hello
i can find a value of AG in any row by using;
Row = sum row number
Worksheets("Sheet1").Cells(Row, 33).Value
But if i was to use;
Worksheets("Sheet1").Range("A6:B6").Offset(2, 0).Value
'.Offset(2, 0) = row 8
how would i then find the value of the 'offset row' (8) in cell AG??
Cheers guys,
Boris
### Excel Facts
Why are there 1,048,576 rows in Excel?
The Excel team increased the size of the grid in 2007. There are 2^20 rows and 2^14 columns for a total of 17 billion cells.
#### Greg Truby
##### MrExcel MVP
Worksheets("Sheet1").Range("A6:B6").Offset(2, 32).Resize(1,1).Value
#### DRJ
##### MrExcel MVP
x = Range("AG" & Range("B6").Offset(2, 0).Row)
#### boris2004
##### New Member
Cheers guys it worked!!!
Cheers,
Boris
Excel contains over 450 functions, with more added every year. That’s a huge number, so where should you start? Right here with this bundle.
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Go back | 579 | 2,045 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2022-27 | latest | en | 0.738673 |
https://nrich.maths.org/public/topic.php?code=-445&cl=3&cldcmpid=866 | 1,586,129,776,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00432.warc.gz | 614,111,496 | 8,826 | # Resources tagged with: Maths Supporting SET
Filter by: Content type:
Age range:
Challenge level:
### There are 66 results
Broad Topics > Cross-curricular Connections > Maths Supporting SET
### Efficient Packing
##### Age 14 to 16 Challenge Level:
How efficiently can you pack together disks?
### Track Design
##### Age 14 to 16 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### Gym Bag
##### Age 11 to 16 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Investigating the Dilution Series
##### Age 14 to 16 Challenge Level:
Which dilutions can you make using only 10ml pipettes?
### Construct the Solar System
##### Age 14 to 18 Challenge Level:
Make an accurate diagram of the solar system and explore the concept of a grand conjunction.
### Witch's Hat
##### Age 11 to 16 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Air Nets
##### Age 7 to 18 Challenge Level:
Can you visualise whether these nets fold up into 3D shapes? Watch the videos each time to see if you were correct.
### Far Horizon
##### Age 14 to 16 Challenge Level:
An observer is on top of a lighthouse. How far from the foot of the lighthouse is the horizon that the observer can see?
### Speedy Sidney
##### Age 11 to 14 Challenge Level:
Two trains set off at the same time from each end of a single straight railway line. A very fast bee starts off in front of the first train and flies continuously back and forth between the. . . .
### How Do You React?
##### Age 14 to 16 Challenge Level:
To investigate the relationship between the distance the ruler drops and the time taken, we need to do some mathematical modelling...
### Chocolate Cake
##### Age 11 to 14 Challenge Level:
If I don't have the size of cake tin specified in my recipe, will the size I do have be OK?
### Mystery Procedure
##### Age 14 to 16 Challenge Level:
Can you work out what this procedure is doing?
### Alternative Record Book
##### Age 14 to 18 Challenge Level:
In which Olympic event does a human travel fastest? Decide which events to include in your Alternative Record Book.
### Genetics
##### Age 14 to 16 Challenge Level:
A problem about genetics and the transmission of disease.
### What's That Graph?
##### Age 14 to 16 Challenge Level:
Can you work out which processes are represented by the graphs?
### Designing Table Mats
##### Age 11 to 16 Challenge Level:
Formulate and investigate a simple mathematical model for the design of a table mat.
### Fill Me up Too
##### Age 14 to 16 Challenge Level:
In Fill Me Up we invited you to sketch graphs as vessels are filled with water. Can you work out the equations of the graphs?
### Guessing the Graph
##### Age 14 to 16 Challenge Level:
Can you suggest a curve to fit some experimental data? Can you work out where the data might have come from?
### Big and Small Numbers in Chemistry
##### Age 14 to 16 Challenge Level:
Get some practice using big and small numbers in chemistry.
##### Age 14 to 18 Challenge Level:
How would you design the tiering of seats in a stadium so that all spectators have a good view?
### Investigating Epidemics
##### Age 11 to 16 Challenge Level:
Simple models which help us to investigate how epidemics grow and die out.
### Make Your Own Pencil Case
##### Age 11 to 14 Challenge Level:
What shape would fit your pens and pencils best? How can you make it?
### 3D Drawing
##### Age 11 to 16
The design technology curriculum requires students to be able to represent 3-dimensional objects on paper. This article introduces some of the mathematical ideas which underlie such methods.
### Pinhole Camera
##### Age 11 to 14 Challenge Level:
Make your own pinhole camera for safe observation of the sun, and find out how it works.
### Elastic Maths
##### Age 14 to 18
How do you write a computer program that creates the illusion of stretching elastic bands between pegs of a Geoboard? The answer contains some surprising mathematics.
### Truth Tables and Electronic Circuits
##### Age 11 to 18
Investigate circuits and record your findings in this simple introduction to truth tables and logic.
### Isometric Drawing
##### Age 11 to 14 Challenge Level:
Explore the properties of isometric drawings.
### Maths Filler
##### Age 11 to 14 Challenge Level:
Imagine different shaped vessels being filled. Can you work out what the graphs of the water level should look like?
##### Age 11 to 14 Challenge Level:
Is it cheaper to cook a meal from scratch or to buy a ready meal? What difference does the number of people you're cooking for make?
### Perspective Drawing
##### Age 11 to 16 Challenge Level:
Explore the properties of perspective drawing.
### Food Chains
##### Age 11 to 14 Challenge Level:
When a habitat changes, what happens to the food chain?
##### Age 11 to 14 Challenge Level:
Can you work out which drink has the stronger flavour?
### Does This Sound about Right?
##### Age 11 to 14 Challenge Level:
Examine these estimates. Do they sound about right?
### Speed-time Problems at the Olympics
##### Age 14 to 16 Challenge Level:
Have you ever wondered what it would be like to race against Usain Bolt?
### Robot Camera
##### Age 14 to 16 Challenge Level:
Could nanotechnology be used to see if an artery is blocked? Or is this just science fiction?
### Triathlon and Fitness
##### Age 11 to 14 Challenge Level:
The triathlon is a physically gruelling challenge. Can you work out which athlete burnt the most calories?
### Electric Kettle
##### Age 14 to 16 Challenge Level:
Explore the relationship between resistance and temperature
### Biology Measurement Challenge
##### Age 14 to 16 Challenge Level:
Analyse these beautiful biological images and attempt to rank them in size order.
### Temperature
##### Age 11 to 14 Challenge Level:
Is there a temperature at which Celsius and Fahrenheit readings are the same?
### Olympic Measures
##### Age 11 to 14 Challenge Level:
These Olympic quantities have been jumbled up! Can you put them back together again?
### Data Matching
##### Age 14 to 18 Challenge Level:
Use your skill and judgement to match the sets of random data.
### Big and Small Numbers in Biology
##### Age 14 to 16 Challenge Level:
Work with numbers big and small to estimate and calulate various quantities in biological contexts.
### Olympic Records
##### Age 11 to 14 Challenge Level:
Can you deduce which Olympic athletics events are represented by the graphs?
### More or Less?
##### Age 14 to 16 Challenge Level:
Are these estimates of physical quantities accurate?
### Bigger or Smaller?
##### Age 14 to 16 Challenge Level:
When you change the units, do the numbers get bigger or smaller?
### Big and Small Numbers in the Physical World
##### Age 14 to 16 Challenge Level:
Work with numbers big and small to estimate and calculate various quantities in physical contexts.
##### Age 14 to 16 Challenge Level:
Which units would you choose best to fit these situations?
### Nutrition and Cycling
##### Age 14 to 16 Challenge Level:
Andy wants to cycle from Land's End to John o'Groats. Will he be able to eat enough to keep him going?
### Counting Dolphins
##### Age 14 to 16 Challenge Level:
How would you go about estimating populations of dolphins?
### A Question of Scale
##### Age 14 to 16 Challenge Level:
Use your skill and knowledge to place various scientific lengths in order of size. Can you judge the length of objects with sizes ranging from 1 Angstrom to 1 million km with no wrong attempts? | 1,692 | 7,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2020-16 | latest | en | 0.856981 |
https://oeis.org/A302703 | 1,653,024,196,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531352.50/warc/CC-MAIN-20220520030533-20220520060533-00477.warc.gz | 498,853,142 | 4,589 | The OEIS is supported by the many generous donors to the OEIS Foundation.
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A302703 G.f. A(x) satisfies: [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n+1))^(n+1) for n>=0. 2
1, 1, 3, 21, 235, 3470, 61933, 1274893, 29423331, 747440115, 20636072811, 613611700946, 19517927805840, 660667692682175, 23699856058131981, 897955765812058192, 35832679277251514074, 1502303284645831488072, 66031982339561373164915, 3036884343153028302140119, 145885192794643951791449387 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 LINKS Paul D. Hanna, Table of n, a(n) for n = 0..300 FORMULA G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies: (1) [x^n] A(x)^(n+1) = [x^n] (1 + x*A(x)^(n+1))^(n+1) for n>=0. (2) A(x) = Sum_{n>=0} b(n) * x^n/A(x)^n, where b(n) = [x^n] (1 + x*A(x)^(n+1))^(n+1) / (n+1). a(n) ~ c * d^n * n! * n^alfa, where d = 2.12460658362428979..., alfa = 2.20132968515..., c = 0.026186121837... - Vaclav Kotesovec, Oct 06 2020 EXAMPLE G.f.: A(x) = 1 + x + 3*x^2 + 21*x^3 + 235*x^4 + 3470*x^5 + 61933*x^6 + 1274893*x^7 + 29423331*x^8 + 747440115*x^9 + 20636072811*x^10 + ... RELATED SERIES. G.f. A(x) = B(x/A(x)) where B(x) = B(x*A(x)) begins: B(x) = 1 + x + 4*x^2 + 31*x^3 + 356*x^4 + 5291*x^5 + 94592*x^6 + 1948763*x^7 + 45025516*x^8 + 1145651239*x^9 + 31696223593*x^10 + ... + b(n)*x^n + ... such that b(n) = [x^n] (1 + x*A(x)^(n+1))^(n+1) / (n+1), as well as b(n) = [x^n] A(x)^(n+1) / (n+1), so that b(n) begin: [1, 2/2, 12/3, 124/4, 1780/5, 31746/6, 662144/7, 15590104/8, ...] ILLUSTRATION OF DEFINITION. The table of coefficients of x^k in A(x)^(n+1) begins: n=0: [1, 1, 3, 21, 235, 3470, 61933, 1274893, ...]; n=1: [1, 2, 7, 48, 521, 7536, 132657, 2704342, ...]; n=2: [1, 3, 12, 82, 867, 12288, 213282, 4304877, ...]; n=3: [1, 4, 18, 124, 1283, 17828, 305056, 6094832, ...]; n=4: [1, 5, 25, 175, 1780, 24271, 409380, 8094540, ...]; n=5: [1, 6, 33, 236, 2370, 31746, 527824, 10326546, ...]; n=6: [1, 7, 42, 308, 3066, 40397, 662144, 12815839, ...]; n=7: [1, 8, 52, 392, 3882, 50384, 814300, 15590104, ...]; ... Compare to the table of coefficients in (1 + x*A(x)^(n+1))^(n+1): n=0: [1, 1, 1, 3, 21, 235, 3470, 61933, ...]; n=1: [1, 2, 5, 18, 114, 1166, 16355, 283142, ...]; n=2: [1, 3, 12, 55, 354, 3372, 44463, 739917, ...]; n=3: [1, 4, 22, 124, 857, 7908, 98244, 1558788, ...]; n=4: [1, 5, 35, 235, 1780, 16501, 195980, 2955095, ...]; n=5: [1, 6, 51, 398, 3321, 31746, 368032, 5294250, ...]; n=6: [1, 7, 70, 623, 5719, 57302, 662144, 9182013, ...]; n=7: [1, 8, 92, 920, 9254, 98088, 1149804, 15590104, ...]; ... to see that the main diagonals of the tables are the same. PROG (PARI) {a(n) = my(A=[1]); for(m=1, n, A=concat(A, 0); A[m+1] = (Vec((1+x*Ser(A)^(m+1))^(m+1))[m+1] - Vec(Ser(A)^(m+1))[m+1])/(m+1) ); A[n+1]} for(n=0, 30, print1(a(n), ", ")) CROSSREFS Cf. A302702. Sequence in context: A078586 A179331 A138903 * A334262 A234855 A058562 Adjacent sequences: A302700 A302701 A302702 * A302704 A302705 A302706 KEYWORD nonn AUTHOR Paul D. Hanna, Apr 16 2018 STATUS approved
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Last modified May 20 01:04 EDT 2022. Contains 353847 sequences. (Running on oeis4.) | 1,626 | 3,391 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2022-21 | latest | en | 0.442108 |
http://clay6.com/qa/7682/what-is-the-value-of-lim-large-frac- | 1,481,198,759,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542588.29/warc/CC-MAIN-20161202170902-00379-ip-10-31-129-80.ec2.internal.warc.gz | 57,640,904 | 25,867 | Browse Questions
# What is the value of $\;$ $\lim_{x\to 0} \large \frac{a^x-1}{x}$?
Answer: $\;$ $\lim_{x\to 0} \large \frac{a^x-1}{x} = \log_e a$ | 67 | 149 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2016-50 | longest | en | 0.374326 |
https://www.minimins.com/threads/just-a-question-about-pointing-foods.11296/ | 1,544,914,765,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376827137.61/warc/CC-MAIN-20181215222234-20181216004234-00151.warc.gz | 991,037,721 | 16,510 | # Just a question about pointing foods
#### Bethany246
##### Banned
Hiya anybody doing weight watchers
Im a little bit confused about the foods in the shopping guide it says For instance Bread white wholemeal or brown 1 med slice 35g = 1 point so does this mean you have to weight the bread what you get till it weighs 35 grams.
And same as
Beef rump steak uncooked 1 medium 225g = 5 points have you to weigh this till it weighs this
And Gammon steak uncooked 1 med 170g =6 points
If you have to weigh things like this it will be boring for me
Hope someone can help
Bethany
#### ANNTEXT
##### Full Member
weighing food is the worst thing about w.w but i recommed that u do -
slice of bread - well u got nvs on the back so u can work it out yourself with out weighing most bread is - ranges from 0.5pts per slice or two for 1.5pts
or 1.5pts and two for 2.5pts
but the meat i would sure say weigh it first before eating else your just go over your points!
if its says 165g yeah just weigh it till it says that much xx hope this helps!
#### Starlight
##### Gold Member
I dont weigh my bread I just take one slice to be one point.
As far as steak etc goes, I normally weigh the portion I have then work out the points.
If its something like mince I weigh it til I get to the amount specified in the points.
I think if you want any diet to succeed, be it CD, WW SW or whatever you have to expect some element of work or sacrifices. If weighing a few things is all it takes to make WW work then its well worth it.
It depends how much you want the weight loss
#### Bethany246
##### Banned
Thanks everyone
Starlight so if you was having say some mince and you bought a pack and it weighed 400g and there was 2 sharing you would just halve it and point it out is that right.
Or same as if you was having a piece of Steak and it was 300g you would just calculate the points for that amount
It says in the shopping guide wafer thin chickenmed portion 30g so that means you weigh this out and that is only 1/2 point.
Same as you say there is always a sacriffice with all of them but it is the after results you want.
I think i will be better when i have finally joined ww meeting then i will know how to go on.
Bethany
#### Starlight
##### Gold Member
Exactly! I usually buy a pack of mince and make bolognese sauce. I divide it into maybe 2 or 3 portions so add the points of the mince and the sauce and divide it by the portions. Its easy, once youve been doing it a while you dont even give it a 2nd thought
#### ANNTEXT
##### Full Member
sorry to butt in --
i would say go carefull and weigh it as sometimes it not right -
i brought meat the other day (chicken) and it said 45g but in fact it was 55g - so i had to have add 0.5pts!
#### Starlight
##### Gold Member
Youre dead right Ann, I always weigh things like my mince etc, unless it comes out of a packet with all the points on it. Guessing is no use, a half point here and there can make all the difference between a loss and a gain
#### Bethany246
##### Banned
Thanks starlight
You have been a great help How many points are you roughly allowed for your breakfast and lunch as i think i will be on 27.
What happens if you go over your points say 3 or 4 would it stop your weightloss for the week.
Do you still keep a check of what you are eating.
How is your new scales going are they better than the original ones like i had
Kind Regards
Bethany
#### ANNTEXT
##### Full Member
thanks starlight was just trying too be helpfull thats all x
#### Starlight
##### Gold Member
Thanks starlight
You have been a great help How many points are you roughly allowed for your breakfast and lunch as i think i will be on 27.
What happens if you go over your points say 3 or 4 would it stop your weightloss for the week.
Do you still keep a check of what you are eating.
How is your new scales going are they better than the original ones like i had
Kind Regards
Bethany
Im on a similar number of points to you Bethany. I usually have about 3 for breakfast, Im not a morning person lol Maybe about 4/6 for lunch, about 10 for dinner and the rest goes on snacks etc but thats just how I work it, how you split your points up is totally your decision, just dont go over your daily allowance
I never go over my points. I try and keep a small 'bank' of points and even on my treat night I work it round my points.
I love my new scales, tho theyre probably prettymuch like yours, they make things SO easy
#### Starlight
##### Gold Member
thanks starlight was just trying too be helpfull thats all x
You are Ann Thats what these forums are all about, for us all to chip in and help each other.
You jump in anytime
#### Bethany246
##### Banned
Hiya did not mean nothing wrong about starlight she has been a good help and so have you Ann
Bethany
#### Bethany246
##### Banned
I would like to say good night to starlight and Ann for being really helpful.
Take Care
Thanks Again both of you
Bethany | 1,231 | 4,976 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2018-51 | latest | en | 0.956861 |
https://www.jiskha.com/display.cgi?id=1268593196 | 1,516,631,290,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891377.59/warc/CC-MAIN-20180122133636-20180122153636-00473.warc.gz | 933,592,246 | 4,072 | # math
posted by .
A new temperature scale has the freezing point of water at 50 degrees and the boiling point of water at 125 degrees. Normal body temperature is 37 degrees C and 98.6 degrees F. Water freezes at 0 degrees C and 32 degrees F and boils at 100 degrees C and 212 degrees F. What is normal body temperature on this new scale?
## Similar Questions
1. ### SCIENCE
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2. ### Technology
Please check this for me: 5. The freezing point of water on the Celsius temperature scale is a. 0 degrees b. 32 degrees c. 100 degrees d. 212 degrees A 8. A measurement for a rectangular box is given as 150 cubic inches. This measurement …
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when 84 joules of heat are added to 2.0 grams of water at 15 degrees C, what will be the final temperature of the water?
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6. ### Calculus
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On planet X, the alien residents use a temperature scale in which they define 0 degrees X to be the melting point of nitrogen and 100 degrees X to be the boiling point of nitrogen. What temperature difference in X degrees does a temperature …
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# The Size of the Sun
Explore the size relationship between the sun and Earth by using tape and stickers. Learners estimate, then place and count the number of one-inch diameter stickers (representing Earths) that would fit across the diameter of a nine-foot circle of... (View More)
# Our Neighborhood in the Universe
Using stickers created from the templates provided, students create a Venn diagram of objects in our solar system, our galaxy and the universe. This short activity can be used as a formative assessment.
# Planet Hunters Educators Guide
Using the 5-E model, these lessons introduce planets, planetary systems, star types, exoplanets, transits, light curves, and the Planet Hunters citizen science project. Supplemental materials include data/image sheets. Next Generation Science... (View More)
# Designing an Open Spectrograph
Learners will build an open spectrograph to calculate the angle the light is transmitted through a holographic diffraction grating. After finding the desired angles, the students will design their own spectrograph using the information learned. The... (View More)
# Using Spectral Data to Explore Saturn and Titan
Learners will compare known elemental spectra with spectra of Titan and Saturn’s rings from a spectrometer aboard the NASA Cassini spacecraft. They identify the elements visible in the planetary and lunar spectra. The activity is part of Project... (View More)
# A Spectral Mystery
Learners will use a spectrograph to gather data about light sources. Using the data they’ve collected, students are able to make comparisons between different light sources and make conjectures about the composition of a mystery light source. The... (View More)
# Patterns and Fingerprints
This is an activity about detecting elements by using light. Learners will develop and apply methods to identify and interpret patterns to the identification of fingerprints. They look at fingerprints of their classmates, snowflakes, and finally... (View More)
# Using a Fancy Spectrograph
Learners will look at various light sources (including glow sticks and Christmas lights) and make conjectures about their composition. The activity is part of Project Spectra, a science and engineering program for middle-high school students,... (View More)
# The Size of Things
In this hands-on activity, learners begin by estimating the size of each planet in our Solar System and Pluto and making each out of playdough or a similar material. Then, learners follow specific instructions to divide a mass of playdough into the... (View More)
# Measuring Solar Rotation - Image Scale Method
This is an activity about the period of the Sun’s rotation. Learners will select images of the Sun from the SOHO spacecraft image archive. Next, they will calculate an image scale for the selected solar images. Then, they will use it to help... (View More)
Audience: Elementary school, Middle school, High school
Materials Cost: Free
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https://www.laopaibet.com/2018/04/16/att/l-shaped-couch-in-small-room-nice-look-7-16-l-shaped-sofa-for-small-living-room-velvet-and-intended-couch-design-14-11937868/ | 1,534,498,924,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221211935.42/warc/CC-MAIN-20180817084620-20180817104620-00045.warc.gz | 887,596,830 | 19,925 | # L Shaped Couch In Small Room Nice Look #7 16 L Shaped Sofa For Small Living Room Velvet And Intended Couch Design 14
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### L
Roman numerals,
• the numerals in the ancient Roman system of notation, still used for certain limited purposes, as in some pagination, dates on buildings, etc. The common basic symbols are I (=1), V (=5), X (=10), L (=50), C (=100), D (=500), and M (=1000). The Roman numerals for one to nine are: I, II, III, IV, V, VI, VII, VIII, IX. A bar over a letter multiplies it by 1000;
thus, X̄ equals 10,000. Integers are written according to these two rules: If a letter is immediately followed by one of equal or lesser value, the two values are added;
thus, XX equals 20, XV equals 15, VI equals 6. If a letter is immediately followed by one of greater value, the first is subtracted from the second;
thus, IV equals 4, XL equals 40, CM equals 900. Examples: XLVII(=47), CXVI(=116), MCXX(=1120), MCMXIV(=1914). Roman numerals may be written in lowercase letters, though they appear more commonly in capitals.
• ### Shaped
1. of a definite form, shape, or character (often used in combination):aU-shaped driveway.
2. designed to fit a particular form, body, or contour: a shaped garment.
3. having other than a plane surface.
### Couch
couch (kouch or, for 6, 15, ko̅o̅ch),USA pronunciation n.
1. a piece of furniture for seating from two to four people, typically in the form of a bench with a back, sometimes having an armrest at one or each end, and partly or wholly upholstered and often fitted with springs, tailored cushions, skirts, etc.;
sofa.
2. a similar article of furniture, with a headrest at one end, on which some patients of psychiatrists or psychoanalysts lie while undergoing treatment.
3. a bed or other place of rest;
a lounge;
any place used for repose.
4. the lair of a wild beast.
5. [Brewing.]the frame on which barley is spread to be malted.
6. [Papermaking.]the board or felt blanket on which wet pulp is laid for drying into paper sheets.
7. a primer coat or layer, as of paint.
8. on the couch, [Informal.]undergoing psychiatric or psychoanalytic treatment.
v.t.
1. to arrange or frame (words, a sentence, etc.);
put into words;
express: a simple request couched in respectful language.
2. to express indirectly or obscurely: the threat couched under his polite speech.
3. to lower or bend down, as the head.
4. to lower (a spear, lance, etc.) to a horizontal position, as for attack.
5. to put or lay down, as for rest or sleep;
cause to lie down.
6. to lay or spread flat.
7. [Papermaking.]to transfer (a sheet of pulp) from the wire to the couch.
8. to embroider by couching.
9. [Archaic.]to hide;
conceal.
v.i.
1. to lie at rest or asleep;
repose;
recline.
2. to crouch;
bend;
stoop.
3. to lie in ambush or in hiding;
lurk.
4. to lie in a heap for decomposition or fermentation, as leaves.
### In
prep.
1. (used to indicate inclusion within space, a place, or limits): walking in the park.
2. (used to indicate inclusion within something abstract or immaterial): in politics; in the autumn.
3. (used to indicate inclusion within or occurrence during a period or limit of time): in ancient times; a task done in ten minutes.
4. (used to indicate limitation or qualification, as of situation, condition, relation, manner, action, etc.): to speak in a whisper; to be similar in appearance.
5. (used to indicate means): sketched in ink; spoken in French.
6. (used to indicate motion or direction from outside to a point within) into: Let's go in the house.
7. (used to indicate transition from one state to another): to break in half.
8. (used to indicate object or purpose): speaking in honor of the event.
9. in that, because;
inasmuch as: In that you won't have time for supper, let me give you something now.
1. in or into some place, position, state, relation, etc.: Please come in.
2. on the inside;
within.
3. in one's house or office.
4. in office or power.
5. in possession or occupancy.
6. having the turn to play, as in a game.
7. [Baseball.](of an infielder or outfielder) in a position closer to home plate than usual;
short: The third baseman played in, expecting a bunt.
8. on good terms;
in favor: He's in with his boss, but he doubts it will last.
9. in vogue;
in style: He says straw hats will be in this year.
10. in season: Watermelons will soon be in.
11. be in for, to be bound to undergo something, esp. a disagreeable experience: We are in for a long speech.
12. in for it, [Slang.]about to suffer chastisement or unpleasant consequences, esp. of one's own actions or omissions: I forgot our anniversary again, and I'll be in for it now.Also,[Brit.,] for it.
13. in with, on friendly terms with;
familiar or associating with: They are in with all the important people.
1. located or situated within;
inner;
internal: the in part of a mechanism.
2. [Informal.]
• in favor with advanced or sophisticated people;
fashionable;
stylish: the in place to dine; Her new novel is the in book to read this summer.
• comprehensible only to a special or ultrasophisticated group: an in joke.
3. well-liked;
included in a favored group.
4. inward;
incoming;
inbound: an in train.
5. plentiful;
available.
6. being in power, authority, control, etc.: a member of the in party.
7. playing the last nine holes of an eighteen-hole golf course (opposed to out): His in score on the second round was 34.
n.
1. Usually, ins. persons in office or political power (distinguished from outs).
2. a member of the political party in power: The election made him an in.
3. pull or influence;
a social advantage or connection: He's got an in with the senator.
4. (in tennis, squash, handball, etc.) a return or service that lands within the in-bounds limits of a court or section of a court (opposed to out).
v.t. Brit. [Dial.]
1. to enclose.
### Small
1. of limited size;
of comparatively restricted dimensions;
not big;
little: a small box.
2. slender, thin, or narrow: a small waist.
3. not large as compared with others of the same kind: a small elephant.
4. (of letters) lower-case (def. 1).
5. not great in amount, degree, extent, duration, value, etc.: a small salary.
6. not great numerically: a small army.
7. of low numerical value;
denoted by a low number.
8. having but little land, capital, power, influence, etc., or carrying on business or some activity on a limited scale: a small enterprise.
9. of minor importance, moment, weight, or consequence: a small problem.
10. humble, modest, or unpretentious: small circumstances.
11. characterized by or indicative of littleness of mind or character;
mean-spirited;
petty: a small, miserly man.
12. of little strength or force: a small effort.
13. (of sound or the voice) gentle;
with little volume.
14. very young: when I was a small boy.
15. diluted;
weak.
16. feel small, to be ashamed or mortified: Her unselfishness made me feel small.
1. in a small manner: They talked big but lived small.
2. into small pieces: Slice the cake small.
3. in low tones;
softly.
n.
1. something that is small: Do you prefer the small or the large?
2. a small or narrow part, as of the back.
3. those who are small: Democracy benefits the great and the small.
4. smalls, small goods or products.
5. smalls, [Brit.]
• underclothes.
• household linen, as napkins, pillowcases, etc.
6. smalls, [Brit. Informal.]the responsions at Oxford University.
7. smalls, coal, ore, gangue, etc., in fine particles.
smallness, n.
### Room
room (ro̅o̅m, rŏŏm),USA pronunciation n.
1. a portion of space within a building or other structure, separated by walls or partitions from other parts: a dining room.
2. rooms, lodgings or quarters, as in a house or building.
3. the persons present in a room: The whole room laughed.
4. space or extent of space occupied by or available for something: The desk takes up too much room.
5. opportunity or scope for something: room for improvement; room for doubt.
6. status or a station in life considered as a place: He fought for room at the top.
7. capacity: Her brain had no room for trivia.
8. a working area cut between pillars.
v.i.
1. to occupy a room or rooms;
lodge.
### Nice
nice (nīs),USA pronunciation adj., nic•er, nic•est.
1. pleasing;
agreeable;
delightful: a nice visit.
2. amiably pleasant;
kind: They are always nice to strangers.
3. characterized by, showing, or requiring great accuracy, precision, skill, tact, care, or delicacy: nice workmanship; a nice shot; a nice handling of a crisis.
4. showing or indicating very small differences;
minutely accurate, as instruments: a job that requires nice measurements.
5. minute, fine, or subtle: a nice distinction.
6. having or showing delicate, accurate perception: a nice sense of color.
7. refined in manners, language, etc.: Nice people wouldn't do such things.
8. virtuous;
respectable;
decorous: a nice girl.
9. suitable or proper: That was not a nice remark.
10. carefully neat in dress, habits, etc.
11. (esp. of food) dainty or delicate.
12. having fastidious, finicky, or fussy tastes: They're much too nice in their dining habits to enjoy an outdoor barbecue.
13. [Obs.]coy, shy, or reluctant.
14. [Obs.]unimportant;
trivial.
15. [Obs.]wanton.
16. make nice, to behave in a friendly, ingratiating, or conciliatory manner.
17. nice and, sufficiently: It's nice and warm in here.
niceness, n.
### L
Roman numerals,
• the numerals in the ancient Roman system of notation, still used for certain limited purposes, as in some pagination, dates on buildings, etc. The common basic symbols are I (=1), V (=5), X (=10), L (=50), C (=100), D (=500), and M (=1000). The Roman numerals for one to nine are: I, II, III, IV, V, VI, VII, VIII, IX. A bar over a letter multiplies it by 1000;
thus, X̄ equals 10,000. Integers are written according to these two rules: If a letter is immediately followed by one of equal or lesser value, the two values are added;
thus, XX equals 20, XV equals 15, VI equals 6. If a letter is immediately followed by one of greater value, the first is subtracted from the second;
thus, IV equals 4, XL equals 40, CM equals 900. Examples: XLVII(=47), CXVI(=116), MCXX(=1120), MCMXIV(=1914). Roman numerals may be written in lowercase letters, though they appear more commonly in capitals.
• ### Shaped
1. of a definite form, shape, or character (often used in combination):aU-shaped driveway.
2. designed to fit a particular form, body, or contour: a shaped garment.
3. having other than a plane surface.
### Sofa
so•fa (sōfə),USA pronunciation n.
1. a long, upholstered couch with a back and two arms or raised ends.
### For
for (fôr; unstressed fər),USA pronunciation prep.
1. with the object or purpose of: to run for exercise.
2. intended to belong to, or be used in connection with: equipment for the army; a closet for dishes.
3. suiting the purposes or needs of: medicine for the aged.
4. in order to obtain, gain, or acquire: a suit for alimony; to work for wages.
5. (used to express a wish, as of something to be experienced or obtained): O, for a cold drink!
6. sensitive or responsive to: an eye for beauty.
7. desirous of: a longing for something; a taste for fancy clothes.
8. in consideration or payment of;
in return for: three for a dollar; to be thanked for one's efforts.
9. appropriate or adapted to: a subject for speculation; clothes for winter.
10. with regard or respect to: pressed for time; too warm for April.
11. during the continuance of: for a long time.
12. in favor of;
on the side of: to be for honest government.
13. in place of;
instead of: a substitute for butter.
14. in the interest of;
on behalf of: to act for a client.
15. in exchange for;
as an offset to: blow for blow; money for goods.
16. in punishment of: payment for the crime.
17. in honor of: to give a dinner for a person.
18. with the purpose of reaching: to start for London.
19. contributive to: for the advantage of everybody.
20. in order to save: to flee for one's life.
21. in order to become: to train recruits for soldiers.
22. in assignment or attribution to: an appointment for the afternoon; That's for you to decide.
23. such as to allow of or to require: too many for separate mention.
24. such as results in: his reason for going.
25. as affecting the interests or circumstances of: bad for one's health.
26. in proportion or with reference to: He is tall for his age.
27. in the character of;
as being: to know a thing for a fact.
28. by reason of;
because of: to shout for joy; a city famed for its beauty.
29. in spite of: He's a decent guy for all that.
30. to the extent or amount of: to walk for a mile.
31. (used to introduce a subject in an infinitive phrase): It's time for me to go.
32. (used to indicate the number of successes out of a specified number of attempts): The batter was 2 for 4 in the game.
33. for it, See in (def. 21).
conj.
1. seeing that;
since.
2. because.
### Small
1. of limited size;
of comparatively restricted dimensions;
not big;
little: a small box.
2. slender, thin, or narrow: a small waist.
3. not large as compared with others of the same kind: a small elephant.
4. (of letters) lower-case (def. 1).
5. not great in amount, degree, extent, duration, value, etc.: a small salary.
6. not great numerically: a small army.
7. of low numerical value;
denoted by a low number.
8. having but little land, capital, power, influence, etc., or carrying on business or some activity on a limited scale: a small enterprise.
9. of minor importance, moment, weight, or consequence: a small problem.
10. humble, modest, or unpretentious: small circumstances.
11. characterized by or indicative of littleness of mind or character;
mean-spirited;
petty: a small, miserly man.
12. of little strength or force: a small effort.
13. (of sound or the voice) gentle;
with little volume.
14. very young: when I was a small boy.
15. diluted;
weak.
16. feel small, to be ashamed or mortified: Her unselfishness made me feel small.
1. in a small manner: They talked big but lived small.
2. into small pieces: Slice the cake small.
3. in low tones;
softly.
n.
1. something that is small: Do you prefer the small or the large?
2. a small or narrow part, as of the back.
3. those who are small: Democracy benefits the great and the small.
4. smalls, small goods or products.
5. smalls, [Brit.]
• underclothes.
• household linen, as napkins, pillowcases, etc.
6. smalls, [Brit. Informal.]the responsions at Oxford University.
7. smalls, coal, ore, gangue, etc., in fine particles.
smallness, n.
### Living
1. having life;
being alive;
2. in actual existence or use;
extant: living languages.
3. active or thriving;
vigorous;
strong: a living faith.
4. burning or glowing, as a coal.
5. flowing freely, as water.
6. pertaining to, suitable for, or sufficient for existence or subsistence: living conditions; a living wage.
7. of or pertaining to living persons: within living memory.
8. lifelike;
true to life, as a picture or narrative.
9. in its natural state and place;
not uprooted, changed, etc.: living rock.
10. very;
absolute (used as an intensifier): to scare the living daylights out of someone.
n.
1. the act or condition of a person or thing that lives: Living is very expensive these days.
2. the means of maintaining life;
livelihood: to earn one's living.
3. a particular manner, state, or status of life: luxurious living.
4. (used with a pl. v.) living persons collectively (usually prec. by the): glad to be among the living.
5. the benefice of a clergyman.
living•ness, n.
### Room
room (ro̅o̅m, rŏŏm),USA pronunciation n.
1. a portion of space within a building or other structure, separated by walls or partitions from other parts: a dining room.
2. rooms, lodgings or quarters, as in a house or building.
3. the persons present in a room: The whole room laughed.
4. space or extent of space occupied by or available for something: The desk takes up too much room.
5. opportunity or scope for something: room for improvement; room for doubt.
6. status or a station in life considered as a place: He fought for room at the top.
7. capacity: Her brain had no room for trivia.
8. a working area cut between pillars.
v.i.
1. to occupy a room or rooms;
lodge.
### Velvet
vel•vet (velvit),USA pronunciation n.
1. a fabric of silk, nylon, acetate, rayon, etc., sometimes having a cotton backing, with a thick, soft pile formed of loops of the warp thread either cut at the outer end or left uncut.
2. something likened to the fabric velvet, as in softness or texture: the velvet of her touch; the velvet of the lawn.
3. the soft, deciduous covering of a growing antler.
4. a very pleasant, luxurious, desirable situation.
• money gained through gambling;
winnings.
• clear gain or profit, esp. when more than anticipated.
1. Also, velvet•ed. made of velvet or covered with velvet.
2. Also, velvet•like′. resembling or suggesting velvet;
smooth;
soft;
velvety: a velvet night; a cat's velvet fur.
### And
and (and; unstressed ənd, ən, or, esp. after a homorganic consonant, n),USA pronunciation conj.
1. (used to connect grammatically coordinate words, phrases, or clauses) along or together with;
as well as;
besides;
also;
moreover: pens and pencils.
plus: 2 and 2 are 4.
3. then: He read for an hour and went to bed.
4. also, at the same time: to sleep and dream.
5. then again;
repeatedly: He coughed and coughed.
6. (used to imply different qualities in things having the same name): There are bargains and bargains, so watch out.
7. (used to introduce a sentence, implying continuation) also;
then: And then it happened.
8. [Informal.]to (used between two finite verbs): Try and do it. Call and see if she's home yet.
9. (used to introduce a consequence or conditional result): He felt sick and decided to lie down for a while. Say one more word about it and I'll scream.
10. but;
on the contrary: He tried to run five miles and couldn't. They said they were about to leave and then stayed for two more hours.
11. (used to connect alternatives): He felt that he was being forced to choose between his career and his family.
12. (used to introduce a comment on the preceding clause): They don't like each other--and with good reason.
13. [Archaic.]if: and you please.Cf. an2.
14. and so forth, and the like;
and others;
et cetera: We discussed traveling, sightseeing, and so forth.
15. and so on, and more things or others of a similar kind;
and the like: It was a summer filled with parties, picnics, and so on.
n.
1. an added condition, stipulation, detail, or particular: He accepted the job, no ands or buts about it.
2. conjunction (def. 5b).
### Couch
couch (kouch or, for 6, 15, ko̅o̅ch),USA pronunciation n.
1. a piece of furniture for seating from two to four people, typically in the form of a bench with a back, sometimes having an armrest at one or each end, and partly or wholly upholstered and often fitted with springs, tailored cushions, skirts, etc.;
sofa.
2. a similar article of furniture, with a headrest at one end, on which some patients of psychiatrists or psychoanalysts lie while undergoing treatment.
3. a bed or other place of rest;
a lounge;
any place used for repose.
4. the lair of a wild beast.
5. [Brewing.]the frame on which barley is spread to be malted.
6. [Papermaking.]the board or felt blanket on which wet pulp is laid for drying into paper sheets.
7. a primer coat or layer, as of paint.
8. on the couch, [Informal.]undergoing psychiatric or psychoanalytic treatment.
v.t.
1. to arrange or frame (words, a sentence, etc.);
put into words;
express: a simple request couched in respectful language.
2. to express indirectly or obscurely: the threat couched under his polite speech.
3. to lower or bend down, as the head.
4. to lower (a spear, lance, etc.) to a horizontal position, as for attack.
5. to put or lay down, as for rest or sleep;
cause to lie down.
6. to lay or spread flat.
7. [Papermaking.]to transfer (a sheet of pulp) from the wire to the couch.
8. to embroider by couching.
9. [Archaic.]to hide;
conceal.
v.i.
1. to lie at rest or asleep;
repose;
recline.
2. to crouch;
bend;
stoop.
3. to lie in ambush or in hiding;
lurk.
4. to lie in a heap for decomposition or fermentation, as leaves.
### Design
de•sign (di zīn),USA pronunciation v.t.
1. to prepare the preliminary sketch or the plans for (a work to be executed), esp. to plan the form and structure of: to design a new bridge.
2. to plan and fashion artistically or skillfully.
3. to intend for a definite purpose: a scholarship designed for foreign students.
4. to form or conceive in the mind;
contrive;
plan: The prisoner designed an intricate escape.
purpose: He designed to be a doctor.
6. [Obs.]to mark out, as by a sign;
indicate.
v.i.
1. to make drawings, preliminary sketches, or plans.
2. to plan and fashion the form and structure of an object, work of art, decorative scheme, etc.
n.
1. an outline, sketch, or plan, as of the form and structure of a work of art, an edifice, or a machine to be executed or constructed.
2. organization or structure of formal elements in a work of art;
composition.
3. the combination of details or features of a picture, building, etc.;
the pattern or motif of artistic work: the design on a bracelet.
4. the art of designing: a school of design.
5. a plan or project: a design for a new process.
6. a plot or intrigue, esp. an underhand, deceitful, or treacherous one: His political rivals formulated a design to unseat him.
7. designs, a hostile or aggressive project or scheme having evil or selfish motives: He had designs on his partner's stock.
8. intention;
purpose;
end.
9. adaptation of means to a preconceived end.
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Category: Couch - Thursday, December 21st, 2017 | 6,404 | 24,513 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2018-34 | latest | en | 0.853441 |
https://math.stackexchange.com/questions/3904314/contractible-manifolds-of-dimension-n-le-2 | 1,723,435,218,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641028735.71/warc/CC-MAIN-20240812030550-20240812060550-00767.warc.gz | 307,943,114 | 39,252 | # Contractible manifolds of dimension $n\le 2$
I have to give a talk about the Whitehead manifold. I would like to stress the fact that it is the first given example of a contractible open (i.e. non compact) manifold which is not homeomorphic to $$\mathbb{R}^n$$ (when $$n=3$$). This happens because the Whitehead manifold is not simply-connected at infinity and as proved by Stallings this is equivalent to not be homeomorphic to $$\mathbb{R}^n$$. The first example of this kind occurs in dimension $$3$$ beacause as I read in many articles the only contractible manifold (up to homeomorphism) of dimension $$n\le 2$$ are: $$\begin{equation*}\{\text{pt.}\} \quad \mathbb{R} \quad \mathbb{R}^2 \end{equation*}$$ I'm thinking about a proof of this classification and I'm having problems with dimension $$2$$. So I have the following questions:
## Question 1
Can you suggest me any reference for the classification in dimension $$2$$ with a detailed proof?
## Question 2
Is the Stalling's result mentioned above valid for a general $$n$$-manifold or do we have to require that it is contractible?
• I am currently thinking of this. If $S$ is an open contractible surface, define $\Sigma$ to be its Alexandrov compactification (its one point compactification). If one can show $\Sigma$ is a topological sphere, then $S$ is homeomorphic to $\mathbb{S}^2\setminus \{pt\}$ which is $\mathbb{R}^2$. Showing $\Sigma$ is the sphere would be equivalent to show it is a simply-connected compact surface, and I think this is not a hard thing to do. Commented Nov 12, 2020 at 10:37
• Is there any relation between the fundamental group of $X$ and the fundamental group of its one point compactification? I know that the one point compactification is compact, so I should only have to prove that it is simply-connected. Commented Nov 12, 2020 at 11:12
• I'm not that sure in the general case. But here, suppose you have a loop in the compactification. If it is not passing through the infinity point, then it is a loop in $S$ and by contractibility, it is homotopic to a point. If the loop passes through the infinity point, I think (this is not a proof!) we can deform it locally to evitate the infinity point, and use the previous work. Commented Nov 12, 2020 at 11:15
• In fact, it seems the hard thing would be to show that the one point compactification will be a topological manifold Commented Nov 12, 2020 at 11:26
• Yes, I think is the hard part of the proof. Commented Nov 12, 2020 at 12:55
## 1 Answer
Forget the one-point compactification, it's a dead end. By the time you prove that a neighborhood of infinity looks like $$[0,\infty) \times S^1$$ (which is what you'd need to see the 1-pt compactification is a manifold) you're more or less already done.
Theorem: If $$\Sigma$$ is an open surface with $$H_1(\Sigma; \Bbb F_2) = 0$$, then $$\Sigma$$ is homeomorphic to $$\Bbb R^2$$.
Sketch of proof.
(1) First prove that $$\Sigma$$ has exactly one end; if $$K \subset \Sigma$$ is a compact subset, then $$\Sigma \setminus \text{int}(K)$$ may have many connected components, but only one of them is noncompact. The proof will be by contrapositive: if $$\Sigma$$ has two ends, show that $$H_1(\Sigma; \Bbb F_2) \neq 0$$. You will need to know that the inclusion $$\partial \Sigma \to \Sigma$$ is nonzero on first homology whenever $$\Sigma$$ is a noncompact surface with boundary.
(2) Use that $$\Sigma$$ has a compact exhaustion --- $$\Sigma = \bigcup \Sigma_n$$, where $$\Sigma_n$$ is a compact surface and $$\Sigma_n \subset \text{int}(\Sigma_{n+1})$$, so that $$\Sigma_{n+1} = \Sigma_n \cup S_n$$, where $$S_n$$ is also a compact surface. This can be justified using the fact that $$\Sigma$$ has a proper smooth function to $$\Bbb R$$, together with Sard's theorem.
(3) Using (1) and (2) together, observe that $$\Sigma \setminus \text{int}(\Sigma_n)$$ only has one noncompact piece. Modify your compact exhaustion so that $$\Sigma \setminus \Sigma_n$$ is connected and so that $$S_n$$ is connected.
(4) Prove that $$\partial \Sigma_n$$ is a single circle, as otherwise $$\Sigma$$ has positive genus (you'll glue on a pair of pants as $$S_n$$, because $$S_n$$ is connected), which would imply $$H_1(\Sigma;\Bbb F_2) \neq 0$$; again this involves some Mayer-Vietoris work.
(5) Prove that each $$\Sigma_1$$ is a disc and each $$S_n$$ is a cylinder. From here it follows that $$\Sigma \cong \Bbb R^2$$.
The details are not completely trivial, in particular on (3). I will not see or respond to comments, so please feel free to edit this answer as you desire. This gives a rough strategy for the general classification of noncompact surfaces without boundary, the ideas simplify when $$\Sigma$$ is contractible like this; similarly you can show that if $$H_1(\Sigma)$$ is finite, then $$\Sigma$$ is obtained by deleting finitely many points from a closed surface.
• Thanks very much for your answer! Could you please add the references for all of this? Commented Nov 12, 2020 at 14:47 | 1,367 | 4,996 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 43, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2024-33 | latest | en | 0.926301 |
http://ficoforums.myfico.com/t5/Credit-Cards/Stupid-Question-1/m-p/1779634 | 1,516,300,566,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084887535.40/warc/CC-MAIN-20180118171050-20180118191050-00138.warc.gz | 120,239,772 | 41,609 | cancel
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## Stupid Question #1
Regular Contributor
## Stupid Question #1
Why is it not good to use what a lender gives you under the terms the lender set? On another thread a poster was asking about possible AA for being maxed out in all of his cards. We all know and accept, albeit reluctantly, that this makes a lender nervous.
But why?
Chase, for example, gives you a \$10k limit and you buy something for \$9,999. When you opened the account Chase told you that you were supposed to make at least minimum payments. You do.
Why do they give you \$10k and then penalize you for using it? If you come to my house and I put a glass of water and three apples in front of you, should I get upset because you eat all the apples and drink all the water?
If Chase wants you to only spend 1/3 of your limit, then shouldn't they give you a limit of \$6,666 instead of \$10,000?
Message 1 of 17
16 REPLIES
Valued Contributor
## Re: Stupid Question #1
McArthur wrote:
Why is it not good to use what a lender gives you under the terms the lender set? On another thread a poster was asking about possible AA for being maxed out in all of his cards. We all know and accept, albeit reluctantly, that this makes a lender nervous.
But why?
Chase, for example, gives you a \$10k limit and you buy something for \$9,999. When you opened the account Chase told you that you were supposed to make at least minimum payments. You do.
Why do they give you \$10k and then penalize you for using it? If you come to my house and I put a glass of water and three apples in front of you, should I get upset because you eat all the apples and drink all the water?
If Chase wants you to only spend 1/3 of your limit, then shouldn't they give you a limit of \$6,666 instead of \$10,000?
I think, in theory, AA is only considered where there is a pattern of up-to-the-limit on several cards, as this is a common sign on financial crisis/pending bankrupcy.
So Chase is saying, "You can spend up to \$10K", with the unstated assumption "providing you don't owe too much to others". If Chase is the last card opened, then they should have some idea of potential obligations, but if it was the first card, it might have been reasonable to give you \$10K, but now I look with increasing anxiety as you pile on inquiries and sometimes successfully get new cards, which you then max out.
Not sure what else I can do other than reduce your CL. I guess ideally the issuer should be able to state the unstated assumption "I am giving you 10K now. Any future cards that you get may reduce your CL in the following way."
Message 2 of 17
Valued Contributor
## Re: Stupid Question #1
McArthur wrote:
Why is it not good to use what a lender gives you under the terms the lender set? On another thread a poster was asking about possible AA for being maxed out in all of his cards. We all know and accept, albeit reluctantly, that this makes a lender nervous.
But why?
Chase, for example, gives you a \$10k limit and you buy something for \$9,999. When you opened the account Chase told you that you were supposed to make at least minimum payments. You do.
Why do they give you \$10k and then penalize you for using it? If you come to my house and I put a glass of water and three apples in front of you, should I get upset because you eat all the apples and drink all the water?
If Chase wants you to only spend 1/3 of your limit, then shouldn't they give you a limit of \$6,666 instead of \$10,000?
Lenders want you to use the available credit offered to you, only they want you to use it responsibly. Maxing out your cards is never a good idea for a long period of time. FICO penializes you for using high utility on your accounts because it is a proven statistic that those that use the majority of their credit are at higher risk of default. Banks know this and are reluctant to grant credit to those that are at a high utility %. Most lenders won't take AA against you for the high utility, but you are limiting your ability to get more credit at a decent rate and gives your lender cause for alarm.
Starting Score: 642
Current Score: EQ 773, EX 780, TU 777 (All FICO)
Goal Score: 800+
Cards: NFCU Flagship 50K, DC 30K, BCP 28.6K, Arrival+ 25K, Citi DP 22.8K, CSR 20.5K, TotalRewards 25K, QuickSilver 20K
Message 3 of 17
Valued Contributor
## Re: Stupid Question #1
YOU CAN!
but if you max out multiple cards at once (which if im think about the same poster as you) it looks like you can't handle credit!
not that im saying you won't pay it back.
but think of if this way.
say i know you personally and i tell you oh i have 10k that i have as a reserve that you can have some and pay it back over time.
he never uses more then let's say 2k at a time. and always pays it back
and then i come to you and say hey i need the whole 10k? but i also take 10k from 3 other people all at the same time.
you would want to know when your going to get some or all of your money back correct?
when people leave a LARGE balance on a credit card some banks tend to balance chase. becuase they want to limit liability.
that's all their doing.
in my short 2 year credit history the largest balance ive carried was probably about 3k. and ive never had a min payment over 30 dollars because i dont carry my balances.
unless there is some formula to calculate min payment vs. balance
Current: Discover Fico 709 3/15 Walmart Fico 743 4/15Inquiries (24 Months): EQ 6 TU 1 EX 6 | Most Recent: 4/09/2015Over 12 Months:9 2015 Goals:Lower UtilityEarn Cash Back Amex Zync(Unicorn) Chase Freedom\$1500Discover IT\$7,400Citi DC \$10,000Citizens Mastercard\$7,000
Message 4 of 17
Regular Contributor
## Re: Stupid Question #1
bs6054 wrote:
McArthur wrote:
Why is it not good to use what a lender gives you under the terms the lender set? On another thread a poster was asking about possible AA for being maxed out in all of his cards. We all know and accept, albeit reluctantly, that this makes a lender nervous.
But why?
Chase, for example, gives you a \$10k limit and you buy something for \$9,999. When you opened the account Chase told you that you were supposed to make at least minimum payments. You do.
Why do they give you \$10k and then penalize you for using it? If you come to my house and I put a glass of water and three apples in front of you, should I get upset because you eat all the apples and drink all the water?
If Chase wants you to only spend 1/3 of your limit, then shouldn't they give you a limit of \$6,666 instead of \$10,000?
I think, in theory, AA is only considered where there is a pattern of up-to-the-limit on several cards, as this is a common sign on financial crisis/pending bankrupcy.
So Chase is saying, "You can spend up to \$10K", with the unstated assumption "providing you don't owe too much to others". If Chase is the last card opened, then they should have some idea of potential obligations, but if it was the first card, it might have been reasonable to give you \$10K, but now I look with increasing anxiety as you pile on inquiries and sometimes successfully get new cards, which you then max out.
Not sure what else I can do other than reduce your CL. I guess ideally the issuer should be able to state the unstated assumption "I am giving you 10K now. Any future cards that you get may reduce your CL in the following way."
Since Chase softs me monthly, they know I have increased my available credit by 500%. Shouldn't they be proactive and CLD me automatically?
I mean, you and I both know the game rules and can skillfully play the game. But most are not as educated as we are. I have seen scores of cases where people have had all their cards maxed out, never missed payments and then had a lender take AA. The resulting chain reaction of over limit fees, account closures, high util, penalty interest and even further AA from additional lenders pushed them to where now they can't make payments.
Sometimes I think lenders are directly responsible for pushing good people into financial ruin via a system of smoke and mirrors.
Message 5 of 17
Super Contributor
## Re: Stupid Question #1
creditnocash wrote:
YOU CAN!
but if you max out multiple cards at once (which if im think about the same poster as you) it looks like you can't handle credit!
not that im saying you won't pay it back.
but think of if this way.
say i know you personally and i tell you oh i have 10k that i have as a reserve that you can have some and pay it back over time.
he never uses more then let's say 2k at a time. and always pays it back
and then i come to you and say hey i need the whole 10k? but i also take 10k from 3 other people all at the same time.
you would want to know when your going to get some or all of your money back correct?
when people leave a LARGE balance on a credit card some banks tend to balance chase. becuase they want to limit liability.
that's all their doing.
in my short 2 year credit history the largest balance ive carried was probably about 3k. and ive never had a min payment over 30 dollars because i dont carry my balances.
unless there is some formula to calculate min payment vs. balance
There is. It is usually listed on the bottom of your T&C sheet somewhere. Something like 2-4% of your balance. But most banks make you pay the whole balance if it is less than a certain amount, I don't think a min payment is ever less than \$25.
Message 6 of 17
Highlighted
Valued Contributor
## Re: Stupid Question #1
I agree with the others. I saw my FICO score drop significantly when I reported a balance on 5 credit cards....at the same time! LOL. I'm able to pay it back...I just wanted to show usage but I didn't realize the affect it would have on my FICO until...well after the fact.
I say that if you do rack up a balance on a card with a 10k limit...have lots of available credit to keep your UTL low. Also pay more than the min so you don't freak the credit card company out thinking you won't pay them back.
As of 2017, rebuilding...
Message 7 of 17
Regular Contributor
## Re: Stupid Question #1
creditnocash wrote:
YOU CAN!
but if you max out multiple cards at once (which if im think about the same poster as you) it looks like you can't handle credit!
not that im saying you won't pay it back.
but think of if this way.
say i know you personally and i tell you oh i have 10k that i have as a reserve that you can have some and pay it back over time.
he never uses more then let's say 2k at a time. and always pays it back
and then i come to you and say hey i need the whole 10k? but i also take 10k from 3 other people all at the same time.
you would want to know when your going to get some or all of your money back correct?
when people leave a LARGE balance on a credit card some banks tend to balance chase. becuase they want to limit liability.
that's all their doing.
in my short 2 year credit history the largest balance ive carried was probably about 3k. and ive never had a min payment over 30 dollars because i dont carry my balances.
unless there is some formula to calculate min payment vs. balance
We are thinking of the same poster.
I agree with you and I also agree with improvingmycredit. We, however, are sophisticated consumers. I'd bet that the vast majority of credit consumers about there are exactly like you-know-who. They've never been presented with Chase's rule book which states what we all here understand. In fact, pretty much all Chase has told them when they apply for the card is the opposite of good credit behavior.
It just seems there is not a level playing field and that creditors are sometimes directly responsibly for the mess people get in.
Something just doesn't seem right with having to push a creditor for CLIs to, say, \$25k just so we can play the UTIL game. I'm guilty. I've played that game. I've got CLs equal to 3x gross income. And all because creditors and FICO force us to play the UTIL game. So how does that help a creditor if I max everything out and walk away?
But if that's the way the system has to work, great. Maybe it should be like getting your first driver license: mandatory training and a credit road test prior to getting any credit.
I bet 90% of the credit population are you-know-whos. Not a pretty picture.
Message 8 of 17
Regular Contributor
## Re: Stupid Question #1
maiden_girl wrote:
I agree with the others. I saw my FICO score drop significantly when I reported a balance on 5 credit cards....at the same time! LOL. I'm able to pay it back...I just wanted to show usage but I didn't realize the affect it would have on my FICO until...well after the fact.
I say that if you do rack up a balance on a card with a 10k limit...have lots of available credit to keep your UTL low. Also pay more than the min so you don't freak the credit card company out thinking you won't pay them back.
Why should the credit card company freak out if I am using exactly what they said I can use and paying back exactly what their agreement said I can pay?
Dont get me wrong. I agree with what you have written. It's just that the vast majority of credit card users are not as sophisticated as we are. Credit card company's should be the prime educators, but they are not. At all.
Message 9 of 17
Valued Contributor
## Re: Stupid Question #1
McArthur wrote:
We are thinking of the same poster.
I agree with you and I also agree with improvingmycredit. We, however, are sophisticated consumers. I'd bet that the vast majority of credit consumers about there are exactly like you-know-who. They've never been presented with Chase's rule book which states what we all here understand. In fact, pretty much all Chase has told them when they apply for the card is the opposite of good credit behavior.
It just seems there is not a level playing field and that creditors are sometimes directly responsibly for the mess people get in.
Something just doesn't seem right with having to push a creditor for CLIs to, say, \$25k just so we can play the UTIL game. I'm guilty. I've played that game. I've got CLs equal to 3x gross income. And all because creditors and FICO force us to play the UTIL game. So how does that help a creditor if I max everything out and walk away?
But if that's the way the system has to work, great. Maybe it should be like getting your first driver license: mandatory training and a credit road test prior to getting any credit.
I bet 90% of the credit population are you-know-whos. Not a pretty picture.
i agree with you completely.
back in school they taught us algebra and geometry but not to balance a check book or handling finances.
As for playing the fico util game really how many people know that? unless looking to improve your credit i never would have found this site.
before i became a normal poster on here, i had no idea that leaving a 0% balance on my best buy card was hurting me so much. (2000/3000) out of total cl.
i paid it off and my score skyrocketed 60 points.
as for large balances when you have credit cards and you get a mortgage cc companies dont just close out your cards or cld them.
Current: Discover Fico 709 3/15 Walmart Fico 743 4/15Inquiries (24 Months): EQ 6 TU 1 EX 6 | Most Recent: 4/09/2015Over 12 Months:9 2015 Goals:Lower UtilityEarn Cash Back Amex Zync(Unicorn) Chase Freedom\$1500Discover IT\$7,400Citi DC \$10,000Citizens Mastercard\$7,000
Message 10 of 17 | 3,801 | 15,478 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2018-05 | longest | en | 0.971808 |
https://cliffsnotes.com/study-guides/algebra/algebra-i/quadratic-equations/solving-quadratic-equations | 1,632,029,961,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056711.62/warc/CC-MAIN-20210919035453-20210919065453-00428.warc.gz | 227,403,574 | 21,944 | A quadratic equation is an equation that could be written as
ax 2 + bx + c = 0
when a 0.
There are three basic methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square.
#### Factoring
To solve a quadratic equation by factoring,
1. Put all terms on one side of the equal sign, leaving zero on the other side.
2. Factor.
3. Set each factor equal to zero.
4. Solve each of these equations.
##### Example 1
Solve x 2 – 6 x = 16.
Following the steps,
x 2 – 6 x = 16 becomes x 2 – 6 x – 16 = 0
Factor.
( x – 8)( x + 2) = 0
Setting each factor to zero,
Then to check,
Both values, 8 and –2, are solutions to the original equation.
##### Example 2
Solve y 2 = – 6 y – 5.
Setting all terms equal to zero,
y 2 + 6 y + 5 = 0
Factor.
( y + 5)( y + 1) = 0
Setting each factor to 0,
To check, y 2 = –6 y – 5
A quadratic with a term missing is called an incomplete quadratic (as long as the ax 2 term isn't missing).
##### Example 3
Solve x 2 – 16 = 0.
Factor.
To check, x 2 – 16 = 0
##### Example 4
Solve x 2 + 6 x = 0.
Factor.
To check, x 2 + 6 x = 0
##### Example 5
Solve 2 x 2 + 2 x – 1 = x 2 + 6 x – 5.
First, simplify by putting all terms on one side and combining like terms.
Now, factor.
To check, 2 x 2 + 2 x – 1 = x 2 + 6 x – 5
Many quadratic equations cannot be solved by factoring. This is generally true when the roots, or answers, are not rational numbers. A second method of solving quadratic equations involves the use of the following formula:
a, b, and c are taken from the quadratic equation written in its general form of
ax 2 + bx + c = 0
where a is the numeral that goes in front of x 2, b is the numeral that goes in front of x, and c is the numeral with no variable next to it (a.k.a., “the constant”).
When using the quadratic formula, you should be aware of three possibilities. These three possibilities are distinguished by a part of the formula called the discriminant. The discriminant is the value under the radical sign, b 2 – 4 ac. A quadratic equation with real numbers as coefficients can have the following:
1. Two different real roots if the discriminant b 2 – 4 ac is a positive number.
2. One real root if the discriminant b 2 – 4 ac is equal to 0.
3. No real root if the discriminant b 2 – 4 ac is a negative number.
##### Example 6
Solve for x: x 2 – 5 x = –6.
Setting all terms equal to 0,
x 2 – 5 x + 6 = 0
Then substitute 1 (which is understood to be in front of the x 2), –5, and 6 for a, b, and c, respectively, in the quadratic formula and simplify.
Because the discriminant b 2 – 4 ac is positive, you get two different real roots.
Example produces rational roots. In Example , the quadratic formula is used to solve an equation whose roots are not rational.
##### Example 7
Solve for y: y 2 = –2y + 2.
Setting all terms equal to 0,
y 2 + 2 y – 2 = 0
Then substitute 1, 2, and –2 for a, b, and c, respectively, in the quadratic formula and simplify.
Note that the two roots are irrational.
##### Example 8
Solve for x: x 2 + 2 x + 1 = 0.
Since the discriminant b 2 – 4 ac is 0, the equation has one root.
The quadratic formula can also be used to solve quadratic equations whose roots are imaginary numbers, that is, they have no solution in the real number system.
##### Example 9
Solve for x: x( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0.
Since the discriminant b 2 – 4 ac is negative, this equation has no solution in the real number system.
But if you were to express the solution using imaginary numbers, the solutions would be .
#### Completing the square
A third method of solving quadratic equations that works with both real and imaginary roots is called completing the square.
1. Put the equation into the form ax 2 + bx = – c.
2. Make sure that a = 1 (if a ≠ 1, multiply through the equation by before proceeding).
3. Using the value of b from this new equation, add to both sides of the equation to form a perfect square on the left side of the equation.
4. Find the square root of both sides of the equation.
5. Solve the resulting equation.
##### Example 10
Solve for x: x 2 – 6 x + 5 = 0.
Arrange in the form of
Because a = 1, add , or 9, to both sides to complete the square.
Take the square root of both sides.
x – 3 = ±2
Solve.
##### Example 11
Solve for y: y 2+ 2 y – 4 = 0.
Arrange in the form of
Because a = 1, add , or 1, to both sides to complete the square.
Take the square root of both sides.
Solve.
##### Example 12
Solve for x: 2 x 2 + 3 x + 2 = 0.
Arrange in the form of
Because a ≠ 1, multiply through the equation by . | 1,342 | 4,619 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2021-39 | latest | en | 0.909428 |
http://quatr.us/math/numbers/decimals.htm | 1,484,703,799,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280133.2/warc/CC-MAIN-20170116095120-00168-ip-10-171-10-70.ec2.internal.warc.gz | 229,919,153 | 32,015 | Decimals and Fractions
# Decimals
Number line
Decimals, like fractions, are a way of describing points on the number line that fall in between the whole numbers. If you were using fractions, you'd say that the point halfway between 1 and 2 was 1 1/2. In decimals, you'd call that same point 1.5.
Each digit after the decimal point represents 1/10 of the one before it. So 1.55 means 1 and 1/2 and 5/100. One common use of decimals is in money, where \$1.55 means a dollar and 55 cents. Another common use of decimals is in percents, where 50% of 200 means .5 times 200, or 100.
Early measuring systems and mathematics didn't use decimals, because they didn't have a good way to write numbers down. After Indian mathematicians invented the numbers we use today, adding the idea of zero about 500 AD, African mathematicians used them to work more with fractions around 1300 AD. But decimals became really common only with the French Revolution about 1800 AD. French mathematicians encouraged people to use decimals after the Revolution as a new, more scientific, easier to understand way of doing math, that would be more available to ordinary people, so everybody could be an educated citizen.
## FractionsNumbersMore MathQuatr.us home
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• Christian Persecution
• Christian Empire
Happy New Year! Welcome back! Get ready for Martin Luther King day with these articles about medieval Africa, slavery, the Civil War, emancipation, the civil rights movement, and Martin Luther King Jr. himself. More about King here... | 506 | 2,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-04 | longest | en | 0.940833 |
http://reference.wolfram.com/mathematica/ref/GraphDistance.html | 1,397,808,779,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609533121.28/warc/CC-MAIN-20140416005213-00633-ip-10-147-4-33.ec2.internal.warc.gz | 197,537,102 | 7,231 | BUILT-IN MATHEMATICA SYMBOL
GraphDistance
GraphDistance[g, s, t]
gives the distance from source vertex s to target vertex t in the graph g.
GraphDistance[g, s]
gives the distance from s to all vertices of the graph g.
Details and OptionsDetails and Options
• GraphDistance[g, s, t] will give the length of the shortest path between s and t.
• The distance is Infinity when there is no path between s and t.
• For a weighted graph, the distance is the minimum of the sum of weights along any path between s and t.
ExamplesExamplesopen allclose all
Basic Examples (1)Basic Examples (1)
Give the distance for a grid graph:
Out[1]=
Out[2]= | 162 | 646 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2014-15 | latest | en | 0.779245 |
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### Removing one set of curly brackets in a cell array
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```Is there a way of changing
{{[1,2]},{'ab'}} to {[1,2],'ab'}, without looking at cells individually, and without using a loop? I have a cell array about 1000 by 1000 to which I want to apply such an operation. The example I give is a 1 by 2 example.
Thanks
David
```
0
See related articles to this posting
```"David Epstein" <David.Epstein.spam@remove.warwick.ac.uk> wrote in message <i67v1o\$qra\$1@fred.mathworks.com>...
> Is there a way of changing
> {{[1,2]},{'ab'}} to {[1,2],'ab'}, without looking at cells individually, and without using a loop? I have a cell array about 1000 by 1000 to which I want to apply such an operation. The example I give is a 1 by 2 example.
>
> Thanks
> David
A = {{[1,2]},{'ab'}}
A =
{1x1 cell} {1x1 cell}
A = [A{:}]
A =
[1x2 double] 'ab'
Oleg
```
0
```I suppose you want to preserve original size of the cell of cells, in which case Oleg's answer is not appropriate, because it works only for 1 by N cells
Do as following :
>> A = { {[1,2]},{'ab'}; {[3,4,5]},{'cde'}; {[6]},{'f'} }
A =
{1x1 cell} {1x1 cell}
{1x1 cell} {1x1 cell}
{1x1 cell} {1x1 cell}
>> B = reshape([A{:}], size(A))
B =
[1x2 double] 'ab'
[1x3 double] 'cde'
[ 6] 'f'
```
0 | 462 | 1,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2013-48 | latest | en | 0.870703 |
http://pharmapdf.com/t/tbp.org1.html | 1,596,732,888,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736972.79/warc/CC-MAIN-20200806151047-20200806181047-00101.warc.gz | 82,708,863 | 4,796 | ## Tbp.org
trait for a total of 8. Since each woman sion as n approaches infinity; when i is 1, this limit is defined as e, the base Perfect
*Beaudet, Paul R.
ra, three traits; A also cannot be rb and rr, for then she would also be ra, three answer is approaching e times 108.
traits. Therefore, A is ra, rb or ra, rr, 4 The distance of the center of the 1-
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Yes,” “Answer No,” and “Have You 5 The young girl should release from
Cheated” cards are selected with equal static position. She will land about 8.7 is 784 for each card. Then 928 – 784 = lum, which the swing is, total energy is difficult regular problem was No. 5 about mg(h – h ), plus kinetic energy, mv2/2, the vertical belt sander. However, of the at any point in its arc. Equating kinetic and potential energy gives v2/2 = gh – gh, so v = √[2g(h h)] = √[2gL(cosθ 3 After one year, you will owe the
cal and L is the length of the chains. Reader’s entries for the Fall problems will is (1 + i/n)n, where i is the annual inter- trajectory of a projectile, which are x 1 Cleo is not remarkably rich. There
est rate and n is the number of times = x + v t and 0 = y + v t – gt2/2. Now, v are four “remarkable” traits—artistic = vcosθ and v = vsinθ. Solving for t (the (ra), beautiful (rb), intelligent (ri), and time it takes the girl to hit the ground) gives t = [v + √(v 2 +2gy )]/g. Substi- tuting into the equation for x gives x = x + (vcosθ/g)[(vsinθ + √(v2sin2θ + 2gy )], where x = Lsinθ and y = L(1 1 Find the smallest positive integer
– cosθ) + h . Using trial and error (with L = 6 ft and h = 2 ft) on the swing’s 5 A sparrow, flying horizontally in a
straight line, is 20 m directly below an B onus. The punter should aim at the
—Christopher R. Oliver, AL E ’08
2 Five pirates find a chest contain-
tion is given by p = [1/(σ√(2π))]exp(- θ2/2σ2), where θ is the deviation from hawk is flying at 30 m/s, how fast is the the aim angle, θ and varies from -45o to +45o, and σ = 7.5 is the standard devia- vote, and if half or more of the pirates B onus. A regular dodecahedron has
After all, this is football!) Pick a yard- line aim point y ; then the aim angle gons. If some, including none or all, of is given by θ = tan-1[(50-y )/(80/3)]. by y = 50-(80/3)tan(θ +θ), except that when y is negative (indicating that the ball crossed the goal line), y = 20. The expected value of the yard line for spot- -45o to +45o of pydθ. It is easy to write a C omputer Bonus. A, B, and C are
program that varies y to find the mini- 3 My wife and I recently attended a
use the digits 1 through 9 once and only fies the equation, A x B = C2, and where well. Program the expressions for y and p into the spreadsheet (with y as a variable), and calculate y for unit increases in θ from -45 to 45. (This is sufficiently accurate for this problem.) with various values of y until the mini- Tau Beta Pi, P. O. Box 2697, Knoxville,
TN 37901-2697 or email to BrainTick-
[email protected] only as plain text. The C omputer Bonus. The next two pal-
with 16 digits! This is quite a distance 4 Professor Asterisk treated his
to the judges who are Dr. H.G. McIl-
abcddcba where a can equal 1, 4, 5, 6, vried III, PA G ’53; F. J. Tydeman, CA
or 9 and b, c, and d can equal 0 through Δ ’73; J. L. Bradshaw, PA A ’82; and the
D. A. Dechman, TX A ’57.
6, 8, 10, 12, 14, and, finally, 16-digit in the range of 29 to 39, inclusive. After
Source: http://www.tbp.org/pubs/BTs/W11.pdf
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# There are n different size pairs of shoes in the box. One
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There are n different size pairs of shoes in the box. One [#permalink] 02 May 2008, 09:04
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There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<2n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?
Last edited by lexis on 06 May 2008, 22:20, edited 1 time in total.
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Re: Math: Probability - n Shoes [#permalink] 05 May 2008, 21:46
Well, no one left comments here
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Re: Math: Probability - n Shoes [#permalink] 06 May 2008, 05:47
lexis wrote:
There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?
I get nC1 * 2n-2C2k-2 / 2nC2k
= (2*k^2-k)/(2*n-1) .
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Re: Math: Probability - n Shoes [#permalink] 07 May 2008, 05:42
lexis; Wats the OA ?
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Re: Math: Probability - n Shoes [#permalink] 07 May 2008, 21:45
farend wrote:
lexis wrote:
There are n different size pairs of shoes in the box. One day, Tan took 2k (2k<n) shoes out of box to clean. What is the probability that only 1 pair of shoes with the same size was taken out?
I get nC1 * 2n-2C2k-2 / 2nC2k
= (2*k^2-k)/(2*n-1) .
u mean nC1=C(n,1)?
U should explain how did U get it. As I see, your answer is not correct.
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Re: Math: Probability - n Shoes [#permalink] 08 May 2008, 04:49
My contribution:
Prob=2k/(2n-1)
Reasoning:
What is the prob of taking the appropriate shoe out once you have taken one out before?
1/(2n-1)
Because you have taken 2k shoes out, thus, 2k/(2n-1)
Regards
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Re: Math: Probability - n Shoes [#permalink] 08 May 2008, 08:58
all too often, I see a problem like this and just draw a blank.
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Re: Math: Probability - n Shoes [#permalink] 08 May 2008, 11:28
JohnLewis1980 wrote:
My contribution:
Prob=2k/(2n-1)
Reasoning:
What is the prob of taking the appropriate shoe out once you have taken one out before?
1/(2n-1)
Because you have taken 2k shoes out, thus, 2k/(2n-1)
Regards
-----------
For example, there are 5 pairs of shoes A,B,C,D,E
probability to take only one pair of shoes is correct and 1 incorrect pair of shoes (mean 2 different shoes)?
Mean:
Let A1, A2 is correct pair (pretend)
C1,D2 is incorrect pair or C2, E2 or... (pretend)
Hope it helps you solve the general puzzle.
@ RyanDe680: Your avatar is so interesting.
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Re: Math: Probability - n Shoes [#permalink] 16 May 2008, 02:19
If no one can take the correct answer, I will leave the OA in next week.
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Re: Math: Probability - n Shoes [#permalink] 16 May 2008, 06:14
lexis wrote:
JohnLewis1980 wrote:
My contribution:
Prob=2k/(2n-1)
Reasoning:
What is the prob of taking the appropriate shoe out once you have taken one out before?
1/(2n-1)
Because you have taken 2k shoes out, thus, 2k/(2n-1)
Regards
-----------
For example, there are 5 pairs of shoes A,B,C,D,E
probability to take only one pair of shoes is correct and 1 incorrect pair of shoes (mean 2 different shoes)?
Mean:
Let A1, A2 is correct pair (pretend)
C1,D2 is incorrect pair or C2, E2 or... (pretend)
Hope it helps you solve the general puzzle.
@ RyanDe680: Your avatar is so interesting.
I'm afraid so
but why?
Don't we agree in the probability to take one complete pair of shoes? i.e. to take the right shoe once you've already take one out?
For me: 1/(2n-1)
Explanation: you take one shoe out, therefore, just 2n-1 shoes remain in the box. The probability to take the right one off is 1/(2n-1), doesn't it?
I need to improve my statistic skill.
Thanks for the explanation
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Re: Math: Probability - n Shoes [#permalink] 16 May 2008, 23:48
You should try ***. It is a really good and comprehensive website for preparing yourself for a GMAT exam. Do try it and see for yourself. Try a 24 hour free trial which you get when you sign up.
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Re: Math: Probability - n Shoes [#permalink] 18 May 2008, 00:52
I agree with alex1234 that *** is very helpful for preparing yourself for the math section of GMAT. I love the practice tests and the way they explain each sum with a video. I found it very useful. It is worth trying.
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Re: Math: Probability - n Shoes [#permalink] 18 May 2008, 09:24
JohnLewis1980 wrote:
My contribution:
Prob=2k/(2n-1)
Reasoning:
What is the prob of taking the appropriate shoe out once you have taken one out before?
1/(2n-1)
Because you have taken 2k shoes out, thus, 2k/(2n-1)
Regards
===============================================
Shoudn't this be solved like this ->
Probability of taking out 2k shoes from total of 2n shoes = 2k/2n
Probability of taking out second compatible shoe = 1/(2n-1)
Hence probability of both events happening together - 2k/2n(2n-1)
Can someone confirm the OA please?
Regards,
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Re: Math: Probability - n Shoes [#permalink] 18 May 2008, 10:29
Expert's post
Interesting problem. +1
My attempt: $$P=\frac{C^n_k*C^{k}_{1}*(C^2_1)^{k-1}*C^{n-k}_{k-1}}{C^{2n}_{2k}}$$
when 2k>n+1 P=0
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Re: Math: Probability - n Shoes [#permalink] 18 May 2008, 10:34
Expert's post
alex1234 wrote:
You should try ***. It is a really good and comprehensive website for preparing yourself for a GMAT exam. Do try it and see for yourself. Try a 24 hour free trial which you get when you sign up.
alexsmith wrote:
I agree with alex1234 that *** is very helpful for preparing yourself for the math section of GMAT. I love the practice tests and the way they explain each sum with a video. I found it very useful. It is worth trying.
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Re: Math: Probability - n Shoes [#permalink] 28 May 2008, 07:46
I'm just giving my views, I'm not advertising.
Is it so that if someone gives their views about a site which I used for my preparation of GMAT is called advertisement??
I benefited from this site and I want everyone else to also.
I'll keep giving my views about the source what I used for my GMAT entrance and will ask all to just try it once.
I'm sure you will get the results that you are dreaming for.
Thats all I want to say.
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Re: Math: Probability - n Shoes [#permalink] 29 May 2008, 05:11
Let C(i, k) = iCk = i!/(k!(k-i)!)
Total number of ways to grab shoes:
C(2n, 2k)
Total number of ways to grab only 1 pair:
C(n, 1) * C(n-1, 2k-2) * 2^(k-2)
C(n, 1) * C(n-1, 2k-2) * 2^(2k-2) / C(2n, 2k)
(unless 2k > n+1, then prob = 0)
This isn't an official question, right? The variables are awkwardly defined.
Either way, can I get the A, B, C, D, E answers/distractors, I would like to give this problem to a friend.
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Re: Math: Probability - n Shoes [#permalink] 30 May 2008, 06:42
Hey friends try *** its really a good site for all people who have problem in GMAT Math and Verbal..
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Re: Math: Probability - n Shoes [#permalink] 12 Jun 2008, 03:44
JingChan wrote:
Let C(i, k) = iCk = i!/(k!(k-i)!)
Total number of ways to grab shoes:
C(2n, 2k)
Total number of ways to grab only 1 pair:
C(n, 1) * C(n-1, 2k-2) * 2^(k-2)
C(n, 1) * C(n-1, 2k-2) * 2^(2k-2) / C(2n, 2k)
(unless 2k > n+1, then prob = 0)
This isn't an official question, right? The variables are awkwardly defined.
Either way, can I get the A, B, C, D, E answers/distractors, I would like to give this problem to a friend.
Congratulation! You're correct. Your math skill is very good!
The tricky is to require ONLY ONE pair of shoes be same size.
To solve it, we separate two sequences:
1st: There is NO pair of shoes
2nd: There is ONLY ONE pair of shoes be same size
==> Combine: ONLY ONE pair of shoes be same size + NO pair of shoes in rest shoes (2n-2)
Re: Math: Probability - n Shoes [#permalink] 12 Jun 2008, 03:44
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### Programming Praxis – Three Binary Algorithms
January 15, 2010
In today’s Programming Praxis we have to implement binary algorithms for multiplying, dividing, and finding the greatest common divisor of two numbers. Let’s get started.
Since all our functions require the Bits typeclass, for which Haskell doesn’t do type defaulting, we use the following language pragma so we don’t have to specify types in the tests.
```{-# LANGUAGE ExtendedDefaultRules #-}
```
We need an import to do bitshifting.
```import Data.Bits
```
And because we’re going to be doing quite a bit of it, two quick convenience convenience functions for doubling and halving numbers:
```left, right :: Bits a => a -> a
left = flip shiftL 1
right = flip shiftR 1
```
Binary multiplication. Piece of cake.
```binmult :: (Bits a, Integral a) => a -> a -> a
binmult 1 b = b
binmult a b = binmult (right a) (left b) + if odd a then b else 0
```
Binary division. By using the until function we don’t have use explicit recursion to find t.
```bindiv :: (Bits a, Ord a) => a -> a -> (a, a)
bindiv n d = f (right \$ until (> n) left d) 0 n where
f t q r | t < d = (q, r)
| t <= r = f (right t) (left q + 1) (r - t)
| otherwise = f (right t) (left q) r
```
Binary gcd. A lot of different conditions, but all very straightforward.
```bingcd :: (Bits a, Integral a) => a -> a -> a
bingcd a 0 = a
bingcd 0 b = b
bingcd a b | even a && even b = 2 * bingcd (right a) (right b)
| even a = bingcd (right a) b
| even b = bingcd a (right b)
| a > b = bingcd (a - b) b
| otherwise = bingcd a (b - a)
```
A quick test shows that everything is working correctly:
```main :: IO ()
main = do print \$ binmult 14 12
print \$ bindiv 837 43
print \$ bingcd 2322 654
```
### Programming Praxis – Two Sub-Quadratic Sorts
October 31, 2009
In yesterday’s Programming Praxis problem we have to implement two sort algorithms. Let’s get started.
First, some imports:
```import Control.Monad
import Data.List
import Data.List.HT
import Data.Array.IO
import Data.Array.MArray```
For the Comb sort algorithm, we’re going to need a function to swap two elements of an array.
```swap :: (MArray a e m, Ix i, Ord e) => i -> i -> a i e -> m ()
swap i j a = do x <- readArray a i
when (y < x) \$ writeArray a i y >> writeArray a j x```
The Comb sort algorithm itself:
```combSort :: Ord a => [a] -> IO [a]
combSort [] = return []
combSort xs = comb (s-1) =<< newListArray (1, s) xs where
comb :: Ord a => Int -> IOArray Int a -> IO [a]
comb 0 a = getElems a
comb n a = mapM_ (\i -> swap i (i+n) a) [1..s-n] >> comb (n-1) a
s = length xs```
We don’t need array access for the Shell sort algorithm, so that saves some code. It’s in the IO monad so we can use the same test function, but the algorithm itself is pure.
```shellSort :: Ord a => [a] -> IO [a]
shellSort [] = return []
shellSort xs = return \$ shell (last . takeWhile (< length xs) \$
iterate (succ . (*3)) 1) xs where
shell 1 = foldr insert []
shell n = shell (div (n-1) 3) . concatMap (shell 1) . sliceHorizontal n```
A little test harness to see of everything’s working:
```test :: ([Int] -> IO [Int]) -> IO ()
test f = do print . null =<< f []
print . (== [1..9]) =<< f [4,7,3,9,1,5,2,6,8]
main :: IO ()
main = do test combSort
test shellSort```
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# How do you solve 2r-1 equals 9?
Wiki User
2013-10-05 01:54:57
2r-1=9 2r =10 r =5
Wiki User
2013-10-05 01:54:57
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https://mooseframework.inl.gov/docs/doxygen/moose/classTrilinearInterpolation.html | 1,556,214,672,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578732961.78/warc/CC-MAIN-20190425173951-20190425195951-00325.warc.gz | 508,383,329 | 7,250 | TrilinearInterpolation Class Reference
This class interpolates a function of three values (f(x,y,z)). More...
#include <TrilinearInterpolation.h>
## Public Member Functions
TrilinearInterpolation (const std::vector< Real > &x, const std::vector< Real > &y, const std::vector< Real > &z, const std::vector< Real > &data)
Constructor initializes data for interpolation. More...
virtual ~TrilinearInterpolation ()=default
Real sample (Real x, Real y, Real z) const
Interpolates for the desired (x,y,z) coordinate and returns the value based on the function values vector. More...
## Protected Member Functions
void getCornerIndices (const std::vector< Real > &v, Real x, int &lower, int &upper, Real &d) const
Finds the indices of the cube that point (x,y,z) is in. More...
Real getCornerValues (int x, int y, int z) const
Searches the function value vector for the value at a given corner coordinate from the getCornerIndices function. More...
## Protected Attributes
std::vector< Real > _x_axis
vector of x-values More...
std::vector< Real > _y_axis
vector of y-values More...
std::vector< Real > _z_axis
vector of z-values More...
std::vector< Real > _fxyz
vector of function values, f(x,y,z) More...
## Detailed Description
This class interpolates a function of three values (f(x,y,z)).
It takes 4 vectors for x, y, z, and function values, f(x,y,z). The vector of function values should be done in the following manner: Function values of constant x and y are written, corresponding with values in vector, z. Function values for the constant x, next y-value, corresponding with values in vector, z. After last y-value, function values for the next x, first y-value, corresponding with values in vector, z.
An example: f(1,4,7) = 10, f(1,4,8) = 11, f(1,4,9) = 12 f(1,5,7) = 13, f(1,5,8) = 14, f(1,5,9) = 15 f(1,6,7) = 16, f(1,6,8) = 17, f(1,6,9) = 18 f(2,4,7) = 20, f(2,4,8) = 21, f(2,4,9) = 22 f(2,5,7) = 23, f(2,5,8) = 24, f(2,5,9) = 25 f(2,6,7) = 26, f(2,6,8) = 27, f(2,6,9) = 28 f(3,4,7) = 30, f(3,4,8) = 31, f(3,4,9) = 32 f(3,5,7) = 33, f(3,5,8) = 34, f(3,5,9) = 35 f(3,6,7) = 36, f(3,6,8) = 37, f(3,6,9) = 38
x = {1, 2, 3}; y = {4, 5, 6}; z = {7, 8, 9}; fxyz = { // fxyz for x = 1 10, 11, 12, 13, 14, 15, 16, 17, 18, // fxyz for x = 2 20, 21, 22, 23, 24, 25, 26, 27, 28, // fxyz for x = 3 30, 31, 32, 33, 34, 35, 36, 37, 38 };
Definition at line 57 of file TrilinearInterpolation.h.
## ◆ TrilinearInterpolation()
TrilinearInterpolation::TrilinearInterpolation ( const std::vector< Real > & x, const std::vector< Real > & y, const std::vector< Real > & z, const std::vector< Real > & data )
Constructor initializes data for interpolation.
Parameters
x vector for x-coordinates y vector for y-coordinates z vector for z-coordinates data vector for function values formatted in the same manner as the example
Definition at line 13 of file TrilinearInterpolation.C.
17 : _x_axis(x), _y_axis(y), _z_axis(z), _fxyz(data)
18 {
19 if (_x_axis.size() < 1)
20 mooseError("x vector has zero elements. At least one element is required.");
21 if (_y_axis.size() < 1)
22 mooseError("y vector has zero elements. At least one element is required.");
23 if (_z_axis.size() < 1)
24 mooseError("z vector has zero elements. At least one element is required.");
25 if (_x_axis.size() * _y_axis.size() * _z_axis.size() != data.size())
26 mooseError("The size of data (",
27 data.size(),
28 ") does not match the supplied dimensions (",
29 _x_axis.size(),
30 ", ",
31 _y_axis.size(),
32 ", ",
33 _z_axis.size(),
34 ")");
35 }
std::vector< Real > _y_axis
vector of y-values
std::vector< Real > _x_axis
vector of x-values
void mooseError(Args &&... args)
Emit an error message with the given stringified, concatenated args and terminate the application...
Definition: MooseError.h:208
std::vector< Real > _fxyz
vector of function values, f(x,y,z)
static PetscErrorCode Vec x
std::vector< Real > _z_axis
vector of z-values
## ◆ ~TrilinearInterpolation()
virtual TrilinearInterpolation::~TrilinearInterpolation ( )
virtualdefault
## ◆ getCornerIndices()
void TrilinearInterpolation::getCornerIndices ( const std::vector< Real > & v, Real x, int & lower, int & upper, Real & d ) const
protected
Finds the indices of the cube that point (x,y,z) is in.
Parameters
[in] v vector to find lower and upper limits for the cube [in] x desired coordinate [out] lower lower limit for cube [out] upper upper limit for cube [out] d ratio of (x - lower) / (upper - lower)
Definition at line 38 of file TrilinearInterpolation.C.
Referenced by sample().
40 {
41 unsigned int N = v.size();
42 if (x < v[0])
43 {
44 lower = 0;
45 upper = 0;
46 }
47 else if (x >= v[N - 1])
48 {
49 lower = N - 1;
50 upper = N - 1;
51 }
52 else
53 {
54 for (unsigned int i = 0; i < N - 1; i++)
55 {
56 if (x > v[i] && x < v[i + 1])
57 {
58 lower = i;
59 upper = i + 1;
60 d = (x - v[lower]) / (v[upper] - v[lower]);
61 break;
62 }
63 else if (x == v[i])
64 {
65 lower = i;
66 upper = i;
67 break;
68 }
69 }
70 }
71 }
PetscInt N
static PetscErrorCode Vec x
## ◆ getCornerValues()
Real TrilinearInterpolation::getCornerValues ( int x, int y, int z ) const
protected
Searches the function value vector for the value at a given corner coordinate from the getCornerIndices function.
Parameters
x index for x-coordinate of corner y index for y-coordinate of corner z index for z-coordinate of corner
Returns
function value for the (x,y,z) coordinate
Definition at line 74 of file TrilinearInterpolation.C.
Referenced by sample().
75 {
76 int nY = _y_axis.size();
77 int nZ = _z_axis.size();
78
79 return _fxyz[x * nY * nZ + y * nZ + z];
80 }
std::vector< Real > _y_axis
vector of y-values
std::vector< Real > _fxyz
vector of function values, f(x,y,z)
static PetscErrorCode Vec x
std::vector< Real > _z_axis
vector of z-values
## ◆ sample()
Real TrilinearInterpolation::sample ( Real x, Real y, Real z ) const
Interpolates for the desired (x,y,z) coordinate and returns the value based on the function values vector.
Parameters
x desired x-coordinate y desired y-coordinate z desired z-coordinate
Returns
interpolated value at coordinate (x,y,z)
Definition at line 83 of file TrilinearInterpolation.C.
84 {
85 int x0 = 0;
86 int y0 = 0;
87 int z0 = 0;
88 int x1 = 0;
89 int y1 = 0;
90 int z1 = 0;
91 Real Dx = 0;
92 Real Dy = 0;
93 Real Dz = 0;
94
95 // find the the indices of the cube, which contains the point
96 getCornerIndices(_x_axis, x, x0, x1, Dx);
97 getCornerIndices(_y_axis, y, y0, y1, Dy);
98 getCornerIndices(_z_axis, z, z0, z1, Dz);
99
100 // find the corresponding function values for the corner indices
101 Real f000 = getCornerValues(x0, y0, z0);
102 Real f001 = getCornerValues(x0, y0, z1);
103 Real f010 = getCornerValues(x0, y1, z0);
104 Real f011 = getCornerValues(x0, y1, z1);
105 Real f100 = getCornerValues(x1, y0, z0);
106 Real f101 = getCornerValues(x1, y0, z1);
107 Real f110 = getCornerValues(x1, y1, z0);
108 Real f111 = getCornerValues(x1, y1, z1);
109
110 // interpolation
111 Real f00 = (f100 - f000) * Dx + f000;
112 Real f10 = (f110 - f010) * Dx + f010;
113 Real f01 = (f101 - f001) * Dx + f001;
114 Real f11 = (f111 - f011) * Dx + f011;
115 Real f0 = (f10 - f00) * Dy + f00;
116 Real f1 = (f11 - f01) * Dy + f01;
117
118 return (f1 - f0) * Dz + f0;
119 }
std::vector< Real > _y_axis
vector of y-values
Real getCornerValues(int x, int y, int z) const
Searches the function value vector for the value at a given corner coordinate from the getCornerIndic...
std::vector< Real > _x_axis
vector of x-values
static PetscErrorCode Vec x
void getCornerIndices(const std::vector< Real > &v, Real x, int &lower, int &upper, Real &d) const
Finds the indices of the cube that point (x,y,z) is in.
std::vector< Real > _z_axis
vector of z-values
## ◆ _fxyz
std::vector TrilinearInterpolation::_fxyz
protected
vector of function values, f(x,y,z)
Definition at line 95 of file TrilinearInterpolation.h.
Referenced by getCornerValues().
## ◆ _x_axis
std::vector TrilinearInterpolation::_x_axis
protected
vector of x-values
Definition at line 86 of file TrilinearInterpolation.h.
Referenced by sample(), and TrilinearInterpolation().
## ◆ _y_axis
std::vector TrilinearInterpolation::_y_axis
protected
vector of y-values
Definition at line 89 of file TrilinearInterpolation.h.
Referenced by getCornerValues(), sample(), and TrilinearInterpolation().
## ◆ _z_axis
std::vector TrilinearInterpolation::_z_axis
protected
vector of z-values
Definition at line 92 of file TrilinearInterpolation.h.
Referenced by getCornerValues(), sample(), and TrilinearInterpolation().
The documentation for this class was generated from the following files: | 2,827 | 8,752 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-18 | latest | en | 0.57369 |
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# Air at pressure P1 is injected to a manometer with a tube size of 2 cm diameter and air at pressure P2 is injected to another manometer with a tube size of 4 cm diameter as shown in the figure. Which of the following options about P1 and P2 is true? A P1 = P2 B 2P1 = P2 C 4P1 = P2 D P1 = 4P2
Asked by rajiv.gupta262 14th February 2021, 1:43 PM
It is assumed that liquids in both manometers are same .
Pressure is directly proportional to height of liquid column .
Hence answer :- (A) P1 = P2
Answered by Expert 5th September 2022, 9:08 AM
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### Free related questions
28th October 2021, 11:47 AM
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## Serre’s open image theorem for products of elliptic curves over function fields via specialization
In Propriétés galoisiennes des points d'ordre fini des courbes elliptiques, Invent. Math. 15, 259--331 (1972), Serre proved the following (Theorem 6 ′′, p. 325):
Let $K$ be a number field and let $K^{cycl}$ be the cyclotomic extension of $K$ generated by all roots of unity. Let $E$ and $E'$ be two elliptic curves such that, over $\bar{K}$,
(i) $E$ and $E'$ have no complex multiplication;
(ii) The $l$-adic representations $(\rho_l)$, $(\rho'_l)$ attached to $E$ and $E'$ don't become isomorphic over any finite extension of $K$.
Then $K(E_{tors}) \cap K(E'_{tors})$ is finite over $K^{cycl}$.
My question is whether this holds for $E$ and $E'$ defined over a function field? If this hasn't already been considered somewhere with an argument specific to the function field case, then maybe a specialization argument might work? Could anyone please provide a reference where there are similar specialization arguments used, or a standard reference for the basic theory of these specialization theorems?
-
If $j,j'$ are non-constant functions on the same algebraic curve, then it's not hard to show that they would have the same poles. Beyond that, I have nothing else to add, except that making trivial edits to keep bumping your question to the front page is not cool. – Felipe Voloch Sep 30 2011 at 17:21 I mean if they don't have the same poles the answer is yes. – Felipe Voloch Sep 30 2011 at 17:24 I think Serre's proof may go through for a finite extension of $\mathbb{Q}(T)$, but I'll have a look at the more geometric approach you suggest also. Thanks for the help - sorry about the edits, won't be doing that again! – Adam Harris Oct 2 2011 at 8:59
If $K$ is a function field over an algebraically closed field and one of your elliptic curves is constant (which does not necessarily violate your hypotheses unless the constant field is the algebraic closure of a finite field) then the answer is no. What kind of constant field are you interested in? You might want to add some non-isotriviality condition. The person to ask is probably Chris Hall, but I don't think he reads MO.
Thanks Felipe - I should have given more information: I specifically am thinking of a situation where I have two non-CM, non-isogenous curves $E$ and $E'$ with $j$-invariants $j(\tau)$ and $j(\tau')$ which are transcendental over $\mathbb{Q}$ but $j(\tau')$ is algebraic over $\mathbb{Q}(j(\tau))$. Also $E$ is defined over $\mathbb{Q}(j(\tau))$ and $E'$ over $\mathbb{Q}(j(\tau'))$. – Adam Harris Sep 28 2011 at 14:47 | 738 | 2,775 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2013-20 | latest | en | 0.880272 |
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By Angelette Perillard at October 25 2018 20:13:15
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The use of math worksheets can help solve numerous arithmetic problems. "Practice makes an individual perfect," is the best motto to be kept in mind while studying math. The motto will help a person to reinforce his desire to better himself in the subject. Without the help of these online resources, one will not be able to achieve the mastery of math. Reading is required in every day in life. Whether it is reading a street sign, reading the news, or reading a menu, comprehension is the fundamental reason for reading. Reading for comprehension is an essential common core skill that can and must be taught to students. | 856 | 4,270 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2019-04 | latest | en | 0.957893 |
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## Presentation on theme: "Multilevel Event History Models with Applications to the Analysis of Recurrent Employment Transitions Fiona Steele."— Presentation transcript:
Multilevel Event History Models with Applications to the Analysis of Recurrent Employment Transitions Fiona Steele
Outline The discrete-time approach Multilevel models and examples for: –Recurrent events –Multiple states Handling large datasets Examples of other applications Estimation/software
Why use discrete-time methods? Events times are often measured in discrete time units, e.g. months or years. Straightforward to allow and test for non-proportional hazards. We can use familiar models for discrete response data. For more complex data structures and processes, we can use existing estimation procedures for multilevel models.
Restructuring data for a discrete-time analysis: Individual-based file
Restructuring data for a discrete-time analysis: Person-period file
Discrete-time hazard function
A simple discrete-time logit model We can fit a logit regression model of the form: The covariates x tj can be constant over time or time-varying. z tj is vector of functions of time (e.g. polynomials or dummy variables) and α T z tj is the logit of the baseline hazard function. Other link functions possible, e.g. clog-log or probit.
Recurrent events Analyse duration of periods of continuous exposure (episodes), e.g. employment episodes, birth intervals, partnerships There may be unobserved individual-specific (i.e. time-invariant) factors which affect the probability of an event for all of an individuals episodes –referred to as unobserved heterogeneity or frailty
Repeated events lead to a two-level hierarchical structure Level 2: Individuals Level 1: Episodes Hierarchical data structure
is probability of event in time interval t during episode i of individual j are covariates which might be time-varying or defined at the episode or individual level 2-level model for recurrent events random effect representing unobserved characteristics of individual j – unobserved heterogeneity or frailty Assume j u
Example: womens employment Duration of non-employment spells; event is (re)entry into employment Data are subsample from British Household Panel Study: 1401 women, 2290 episodes and 15314 person-year records Employment, birth and union histories collected retrospectively at wave 2. These were linked to subsequent panel data to form continuous histories Focus on effects of duration non-employed and time-varying indicators of number and age of children, but also adjust for age, characteristics of previous job (if any)
Unobserved individual heterogeneity Estimated standard deviation of woman-level random effect is 0.65 (se=0.09) –significant variation between women in log-odds of entering employment due to unmeasured time-invariant characteristics Failure to account for unobserved heterogeneity (UH) leads to overstatement of negative duration effects and understatement of positive duration effects After accounting for UH, effects of time-varying covariates (e.g. duration and number/age children) are subject-specific, i.e. within-woman effects
Duration effects before and after allowing for unobserved heterogeneity Duration non-employed (ref is < 1 yr)BeforeAfter [1,2) years-0.788*-0.648* [2,3)-1.133*-0.927* [3,4)-1.489*-1.225* [4,5)-1.372*-1.077* [5,6)-1.256*-0.930* [6,7)-1.353*-1.025* [7,8)-1.575*-1.248* [8,9)-1.688*-1.350* 9+ years-2.130*-1.765* * p<0.05
Estimates from multilevel logit model of entry into employment Child indicatorEst.(SE) Imminent birth (within 1 year)-0.836*(0.124) No. children age <=5 years (ref = 0) 1 child-0.204*(0.096) 2 -0.356*(0.142) No. children age > 5 years (ref = 0) 1 child 0.244*(0.117) 2 0.428*(0.115) * p<0.05
Modelling transitions between multiple states An individual may pass through various states, e.g. employment and non-employment. Suppose there are 2 states, and denote by p stij the probability of a transition from state s. where (u 1j, u 2j ) ~ bivariate normal Note: Generalises to multinomial logit for > 2 states
Multiple states: data structure (1) Start with an episode-based file, e.g. jiState ij t ij EVENT ij Age ij 11E3116 12NE2019 States are employment (E) and non-employment (NE) Notes: (i) t in years; (ii) EVENT ij =1 if uncensored, 0 if censored; (iii) age, in years, at start of episode.
Multiple states: data structure (2) Convert to discrete-time format: ty tij E ij NE ij E ij *Age ij NE ij *Ag e ij 1010160 2010 0 3110 0 1001019 20010 E ij dummy for Employment, NE ij dummy for Non-Employment
Example: transitions between employment and non-employment corr( u 1j, u 2j )=0.58, se=0.13, so large positive residual correlation between ENE and NEE –Women with high (low) chance of entering E tend to have a high (low) chance of leaving E –Positive correlation arises from two sub-groups: short spells of E and NE, and longer spells of both types BUT little impact on estimates for child indicators on (re)entry into employment
Handling large datasets Although flexible, a drawback of the discrete-time approach is that the analysis file can be very large. This is a particular problem when we wish to fit complex models with multiple correlated random effects. Two possible approaches: –Group time intervals –More efficient algorithms, e.g. reparameterisation in MCMC estimation (Browne et al. 2009)
Grouped time intervals Suppose we analyse 6-month rather than monthly intervals. Need to allow for different lengths of exposure time. In any 6-month interval, some will have the event or be censored after 1 st month while others will be exposed for full 6 months. Denote by n tij exposure time in grouped interval t. Estimate binomial logit model with response y tij and denominator n tij Note: intervals do not need to be the same width.
Example of grouped time intervals Suppose an individual is observed to have an event during the 17 th month, and we wish to group durations into 6-month intervals (t). jitn tij y tij y* tij 111600 112600 113510.2
Implications of aggregation Need to assume that hazard function is constant within the grouped intervals. Need to fix values of time-varying covariates within intervals, e.g. value at start. In practice, aggregation has little impact on estimated baseline hazard or effects of episode/individual-level covariates. But impact on coefficients of time-varying covariates can be substantial.
Examples of other applications Hospital admissions: length of stay or duration between admissions –Repeated episodes nested within patients if multiple admissions –Hospital and GP effects using cross-classified multilevel model (GPs refer to multiple hospitals, and hospitals take patients from multiple GPs) Area effects on mortality or fertility –Repeated birth intervals (for fertility) for individuals nested within areas
Area effects on mortality: alternative approaches As in employment example, set up person-period file with multiple records per person, e.g. Kravdal (2006) Define a single binary response for each person and include number of years of exposure as offset in a Poisson regression, e.g. Tarkiainen et al. (2009). Could also treat as binomial response (as for grouped time intervals). If few, categorical covariates apply Poisson regression to aggregate data (1 record for each combination of t and covariate values)
Area effects on mortality: Multilevel Poisson modelling of aggregate data (1) Suppose we want to estimate effect of age, sex and area characteristics on individual mortality risk Suppose we group age into four 5-year age categories. Then for each area define 8 cells, one for each age-sex combination For area j denote by y ij the observed number of deaths for age-sex cell i Denote the total population at risk of mortality in cell i of area j by n ij, or might use expected number of deaths E ij
Area effects on mortality: Multilevel Poisson modelling of aggregate data (2) Analyse (y ij, n ij ) using 2-level Poisson model Define age and sex dummies characterising cells and include these and area-level variables as predictors Application to cancer mortality: Langford and Day (2001) - No. deaths for small areas (i) within regions (j) within EC nations (k). Covariates at regional level Application to teenage conception: Diamond et al. (2002) –No. conceptions for age-year cell (i) within electoral wards (j). Deprivation indicators at ward level
Software Recurrent events and multiple states. Any software for multilevel binary responses Binomial models for grouped intervals. GLLAMM, MLwiN, WinBUGS Simultaneous equations models for correlated processes. aML, GLLAMM, MLwiN, Sabre, WinBUGS. aML is the most general (mixed response types at different levels)
Browne, W. J., Steele, F., Golalizadeh, M. & Green, M. (2009). The use of simple reparameterisations in MCMC estimation of multilevel models with applications to discrete- time survival models. JRSS A, 172, 579-598. Diamond, I., Clements, S., Stone, N. and Ingham, R. (2002) Spatial variation in teenage conceptions in south and west England. Journal of the Royal Statistical Society, Series A, 162: 273-289. Goldstein, H., Pan, H. and Bynner, J. (2004) A flexible procedure for analysing longitudinal event histories using a multilevel model. Understanding Statistics, 3: 85-99. Kravdal, Ø (2006) Does place matter for cancer survival in Norway? A multilevel analysis of the importance of hospital affiliation and municipality socio-economic resources. Health and Place, 12: 527-537. Langford, I. H. and Day, R.J. (2001) Poisson Regression. In A.H. Leyland and H. Goldstein (ed) Multilevel Modelling of Health Statistics. London: Wiley. Chapter 4. References
Steele, F., Goldstein, H. and Browne, W. (2004) A general multistate competing risks model for event history data, with an application to a study of contraceptive use dynamics. Statistical Modelling, 4: 145-159. Steele, F. (2011) Multilevel discrete-time event history models with applications to the analysis of recurrent employment transitions (with discussion). Australian and New Zealand Journal of Statistics (to appear). Tarkiainen, L., Martikainen, P., Laaksonen, M. and Leyland, A.H. (2009) Comparing the effects of neighbourhood characteristics on all-cause mortality using two hierarchical areal units in the capital region of Helsinki. Health and Place, 16: 409-412. See also downloadable materials: http://www.cmm.bris.ac.uk/MLwiN/tech-support/workshops/materials/models.shtml http://www.cmm.bris.ac.uk/MLwiN/tech-support/workshops/materials/eha.shtml
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# FINANCIAL STATEMNET ANALYSIS
The statements prepared and presented by a business enterprise at the end of accounting year i.e. after the balance sheet and profit and loss account are declared it is called financial statements. With the help of these statements we can get the true picture of the current business with the comparing four - five year back financial result. This statement is also helpful to the money lender or the trade creditors for credit periods. These statements are also helpful to the management committee to take some important decisions. In these statements some of the following analysis or statements are prepared like, Trend Percentage Analysis, Common Size Statements and Fund Flow & Cash Flow Statements. In the Trend Percentage Analysis, the financial statements of past few years are also presented that the percentage relation of various items of the statements with some important base item is shown. The Common Size Statements are used to provide common base for comparison. For Example, the total of the balance sheet is taken as 100 and various items in the balance sheet are stated as a percentage of the total of 100.
## COMMON SIZE STATEMENTS
Common Size Statement is different from the Trend Percentage Method. In the common size statement we decide the base year and each and every years total are divided with that figure but in Trend Percentage Method particular items base years balance is taken as denominator and the other years figure are taken as numerator. ANAND MILK UNION LIMITED
1
LIABLITIES:
(The entire figures are calculated after taking the base of total Liabilities of that particular year) PARTICULARS Share Capital Reserve & Surplus Red. Debentures Loans Fixed Deposit C.L & Provision Net Profit Total 03-04 Rs 1213.04 7234.43 1187.68 7150.00 2502.48 6786.00 252.46 26326.09 03-04 04.61 27.48 04.51 27.16 09.51 25.77 00.96 100 04-05 05.52 16.69 04.65 29.00 11.31 31.60 01.23 100 All figures are in % 05-06 06.50 16.24 04.73 21.51 13.27 36.43 01.32 100 06-07 07.69 15.14 04.45 09.81 14.31 47.00 01.60 100 07-08 06.22 10.92 02.90 27.89 12.08 38.73 01.26 100
INTERPRETATION:
2003-04, 2004-05 and 2005-06:
## The percentage of Current
Liability, Loans and Reserve and Surplus covers more than 75% of total Liability; the reason of lower percentage of Net Profit is higher amount of Loans, Debentures, Current Liability and Fixed Deposits.
2006-07:
## In this year the amount of Loan is decreased
by 50% from last year. And in compare the percentage of Current Liability increased and covered 47% of Total Liability. So in that year the percentage of Net Profit is slightly higher than the last three years.
2007-08:
## In this year the amount of Loan are increased
by more than 350%. Increase in amount of Loan, Current Liability and Fixed Deposit is reason for decline in percentage of Net Profit.
ASSETS:
2
## ANAND MILK UNION LIMITED
(The entire figures are calculated after taking the base of total Assets of that particular year) PARTICULARS Fixed Assets Capital W.I.P Investment Stock Adv. & Debtors Cash & Bank Bal. Deferred Expenses Total 03-04 Rs 9588.26 0000.32 0631.31 4588.81 5809.35 5708.04 00.00 26326.09 03-04 36.42 00.00 02.40 17.43 22.07 21.68 00.00 100 04-05 24.16 00.00 02.27 30.31 34.48 08.78 00.00 100 All figures are in % 05-06 20.20 00.21 02.09 39.32 32.22 05.67 00.29 100 06-07 20.85 00.00 01.68 35.24 27.44 14.47 00.32 100 07-08 17.07 00.35 01.44 43.88 27.73 09.23 00.30 100
INTERPRETATION:
From the above table we can say that more than 75% of Total
## Assets are covered in Fixed Assets, Stock and Debtors.
The percentage of Investment covered more than 2% in first three
## year, and 1.5% in last two year, of Total Assets.
The percentage of Cash and Bank Balance covered 21% in 2003-
## 04 and than increased averagely by 9% of Total Assets.
INCOME:
(The entire figures are calculated after taking the base of total Income of that particular year) PARTICULARS All figures are in %
3
## ANAND MILK UNION LIMITED
Net Sales Interest Income Dividend Income Sundry Income Prior Period Income Closing Stock Total
## 88.03 00.21 00.06 00.62 00.01 11.07 100
INTERPRETATION:
From the above table we can say that every year the percentage
of Sales covers total income averagely by 90% and the other amount are covered by other incomes.
From the table we conclude that the percentage of Closing Stock
## is increased every year in percentage of total income.
EXPENDITURE:
(The entire figures are calculated after taking the base of total Expenditure of that particular year) PARTICULARS Opening Stock Milk Purchase R.M. Consumption Research & Ext.Exp Processing Expenses
4
All figures are in % 04-05 05.11 60.03 15.11 00.88 00.45 05-06 08.12 58.94 13.77 00.91 00.72 06-07 08.94 58.54 13.11 01.07 00.80 07-08 05.71 64.03 12.92 00.97 00.90
## ANAND MILK UNION LIMITED
Packaging Expenses Power and Fuel Exp. Salary and Wages P.F & Gratuity Repairs and Main. Freight & For. Exp. Marketing Expenses Postage & Telegram Insurance premium Rent Rates & Taxes Audit Fees Administrative Exp. Int. & Bank Commi. Depreciation Decline in provision Prior period Exp. Net profit Total
03385.08 02150.30 01241.34 00303.53 00494.12 00269.67 00035.46 00029.92 00025.02 00010.25 00065.55 00040.94 01099.22 01281.32 00000.0
05.82 03.70 02.09 00.54 00.81 00.46 00.06 00.05 00.04 00.02 00.11 00.12 01.89 02.20 00.00
06.50 03.50 02.12 00.56 01.01 00.46 00.10 00.07 00.07 00.04 00.11 00.17 01.28 01.82 00.07
06.67 03.24 01.75 00.61 00.92 00.58 00.11 00.07 00.09 00.05 00.10 00.17 01.00 01.46 00.06
07.21 03.23 01.53 00.53 00.95 00.78 00.12 00.06 00.09 00.05 00.10 00.15 00.79 00.75 00.03
06.87 03.13 01.38 00.39 00.87 00.83 00.09 00.05 00.05 00.04 00.08 00.14 00.67 00.47 00.04
0 00020.20 00.03 00.07 00.12 00.03 00.04 00252.46 00.43 00.47 00.42 00.46 00.37 58187.06 100 100 100 100 100
INTERPRETATION:
From the above table we conclude that in every year more than 70% of total expenses are made for Purchase of Milk and Raw Materials.
In first three years, more than 15% and 12% in last two years of
the total expenses are made for Packaging, Power and Fuel, Salary and Wages, Depreciation and Interest and Bank Commission. More than 10% to 12% amounts of total expenses are made for other expenses mentioned above in table.
## TREND PERCENTAGE ANALYSIS
For studying the Trend of various items of financial statement, figure of a single year are not enough, comparative figure for some more years are significantly required. Such comparative figures may be either absolute figure or may be presented in percentage form. In this method we assume one year as a base year, and compare that figure with similar items of some other years. In this method all comparative figure are shown in the percentage so that it is called percentage or Trend Ratio Analysis.
LIABLITIES:
(The entire figures are calculated after taking the base year 2003-04) PARTICULARS 03-04 Rs Share Capital 1213.04 Reserve & Surplus Red. Debentures Loans Fixed Deposit
6
All figures are in % 03-04 100 100 100 100 100 04-05 115.0 3 058.3 3 098.9 4 102.5 2 114.2 05-06 06-07 07-08 131.8 163.21 183.75 7 055.2 053.91 054.13
8 147.39 173.18
## C.L & Provision Net Profit
6786.00 252.46
100 100
1 117.7 2 123.2 8
2 132.0 5 128.2 3
168.4 7 163.0 0
204.72 178.84
INTERPRETATION:
From the above table we conclude that
SHARE CAPITAL:
## The percentage of paid up Share Capital
increased averagely by 15% for first three year and than increased by more than 25% because of increase in Authorize Share Capital by 100% from 2006-07.
FIXED DEPOSIT AND CURRENT LIABLITIES:
The
percentage of Fixed Deposits increased averagely by 10% to 12% in whole five years. Current Liability averagely increased by 30%. And it becomes more than double from the base year.
DEBENTURES, LOANS AND NET PROFIT: The
percentages of Debentures are declining every year. The percentage of Loan are declined up to 2006-07 and increased in 2007-08. The percentages of Net Profit are increased every year.
ASSETS:
(The entire figures are calculated after taking the base year 2003-04) PARTICULARS Fixed Assets Investment 03-04 Rs 9588.26 0631.31 03-04 100 100 04-05 063.7 0 090.8 All figures are in % 05-06 051.8 2 081.3 06-07 07-08 056.0 063.85 2 068.5 081.54
7
## 342.96 171.22 058.00
INTERPRETATION:
From the above table we conclude that
FIXED ASSETS INVESTMENT AND DEBTORS:
The
percentage of Fixed Assets, Investment and Debtors are decreasing up to 2006-07 and increase in Fixed Assets by 7.5%, Investment by 13% and Debtors by 50% in compare to last year i.e. 2006-07.
STOCK:
## 06 and 200% in 2007-08 from the base year.
CASH AND BANK BALANCE:
Balance is decreased in first two year from the base year and increased in last two year from the 2005-06.
INCOME:
(The entire figures are calculated after taking the base year 2003-04) PARTICULARS
8
## ANAND MILK UNION LIMITED
Net Sales Interest Income Dividend Income Sundry Income Prior Period Income Closing Stock
## 7 194.77 339.02 421.2 4 204.9 4 058.97 397.25
INTERPRETATION:
From the above table we conclude that
NET SALES AND SUNDRY INCOME:
The percentage of
Net Sales is increasing every year and becomes double in 200708. Sundry Income is also increased every year from the base year and becomes double in 2005-2006 and becomes triple in 2007-08.
INTREST AND DIVIDENED INCOME: The
percentage of
Interest Income is decreasing every year from the base year. And the Dividend Income is decreased in first two year and increased in last two year from the base year.
## ANAND MILK UNION LIMITED
EXPENDITURE:
(The entire figures are calculated after taking the base year 2003-04) PARTICULARS Opening Stock Milk Purchase R.M. Consumption Research & Ext.Exp Processing Expenses Packaging Expenses Power and Fuel Exp. Salary and Wages P.F & Gratuity Repairs and Main. Freight & For. Exp. Marketing Expenses Postage & Telegram Insurance premium Rent Rates & Taxes
10
All figures are in % 03-04 Rs 05594.5 0 31903.37 09285.2 2 00526.3 9 00173.2 0 03385.0 8 02150.3 0 01241.3 4 00303.5 3 00494.1 2 00269.6 7 00035.4 6 00029.9 2 00025.0 2 00010.2 03-04 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 04-05 060.6 3 125.0 3 108.1 3 110.7 4 173.0 8 127.53 108.1 0 113.65 121.4 1 136.42 113.47 05-06 114.5 4 145.7 5 117.0 9 138.1 8 329.20 155.6 2 119.24 111.0 5 158.7 3 149.0 0 172.7 06-07 07-08 142.8 124.27 1 163.97 244.38 126.2 0 181.3 6 410.8 169.35 224.66 628.85
## 7 154.96 158.17 172.6 2 257.8 215.24 376.44
1 0 196.61 275.29 299.40 310.18 150.5 3 175.33 227.1 194.1 8 304.1 5 458.6 188.8 187.63
## ANAND MILK UNION LIMITED
Audit Fees Administrative Exp. Int. & Bank Commi. Depreciation Prior period Exp. Net profit
## 100 100 100 100 100 100
2 110.0
3 126.5
9 141.6
152.03
5 5 7 298.43 312.23 326.11 407.35 077.2 0 094.4 071.7 9 090.5 064.1 4 052.2 074.12 044.77
## 4 8 2 234.25 427.32 137.77 236.18 123.27 128.2 3 162.99 178.84
INTERPRETATION:
From the above table we conclude that Raw Material Consumption and Milk Purchases are increased by .69 times and 2.5 times.
Interest & Bank Commission and Depreciation both are declining
every year from the base year. The percentage of Depreciation is decreased by 0.50 times in 2007-08 from the base year.
Processing Expenses is increased by 6 times, Rent Rates and
Administrative Expense is increased by 4 times, Marketing Expense is increased by 3 times, Packaging Expense, Insurance Premium and Research Expense is nearly increasing by 2.5 times, Repairs and Maintenance is increased by 2 times in 2007-08 from the base year.
## ANAND MILK UNION LIMITED
11
RATIO ANALYSIS
The most important task of a financial manager is to interpret the financial information in such a manner, that it can be well understood by the people, who are not well versed in financial information figures. The technique, by which it is to be calculated, is known as Ratio Analysis. 1) Percentage 2) Rate 3) Proportion Ratio Analysis is an important technique of financial analysis. It depicts the efficiency or shortfall of the organization in the form of trend Analysis. Different ratio appeal to different people managements, having the task of running business efficiency, will interest in all ratios.
12
## ANAND MILK UNION LIMITED
A Supplier of goods on credit will be partially interested in liquidity ratios, which indicate the ability of the business to pay its bills. Existing and future shareholders will indicate the ability of business to purchase. Existing and future shareholders will interest in investment ratios, which indicate the level of return that can be expected on an investment in business. Major customers, intent on having a continuing source of supply, will be interested in the financial stability, as reveled by the capital structure, liquidity and profitability ratios. Debenture and loan stock holders will be interested in ability of a business will be interested in the ability of a business to pay interest, and ultimately to repay capital.
A banker, gibing only short-term loans, will be interested mainly in the liquidity of the business, and its ability to repay those loans.
## STEPS IN RATIO ANALYSIS
Collection of information, which are relevant from the financial statements and then to calculate different ratios accordingly. Comparison of computed ratios of the same organization or with the industry ratios.
Interpretation, drawing of the inference and report-writing.
RATIO ANALYSIS
Ratio analyses are a powerful tool of financial analysis. A ratio is defined as the indicated quotient of two mathematical expressions and ANAND MILK UNION LIMITED
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as the relationship between two or more things. In financial analysis a ratio is used as a benchmark for evaluating the financial position and performance of a firm. The relationship between two accounting figures, expressed mathematically, is known as a financial ratio.
TYPES OF RATIOS
Several ratios; calculated from the accounting data, can be grouped into various classes according to financial activity or function to be evaluated. We may classify the ratios into the following categories. Liquidity ratios Leverage ratios Profitability ratios Capital Structure Ratio Other Ratio
LIQUIDITY RATIO
It is extremely essential for a firm to be able to meet its obligations as they become due. Liquidity ratios measure the ability of the firm to meet its current obligations. In fact, analyses of liquidity needs the preparation of cash budgets and cash and fund flow statements but liquidity ratios by establishing a relationship between cash and other current assets to current obligations, provide a quick current assets to current obligations, provide a quick measure of liquidity. A firm should ensure that it should not suffer from lack of liquidity, and also that it does not have excess liquidity. The failure of company to meet its current obligation due to lack of sufficient liquidity, will result in poor credit worthless, loss of creditors
14
## ANAND MILK UNION LIMITED
confidence for even in legal tangles resulting in the closure of the company. A very high degree of liquidity is also bed. the firms fund will be unnecessarily tied up in current assets therefore, it is necessary to strike a proper balance between high liquidity and lack of liquidity. The most common ratios, which indicate the extent of liquidity, are Current ratio Quick ratio
CURRENT RATIO
Current ratio is the ratio of total current assets to total current liabilities. Current assets of a firm represent those assets which can be in ordinary course of business converted into cash with in short period of time and current liabilities defined as liabilities which are short term manufacturing obligation to meet current assets. To measure the financial liquidity of Amul Current assets = Stock, Advance & debtors, Cash & Bank Balance. Current liabilities = Deposits, Due to societies, O/s against Expenses and Purchases, Sundry Creditors, Provisions. Current assets Current Ratio = ________________ ANAND MILK UNION LIMITED
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Current liabilities
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
CURRENT ASSETS
16102.2 18596.49 18990.94 19874.21 28995.9
CURRENT LIABILITY
6786 8988.56 9460.79 12106.54 23992.01
RATIO (C.A/C.L)
2.37 2.07 2.12 1.64 1.21
CURRENT RATIO 2.5 2 TIMES 1.5 1 0.5 0 current ratio 2003-04 2004-05 2005-06 2006-07 2007-08 2.37 2.07 2.12 1.64 1.21
YEAR
INTERPRETATION
The ideal Current Ratio of any firm is 2:1. In AMUL first three year the ratio is more than 2, it indicates good financial ability of the sector. But after that the ratio is declining because of the increase in Current Liability. It indicates that day by day the amounts of creditors are increasing which is not good for the sector.
16
## ANAND MILK UNION LIMITED
QUICK RATIO
Quick ratio is also called acid test ratio. It is the ratio between quick current assets and current liabilities. It is calculated by dividing the quick assets by current claim. Quick ratio is the measurement of firms ability to convert its current assets quickly into cash in order to meet its current claim. The term quick assets refers to current assets which can be converted into cash immediately or at a short notice without reduction in value of quick ratio. Quick Assets = Stock, due from societies, Advances, trade and Sundry Debtors Cash and Bank Balance Quick assets Quick Ratio = ______________ Current liabilities ANAND MILK UNION LIMITED
17
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
QUICK ASSETS
11365.37 10686.31 9078.21 10533.65 12952.44
CURRENT LIABILITY
6786 8988.56 9460.79 12106.54 23992.01
RATIO (Q.A/C.L)
1.67 1.19 0.96 0.87 0.54
LIQUID RATIO 1.8 1.6 1.4 TIMES 1.2 1 0.8 0.6 0.4 0.2 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 1.67 1.19 0.96 0.87 0.54
INTERPRETATION
The ideal Quick Ratio is 1:1. In AMUL the Quick Ratio is more than 1 in 2003-04 and 2004-05. But than after it started declining and reached below 1 for the next three years. The reason is continuous increase in the current liability.
18
## ANAND MILK UNION LIMITED
LEVERAGE RATIO
In the short term creditors like bankers and suppliers of raw material; are more concerned with firms current debt-paying ability, on the other hand, long-term creditors like debenture holders, financial institution are more concerned with the firms long term financial strength In fact a firm should have short as well as long term financial position. To judge the long-term financial position of the firms financial leverage or capital structure ratios are calculated. These ratios indicate funds provided by owners and lenders. As a general rule, there should be an appropriate mix of debt and owners equity in financing the firms assets.
## STOCK TURNOVER RATIO
This ratio indicates the efficiency of the firm in selling its product. It is calculated by dividing the cost of good sold by average inventory. Average Stock Inventory turnover = ___________________ 300 Cost of Goods Sold ANAND MILK UNION LIMITED
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## Cost of good sold =
Opening stock + Purchase of Milk + Purchase of Raw Material + Purchase expenses Closing stock
YEAR
## AVGERAGE COST STOCK GOODS SOLD
OF
RATIO (A.S./C.O.G.S)*300
31 days 31 days 38 days 34 days 35 days
STOCKS TURNOVER
38 34
35
2005-06 YEARS
2006-07
2007-08
20
## ANAND MILK UNION LIMITED
INTERPRETATION
From the above ratio we can say that AMUL is turning its inventory of finished good into sales in 31 days in 2003-04 and 2004-05, 38 days in 2005-06, 34 days and 35 days in 2006-07 and 2007-08 respectively. It is good for any co-operative sector.
## DEBTORS TURNOVER RATIO
A firm sells goods for cash and credit is used as a marketing tool by a number of companies. When the firm extends credit to its customer. Book debts (debtors or receivables) are created in the firms accounts. Book debts are expected to be converting into cash over a short period and, therefore are included in current assets. The liquidity position of the firm depends on the quality of debtors to a great extent. Financial analyses apply three ratios to judge the quality or liquidity debtors: (a) debtor turnover (b) Collection period and (c) again schedule of debtors. Debtors turnover Ratio can be found out dividing Average Debtors by Sales Sales = Net Sales of Milk and Milk products Debtors = Trade Debtors, Sundry debtors, Debtors Debtors Turnover Ratio = ________ 300 Sales
## ANAND MILK UNION LIMITED
21
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
## AVGERAGE SALES DEBTORS
6690.15 6797.52 7679.62 6759.25 7625.77 54088.29 59459.07 70206.23 81631.69 107187.29
RATIO (A.D./SALES)*300
37 days 34 days 33 days 25 days 21 days
## DEBTORS TURNOVER 40 35 30 DAYS 25 20 15 10 5 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 37
34
33 25 21
INTERPRETATION
From the above ratios we can say that AMULS Debtors remain outstanding for 37 days in 2003-04, 34 days in 2004-05, 33 days, 25
22
## ANAND MILK UNION LIMITED
days, 21 days in 2005-06, 2006-07 and 2007-08 respectively. The Collection period of Debtors is decreasing day by day. It is good sign for AMUL.
## CREDITOR TURNOVER RATIO
Creditor turnover ratio is computed by net credit purchases to average creditors. Creditors = Outstanding against Purchases + Societies + Outstanding against Expenses + Sundry Creditors Average Creditors Creditor Turnover Ratio = __________________ 300 Net Purchase
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
## AVGERAGE NET CREDITORS PURCHASE
5930.37 7098.06 8095.97 9798.1 12312.25 41188.59 49931.03 57374.94 64034.16 93690.87
## RATIO (A.C. /N.P.)*300
43 days 43 days 42 days 46 days 39 days
## ANAND MILK UNION LIMITED
23
CREDITORS TURNOVER
48 46 44 DAYS 42 40 38 36 34 2003-04 2004-05 2005-06 YEARS 2006-07 2007-08 39 43 43 42 46
INTERPRETATION
From the above we can say that the payment of Creditors is outstanding by AMUL. 43 days in 2003-04 and 2004-05, 42 days, 46 days, 39 days in 2005-06, 2006-07 and 2007-08 respectively. The payment period to Creditor remain same in every year. It is good for AMUL.
24
## FIXED ASSETS TURNOVER RATIO
Firm may wish to know its efficiency of utilizing fixed assets separately for AMUL the fixed assets turnover ratio is Sales Fixed Assets Turnover = __________________ Net Fixed Assets
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
SALES
54088.29 59459.07 70206.23 81631.69 107187.29
## NET FIXED ASSETS
9588.26 6107.85 4968.66 5371.69 6122.97
RATIO (Sales/N.F.A)
5.64 9.73 14.13 15.20 17.51
25
## 17.51 14.13 9.73 5.64 15.2
TIMES
2003-04
2004-05
2005-06 YEAR
2006-07
2007-08
INTERPRETATION
From the above ratio we can say that in AMUL Fixed Assets is recovered in 5.64 times in 2003-04, 9.73 times in 2004-05, 14.13 times, 15.2 times, 17.51 times in 2005-06, 2006-07 and 2007-08 respectively. We can say that the total Fixed Assets Turnover is increasing day by day. It is good sign for AMUL.
26
## WORKING CAPITAL TURNOVER RATIO
A firm may also like to relate net working capital to sales. It may thus computer net working capital turnover by dividing sales by net working capital turnover for Amul the ratio is Sales _______________ Working Capital
## Net Current assets Turnover =
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
SALES
54088.29 59459.07 70206.23 81631.69 107187.29
WORKING CAPITAL
9320.2 9607.93 9530.15 7767.67 5033.89
RATIO (Sales/W.C.)
5.80 6.18 7.37 10.51 21.42
## ANAND MILK UNION LIMITED
27
WORKING CAPITAL TURNOVER 25 20 TIMES 15 10.51 10 5 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 5.8 6.18 7.37
21.42
INTERPRETATION
From the above we can say that AMUL is able to recover its Working Capital 5.8 times in 2003-04, 6.18 times in 2004-05, 7.37 times in 200506, 10.51 times 2006-07 and 21.42 times in 2007-08. Working Capital Turnover is also increasing day by day. It is good and from it AMUL can generate more and more sales.
28
## TOTAL ASSETS TURNOVER RATIO
Total Assets Turnover Ratio shows the firms ability in generating sales from all financial sources committed to total assets thus Sales Total Assets Turnover = _____________ Total Assets Total assets (TA) include net fixed assets (NFA) and current assets (CA) for Amul the ratio is
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
SALES
54088.29 59459.07 70206.23 81631.69 107187.29
TOTAL ASSETS
26326.08 25277.94 24595.17 25761.82 35864.51
RATIO (Sales/T.A.)
2.05 2.35 2.85 3.17 2.99
## ANAND MILK UNION LIMITED
29
TOTAL ASSET TURNOVER 3.5 3 2.5 TIMES 2 1.5 1 0.5 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 2.35 2.05 2.85 3.17
2.99
INTERPRETATION
From the above data we can say that in AMUL Total Asset Turnover is recovered 2.05 times in 2003-04, 2.35 times in 2004-05, 2.85 times, 3.17 times, 2.99 times in 2005-06, 2006-07 and 2007-08 respectively. Till 2007-08 the Total Asset Turnover Ratio is increasing because the total asset is quiet same in every year. But in 2007-08 the Total Assets is increasing by 40% from 2006-07. So the turnover ratio is declining in that year.
30
## NET ASSETS TURNOVERS
The firm can compute net assets turnover simply by dividing sales by net assets (NA) net assets turnover may also called capital employed turnover for Amul the ratio is Net Assets = Net Fixed Asset + Current Asset after deducting Current Liability Sales Net assets turnover = ____________ Net Assets
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
SALES
54088.29 59459.07 70206.23 81631.69 107187.29
NET ASSETS
18908.46 9607.93 14498.81 13139.36 11126.86
RATIO (Sales/N.A.)
2.86 6.18 4.84 6.21 9.63
## ANAND MILK UNION LIMITED
31
ASSETS TURNOVER 12 10 8 TIMES 6 4 2 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 2.86 6.18 4.84 6.21 9.63
INTERPRETATION
From the above data we can say that Assets Turnover in AMUL during 2003-04 is 2.86 times, 6.15 times in 2004-05, 4.84 in 2005-06, 6.21 times in 2006-07 and 9.63 times in 2007-08 respectively. The Net Asset Turnover is increasing day by day. But the total Current Liability is also increasing and it is directly affected to Net Asset Turnover. If the Current Liability is decreased than the Net Asset can be Turnover by more than this ratio.
32
## ANAND MILK UNION LIMITED
PROFITABILITY RATIOS
A company should earn profit to survive and grow over a long period of time. Profit are essential but it would be wrong to assume that every action initiated by management of a company should be aimed at maximizing profits irrespective of social consequences and profit is looked upon as a term of above since some firms always want to maximize profits at due cost of employees, customers, and society. Except such infrequent cases, it is fact profit must be earned to sustain the operation of the business to be able to obtain funds from investor for expansion and growth and to contribute towards the social overhead for the welfare of society. Profit is the difference between revenues and expenses over a period of time. Profit is the ultimate output of the company; and it will have no future if it fails to make sufficient profits. There fore financial manager should continuously evaluate the efficiency of its company in term of profits. Generally two types of profitability ratios are calculated. Profitability in relation to sales Profitability in relation to investment Measures of Profit Profit can be measured in various ways 1) Gross Profit (2) Net Profit
33
## GROSS PROFIT TO SALES RATIO
Gross profit ratio is calculated by dividing Gross Profit by sales. Here gross profit is the different between sales and the manufacturing cost of good sold. Sales Cost of Good sold ______________________ 100 Sales
## Gross profit margin =
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
GROSS PROFIT
10428.01 12235.94 13946.75 15869.49 19005.32
SALES
54088.29 59459.07 70206.23 81631.69 107187.29
RATIO (G.P./Sales)*100
19.27 20.58 19.87 19.44 17.73
34
## ANAND MILK UNION LIMITED
GROSS PROFIT RATIO 21 20.5 20 19.5 19 18.5 18 17.5 17 16.5 16 20.58 19.87 19.27 19.44
PERCENTAGE
17.73
2003-04
2004-05
2005-06 YEAR
2006-07
2007-08
INTERPRETATION
From the above ratio we can say that Gross Profit Ratio in 2003-04 is 19.27%, 20.58% in 2004-05 20.58%, 19.87% in 2005-06, 19.44% in 2006-07 and 17.73% in 2007-08 respectively. The total amount of Gross Profit is increasing every year. But the ratio is decreasing; the main reason is increase in the Purchase Price Milk and Raw Material.
35
## NET PROFIT RATIO
Net Profit is obtained when operating expenses; interest and taxes are subtracted from gross-profit. The net profit margin ratio is measured by dividing profit after tax by sales. Net profit Net profit margin = __________ 100 Sales
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
NET PROFIT
252.46 311.23 323.74 411.50 451.51
SALES
54088.29 59459.07 70206.23 81631.69 107187.29
RATIO (N.P./Sales)
0.46 0.52 0.46 0.50 0.42
36
## ANAND MILK UNION LIMITED
NET PROFIT 0.6 0.5 PERCENTAGE 0.4 0.3 0.2 0.1 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 0.46
0.52 0.46
0.5 0.42
INTERPRETATION
From the above figure we can say that the percentage of Net Profit is 0.46% in 2003-04 and 2005-06. In 2004-05 it is 0.52%, 0.50% in 200607 and 0.42% in 2007-08 respectively. The total amount of sales is increased every year but at the other side total operating expenses is also increased day by day. So it directly affect to Net Profit Ratio of AMUL.
37
## OPERATING EXPENSES RATIO
Operating Expenses ratio is found out by dividing operating cost by sales. Operating Cost = Total Cost Interest Provision Closing Stock
Operating Cost Operating Cost Ratio = __________ Sales
100
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
OPERATIN G COST
53422.7 58785.45 69636.1 80627.15 106956.62
SALES
54088.29 59459.07 70206.23 81631.69 107187.29
RATIO (O.C./Sales)
98 98 99 98 99
38
## ANAND MILK UNION LIMITED
OPERATING EXPENSE 99.2 99 PERCENTAGE 98.8 98.6 98.4 98.2 98 97.8 97.6 97.4 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 98 98 98
99
99
INTERPRETATION
From the above data we can say that Operating Expense remains same during the 2003-04, 2004-05, 2006-07 i.e.98 %. In 2005-06 and 200708 it was 99%. The Operating Expenses of AMUL is increasing out of 1 Rs of sales 98 and 99 paisa is consumed in operating Expenses. The main reason is increase in Marketing, Packaging, and Processing Expenses and the price of Raw Material and Milk.
39
## RETURN ON CAPITAL EMPLOYEED
Return on Capital Employed Ratio is found out by dividing Net Profit by Capital Employed Capital Employed = Total Assets Misc. Expenses Current Liability Net profit Return on Capital Employed = __________
100
Capital Employed
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
NET PROFIT
252.46 311.23 323.74 411.50 451.51
CAPITAL EMPLOYED
19540.09 16289.38 15063.48 13572.19 11767.44
RATIO (N.P./C.E.)
1.29 1.91 2.15 3.03 3.84
40
## ANAND MILK UNION LIMITED
RETURN ON CAPITAL EMPLOYED 4.5 4 PERCENTAGE 3.5 3 2.5 2 1.5 1 0.5 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 1.91 1.29 2.15 3.03
3.84
INTERPRETATION
From the above data we can say that Return on Capital Employed during 2003-04 is 1.29%. It was 1.91% in 2004-05, 2.15% in 2005-06, 3.03% in 2006-07 and 3.84% in 2007-08 respectively. Return on Capital Employed is increasing day by day. The recover Ratio of Capital Employed in the business is increasing it is good sign for AMUL.
41
## RETURN ON SHAREHOLDER FUND
Return on Shareholder Fund Ratio is found out by dividing Net Profit by Shareholders Fund. Shareholders Fund = Equity Share Capital + Reserve and Surplus Misc. Expenses
## Net profit Return on Shareholders Fund = __________
100
Shareholders Fund
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
NET PROFIT
252.46 311.23 323.74 411.50 451.51
SHAREHOLDERS FUND
3201.91 3452.65 3683.51 4138.50 4486.30
## RATIO (N.P. /S.F.)
7.88 9.01 8.78 9.94 10.06
42
## ANAND MILK UNION LIMITED
RETURN ON SHARE HOLDER FUND 12 10 PERCENTAGE 8 6 4 2 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 7.88 9.01 9.94 8.78
10.06
INTERPRETATION
From the above ratio we can say that the Return on Shareholder Fund is 7.88% in 2003-04, 9.01% in 2004-05, 8.78% in 2005-06, 9.94% in 2006-07, and 10.06% in 2007-08 respectively. The Return on Shareholders Fund is increasing every year. So it is good sign that the recover ratio of amount invested is increasing.
43
## EARNING PER SHARE
The profitability of the common shareholders investment can also be measured in many other ways on such measure is to calculate the earning per share. The earning per share is calculated by dividing the profit after taxes by the total number of common (ordinary) shares outstanding. EPS calculation made over years indicate whether or net the firms earning power on per-share basis has changed over that period. Earning per share of the company should be compared with the industry average and the earning per share of other firms. Earning per share simply shows the profitability of the firm on per share basis. Net Profit Earning per share = __________________________________ Number of common shares outstanding
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
NET PROFIT
252.46 311.23 323.74 411.50 451.51
NO. OF
12.13 13.95 15.97 19.80 22.29
RATIO
20.81 22.31 20.25 20.78 20.26
44
## ANAND MILK UNION LIMITED
EARNING PER SHARE 22.5 22 21.5 IN RS 21 20.5 20 19.5 19 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 20.81 20.25 20.78 20.26 22.31
INTERPRETATION
From the above data we can say that Earning per Share is 20.81Rs in 2003-04, 22.31 Rs in 2004-05, 20.25 Rs in 2005-06, 20.78 Rs in 200607 and 20.26 Rs in 2007-08. Earning per Share ratio is comparatively better for AMUL. The shares of AMUL are distributed only to the Societies, so the main earning is distributed to its Societies.
45
## DIVIDEND PAYOUT RATIO
Dividend payout ratio is known as a payout ratio. It measures the relationship between the earning belonging to the ordinary shareholders and the dividend paid to them. In other words, the D/P ratio shows what percentage share the net profit after taxes and preference dividend is paid out as a dividend to the equity holders. It can be calculated by dividing the total dividend paid to owners by the total profits/earnings available to them. Alternatively it can be found out by dividing the DPS by the EPS. The rate of dividend per ordinary share = 15 % Amt. provided for Dividend ________________ 100 Earning per share
(1) Dividend =
## Dividend per share ________________ 100 Earning per share
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
DIVIDENED
14.53 14.53 14.60 14.50 14.60
## EARNING PER SHARE
20.81 22.31 20.25 20.78 20.26
RATIO (Div./E.P.S)
69.82 64.95 72.10 69.78 72.06
46
## ANAND MILK UNION LIMITED
DIVIDENDED PAY OUT RATIO 74 72 PERCENTAGE 70 68 66 64 62 60 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 64.95 69.82
72.1 69.78
72.06
INTERPRETATION
From the above ratio we can say that Dividend Pay Out Ratio is 69.82% in 2003-04, 64.95% in 2004-05, 72.10% in 2005-06, 69.78% in 2006-07 and 72.06 in 2007-08. From the Total Earning per Share averagely 70% amount is distributed to the Shareholders.
## CAPITAL STRUCTURE RATIO
ANAND MILK UNION LIMITED
47
In this type of Ratio the comparison made for Capital Structure. In this Ratio the proportion is to be found out between different types of long term capital. In this type of ratios we can find out following type of ratios Debt Equity Ratio Proprietary Ratio
## DEBT EQUITY RATIO
Debt-Equity Ratio Compute by dividing Total Debt to Net Worth. Total Debt = Debentures + Deposits + Long Term Loans Net Worth = Equity Share Capital + Reserve & Surplus Total Debt Debt-Equity Ratio = ____________ Net Worth
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
TOTAL DEBT
10840.16 10363.16 9216.63 7363.71 5874.95
NET WORTH
3201.91 3452.65 3754.41 4221.59 4591.36
RATIO (T.D./N.W)
3.38 3.00 2.45 1.74 1.28
48
## ANAND MILK UNION LIMITED
DEBT-EQUITY RATIO 4 3.5 3 TIMES 2.5 2 1.5 1 0.5 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08
## 3.38 3 2.45 1.74 1.28
INTERPRETATION
From the above ratio it is clear that Debt-Equity Ratio in 2003-04 is 3.38 times. It was 3.0 times in 2004-05, 2.45 times in 2005-06, 1.74 times in 2006-07 and 1.28 times in 2007-08. The ideal Debt-Equity Ratio is 2:1, in AMUL 2003-04, 2004-05 and 2005-06 the Ratio is more than 2, because of the higher amount of Long Term Debt but than after it is declining, so it shows that Total Long Term Debt is decreasing. It is good sign for AMUL.
## ANAND MILK UNION LIMITED
49
PROPRIETARY RATIO
Proprietary Ratio is found out by dividing Proprietary Fund by Total Assets. Proprietary Fund = Equity Share Capital + Reserve and Surplus Proprietary fund Proprietary Ratio = ____________ Total Assets
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
PROPRIETAR Y FUND
3201.01 3452.65 3754.41 4221.59 4591.36
TOTAL ASSETS
26326.08 25277.94 24595.17 25761.82 35864.51
RATIO (Sales/T.A.)
12.16 13.66 15.26 16.39 12.80
50
## ANAND MILK UNION LIMITED
PROPRIETORY RATIO 18 16 PERCENTAGE 14 12 10 8 6 4 2 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 12.16 13.66 16.39 12.8
15.26
INTERPRETATION
From the above ratio it is clear that Proprietary Ratio for the year 200304 is 12.16%. In 2004-05 it is 13.66%, in 2005-06 it is 15.26%, in 2006-07 it is 16.39% and in 2007-08 it is 12.8%. Out of total Assets the above percentage is invested by Proprietor and it is not better but we can say it is good for any Co-Operative Society.
## ANAND MILK UNION LIMITED
51
OTHER RATIOS
In this Ratio we find following types of Ratio Debt Ratio Debtors Asset Ratio
DEBT RATIO:
This Ratio can be found out by dividing Total Debt by Net Asset. Debt = Loans + Debentures + Fixed Deposit Net Asset = Total Assets Current Liability
Debt Ratio =
## Total Debt ____________ Net Assets
100
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
TOTAL DEBT
10840.16 11363.16 9716.63 7363.71 15374.95
NET ASSETS
19540.09 16289.38 15063.48 13572.19 11767.44
RATIO (Debt./N.A.)
55.48 69.75 64.50 54.25 130.65
52
## ANAND MILK UNION LIMITED
DEBT RATIO 140 120 PERCENTAGE 100 80 60 40 20 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 55.48 69.75 64.5 54.25 130.65
INTERPRETATION
From the above ratio we can say that Debt Ratio in 2003-04 is 55.48%. It was 69.75% in 2004-05, 64.5% in 2005-06, 54.25% in 2006-07 and 130.65% in 2007-08. It is comparatively good, except 2007-08.
53
## TOTAL LIABILITY TO ASSET RATIO
This Ratio can be found out by dividing Total Liability by Total Asset Total Liability = Debentures + Loans + Fixed Deposit + Current Liability Total Liability ____________ Total Assets
## Total Liability to Total Assets Ratio =
100
YEAR
2003-04 2004-05 2005-06 2006-07 2007-08
TOTAL LIABILITY
17618.62 19296.36 18608.22 18796.43 28947.46
TOTAL ASSETS
26326.08 25277.94 24595.17 25761.82 35864.51
RATIO (Sales/T.A.)
67.00 76.00 76.00 73.00 81.00
54
## TOTAL LIABILITY TO TOTAL ASSET RATIO
90 80 PERCENTAGE 70 60 50 40 30 20 10 0 2003-04 2004-05 2005-06 YEAR 2006-07 2007-08 67 81 73
76
76
INTERPRETATION
From the above ratio we can say that Total Liability to Total Asset Ratio in 2003-04 is 67%. In 2004-05 and 2005-06 it is 76%, in 2006-07 it is 73% and in 2007-08 it is 81%. From the above ratio we interpret that in compare of Total Assets. The percentage of Total Liability is comparatively less so it indicate good sign for AMUL.
## CHALLENGES TO BE MET ANAND MILK UNION LIMITED
55
Expansion upgrading of plant and equipment to met increasing demanded for quality and quantity with the help of betterqualified personnel. Rapid increase in productivity while respecting the basic man land animal dynamic that is control to dairy and agriculture development in India Development of new markets and expansion of old ones replacing additional system with quality packaged milk products and vegetable. Creating a national information network to ensure that accurate timely information is available to all who need it. Rapid progress towards the highest qualify standard strengthens institutions leaders, managers and members.
56
## ANAND MILK UNION LIMITED
PRODUCTION DEPARTMENT Strengthen the relationship with villagers. As the dairy industry depends on agriculture & veterinary Amul Dairy should have to concentrate on making strong relationship with villagers by providing societal services like health and safety (medical facility) to the farmers, educational facility to their children, insurance facilities etc. Though milk is perishable item durability of milk should be increase by increase in number of chilling centers in all over the Gujarat. HUMAN RESOURCE DEPARTMENT Working system in many departments is obsolescing. So, union has to adopt new operating systems to work like increase in no. of computers in administrative department, salary department etc. So that work can be done speedier and efficiently and newly recruited employees can be motivated.
SUGGESTION
ANAND MILK UNION LIMITED
57
After analyzing the ratio and watching the whole condition business. I would like to give suggestion as bellow.
of
The operating expenses are very high. Such has a raw material,
research & extension, packaging, processing, power & fuel etc. In this expenses raw material consumption is most and company can never controlled. Such expenses. Because Amul is expanding its dairy network year to year. So plant machinery consumption, operation expanses and other expenses will be increase in future. So firm should more focus to increase in sales. Because selling is being increased more and more company can achieve such an expenses. So increasing in the selling of milk products is necessary. Average collection period of Amul is not covered with in credit period, which is granted by firm to its debtor with in month. So firm should try to collect amount from debtors with in month. To purchase milk is most important factor of Amul because firm milk as a raw material which is perishable. So firm should use more and more high speed vehicles like tanker, railways etc for to collect raw-material and arrangement of more calling centers are necessary to protect the raw-material. Because Wastage of milk can effect on the expected production or on the profit also.
Firm should
58
## places is very costly and its also may be increased. So
Amul should establish satellite dairies on such a far places. Because such this steps the production and sales will be increased and market of the private competitor will be decreased.
In the globalization world when cutthroat competition is
prevailing in all developed or developing country so to survive in the market, high qualified persons are more necessary for every firm. Due to lack of salary and other. Facilities qualified person cannot stay in the particular firm. So for Amul should take necessary steps for this. Firm should increase extra facility or high salary.
In the marketing field to increase the market of new products.
Firm should use more and more distribution channels like Amul parlor, retailer, and direct selling and media like more advertisement in television, news-paper posters, board and many more.
In the matter of the Amul products market of Amul chocolate is
not very well because its quality is very soft and it melts in very short time in the normal hot. So firm should improve it.
## ANAND MILK UNION LIMITED
59
CONCLUSION
The very concept of Kaira Unions system of co-operative dairying was destined to become priceless for millions of farmers. The Kaira District Co-operative Milk Producers Union is the pioneer. And it succeeded. It came to be regarded as a model. The Anand pattern of co-operative dairying is now being replicated in all the states and union territories of India. Amul itself is also growing day by day. In the world of liberalization, globalization and competition in international market, Amul tries best to compete with dairy industry of the world. Union maintains a constant link with rural producers. All the necessary activities related to Animal Husbandry are seriously undertaken to increase the production. Welfare of milk producers is the only aim of Amul. Amuls motto is to give maximum possible returns to milk producers. Various welfare and educational activities are also arranged by the union for the betterment of rural people. This way, Amul is doing a type of social work in our country. It can be seen from above figures that growing rate of Amul has increased over the period of time. Though profit making is not a goal of Amul, its profit is also increasing year by year. This shows the efficiency of production methods. Amul has reached the turnover of 600 crores this year. Profit of this year is 3.10 crores. In comparison to the sales, profit is less because Amul believes in maximum welfare of milk producers. So it gives maximum price to milk producers and charges minimum for cattle feed and other products.
60
## ANAND MILK UNION LIMITED
FUTURE CHALLENGES
The future of any institution is a subject, which requires constant attention. The future is perceived as one embroiled with hardship. Hardships may surface in many forms as global demands and changes, foreign affiliations, competition, liberalization, changing values, urban shifting etc. are some of them to name which we foresee and union has to cope with these. In future union should adopt latest technology. Union should not sit with foremost quality of product and co-placement with existing product range but think in a more innovative way. To stay ahead research and development unit has to be strengthened. Union also thinks of price to sustain leading position in the market. Prices will have to remain steady and so union should concentrate on reducing maintenance expenses rather than proposing increasing product price. In comparison with other countries, which are in leading position in dairy industry, the milk-producing animals in our country yield 4 times less milk. Hence, it is the time that union should seriously review enhancement of milk production by adopting better animal husbandry practices.
## ANAND MILK UNION LIMITED
61
REFERENCES
Annual Report of Amul. www.amulindia.com www.google.com.
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https://programs.programmingoneonone.com/2021/10/hackerearth-car-names-problem-solution.html | 1,713,927,726,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296818999.68/warc/CC-MAIN-20240424014618-20240424044618-00152.warc.gz | 425,337,796 | 33,393 | HackerEarth Car Names problem solution
In this HackerEarth Car Names, problem-solution Brian built his own car and was confused about what name he should keep for it. He asked Roman for help. Roman being his good friend, suggested a lot of names.
Brian only liked names that:
- Consisted of exactly three distinct characters, say C1, C2, and C3
- Satisfied the criteria that the string was of the form - C1n C2n C3n: This means, first C1 occurs n times, then C2 occurs n times and then C3 occurs n times. For example, xyz, ccaarr, mmmiiiaaa satisfy the criteria, but xyzw, aabbbcccc don't.
Given N names suggested by Roman, print "OK" if Brian likes the name and "Not OK" if he doesn't.
HackerEarth Car Names problem solution.
`#include <bits/stdc++.h>using namespace std;int main(){ int N; cin>>N; string s; while(N--) { cin>>s; int num1[1000]={0}; int k=0; long long cnt=1; int num=0; for(int i=1;i<s.size();i++) { if(s[i]==s[i-1])cnt++; else { num1[k++]=cnt; if(k>3) break; num++; cnt=1; } } if(cnt>=1) { num1[k++]=cnt; num++; } if(num==3&&num1[0]==num1[1]&&num1[1]==num1[2] && s[0] != s[s.size() - 1]) cout<<"OK"<<endl; else cout <<"Not OK"<<endl; }}`
Second solution
`#include <bits/stdc++.h>#define F first#define S second#define pb push_back#define FOR(i,lb,ub) for(i=lb;i<=ub;i++)#define RFOR(i,ub,lb) for(i=ub;i>=lb;i--)#define FORS(it,v) for(it=v.begin();it!=v.end();it++)#define int long long#define st_clk double st=clock();#define end_clk double en=clock();#define show_time cout<<"\tTIME="<<(en-st)/CLOCKS_PER_SEC<<endl;int gcd(int a, int b) { return b?gcd(b,a%b):a; }using namespace std;void solve(){ int i,j,k,n; cin>>n; assert(n>=1 && n<=100); string s; FOR(i,0,n-1) { cin>>s; set<char> sc; int ok = true, sz = s.size(); if (sz%3 != 0) ok = false; else { int j=0; sc.insert(s[j]); for (; j<sz/3-1; j++) { if (s[j]!=s[j+1]) ok = false; } j++; sc.insert(s[j]); for (; j<2*sz/3 - 1; j++) { if (s[j]!=s[j+1]) ok = false; } j++; sc.insert(s[j]); for (; j<sz-1; j++) { if (s[j]!=s[j+1]) ok = false; } } if (sc.size()!=3) ok = false; if (!ok) { cout<<"Not OK\n"; } else cout<<"OK\n"; }}main(){ st_clk ios_base::sync_with_stdio(0); cin.tie(0); int t,i,j; solve(); return 0;}` | 814 | 2,495 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-18 | latest | en | 0.630155 |
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Post subject: [原创教程] 汇编寻址与高级代码的对应关系Posted: Fri Oct 02, 2009 2:39 am
Joined: Sun Feb 15, 2009 8:23 am
Posts: 27
6502资料可以看"图解6502 指令集.pdf"或"6502编程大奥秘.chm(前三章节就够)"。我个人比较喜欢看后者。
6502的寻址有13个。
1。立即寻址
LDA #03 //就是将03传入A寄存器。
STA W //W为一个变量，这里用了“直接寻址”，下面有说明，这一行的作用是将A寄存器的值传到W上。
2。直接寻址
2.1
LDA #03
STA W //W为一个变量，W也是一个地址。高级代码不直接写地址而是写成变量。
2.2
2.3
LDA #03
STA \$0300
2.4
N=W
LDA W
STA N
3。零页寻址
4。累加器寻址
LDA W
ASL //操作数就在A中
STA N
N = W << 1
5。隐含寻址
6。直接X变址
6.1
LDX #03
LDA W,X
STA N
6.2
LDX #03
LDA \$0300,X
STA \$0200
LDA \$0303
STA \$0200
7。直接Y变址
8。零页X变址，这个与“直接X变址”相近，就是局限在零页，省略高位。
9。零页Y变址，这个与“直接Y变址”相近，就是局限在零页，省略高位。
10。间接寻址
JMP (P) //这个不是跳到P地址上，而是跳到P所指向的地址。
11。先变址X后间接寻址
ldx #03
lda (P,X)
sta W
W=*P(3)
12。先间址后变址Y
P记录着这个数组的首地址。即P=&W
ldy #03
lda (P),Y
sta N
P=&W
...
N=*P(3)
13。相对寻址
while....
{...
}
if .... then ... else ....
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posted by .
write the equation in slope intercept form and identify the slope and y-intercept 4x+9y=4
• Algebra -
to change this into the required form takes two steps
1. keep the y term on the left, everybody else goes to the right.
9y = - 4x + 4
2. divide each term by the coefficient of the y term
y = (-4/9)x + 4/9
All done, the slope and y-intercept are now staring at you.
• Algebra -
Thank You so much
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More Similar Questions | 763 | 2,506 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-34 | latest | en | 0.899854 |
https://www.supermoney.com/encyclopedia/fibonacci-retracement-levels/ | 1,695,483,036,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233506481.17/warc/CC-MAIN-20230923130827-20230923160827-00597.warc.gz | 1,138,450,002 | 28,201 | # Understanding Fibonacci Retracement Levels in Financial Trading
Summary:
Fibonacci retracement levels are horizontal lines drawn on a price chart to identify potential support and resistance levels during market corrections. These levels, based on the Fibonacci sequence, include 23.6%, 38.2%, 50%, 61.8%, and 78.6%. Traders can use Fibonacci retracement levels to identify entry and exit points, confirm trends, manage risk, and benefit from their widespread recognition. However, it’s important to consider subjectivity and the tool’s limitations when incorporating them into trading strategies.
In the world of financial trading, there are various tools and techniques that traders utilize to analyze price movements and identify potential trends. One such tool is Fibonacci retracement levels. While the name might sound intimidating, understanding Fibonacci retracement levels can provide valuable insights for traders. In this article, we will delve into the concept of Fibonacci retracement levels, explore how they work, and discuss their significance in financial markets.
## What are Fibonacci retracement levels?
Fibonacci retracement levels are based on the mathematical sequence discovered by an Italian mathematician named Leonardo Fibonacci. This sequence, known as the Fibonacci sequence, starts with 0 and 1, and each subsequent number is the sum of the two preceding numbers (0, 1, 1, 2, 3, 5, 8, 13, and so on).
Fibonacci retracement levels are horizontal lines that are drawn on a price chart to identify potential support and resistance levels during market retracements. These levels are derived from the Fibonacci sequence and are expressed as percentages: 23.6%, 38.2%, 50%, 61.8%, and 78.6%. Traders believe that these levels represent areas where the price of an asset is likely to reverse or continue its trend.
## How do Fibonacci retracement levels work?
Fibonacci retracement levels work by identifying potential areas of price retracement within a broader trend. The basic idea is that after an asset’s price makes a significant move in one direction, it is likely to retrace or pull back before continuing its original trend. Traders use Fibonacci retracement levels to determine where these pullbacks may occur and to identify potential buying or selling opportunities.
To apply Fibonacci retracement levels, a trader selects a significant high and low on a price chart. These points represent the starting and ending points of the trend being analyzed. The Fibonacci retracement levels are then drawn based on these two points, creating horizontal lines at the specified percentages.
## Significance of Fibonacci retracement levels
Fibonacci retracement levels are considered significant for several reasons:
### Support and resistance
The retracement levels act as potential support or resistance areas where traders expect the price to react. If the price of an asset reaches one of these levels, it is likely to bounce off or reverse its direction.
### Psychological factors
Fibonacci retracement levels are widely used by traders, which increases their significance. As more traders observe and act upon these levels, they can become self-fulfilling prophecies, reinforcing their importance in the market.
### Confluence with other indicators
Fibonacci retracement levels are often used in conjunction with other technical analysis tools such as trendlines, moving averages, or candlestick patterns. When multiple indicators align with Fibonacci levels, it strengthens their relevance and increases the likelihood of accurate predictions.
## Limitations of Fibonacci retracement levels
While Fibonacci retracement levels can be a valuable tool, it’s essential to recognize their limitations:
Subjectivity: Selecting the appropriate high and low points to draw Fibonacci retracement levels can be subjective. Different traders may choose different points, leading to variations in the levels drawn.
Market noise: In volatile markets or during news events, price movements may not adhere strictly to Fibonacci retracement levels. Sudden market reactions can override the expected retracement patterns.
The supplementary analysis required: Fibonacci retracement levels should not be relied upon solely for trading decisions. They are most effective when used in conjunction with other technical indicators, fundamental analysis, and risk management strategies.
Fibonacci retracement levels are a popular tool among traders for identifying potential areas of a price retracement. By understanding these levels and their significance, traders can make more informed decisions based on historical price patterns. However, it’s important to remember that no trading tool is foolproof, and proper risk management and supplementary analysis are crucial for successful trading.
### How do Fibonacci retracement levels work?
Fibonacci retracement levels work by identifying potential areas of price retracement within a broader trend. The basic idea is that after an asset’s price makes a significant move in one direction, it is likely to retrace or pull back before continuing its original trend. Traders use Fibonacci retracement levels to determine where these pullbacks may occur and to identify potential buying or selling opportunities.
### How do I identify the high and low points to draw Fibonacci retracement levels?
To identify the high and low points for drawing Fibonacci retracement levels, you should look for significant price swings or trends on a price chart. The high point represents the peak of an upward move, while the low point represents the bottom of a downward move. These points should be selected based on the specific trend you want to analyze.
### Can Fibonacci retracement levels be used in all financial markets?
Yes, Fibonacci retracement levels can be used in various financial markets, including stocks, forex, commodities, and cryptocurrencies. The underlying principle of price retracement exists in all markets, making Fibonacci retracement levels applicable across different asset classes.
### How can Fibonacci retracement levels be used to determine entry and exit points?
Fibonacci retracement levels can help traders identify potential entry and exit points. Traders often look for price reactions or patterns near the Fibonacci levels, such as a bounce off a retracement level indicating a potential entry point, or a breach of a retracement level suggesting an exit point.
### Are Fibonacci retracement levels a standalone trading strategy?
Fibonacci retracement levels are not a standalone trading strategy. They are most effective when used in conjunction with other technical analysis tools, such as trendlines, support and resistance levels, and indicators. It’s important to consider multiple factors and conduct a comprehensive analysis to make well-informed trading decisions.
## Key takeaways
• Fibonacci retracement levels are derived from the Fibonacci sequence and represent potential areas of price retracement.
• Traders use Fibonacci retracement levels to identify support and resistance levels during market retracements.
• These levels are commonly used in conjunction with other technical analysis tools for more accurate predictions.
• Fibonacci retracement levels can act as self-fulfilling prophecies due to their widespread usage among traders.
• Selecting the appropriate high and low points for drawing Fibonacci retracement levels can be subjective and may vary among traders.
• Fibonacci retracement levels should be used as part of a comprehensive trading strategy and not relied upon as the sole basis for trading decisions.
###### View Article Sources
1. Energy cryptocurrencies and leading U.S. energy stock prices? – NCBI
2. Demystifying EQA statistics and reports – NCBI
3. Variance and Standard Deviation – CUNY
4. Scarcity in Economics — SuperMoney
5. Demand Curve — SuperMoney
6. Variance — SuperMoney | 1,492 | 7,946 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2023-40 | latest | en | 0.929451 |
https://www.geeksforgeeks.org/prims-mst-for-adjacency-list-representation-greedy-algo-6/ | 1,695,776,428,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510238.65/warc/CC-MAIN-20230927003313-20230927033313-00038.warc.gz | 847,248,454 | 58,684 | Open In App
# Prim’s MST for Adjacency List Representation | Greedy Algo-6
We recommend reading the following two posts as a prerequisite to this post.
We have discussed Prim’s algorithm and its implementation for adjacency matrix representation of graphs. The time complexity for the matrix representation is O(V^2). In this post, O(ELogV) algorithm for adjacency list representation is discussed.
As discussed in the previous post, in Prim’s algorithm, two sets are maintained, one set contains list of vertices already included in MST, other set contains vertices not yet included. With adjacency list representation, all vertices of a graph can be traversed in O(V+E) time using BFS. The idea is to traverse all vertices of graph using BFS and use a Min Heap to store the vertices not yet included in MST. Min Heap is used as a priority queue to get the minimum weight edge from the cut. Min Heap is used as time complexity of operations like extracting minimum element and decreasing key value is O(LogV) in Min Heap.
Following are the detailed steps.
1. Create a Min Heap of size V where V is the number of vertices in the given graph. Every node of min heap contains vertex number and key value of the vertex.
2. Initialize Min Heap with first vertex as root (the key value assigned to first vertex is 0). The key value assigned to all other vertices is INF (infinite).
3. While Min Heap is not empty, do following
1. Extract the min value node from Min Heap. Let the extracted vertex be u.
2. For every adjacent vertex v of u, check if v is in Min Heap (not yet included in MST). If v is in Min Heap and its key value is more than weight of u-v, then update the key value of v as weight of u-v.
Let us understand the above algorithm with the following example:
Initially, key value of first vertex is 0 and INF (infinite) for all other vertices. So vertex 0 is extracted from Min Heap and key values of vertices adjacent to 0 (1 and 7) are updated. Min Heap contains all vertices except vertex 0.
The vertices in green color are the vertices included in MST.
Since key value of vertex 1 is minimum among all nodes in Min Heap, it is extracted from Min Heap and key values of vertices adjacent to 1 are updated (Key is updated if the a vertex is in Min Heap and previous key value is greater than the weight of edge from 1 to the adjacent). Min Heap contains all vertices except vertex 0 and 1.
Since key value of vertex 7 is minimum among all nodes in Min Heap, it is extracted from Min Heap and key values of vertices adjacent to 7 are updated (Key is updated if the a vertex is in Min Heap and previous key value is greater than the weight of edge from 7 to the adjacent). Min Heap contains all vertices except vertex 0, 1 and 7.
Since key value of vertex 6 is minimum among all nodes in Min Heap, it is extracted from Min Heap and key values of vertices adjacent to 6 are updated (Key is updated if the a vertex is in Min Heap and previous key value is greater than the weight of edge from 6 to the adjacent). Min Heap contains all vertices except vertex 0, 1, 7 and 6.
The above steps are repeated for rest of the nodes in Min Heap till Min Heap becomes empty
Implementation:
## C++
`// C / C++ program for Prim's MST for adjacency list``// representation of graph` `#include ``#include ``#include ` `// A structure to represent a node in adjacency list``struct` `AdjListNode {`` ``int` `dest;`` ``int` `weight;`` ``struct` `AdjListNode* next;``};` `// A structure to represent an adjacency list``struct` `AdjList {`` ``struct` `AdjListNode*`` ``head; ``// pointer to head node of list``};` `// A structure to represent a graph. A graph is an array of``// adjacency lists. Size of array will be V (number of``// vertices in graph)``struct` `Graph {`` ``int` `V;`` ``struct` `AdjList* array;``};` `// A utility function to create a new adjacency list node``struct` `AdjListNode* newAdjListNode(``int` `dest, ``int` `weight)``{`` ``struct` `AdjListNode* newNode`` ``= (``struct` `AdjListNode*)``malloc``(`` ``sizeof``(``struct` `AdjListNode));`` ``newNode->dest = dest;`` ``newNode->weight = weight;`` ``newNode->next = NULL;`` ``return` `newNode;``}` `// A utility function that creates a graph of V vertices``struct` `Graph* createGraph(``int` `V)``{`` ``struct` `Graph* graph`` ``= (``struct` `Graph*)``malloc``(``sizeof``(``struct` `Graph));`` ``graph->V = V;` ` ``// Create an array of adjacency lists. Size of array`` ``// will be V`` ``graph->array = (``struct` `AdjList*)``malloc``(`` ``V * ``sizeof``(``struct` `AdjList));` ` ``// Initialize each adjacency list as empty by making`` ``// head as NULL`` ``for` `(``int` `i = 0; i < V; ++i)`` ``graph->array[i].head = NULL;` ` ``return` `graph;``}` `// Adds an edge to an undirected graph``void` `addEdge(``struct` `Graph* graph, ``int` `src, ``int` `dest,`` ``int` `weight)``{`` ``// Add an edge from src to dest. A new node is added to`` ``// the adjacency list of src. The node is added at the`` ``// beginning`` ``struct` `AdjListNode* newNode`` ``= newAdjListNode(dest, weight);`` ``newNode->next = graph->array[src].head;`` ``graph->array[src].head = newNode;` ` ``// Since graph is undirected, add an edge from dest to`` ``// src also`` ``newNode = newAdjListNode(src, weight);`` ``newNode->next = graph->array[dest].head;`` ``graph->array[dest].head = newNode;``}` `// Structure to represent a min heap node``struct` `MinHeapNode {`` ``int` `v;`` ``int` `key;``};` `// Structure to represent a min heap``struct` `MinHeap {`` ``int` `size; ``// Number of heap nodes present currently`` ``int` `capacity; ``// Capacity of min heap`` ``int``* pos; ``// This is needed for decreaseKey()`` ``struct` `MinHeapNode** array;``};` `// A utility function to create a new Min Heap Node``struct` `MinHeapNode* newMinHeapNode(``int` `v, ``int` `key)``{`` ``struct` `MinHeapNode* minHeapNode`` ``= (``struct` `MinHeapNode*)``malloc``(`` ``sizeof``(``struct` `MinHeapNode));`` ``minHeapNode->v = v;`` ``minHeapNode->key = key;`` ``return` `minHeapNode;``}` `// A utility function to create a Min Heap``struct` `MinHeap* createMinHeap(``int` `capacity)``{`` ``struct` `MinHeap* minHeap`` ``= (``struct` `MinHeap*)``malloc``(``sizeof``(``struct` `MinHeap));`` ``minHeap->pos = (``int``*)``malloc``(capacity * ``sizeof``(``int``));`` ``minHeap->size = 0;`` ``minHeap->capacity = capacity;`` ``minHeap->array = (``struct` `MinHeapNode**)``malloc``(`` ``capacity * ``sizeof``(``struct` `MinHeapNode*));`` ``return` `minHeap;``}` `// A utility function to swap two nodes of min heap. Needed``// for min heapify``void` `swapMinHeapNode(``struct` `MinHeapNode** a,`` ``struct` `MinHeapNode** b)``{`` ``struct` `MinHeapNode* t = *a;`` ``*a = *b;`` ``*b = t;``}` `// A standard function to heapify at given idx``// This function also updates position of nodes when they``// are swapped. Position is needed for decreaseKey()``void` `minHeapify(``struct` `MinHeap* minHeap, ``int` `idx)``{`` ``int` `smallest, left, right;`` ``smallest = idx;`` ``left = 2 * idx + 1;`` ``right = 2 * idx + 2;` ` ``if` `(left < minHeap->size`` ``&& minHeap->array[left]->key`` ``< minHeap->array[smallest]->key)`` ``smallest = left;` ` ``if` `(right < minHeap->size`` ``&& minHeap->array[right]->key`` ``< minHeap->array[smallest]->key)`` ``smallest = right;` ` ``if` `(smallest != idx) {`` ``// The nodes to be swapped in min heap`` ``MinHeapNode* smallestNode`` ``= minHeap->array[smallest];`` ``MinHeapNode* idxNode = minHeap->array[idx];` ` ``// Swap positions`` ``minHeap->pos[smallestNode->v] = idx;`` ``minHeap->pos[idxNode->v] = smallest;` ` ``// Swap nodes`` ``swapMinHeapNode(&minHeap->array[smallest],`` ``&minHeap->array[idx]);` ` ``minHeapify(minHeap, smallest);`` ``}``}` `// A utility function to check if the given minHeap is empty``// or not``int` `isEmpty(``struct` `MinHeap* minHeap)``{`` ``return` `minHeap->size == 0;``}` `// Standard function to extract minimum node from heap``struct` `MinHeapNode* extractMin(``struct` `MinHeap* minHeap)``{`` ``if` `(isEmpty(minHeap))`` ``return` `NULL;` ` ``// Store the root node`` ``struct` `MinHeapNode* root = minHeap->array[0];` ` ``// Replace root node with last node`` ``struct` `MinHeapNode* lastNode`` ``= minHeap->array[minHeap->size - 1];`` ``minHeap->array[0] = lastNode;` ` ``// Update position of last node`` ``minHeap->pos[root->v] = minHeap->size - 1;`` ``minHeap->pos[lastNode->v] = 0;` ` ``// Reduce heap size and heapify root`` ``--minHeap->size;`` ``minHeapify(minHeap, 0);` ` ``return` `root;``}` `// Function to decrease key value of a given vertex v. This``// function uses pos[] of min heap to get the current index``// of node in min heap``void` `decreaseKey(``struct` `MinHeap* minHeap, ``int` `v, ``int` `key)``{`` ``// Get the index of v in heap array`` ``int` `i = minHeap->pos[v];` ` ``// Get the node and update its key value`` ``minHeap->array[i]->key = key;` ` ``// Travel up while the complete tree is not heapified.`` ``// This is a O(Logn) loop`` ``while` `(i`` ``&& minHeap->array[i]->key`` ``< minHeap->array[(i - 1) / 2]->key) {`` ``// Swap this node with its parent`` ``minHeap->pos[minHeap->array[i]->v] = (i - 1) / 2;`` ``minHeap->pos[minHeap->array[(i - 1) / 2]->v] = i;`` ``swapMinHeapNode(&minHeap->array[i],`` ``&minHeap->array[(i - 1) / 2]);` ` ``// move to parent index`` ``i = (i - 1) / 2;`` ``}``}` `// A utility function to check if a given vertex``// 'v' is in min heap or not``bool` `isInMinHeap(``struct` `MinHeap* minHeap, ``int` `v)``{`` ``if` `(minHeap->pos[v] < minHeap->size)`` ``return` `true``;`` ``return` `false``;``}` `// A utility function used to print the constructed MST``void` `printArr(``int` `arr[], ``int` `n)``{`` ``for` `(``int` `i = 1; i < n; ++i)`` ``printf``(``"%d - %d\n"``, arr[i], i);``}` `// The main function that constructs Minimum Spanning Tree``// (MST) using Prim's algorithm``void` `PrimMST(``struct` `Graph* graph)``{`` ``int` `V = graph->V; ``// Get the number of vertices in graph`` ``int` `parent[V]; ``// Array to store constructed MST`` ``int` `key[V]; ``// Key values used to pick minimum weight`` ``// edge in cut` ` ``// minHeap represents set E`` ``struct` `MinHeap* minHeap = createMinHeap(V);` ` ``// Initialize min heap with all vertices. Key value of`` ``// all vertices (except 0th vertex) is initially`` ``// infinite`` ``for` `(``int` `v = 1; v < V; ++v) {`` ``parent[v] = -1;`` ``key[v] = INT_MAX;`` ``minHeap->array[v] = newMinHeapNode(v, key[v]);`` ``minHeap->pos[v] = v;`` ``}` ` ``// Make key value of 0th vertex as 0 so that it`` ``// is extracted first`` ``key[0] = 0;`` ``minHeap->array[0] = newMinHeapNode(0, key[0]);`` ``minHeap->pos[0] = 0;` ` ``// Initially size of min heap is equal to V`` ``minHeap->size = V;` ` ``// In the following loop, min heap contains all nodes`` ``// not yet added to MST.`` ``while` `(!isEmpty(minHeap)) {`` ``// Extract the vertex with minimum key value`` ``struct` `MinHeapNode* minHeapNode`` ``= extractMin(minHeap);`` ``int` `u`` ``= minHeapNode`` ``->v; ``// Store the extracted vertex number` ` ``// Traverse through all adjacent vertices of u (the`` ``// extracted vertex) and update their key values`` ``struct` `AdjListNode* pCrawl = graph->array[u].head;`` ``while` `(pCrawl != NULL) {`` ``int` `v = pCrawl->dest;` ` ``// If v is not yet included in MST and weight of`` ``// u-v is less than key value of v, then update`` ``// key value and parent of v`` ``if` `(isInMinHeap(minHeap, v)`` ``&& pCrawl->weight < key[v]) {`` ``key[v] = pCrawl->weight;`` ``parent[v] = u;`` ``decreaseKey(minHeap, v, key[v]);`` ``}`` ``pCrawl = pCrawl->next;`` ``}`` ``}` ` ``// print edges of MST`` ``printArr(parent, V);``}` `// Driver program to test above functions``int` `main()``{`` ``// Let us create the graph given in above figure`` ``int` `V = 9;`` ``struct` `Graph* graph = createGraph(V);`` ``addEdge(graph, 0, 1, 4);`` ``addEdge(graph, 0, 7, 8);`` ``addEdge(graph, 1, 2, 8);`` ``addEdge(graph, 1, 7, 11);`` ``addEdge(graph, 2, 3, 7);`` ``addEdge(graph, 2, 8, 2);`` ``addEdge(graph, 2, 5, 4);`` ``addEdge(graph, 3, 4, 9);`` ``addEdge(graph, 3, 5, 14);`` ``addEdge(graph, 4, 5, 10);`` ``addEdge(graph, 5, 6, 2);`` ``addEdge(graph, 6, 7, 1);`` ``addEdge(graph, 6, 8, 6);`` ``addEdge(graph, 7, 8, 7);` ` ``PrimMST(graph);` ` ``return` `0;``}`
## Java
`// java program for Prim's MST for adjacency list``// representation of graph` `import` `java.util.ArrayList;``import` `java.util.PriorityQueue;` `public` `class` `Main {`` ``static` `class` `Graph {`` ``int` `V;`` ``ArrayList> adj;` ` ``// Inner class to represent an edge (destination and weight)`` ``static` `class` `Node {`` ``int` `dest;`` ``int` `weight;` ` ``Node(``int` `dest, ``int` `weight) {`` ``this``.dest = dest;`` ``this``.weight = weight;`` ``}`` ``}` ` ``Graph(``int` `V) {`` ``this``.V = V;`` ``adj = ``new` `ArrayList<>(V);`` ``for` `(``int` `i = ``0``; i < V; i++)`` ``adj.add(``new` `ArrayList<>());`` ``}` ` ``// Function to add an undirected edge between two vertices with given weight`` ``void` `addEdge(``int` `src, ``int` `dest, ``int` `weight) {`` ``adj.get(src).add(``new` `Node(dest, weight));`` ``adj.get(dest).add(``new` `Node(src, weight));`` ``}` ` ``// Function to find the Minimum Spanning Tree using Prim's algorithm`` ``void` `primMST() {`` ``int``[] parent = ``new` `int``[V];`` ``int``[] key = ``new` `int``[V];`` ``boolean``[] inMST = ``new` `boolean``[V];` ` ``for` `(``int` `i = ``0``; i < V; i++) {`` ``parent[i] = -``1``; ``// Array to store the parent node of each vertex in the MST`` ``key[i] = Integer.MAX_VALUE; ``// Array to store the minimum key value for each vertex`` ``inMST[i] = ``false``; ``// Array to track if the vertex is in the MST or not`` ``}` ` ``PriorityQueue minHeap = ``new` `PriorityQueue<>((a, b) -> a.weight - b.weight);` ` ``key[``0``] = ``0``; ``// Start the MST from vertex 0`` ``minHeap.add(``new` `Node(``0``, key[``0``]));` ` ``while` `(!minHeap.isEmpty()) {`` ``Node u = minHeap.poll(); ``// Extract the node with the minimum key value`` ``int` `uVertex = u.dest;`` ``inMST[uVertex] = ``true``;` ` ``// Traverse through all adjacent vertices of u (the extracted vertex) and update their key values`` ``for` `(Node v : adj.get(uVertex)) {`` ``int` `vVertex = v.dest;`` ``int` `weight = v.weight;` ` ``// If v is not yet included in MST and weight of u-v is less than key value of v, then update key value and parent of v`` ``if` `(!inMST[vVertex] && weight < key[vVertex]) {`` ``parent[vVertex] = uVertex;`` ``key[vVertex] = weight;`` ``minHeap.add(``new` `Node(vVertex, key[vVertex]));`` ``}`` ``}`` ``}` ` ``printMST(parent);`` ``}` ` ``// Function to print the edges of the Minimum Spanning Tree`` ``void` `printMST(``int``[] parent) {`` ``System.out.println(``"Edges of Minimum Spanning Tree:"``);`` ``for` `(``int` `i = ``1``; i < V; i++) {`` ``System.out.println(parent[i] + ``" - "` `+ i);`` ``}`` ``}`` ``}` ` ``public` `static` `void` `main(String[] args) {`` ``int` `V = ``9``;`` ``Graph graph = ``new` `Graph(V);`` ``graph.addEdge(``0``, ``1``, ``4``);`` ``graph.addEdge(``0``, ``7``, ``8``);`` ``graph.addEdge(``1``, ``2``, ``8``);`` ``graph.addEdge(``1``, ``7``, ``11``);`` ``graph.addEdge(``2``, ``3``, ``7``);`` ``graph.addEdge(``2``, ``8``, ``2``);`` ``graph.addEdge(``2``, ``5``, ``4``);`` ``graph.addEdge(``3``, ``4``, ``9``);`` ``graph.addEdge(``3``, ``5``, ``14``);`` ``graph.addEdge(``4``, ``5``, ``10``);`` ``graph.addEdge(``5``, ``6``, ``2``);`` ``graph.addEdge(``6``, ``7``, ``1``);`` ``graph.addEdge(``6``, ``8``, ``6``);`` ``graph.addEdge(``7``, ``8``, ``7``);` ` ``graph.primMST();`` ``}``}`
## Python3
`# A Python program for Prims's MST for``# adjacency list representation of graph` `from` `collections ``import` `defaultdict``import` `sys` `class` `Heap():` ` ``def` `__init__(``self``):`` ``self``.array ``=` `[]`` ``self``.size ``=` `0`` ``self``.pos ``=` `[]` ` ``def` `newMinHeapNode(``self``, v, dist):`` ``minHeapNode ``=` `[v, dist]`` ``return` `minHeapNode` ` ``# A utility function to swap two nodes of`` ``# min heap. Needed for min heapify`` ``def` `swapMinHeapNode(``self``, a, b):`` ``t ``=` `self``.array[a]`` ``self``.array[a] ``=` `self``.array[b]`` ``self``.array[b] ``=` `t` ` ``# A standard function to heapify at given idx`` ``# This function also updates position of nodes`` ``# when they are swapped. Position is needed`` ``# for decreaseKey()`` ``def` `minHeapify(``self``, idx):`` ``smallest ``=` `idx`` ``left ``=` `2` `*` `idx ``+` `1`` ``right ``=` `2` `*` `idx ``+` `2` ` ``if` `left < ``self``.size ``and` `self``.array[left][``1``] < \`` ``self``.array[smallest][``1``]:`` ``smallest ``=` `left` ` ``if` `right < ``self``.size ``and` `self``.array[right][``1``] < \`` ``self``.array[smallest][``1``]:`` ``smallest ``=` `right` ` ``# The nodes to be swapped in min heap`` ``# if idx is not smallest`` ``if` `smallest !``=` `idx:` ` ``# Swap positions`` ``self``.pos[``self``.array[smallest][``0``]] ``=` `idx`` ``self``.pos[``self``.array[idx][``0``]] ``=` `smallest` ` ``# Swap nodes`` ``self``.swapMinHeapNode(smallest, idx)` ` ``self``.minHeapify(smallest)` ` ``# Standard function to extract minimum node from heap`` ``def` `extractMin(``self``):` ` ``# Return NULL wif heap is empty`` ``if` `self``.isEmpty() ``=``=` `True``:`` ``return` ` ``# Store the root node`` ``root ``=` `self``.array[``0``]` ` ``# Replace root node with last node`` ``lastNode ``=` `self``.array[``self``.size ``-` `1``]`` ``self``.array[``0``] ``=` `lastNode` ` ``# Update position of last node`` ``self``.pos[lastNode[``0``]] ``=` `0`` ``self``.pos[root[``0``]] ``=` `self``.size ``-` `1` ` ``# Reduce heap size and heapify root`` ``self``.size ``-``=` `1`` ``self``.minHeapify(``0``)` ` ``return` `root` ` ``def` `isEmpty(``self``):`` ``return` `True` `if` `self``.size ``=``=` `0` `else` `False` ` ``def` `decreaseKey(``self``, v, dist):` ` ``# Get the index of v in heap array` ` ``i ``=` `self``.pos[v]` ` ``# Get the node and update its dist value`` ``self``.array[i][``1``] ``=` `dist` ` ``# Travel up while the complete tree is not`` ``# heapified. This is a O(Logn) loop`` ``while` `i > ``0` `and` `self``.array[i][``1``] < \`` ``self``.array[(i ``-` `1``) ``/``/` `2``][``1``]:` ` ``# Swap this node with its parent`` ``self``.pos[``self``.array[i][``0``]] ``=` `(i``-``1``)``/``2`` ``self``.pos[``self``.array[(i``-``1``)``/``/``2``][``0``]] ``=` `i`` ``self``.swapMinHeapNode(i, (i ``-` `1``)``/``/``2``)` ` ``# move to parent index`` ``i ``=` `(i ``-` `1``) ``/``/` `2` ` ``# A utility function to check if a given vertex`` ``# 'v' is in min heap or not`` ``def` `isInMinHeap(``self``, v):` ` ``if` `self``.pos[v] < ``self``.size:`` ``return` `True`` ``return` `False` `def` `printArr(parent, n):`` ``for` `i ``in` `range``(``1``, n):`` ``print``(``"% d - % d"` `%` `(parent[i], i))` `class` `Graph():` ` ``def` `__init__(``self``, V):`` ``self``.V ``=` `V`` ``self``.graph ``=` `defaultdict(``list``)` ` ``# Adds an edge to an undirected graph`` ``def` `addEdge(``self``, src, dest, weight):` ` ``# Add an edge from src to dest. A new node is`` ``# added to the adjacency list of src. The node`` ``# is added at the beginning. The first element of`` ``# the node has the destination and the second`` ``# elements has the weight`` ``newNode ``=` `[dest, weight]`` ``self``.graph[src].insert(``0``, newNode)` ` ``# Since graph is undirected, add an edge from`` ``# dest to src also`` ``newNode ``=` `[src, weight]`` ``self``.graph[dest].insert(``0``, newNode)` ` ``# The main function that prints the Minimum`` ``# Spanning Tree(MST) using the Prim's Algorithm.`` ``# It is a O(ELogV) function`` ``def` `PrimMST(``self``):`` ``# Get the number of vertices in graph`` ``V ``=` `self``.V` ` ``# key values used to pick minimum weight edge in cut`` ``key ``=` `[]` ` ``# List to store constructed MST`` ``parent ``=` `[]` ` ``# minHeap represents set E`` ``minHeap ``=` `Heap()` ` ``# Initialize min heap with all vertices. Key values of all`` ``# vertices (except the 0th vertex) is initially infinite`` ``for` `v ``in` `range``(V):`` ``parent.append(``-``1``)`` ``key.append(``1e7``)`` ``minHeap.array.append(minHeap.newMinHeapNode(v, key[v]))`` ``minHeap.pos.append(v)` ` ``# Make key value of 0th vertex as 0 so`` ``# that it is extracted first`` ``minHeap.pos[``0``] ``=` `0`` ``key[``0``] ``=` `0`` ``minHeap.decreaseKey(``0``, key[``0``])` ` ``# Initially size of min heap is equal to V`` ``minHeap.size ``=` `V` ` ``# In the following loop, min heap contains all nodes`` ``# not yet added in the MST.`` ``while` `minHeap.isEmpty() ``=``=` `False``:` ` ``# Extract the vertex with minimum distance value`` ``newHeapNode ``=` `minHeap.extractMin()`` ``u ``=` `newHeapNode[``0``]` ` ``# Traverse through all adjacent vertices of u`` ``# (the extracted vertex) and update their`` ``# distance values`` ``for` `pCrawl ``in` `self``.graph[u]:` ` ``v ``=` `pCrawl[``0``]` ` ``# If shortest distance to v is not finalized`` ``# yet, and distance to v through u is less than`` ``# its previously calculated distance`` ``if` `minHeap.isInMinHeap(v) ``and` `pCrawl[``1``] < key[v]:`` ``key[v] ``=` `pCrawl[``1``]`` ``parent[v] ``=` `u` ` ``# update distance value in min heap also`` ``minHeap.decreaseKey(v, key[v])` ` ``printArr(parent, V)` `# Driver program to test the above functions``graph ``=` `Graph(``9``)``graph.addEdge(``0``, ``1``, ``4``)``graph.addEdge(``0``, ``7``, ``8``)``graph.addEdge(``1``, ``2``, ``8``)``graph.addEdge(``1``, ``7``, ``11``)``graph.addEdge(``2``, ``3``, ``7``)``graph.addEdge(``2``, ``8``, ``2``)``graph.addEdge(``2``, ``5``, ``4``)``graph.addEdge(``3``, ``4``, ``9``)``graph.addEdge(``3``, ``5``, ``14``)``graph.addEdge(``4``, ``5``, ``10``)``graph.addEdge(``5``, ``6``, ``2``)``graph.addEdge(``6``, ``7``, ``1``)``graph.addEdge(``6``, ``8``, ``6``)``graph.addEdge(``7``, ``8``, ``7``)``graph.PrimMST()` `# This code is contributed by Divyanshu Mehta`
## C#
`using` `System;``using` `System.Collections.Generic;` `// A structure to represent a node in adjacency list``class` `AdjListNode``{`` ``public` `int` `dest;`` ``public` `int` `weight;`` ``public` `AdjListNode next;``}` `// A structure to represent an adjacency list``class` `AdjList``{`` ``public` `AdjListNode head;``}` `// A structure to represent a graph. A graph is an array of``// adjacency lists. Size of array will be V (number of``// vertices in graph)``class` `Graph``{`` ``public` `int` `V;`` ``public` `AdjList[] array;``}` `class` `Program``{`` ``// A utility function to create a new adjacency list node`` ``static` `AdjListNode newAdjListNode(``int` `dest, ``int` `weight)`` ``{`` ``AdjListNode newNode = ``new` `AdjListNode();`` ``newNode.dest = dest;`` ``newNode.weight = weight;`` ``newNode.next = ``null``;`` ``return` `newNode;`` ``}` ` ``// A utility function that creates a graph of V vertices`` ``static` `Graph createGraph(``int` `V)`` ``{`` ``Graph graph = ``new` `Graph();`` ``graph.V = V;`` ``graph.array = ``new` `AdjList[V];` ` ``for` `(``int` `i = 0; i < V; ++i)`` ``graph.array[i] = ``new` `AdjList();` ` ``return` `graph;`` ``}` ` ``// Adds an edge to an undirected graph`` ``static` `void` `addEdge(Graph graph, ``int` `src, ``int` `dest, ``int` `weight)`` ``{`` ``AdjListNode newNode = newAdjListNode(dest, weight);`` ``newNode.next = graph.array[src].head;`` ``graph.array[src].head = newNode;` ` ``newNode = newAdjListNode(src, weight);`` ``newNode.next = graph.array[dest].head;`` ``graph.array[dest].head = newNode;`` ``}` ` ``// Structure to represent a min heap node`` ``class` `MinHeapNode`` ``{`` ``public` `int` `v;`` ``public` `int` `key;`` ``}` ` ``// Structure to represent a min heap`` ``class` `MinHeap`` ``{`` ``public` `int` `size;`` ``public` `int` `capacity;`` ``public` `int``[] pos;`` ``public` `MinHeapNode[] array;`` ``}` ` ``// A utility function to create a new Min Heap Node`` ``static` `MinHeapNode newMinHeapNode(``int` `v, ``int` `key)`` ``{`` ``MinHeapNode minHeapNode = ``new` `MinHeapNode();`` ``minHeapNode.v = v;`` ``minHeapNode.key = key;`` ``return` `minHeapNode;`` ``}` ` ``// A utility function to create a Min Heap`` ``static` `MinHeap createMinHeap(``int` `capacity)`` ``{`` ``MinHeap minHeap = ``new` `MinHeap();`` ``minHeap.pos = ``new` `int``[capacity];`` ``minHeap.size = 0;`` ``minHeap.capacity = capacity;`` ``minHeap.array = ``new` `MinHeapNode[capacity];`` ``return` `minHeap;`` ``}` ` ``// A utility function to swap two nodes of min heap`` ``static` `void` `swapMinHeapNode(``ref` `MinHeapNode a, ``ref` `MinHeapNode b)`` ``{`` ``MinHeapNode t = a;`` ``a = b;`` ``b = t;`` ``}` ` ``// A standard function to heapify at given idx`` ``// This function also updates position of nodes when they`` ``// are swapped. Position is needed for decreaseKey()`` ``static` `void` `minHeapify(MinHeap minHeap, ``int` `idx)`` ``{`` ``int` `smallest, left, right;`` ``smallest = idx;`` ``left = 2 * idx + 1;`` ``right = 2 * idx + 2;` ` ``if` `(left < minHeap.size && minHeap.array[left].key < minHeap.array[smallest].key)`` ``smallest = left;` ` ``if` `(right < minHeap.size && minHeap.array[right].key < minHeap.array[smallest].key)`` ``smallest = right;` ` ``if` `(smallest != idx)`` ``{`` ``MinHeapNode smallestNode = minHeap.array[smallest];`` ``MinHeapNode idxNode = minHeap.array[idx];` ` ``minHeap.pos[smallestNode.v] = idx;`` ``minHeap.pos[idxNode.v] = smallest;` ` ``swapMinHeapNode(``ref` `minHeap.array[smallest], ``ref` `minHeap.array[idx]);` ` ``minHeapify(minHeap, smallest);`` ``}`` ``}` ` ``// A utility function to check if the given minHeap is empty or not`` ``static` `bool` `isEmpty(MinHeap minHeap)`` ``{`` ``return` `minHeap.size == 0;`` ``}` ` ``// Standard function to extract minimum node from heap`` ``static` `MinHeapNode extractMin(MinHeap minHeap)`` ``{`` ``if` `(isEmpty(minHeap))`` ``return` `null``;` ` ``MinHeapNode root = minHeap.array[0];` ` ``MinHeapNode lastNode = minHeap.array[minHeap.size - 1];`` ``minHeap.array[0] = lastNode;` ` ``minHeap.pos[root.v] = minHeap.size - 1;`` ``minHeap.pos[lastNode.v] = 0;` ` ``--minHeap.size;`` ``minHeapify(minHeap, 0);` ` ``return` `root;`` ``}` ` ``// Function to decrease key value of a given vertex v.`` ``// This function uses pos[] of min heap to get the current index`` ``// of node in min heap`` ``static` `void` `decreaseKey(MinHeap minHeap, ``int` `v, ``int` `key)`` ``{`` ``int` `i = minHeap.pos[v];` ` ``minHeap.array[i].key = key;` ` ``while` `(i > 0 && minHeap.array[i].key < minHeap.array[(i - 1) / 2].key)`` ``{`` ``minHeap.pos[minHeap.array[i].v] = (i - 1) / 2;`` ``minHeap.pos[minHeap.array[(i - 1) / 2].v] = i;`` ``swapMinHeapNode(``ref` `minHeap.array[i], ``ref` `minHeap.array[(i - 1) / 2]);` ` ``i = (i - 1) / 2;`` ``}`` ``}` ` ``// A utility function to check if a given vertex 'v' is in min heap or not`` ``static` `bool` `isInMinHeap(MinHeap minHeap, ``int` `v)`` ``{`` ``return` `minHeap.pos[v] < minHeap.size;`` ``}` ` ``// A utility function used to print the constructed MST`` ``static` `void` `printArr(``int``[] arr, ``int` `n)`` ``{`` ``for` `(``int` `i = 1; i < n; ++i)`` ``Console.WriteLine(arr[i] + ``" - "` `+ i);`` ``}` ` ``// The main function that constructs Minimum Spanning Tree (MST) using Prim's algorithm`` ``static` `void` `PrimMST(Graph graph)`` ``{`` ``int` `V = graph.V;`` ``int``[] parent = ``new` `int``[V];`` ``int``[] key = ``new` `int``[V];` ` ``MinHeap minHeap = createMinHeap(V);` ` ``for` `(``int` `v = 1; v < V; ++v)`` ``{`` ``parent[v] = -1;`` ``key[v] = ``int``.MaxValue;`` ``minHeap.array[v] = newMinHeapNode(v, key[v]);`` ``minHeap.pos[v] = v;`` ``}` ` ``key[0] = 0;`` ``minHeap.array[0] = newMinHeapNode(0, key[0]);`` ``minHeap.pos[0] = 0;` ` ``minHeap.size = V;` ` ``while` `(!isEmpty(minHeap))`` ``{`` ``MinHeapNode minHeapNode = extractMin(minHeap);`` ``int` `u = minHeapNode.v;` ` ``AdjListNode pCrawl = graph.array[u].head;`` ``while` `(pCrawl != ``null``)`` ``{`` ``int` `v = pCrawl.dest;` ` ``if` `(isInMinHeap(minHeap, v) && pCrawl.weight < key[v])`` ``{`` ``key[v] = pCrawl.weight;`` ``parent[v] = u;`` ``decreaseKey(minHeap, v, key[v]);`` ``}`` ``pCrawl = pCrawl.next;`` ``}`` ``}` ` ``printArr(parent, V);`` ``}` ` ``static` `void` `Main(``string``[] args)`` ``{`` ``int` `V = 9;`` ``Graph graph = createGraph(V);`` ``addEdge(graph, 0, 1, 4);`` ``addEdge(graph, 0, 7, 8);`` ``addEdge(graph, 1, 2, 8);`` ``addEdge(graph, 1, 7, 11);`` ``addEdge(graph, 2, 3, 7);`` ``addEdge(graph, 2, 8, 2);`` ``addEdge(graph, 2, 5, 4);`` ``addEdge(graph, 3, 4, 9);`` ``addEdge(graph, 3, 5, 14);`` ``addEdge(graph, 4, 5, 10);`` ``addEdge(graph, 5, 6, 2);`` ``addEdge(graph, 6, 7, 1);`` ``addEdge(graph, 6, 8, 6);`` ``addEdge(graph, 7, 8, 7);` ` ``PrimMST(graph);`` ``}``}`
## Javascript
``
Output
```0 - 1
5 - 2
2 - 3
3 - 4
6 - 5
7 - 6
0 - 7
2 - 8
```
Time Complexity:
The time complexity of the above code/algorithm looks O(V^2) as there are two nested while loops. If we take a closer look, we can observe that the statements in inner loop are executed O(V+E) times (similar to BFS). The inner loop has decreaseKey() operation which takes O(LogV) time. So overall time complexity is O(E+V)*O(LogV) which is O((E+V)*LogV) = O(ELogV) (For a connected graph, V = O(E))
Space Complexity:
The space complexity of the above code/algorithm is O(V + E) because we need to store the adjacency list representation of the graph, the heap data structure, and the MST. | 11,118 | 33,709 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2023-40 | latest | en | 0.911239 |
https://cg.cs.uni-bonn.de/en/teaching/2018-ws/lecture-foundations-of-graphics/ | 1,618,725,167,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038468066.58/warc/CC-MAIN-20210418043500-20210418073500-00511.warc.gz | 276,008,631 | 6,428 | Lecture: Foundations of Graphics
Course
• Lecturer(s):
• Start: October 9
• Dates: Tue. 14:15 - 15:45, HS7
• Course number: MA-INF 2111
• Curriculum: Master
• Exams: tba
Exercises
• Tutor(s):
• Start: October 16
• Dates: Tue. 10:15 - 11:45, 12:15 - 13:45, Room 3.035b
Description
This is an introductory course to prepare you for master's projects and advanced courses in computer graphics. Basic math and programming knowledge is needed to follow this course; exercises will include coding problems in Python.
• An elementary introduction to kinematics and ordinary differential equations for Physics based modelling (rigid body motion, etc.)
• Basics of numeric integration
• Clustering algorithms for data compression
• Robust statistics (RANSAC, etc)
• Optimization techniques
• Foundations of Signal processing
• Monte Carlo methods with applications to photorealistic rendering
• Polynomials and subdivision schemes for curves and surfaces.
Welcome to the lecture
The programming exercises in this course will be in Python. In case you have only little knowledge in this language or programming in general, we advise you to have a look at this tutorial. We wish everyone a good start!
Assignment Sheets
Exercise 0: Python Assignment sheet (PDF document, 116 KB) Exercise 1: Transformations Assignment sheet (PDF document, 130 KB) 01-Transformations-framework.zip (ZIP archive, 26 KB) Exercise 2: Physics Assignment sheet (PDF document, 188 KB) 02-Physics-framework.zip (ZIP archive, 854 Bytes) Exercise 3: Raytracing Assignment sheet (PDF document, 347 KB) 03-Raytracing-framework.zip (ZIP archive, 33 KB) Exercise 4: Monte-Carlo-Integration Assignment sheet (PDF document, 166 KB) ImportanceSampling.py (Python script, 1.4 KB) ImportanceSampling_Convergence.png (PNG image, 44 KB) ImportanceSampling_Functions.png (PNG image, 40 KB) ImportanceSampling_Normal_Distribution.png (PNG image, 33 KB) ImportanceSampling_Polynomial.png (PNG image, 34 KB) MonteCarlo.py (Python script, 3.2 KB) Exercise 5: Circle-fitting Assignment sheet (PDF document, 1.0 MB) coins.png (PNG image, 37 KB) hough_circle_fitting.py (Python script, 6.2 KB) ransac_circle_fitting.py (Python script, 6.1 KB) Exercise 6: Segmentation Assignment sheet (PDF document, 541 KB) 06-Segmentation-framework.zip (ZIP archive, 514 KB) Exercise 7: FourierAnalysis Assignment sheet (PDF document, 162 KB) 07-Fourier-framework.zip (ZIP archive, 46 KB) Exercise 8: PCA-and-SVD Assignment sheet (PDF document, 168 KB) framework.zip (ZIP archive, 21 KB) Exercise 9: Parametric-Curves Assignment sheet (PDF document, 121 KB) 09-Curves-framework.zip (ZIP archive, 625 Bytes) Exercise 10: Colors Assignment sheet (PDF document, 573 KB) 10-Colors-framework.zip (ZIP archive, 283 KB) Exercise 98: TestExam Assignment sheet (PDF document, 519 KB) Exercise 98: TestExamSolution Assignment sheet (PDF document, 653 KB) | 768 | 2,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-17 | latest | en | 0.651734 |
http://de.metamath.org/mpegif/clwwnisshclwwn.html | 1,582,301,051,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145533.1/warc/CC-MAIN-20200221142006-20200221172006-00027.warc.gz | 38,651,512 | 10,956 | Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > clwwnisshclwwn Structured version Unicode version
Theorem clwwnisshclwwn 24632
Description: Cyclically shifting a closed walk as word of fixed length results in a closed walk as word of the same length (in an undirected graph). (Contributed by Alexander van der Vekens, 10-Jun-2018.)
Assertion
Ref Expression
clwwnisshclwwn ClWWalksN cyclShift ClWWalksN
Proof of Theorem clwwnisshclwwn
StepHypRef Expression
1 clwwlknprop 24595 . . . . . . 7 ClWWalksN Word
2 simpl 457 . . . . . . . . . . 11
32anim2i 569 . . . . . . . . . 10
4 df-3an 975 . . . . . . . . . 10
53, 4sylibr 212 . . . . . . . . 9
653adant2 1015 . . . . . . . 8 Word
7 clwwlkisclwwlkn 24614 . . . . . . . 8 ClWWalksN ClWWalks
86, 7syl 16 . . . . . . 7 Word ClWWalksN ClWWalks
91, 8mpcom 36 . . . . . 6 ClWWalksN ClWWalks
109adantl 466 . . . . 5 ClWWalksN ClWWalks
1110adantr 465 . . . 4 ClWWalksN ClWWalks
12 eqcom 2476 . . . . . . . . . . . 12
1312biimpi 194 . . . . . . . . . . 11
1413adantl 466 . . . . . . . . . 10
15143ad2ant3 1019 . . . . . . . . 9 Word
161, 15syl 16 . . . . . . . 8 ClWWalksN
1716adantl 466 . . . . . . 7 ClWWalksN
1817oveq2d 6311 . . . . . 6 ClWWalksN
1918eleq2d 2537 . . . . 5 ClWWalksN
2019biimpa 484 . . . 4 ClWWalksN
21 clwwisshclwwn 24631 . . . 4 ClWWalks cyclShift ClWWalks
2211, 20, 21syl2anc 661 . . 3 ClWWalksN cyclShift ClWWalks
23 elfzelz 11700 . . . . . . . . . . 11
2423anim2i 569 . . . . . . . . . 10 Word Word
25 cshwlen 12750 . . . . . . . . . 10 Word cyclShift
2624, 25syl 16 . . . . . . . . 9 Word cyclShift
2726ex 434 . . . . . . . 8 Word cyclShift
28273ad2ant2 1018 . . . . . . 7 Word cyclShift
29 oveq2 6303 . . . . . . . . . . . 12
3029eleq2d 2537 . . . . . . . . . . 11
31 eqeq2 2482 . . . . . . . . . . 11 cyclShift cyclShift
3230, 31imbi12d 320 . . . . . . . . . 10 cyclShift cyclShift
3332eqcoms 2479 . . . . . . . . 9 cyclShift cyclShift
3433adantl 466 . . . . . . . 8 cyclShift cyclShift
35343ad2ant3 1019 . . . . . . 7 Word cyclShift cyclShift
3628, 35mpbird 232 . . . . . 6 Word cyclShift
371, 36syl 16 . . . . 5 ClWWalksN cyclShift
3837adantl 466 . . . 4 ClWWalksN cyclShift
3938imp 429 . . 3 ClWWalksN cyclShift
401simp1d 1008 . . . . . . . 8 ClWWalksN
4140anim1i 568 . . . . . . 7 ClWWalksN
4241, 4sylibr 212 . . . . . 6 ClWWalksN
4342ancoms 453 . . . . 5 ClWWalksN
4443adantr 465 . . . 4 ClWWalksN
45 isclwwlkn 24592 . . . 4 cyclShift ClWWalksN cyclShift ClWWalks cyclShift
4644, 45syl 16 . . 3 ClWWalksN cyclShift ClWWalksN cyclShift ClWWalks cyclShift
4722, 39, 46mpbir2and 920 . 2 ClWWalksN cyclShift ClWWalksN
4847ex 434 1 ClWWalksN cyclShift ClWWalksN
Colors of variables: wff setvar class Syntax hints: wi 4 wb 184 wa 369 w3a 973 wceq 1379 wcel 1767 cvv 3118 cfv 5594 (class class class)co 6295 cc0 9504 cn0 10807 cz 10876 cfz 11684 chash 12385 Word cword 12515 cyclShift ccsh 12739 ClWWalks cclwwlk 24571 ClWWalksN cclwwlkn 24572 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1601 ax-4 1612 ax-5 1680 ax-6 1719 ax-7 1739 ax-8 1769 ax-9 1771 ax-10 1786 ax-11 1791 ax-12 1803 ax-13 1968 ax-ext 2445 ax-rep 4564 ax-sep 4574 ax-nul 4582 ax-pow 4631 ax-pr 4692 ax-un 6587 ax-cnex 9560 ax-resscn 9561 ax-1cn 9562 ax-icn 9563 ax-addcl 9564 ax-addrcl 9565 ax-mulcl 9566 ax-mulrcl 9567 ax-mulcom 9568 ax-addass 9569 ax-mulass 9570 ax-distr 9571 ax-i2m1 9572 ax-1ne0 9573 ax-1rid 9574 ax-rnegex 9575 ax-rrecex 9576 ax-cnre 9577 ax-pre-lttri 9578 ax-pre-lttrn 9579 ax-pre-ltadd 9580 ax-pre-mulgt0 9581 ax-pre-sup 9582 This theorem depends on definitions: df-bi 185 df-or 370 df-an 371 df-3or 974 df-3an 975 df-tru 1382 df-ex 1597 df-nf 1600 df-sb 1712 df-eu 2279 df-mo 2280 df-clab 2453 df-cleq 2459 df-clel 2462 df-nfc 2617 df-ne 2664 df-nel 2665 df-ral 2822 df-rex 2823 df-reu 2824 df-rmo 2825 df-rab 2826 df-v 3120 df-sbc 3337 df-csb 3441 df-dif 3484 df-un 3486 df-in 3488 df-ss 3495 df-pss 3497 df-nul 3791 df-if 3946 df-pw 4018 df-sn 4034 df-pr 4036 df-tp 4038 df-op 4040 df-uni 4252 df-int 4289 df-iun 4333 df-br 4454 df-opab 4512 df-mpt 4513 df-tr 4547 df-eprel 4797 df-id 4801 df-po 4806 df-so 4807 df-fr 4844 df-we 4846 df-ord 4887 df-on 4888 df-lim 4889 df-suc 4890 df-xp 5011 df-rel 5012 df-cnv 5013 df-co 5014 df-dm 5015 df-rn 5016 df-res 5017 df-ima 5018 df-iota 5557 df-fun 5596 df-fn 5597 df-f 5598 df-f1 5599 df-fo 5600 df-f1o 5601 df-fv 5602 df-riota 6256 df-ov 6298 df-oprab 6299 df-mpt2 6300 df-om 6696 df-1st 6795 df-2nd 6796 df-recs 7054 df-rdg 7088 df-1o 7142 df-oadd 7146 df-er 7323 df-map 7434 df-pm 7435 df-en 7529 df-dom 7530 df-sdom 7531 df-fin 7532 df-sup 7913 df-card 8332 df-pnf 9642 df-mnf 9643 df-xr 9644 df-ltxr 9645 df-le 9646 df-sub 9819 df-neg 9820 df-div 10219 df-nn 10549 df-2 10606 df-n0 10808 df-z 10877 df-uz 11095 df-rp 11233 df-fz 11685 df-fzo 11805 df-fl 11909 df-mod 11977 df-hash 12386 df-word 12523 df-lsw 12524 df-concat 12525 df-substr 12527 df-csh 12740 df-clwwlk 24574 df-clwwlkn 24575 This theorem is referenced by: clwwlknscsh 24642
Copyright terms: Public domain W3C validator | 2,698 | 5,279 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2020-10 | latest | en | 0.136455 |
https://www.sawaal.com/square-roots-and-cube-roots-questions-and-answers/a-group-of-students-decided-to-collect-as-many-paise-from-each-member-of-group-as-is-the-number-of-m_5247 | 1,532,160,452,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676592420.72/warc/CC-MAIN-20180721071046-20180721091046-00015.warc.gz | 967,114,394 | 15,436 | 18
Q:
# A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:
A) 57 B) 67 C) 77 D) 87
Explanation:
Money collected = (59.29 x 100) paise = 5929 paise.
Number of members = $5929$ = 77.
Q:
Cube root of 729 then square it
A) 9 B) 36 C) 81 D) 144
Explanation:
729 = 9 x 9 x 9
=> Cube root of 729 = 9
Now, required square of 9 = 9 x 9 = 81.
0 85
Q:
If (89)2 is added to the square of a number, the answer so obtained is 16202. What is the (1/26) of that number?
A) 5.65 B) 2.7 C) 3.5 D) 6.66
Explanation:
Let the number is = x
(89)2 + x2 = 16202
x2 = 8281
x = 91
=> (1/26) of 91 = 3.5
10 575
Q:
What should come in place of x in the following equation?
A) 13 B) 12 C) 17 D) 16
Explanation:
Then x2 = $162x128$
= Sqrt of 82 x 62 x 32
= 8 x 6 x 3
x2 = 144.
x = 12.
5 507
Q:
Find the value of $1.21x0.91.1x0.11$?
A) 0 B) 1 C) 3 D) 5
Explanation:
Given $1.21x0.91.1x0.11$
$1.21x100x0.9x101.1x10x0.11x100$
$9$
= 3.
5 788
Q:
If then find the value of $x3-6x2+6x$ ?
A) 1 B) 2 C) 3 D) 4
Explanation:
Given that
So x - 2 =
Now cubing on both sides, we get
=>
Therefore,
5 580
Q:
If
A) 3/4 B) 4/3 C) 3/5 D) 5/3
Explanation:
By simplifying, we get 4/3
3 1081
Q:
If x =
A) 3sqrt{3} B) 8sqrt{3} C) 14 D) 14+8sqrt{3}
Explanation: | 632 | 1,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 7, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-30 | latest | en | 0.716051 |
https://byjus.com/question-answer/dfrac-sin-quad-7x-sin-quad-5x-sin-quad-9x-sin-quad-3x-cos-quad/ | 1,720,838,358,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514484.89/warc/CC-MAIN-20240713020211-20240713050211-00362.warc.gz | 129,646,123 | 33,665 | 1
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Question
# (sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x)=tan6x
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Solution
## L.H.S=(sin7x+sin5x)+(sin9x+sin3x)(cos7x+cos5x)+(cos9x+cos3x)=2sin(7x+5x2)cos(7x−5x2)+2sin(9x+3x2)cos(9x−3x2)2cos(7x+5x2)cos(7x−5x2)+2cos(9x+3x2)cos(9x−3x2)=sin(6x)cos(x)+sin(6x)cos(3x)cos(6x)cos(x)+cos(6x)cos(3x)=sin6x(cosx+cos3x)cos6x(cosx+cos3x)=sin6xcos6x=tan6x=R.H.SHence proved.
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# Content
`Couple Decomposition Couple Addition`
# Couple Decomposition
Forces in a couple can be decomposed also. Couple in space can therefore resolved into three couple components of forces, Fx , Fy and Fz without change its moment arm at the point of application.
Therefore, the moment M of couple of force, Fa can be expressed as:
Although the couple of force, Fa can be resolved into three couple components of rectangular force, Fx Fy and Fz, the moments of the three couple components, M1, M2 and M3 may not be the rectangular couple components because the plane containing the force couple may be at an angle to the coordinate planes. However, the vector sum of M1, M2 and M3 is always equal to moment M.
The rectangular components Mx, My and Mz of moment M can be determined by resolving the moment arm vector ra into rectangular components, imply
For the addition of couples at their intersection line, forces at the point of application can be represnted by a resultant force. The resultant forces R and -R of force couples of forces Fa and Fb also form a couple of force R.
Since all force couples can be translated to the same points of application with common moment arm, the resultant moment M of force couples can be expressed as:
Therefore the resultant moment M of force couples is the vector sum of moments Ma and Mb of forces Fa and Fb.
ID: 110700004 Last Updated: 7/5/2011 Revision: 0 Ref:
References
1. I.C. Jong; B.G. rogers, 1991, Engineering Mechanics: Statics and Dynamics
2. F.P. Beer; E.R. Johnston,Jr.; E.R. Eisenberg, 2004, Vector Mechanics for Engineers: Statics
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Electric 27 | 538 | 2,102 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2021-39 | latest | en | 0.808872 |
https://www.dezyre.com/recipes/find-most-frequent-value-array | 1,627,654,470,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153966.60/warc/CC-MAIN-20210730122926-20210730152926-00688.warc.gz | 738,793,561 | 25,687 | How to find the most frequent value in an array?
MACHINE LEARNING RECIPES DATA CLEANING PYTHON DATA MUNGING PANDAS CHEATSHEET ALL TAGS
# How to find the most frequent value in an array?
This recipe helps you find the most frequent value in an array
## Recipe Objective
We do handle varies size of arrays. To perform statistical operation, numpy has predefined function to handle such problems.
So this recipe is a short example on how to find the most frequent values in an array. Let's get started.
## Step 1 - Import the library
``` import numpy as np ```
Let's pause and look at these imports. Numpy is generally helpful in data manipulation while working with arrays. It also helps in performing mathematical operation.
## Step 2 - Defining random array
``` a= np.array([0,1,2,3,1,2,1,1,1,3,2,2]) ```
We have a random array having same values multiple times.
## Step 3 - Finding mode of array
``` counts = np.bincount(a) print(np.argmax(counts)) ```
We have firstly used bincount to count the number of times each element is present. Later using argmax, found the argument for which counts has maximum frequency.
## Step 4 - Lets look at our dataset now
Once we run the above code snippet, we will see:
```1
```
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In this deep learning project, you will learn to implement Unet++ models for medical image segmentation to detect and classify colorectal polyps. | 753 | 3,727 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2021-31 | latest | en | 0.8525 |
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View Full Version : hope you can give an answer!
xwang98
03-14-2004, 01:44 AM
hello everyone,
sorry to trouble you again,I had posted a mail almost the same content,but get no reply.I doubt whether the net don't work well,so I tried this again.
I have a simple and basic question to ask. anybody who will answer me as soon as possbile will be very very appreciated!
my confusion is about the joint rotation description based on skin-markers.
I have read a paper by Ralf Schmidt et al----- a maker-based measurement proceduce for unconstrainted wrist and elbow motion. I tried to email prof. Ralf Schmidt , but the email address can't accessed. so I trouble you,if you can correct or clear my confusion,I will be appreictted very much!
In that paper,the center of shoulder is assumed to 7cm inferior to the acromion maker which is the average of visually deterimined distance.
question:
Are there some better ways to determine the rotation center of shoulder. are they easy to be used in actual calculation ? has relative programme code of them has been present?
Mr. Ralf Schmidt use a static refernce measurement (at time t1) of "joint marker" to maker sure the original center coordinates. then use marker on segment to calculate the homogenous transformation matrix at time t1, t2 respectively,where t2 stands for another time instant , and then use these matrixes and the origianl center coordiante to get the rotation center coordiante at time t2.
my confusion:
to get a tranformation matix using the least squares algorithm presented by veldpaus et al (1988). two group of coordinates of 3 maker at least on segment at two position are needed. from my understanding, the corrdinates of makers at time t1,t2 are with respect to a gobal reference frame. how to get the transformation matrix at time t1 and t2.
I think we can calculate a homogenous tranformation matrix describe the mapping relation of coordinates at time t1 and t2.
I think that the basic realtion to calculate the transformation is :
X2=RX1+d, =TX1---(1)
where R is rotation matrix ,d is translation vector.T is the homogenous matrix.
my confusion in detail is:
1 the homogenous tansformation matrix T should include both R and d. right?I think so!
2 take the time t1 as example, (the static reference time), how to get the homogenous transformation matix according to (1) using the least square algorithm?
I am very confused at this point.
any noe who ever read this paper can answer me will be appreciated!
finally ,i would see if anyone know the eamil contact ways of Prof. Ralf Schmidt ,please tell me.
sorry to trouble you!
hope you can help me, hope you can answer me if you receive this letter!
thankyou
xinting wang
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https://www.tensorflow.org/api_docs/python/tf/contrib/bayesflow/monte_carlo/expectation?hl=pt-br | 1,561,045,364,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999261.43/warc/CC-MAIN-20190620145650-20190620171650-00106.warc.gz | 955,078,072 | 56,863 | tf.contrib.bayesflow.monte_carlo.expectation
tf.contrib.bayesflow.monte_carlo.expectation(
f,
samples,
log_prob=None,
use_reparametrization=True,
axis=0,
keep_dims=False,
name=None
)
Computes the Monte-Carlo approximation of $$E_p[f(X)]$$. (deprecated)
This function computes the Monte-Carlo approximation of an expectation, i.e.,
$$E_p[f(X)] \approx= m^{-1} sum_i^m f(x_j), x_j\ ~iid\ p(X)$$
where:
• x_j = samples[j, ...],
• log(p(samples)) = log_prob(samples) and
• m = prod(shape(samples)[axis]).
Tricks: Reparameterization and Score-Gradient
When p is "reparameterized", i.e., a diffeomorphic transformation of a parameterless distribution (e.g., Normal(Y; m, s) <=> Y = sX + m, X ~ Normal(0,1)), we can swap gradient and expectation, i.e., grad[ Avg{ $$s_i : i=1...n$$ } ] = Avg{ grad[$$s_i$$] : i=1...n } where S_n = Avg{$$s_i$$}and$$s_i = f(x_i), x_i ~ p$$.
However, if p is not reparameterized, TensorFlow's gradient will be incorrect since the chain-rule stops at samples of non-reparameterized distributions. (The non-differentiated result, approx_expectation, is the same regardless of use_reparametrization.) In this circumstance using the Score-Gradient trick results in an unbiased gradient, i.e.,
grad[ E_p[f(X)] ]
= grad[ int dx p(x) f(x) ]
= int dx grad[ p(x) f(x) ]
= int dx [ p'(x) f(x) + p(x) f'(x) ]
= int dx p(x) [p'(x) / p(x) f(x) + f'(x) ]
= int dx p(x) grad[ f(x) p(x) / stop_grad[p(x)] ]
= E_p[ grad[ f(x) p(x) / stop_grad[p(x)] ] ]
Unless p is not reparametrized, it is usually preferable to use_reparametrization = True.
Example Use:
import tensorflow_probability as tfp
tfd = tfp.distributions
# Monte-Carlo approximation of a reparameterized distribution, e.g., Normal.
num_draws = int(1e5)
p = tfd.Normal(loc=0., scale=1.)
q = tfd.Normal(loc=1., scale=2.)
exact_kl_normal_normal = tfd.kl_divergence(p, q)
# ==> 0.44314718
approx_kl_normal_normal = tfp.monte_carlo.expectation(
f=lambda x: p.log_prob(x) - q.log_prob(x),
samples=p.sample(num_draws, seed=42),
log_prob=p.log_prob,
use_reparametrization=(p.reparameterization_type
== distribution.FULLY_REPARAMETERIZED))
# ==> 0.44632751
# Relative Error: <1%
# Monte-Carlo approximation of non-reparameterized distribution, e.g., Gamma.
num_draws = int(1e5)
p = ds.Gamma(concentration=1., rate=1.)
q = ds.Gamma(concentration=2., rate=3.)
exact_kl_gamma_gamma = tfd.kl_divergence(p, q)
# ==> 0.37999129
approx_kl_gamma_gamma = tfp.monte_carlo.expectation(
f=lambda x: p.log_prob(x) - q.log_prob(x),
samples=p.sample(num_draws, seed=42),
log_prob=p.log_prob,
use_reparametrization=(p.reparameterization_type
== distribution.FULLY_REPARAMETERIZED))
# ==> 0.37696719
# Relative Error: <1%
# For comparing the gradients, see monte_carlo_test.py.
approx_kl_p_q = tfp.vi.monte_carlo_csiszar_f_divergence(
f=bf.kl_reverse,
p_log_prob=q.log_prob,
q=p,
num_draws=num_draws)
Args:
• f: Python callable which can return f(samples).
• samples: Tensor of samples used to form the Monte-Carlo approximation of $$E_p[f(X)]$$. A batch of samples should be indexed by axis dimensions.
• log_prob: Python callable which can return log_prob(samples). Must correspond to the natural-logarithm of the pdf/pmf of each sample. Only required/used if use_reparametrization=False. Default value: None.
• use_reparametrization: Python bool indicating that the approximation should use the fact that the gradient of samples is unbiased. Whether True or False, this arg only affects the gradient of the resulting approx_expectation. Default value: True.
• axis: The dimensions to average. If None, averages all dimensions. Default value: 0 (the left-most dimension).
• keep_dims: If True, retains averaged dimensions using size 1. Default value: False.
• name: A name_scope for operations created by this function. Default value: None (which implies "expectation").
Returns:
• approx_expectation: Tensor corresponding to the Monte-Carlo approximation of $$E_p[f(X)]$$.
Raises:
• ValueError: if f is not a Python callable.
• ValueError: if use_reparametrization=False and log_prob is not a Python callable. | 1,170 | 4,083 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2019-26 | latest | en | 0.553753 |
https://www.physicsforums.com/threads/diffraction-gratings-calculating-the-highest-order-maximum.628227/ | 1,542,617,781,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039745522.86/warc/CC-MAIN-20181119084944-20181119110944-00484.warc.gz | 936,152,328 | 12,326 | # Homework Help: Diffraction gratings - calculating the highest order maximum
1. Aug 14, 2012
### Marcargo
1. The problem statement, all variables and given/known data
Monochromatic light of wavelength 694. nm is incident on a diffraction grating with 2192 lines per centimetre. What is the highest order maximum that can be observed?
2. Relevant equations
dsinθ = mλ
3. The attempt at a solution
d = 2192/cm = 21.92/m
λ = 694 x 10 -9m
sinθ = ??????????????????
m = dsinθ/λ
= (2192 x sinθ)/(694 x 10-9
I think that this is the equation to use, but I'm not sure what angle to use. I initially tried 90 for θ, but then logically thinking about it, that wouldn't work as light travels only in straight lines. Not sure how to carry on from here. Some insight would help!
Thanks | 219 | 783 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2018-47 | latest | en | 0.925936 |
http://southernbellsinc.com/taco/how-many-pounds-of-taco-meat-would-i-need-for-30-people.html | 1,670,362,105,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711114.3/warc/CC-MAIN-20221206192947-20221206222947-00823.warc.gz | 46,577,217 | 19,924 | # How Many Pounds Of Taco Meat Would I Need For 30 People?
## How many tacos will 2 lbs of meat make?
We had 2 lbs ground beef and each had 2 tacos, so I would say it would make around 8.
## How many tacos will 4 lbs of meat make?
The “conservative” amount of ground beef in a typical taco is 2.5 ounces, but because of the other taco fillings we can estimate 2 ounces of beef for the average taco. So that being said, probably about 8 tacos per pound of ground beef.
## How many pounds of meat are in a street taco?
Although the standard is 3 tacos per person, folks tend to be a bit generous when filling their own tacos so caterers calculate 2 tacos per person for taco bars. One pound (16 ounces) of 80/20 raw ground beef, pan browned and drained, will equal 12 ounces of cooked ground beef.
## What’s a good side for tacos?
Whether you prefer the classic Mexican soft taco or the crunchy American hard shell taco, these sides will not disappoint.
• Chips and Salsa.
• Corn Pudding.
• Mexican Coleslaw.
• Jicama with Mango Slaw.
• Mexican Sweet Potatoes.
You might be interested: Question: How Long Does It Take To Make Taco Meat?
## How many tacos should you eat?
Our taco algorithms show that people will eat approximately 4-5 tacos. Keep in mind that some will eat less, and some will eat more, but it will average out to approximately 4-5 tacos per person.
## How many tacos will a pound of chicken make?
Yep, they are that good. I typically start these tacos with 12 to 16 ounces of frozen boneless, skinless chicken, counting on about 4 ounces of chicken per person. That’s between 3/4 to 1 pound for 3 – 4 people.
## How much rice do you need for a taco bar?
Add 2 cups uncooked white long grain rice, along with 1/2 Tbsp chili powder, 1/2 tsp cumin, 1/4 tsp oregano, and 1 tsp salt. Continue to stir and cook the rice and spices for about 2 minutes. This will toast the rice and spices and enhance their flavor.
## How much taco seasoning does it take to make a pound of meat?
Use 2 Tablespoons of taco seasoning for every pound of ground beef. Add the spices to browned meat with 1/2 cup of water.
## What goes with tacos other than rice?
20 Easy Side Dishes for Taco Tuesday Besides Rice and Beans
• Cilantro Lime Rice.
• Mexican Sweet Potatoes.
• Mexican Rice.
• Pico De Gallo.
• Salsa Verde.
## What do you serve with walking tacos?
30 Sides for Tacos to Serve on Taco Tuesday (or Any Day, Really)
• Mango Guacamole.
• Seven-Layer Dip.
• 15-Minute Gazpacho with Cucumber, Red Pepper and Basil.
• Zucchini Rice.
• Light and Tangy Coleslaw.
• Mexican Street Corn Deviled Eggs. | 666 | 2,614 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.828125 | 3 | CC-MAIN-2022-49 | longest | en | 0.944176 |
https://www.examguruadda.in/2017/01/quant-quiz-for-ibps-2017-india-post.html | 1,580,100,347,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694176.67/warc/CC-MAIN-20200127020458-20200127050458-00015.warc.gz | 834,731,260 | 19,088 | ### Quant Quiz for IBPS 2017, India Post
1).The incomes of A and B are in the ratio of 2:3 and their expenditures are in the ratio of 1:3. If each of them saves RS.12000, find A’s income.
a) RS.20000
b) RS.24000
c) RS.8000
d) RS.16000
e) RS.12000
2).The length of a tree increases at the rate of 5% every year. If it is 20 feet high now, how much will it increase after three years?
a) 23.1525 ft
b) 4.1525 ft
c) 3.1525 ft
d) 6.2515 ft
e) None of these
3).If A exceeds B by 80% and B is less than C by 20%, find the ratio of A to C.
a) 36:25
b) 25:29
c) 25:36
d) 29:36
e) 36.29
4).A man sells two articles for the same price-one on 25% gain and the other on 20% gain. If the first article costs RS.8000 then what will be the cost price of the second article?
a) RS.7330.63
b) RS.8333.33
c) RS.6333.33
d) RS.9333.33
e) RS.9000.53
5).The difference between a 40% discount on RS.1000 and successive discounts of 30% and 5% on the same amount is
a) RS. 31
b) RS. 32
c) RS. 30
d) RS. 40
e) RS. 65
6).A man takes a loan of RS.20000 partly from a bank at 8% pa and the remaining from another bank at 10% pa. He pays a total interest of RS.1900 per annum. The amount of loan taken from the first bank is
a) RS.7000
b) RS.10000
c) RS.15000
d) RS.7500
e) RS.5000
7).If the interest is calculated half-yearly, the compound interest on RS.18000 at10% per annum at the end of 1(1/2) years is
a) RS.2700
b) RS.2037.5
c) RS.2043.7
d) RS.2837.5
e) RS.2187.5
8).30 men working 16 hours per day make a dam in 20days. In how many days will 24 men working 20 hours per day make it?
a) 5 days
b) 10 days
c) 20 days
d) 25 days
e) 40 days
9).There are two leakages in the bottom of a cistern. Both together empty the cistern in 12 hours. If the first leakage alone empties it in 30 hours, then in how many hours will the second leakage alone empty it?
a) 20 hours
b) 25 hours
c) 30 hours
d) 15 hours
e) 12 hours
10).The diameter of a wheel of a vehicle is 56 cm. It completes 1000 rotations in 1 minute. Find the speed (in m/s) of the vehicle.
a) 29.33 m/s
b) 59.33 m/s
c) 58.66 m/s
d) 70 m/s
e) 29.33 m/s
Solutions
1). Let the incomes of A and B be 2x and 3x respectively. Each of them saves RS.12000.
Now, (2x-12000)/ (3x-12000) =1/3
Or, 6x-36000=3x-12000 or, 3x=24000
∴x=RS.8000
Hence income of A=2x=8000×2= RS.16000
2). Height of the tree now=20 feet Height of the tree after 3 years at the rate of 5% increase per year=20×(105/100)×(105/100)×(105/100) =23.1525 feet
∴ Required increase= (23.1525-20) feet=3.1525 feet
3). Let C be 100, then B=80
∴A= (80×180)/100=144
∴ Required ratio=144:100=36:25
4). If two articles bought on different prices, are sold for the same price at % gain or % loss, then
(CP of First article / CP of Second article)
= (100±% gain or loss on the second article / 100±% gain or loss on the first article)
Or, 8000 / x = (100+20) / (100+25) or, 8000/x = 120/125
∴ x= (8000×125) / 120
∴ x=RS.8333.33
5). CP=RS.1000
SP at 40% discount= (100×60)/100=RS.600
∴ Discount=1000-600=400
SP at two successive discounts of 30% and 5%
=1000×(70/100)×(95/100) =RS.665
Discount=1000-665=335
∴ Required difference=400-335=RS.65
6). Let the loan taken from the first bank be RS. x.
Loan from the second bank= RS.(20000-x)
Interest of the first bank+ interest of the second bank=1900
Now,(x×8×1) / 100 + [(20000-x)×10×1] / 100=1900
Or, 8x+200000-10x =190000
or, 2x=10000
∴ x=RS.5000
7). P=RS.18000
R=10% pa=(10/2)% half-yearly
∴r=5% half-yearly
Time will be double when interest is calculated half-yearly, ie 1(1/2) years × 2 = 3 years
∴Amount=18000(I+(r/100))t
=18000(1+[5/100)]3
=18000×(105/100)×(105/100)×(105/100)
=RS.20837.25
∴CI=20837.25-18000=RS.2837.5
8). M1D1H1= M2D2H2
Now, 30×16×20=24×D×20
Or, D= (30×16×20) / (20×24)
∴ D =20 days
9). Required time= (Product of both times / Difference of both times)
= (12×30) / (30-12)
= (12×30)/18 =20 hours
10). Circumference of the wheel=2πr
=2× (22/7)× 28cm =176 cm
The distance covered by the wheel in 1 minute= (176×1000) cm =176000cm
Speed of the wheel=176000 / (60×100) m/s
≈29.33 m/s | 1,634 | 4,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2020-05 | latest | en | 0.820537 |
http://singaporemathguru.com/question/primary-6-problem-sums-word-problems-speed-getting-there-speed-exercise-1-1262 | 1,575,813,984,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540510336.29/warc/CC-MAIN-20191208122818-20191208150818-00014.warc.gz | 130,897,073 | 11,228 | ### Primary 6 Problem Sums/Word Problems - Try FREE
Score :
(Single Attempt)
#### Question
In a laboratory, a robot took 8 h to travel from point A to point B at an average speed of 90 m/h.
Its speed was then reduced by 10 m/h and it travelled for another 6 h to reach point C.
Find the total distance covered by the robot.
The correct answer is : 1200
m | 94 | 360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2019-51 | latest | en | 0.937797 |
https://www.thestudentroom.co.uk/showthread.php?t=3090299 | 1,521,701,967,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647777.59/warc/CC-MAIN-20180322053608-20180322073608-00000.warc.gz | 853,835,254 | 42,458 | x Turn on thread page Beta
You are Here: Home >< Maths
# C4 Integration watch
1. I think integration is a decent topic but I have no idea how to integrate the following:
• 4Cos3xSin2x
• Sec2xtan2x
Guides on how to integrate these would be greatly appreciated,
Thanks
2. (Original post by Leking9)
I think integration is a decent topic but I have no idea how to integrate the following:
• 4Cos3xSin2x
• Sec2xtan2x
Guides on how to integrate these would be greatly appreciated,
Thanks
Do you know how to integrate CosxSinnx
3. (Original post by Leking9)
I think integration is a decent topic but I have no idea how to integrate the following:
• 4Cos3xSin2x
• Sec2xtan2x
Guides on how to integrate these would be greatly appreciated,
Thanks
For the second one recall what the differential of Tanx is
4. (Original post by TenOfThem)
Do you know how to integrate CosxSinnx
Not really, the only way I know is to do it by parts but that's extremely long
5. (Original post by TenOfThem)
For the second one recall what the differential of Tanx is
sec2x
6. (Original post by Leking9)
Not really, the only way I know is to do it by parts but that's extremely long
(Original post by Leking9)
sec2x
Have you done inverse chain rule .... If so these are both easy
If not substitution ... U=sinx and u=tanx
7. (Original post by TenOfThem)
Have you done inverse chain rule .... If so these are both easy
If not substitution ... U=sinx and u=tanx
"Inverse chain rule" I've been taught according to what's inside the C4 edexcel book and for some reason I can't see any exercises on it nor has my teacher gone over it
8. (Original post by Leking9)
"Inverse chain rule" I've been taught according to what's inside the C4 edexcel book and for some reason I can't see any exercises on it nor has my teacher gone over it
It is not called that in the book
It is just by recognition
If you know what sinx differentiates to, then you know what (sinx)^n differentiates to
Therefore you can integrate cosx (sinx)^n
Or use the substitutions that I pointed out
9. (Original post by Leking9)
"Inverse chain rule" I've been taught according to what's inside the C4 edexcel book and for some reason I can't see any exercises on it nor has my teacher gone over it
It's called integration by parts and you definitely should have covered it?
10. (Original post by TVIO)
It's called integration by parts and you definitely should have covered it?
These questions do not require IBP
11. (Original post by Leking9)
"Inverse chain rule" I've been taught according to what's inside the C4 edexcel book and for some reason I can't see any exercises on it nor has my teacher gone over it
It's noting that the form of the integral is in , and answering the question via recognition. In reality it's just a simple substitution, but with time you'll be able to do them by recognition.
Thus the integral becomes:
12. (Original post by TenOfThem)
These questions do not require IBP
could use sin(a+-b) formulas
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13. (Original post by physicsmaths)
could use sin(a+-b) formulas
Posted from TSR Mobile
Could you?
Since both of these integrals are very straightforward I would be wary of introducing anything new
14. (Original post by TenOfThem)
Could you?
Since both of these integrals are very straightforward I would be wary of introducing anything new
That's how I always done the sinnxcosmx ones. It's just using the trig formulas to cancel into a very simple integral.
Posted from TSR Mobile
15. 4cos(3x)sin(2x)= 2sin(5x) -2sinx
Posted from TSR Mobile
16. (Original post by physicsmaths)
That's how I always done the sinnxcosmx ones. It's just using the trig formulas to cancel into a very simple integral.
Posted from TSR Mobile
Those questions .... Yes
Not the ones in the OP though
17. (Original post by TenOfThem)
Those questions .... Yes
Not the ones in the OP though
Oh, is that raised to the power 3? I thought it was cos(3x)sin(2x)
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Top tips from students who have already aced their exams | 1,153 | 4,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2018-13 | latest | en | 0.942894 |
https://www.mathematics-monster.com/lessons/how_to_find_the_least_common_multiple.html | 1,675,558,860,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500158.5/warc/CC-MAIN-20230205000727-20230205030727-00302.warc.gz | 890,107,649 | 4,342 | ## The Lesson
The least common multiple of two or more numbers can be found. Imagine we wanted to find the least common multiple of 3 and 4.
## How to Find the Least Common Multiple
Finding the least common multiple is easy.
## Question
What is the least common multiple of 3 and 4?
# 1
List the multiples of 3.
Multiples of 3 = 3, 6, 9, 12, 15...
Don't forget: The multiples of 3 are the result of multiplying 3 by successive integers: 3 × 1 = 3, 3 × 2 = 6...
# 2
List the multiples of 4.
Multiples of 4 = 4, 8, 12, 16, 20...
# 3
The smallest multiple that appears in both lists is the least common multiple.
Multiples of 3 = 3, 6, 9, 12, 15... Multiples of 4 = 4, 8, 12, 16, 20...
12 is the least common multiple of 3 and 4.
## Lesson Slides
The slider below gives another example of finding the least common multiple and an example of why it is useful. Open the slider in a new tab
## What's in a Name?
The least common multiple (LCM) is so called because it is the smallest (least) multiple that is the same (common) to the numbers.
## What Is a Multiple?
A multiple is the result of multiplying a number by an integer (a whole number). For example, the multiples of 3 are:
3, 6, 9, 12, 15
• 3 is the result of multiplying 3 by 1.
• 6 is the result of multiplying 3 by 2.
• 9 is the result of multiplying 3 by 3.
• 12 is the result of multiplying 3 by 4.
• 15 is the result of multiplying 3 by 5.
## What Is the Least Common Multiple?
The Least Common Multiple is the smallest multiple that is common to two or more numbers. | 456 | 1,548 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2023-06 | longest | en | 0.922807 |
https://stackoverflow.com/questions/2356501/how-do-you-round-up-a-number-in-python/43795978 | 1,582,480,750,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875145818.81/warc/CC-MAIN-20200223154628-20200223184628-00538.warc.gz | 560,783,412 | 42,915 | # How do you round UP a number in Python?
This problem is killing me. How does one roundup a number UP in Python?
I tried round(number) but it round the number down. Example:
``````round(2.3) = 2.0 and not 3, what I would like
``````
The I tried int(number + .5) but it round the number down again! Example:
``````int(2.3 + .5) = 2
``````
Then I tried round(number + .5) but it won't work in edge cases. Example:
``````WAIT! THIS WORKED!
``````
Please advise.
• `round(number + .5)` doesn't work if the number is integer. `round(3+.5) == 4`, when you actually want `3`. – Nearoo Jan 27 '19 at 16:47
## 24 Answers
The ceil (ceiling) function:
``````import math
print(math.ceil(4.2))
``````
• Elaboration: math.ceil returns the smallest integer which is greater than or equal to the input value. This function treats the input as a float (Python does not have strongly-typed variables) and the function returns a float. If you want an int, you can construct an int from the return value, i.e., `int(math.ceil(363))` – R. W. Sinnet Aug 26 '15 at 23:37
• @Sinnet: Actually one could say that python is strongly typed stackoverflow.com/a/11328980/5069869 – Bernhard Jan 8 '16 at 12:54
• @TheEspinosa: Yes, python is definitely strongly typed, its just that many functions ask questions about the type of some parameters and execute different code depending on the answer. – quamrana Dec 17 '17 at 12:27
• @R.W.Sinnet In Python 3, `math.ceil` returns an actual integer object, not just floating object with integer value. – Arthur Tacca May 2 '18 at 15:50
• Take care of float precision, due to `10000000 * 0.00136 = 13600.000000000002` ceil can increase a lot `math.ceil(10000000 * 0.00136) = 13601.0` – ofthestreet Sep 10 '19 at 8:23
Interesting Python 2.x issue to keep in mind:
``````>>> import math
>>> math.ceil(4500/1000)
4.0
>>> math.ceil(4500/1000.0)
5.0
``````
The problem is that dividing two ints in python produces another int and that's truncated before the ceiling call. You have to make one value a float (or cast) to get a correct result.
In javascript, the exact same code produces a different result:
``````console.log(Math.ceil(4500/1000));
5
``````
• In Python 2.x : int/int --> int and int/float --> float In Python 3.x : int/int can result in a float – gecco Oct 31 '11 at 6:46
• you can get the Python 3.x on behavior on certain versions of Python 2.x by enabling "true division" as shown here – Rob Dennis Oct 22 '13 at 17:54
I know this answer is for a question from a while back, but if you don't want to import math and you just want to round up, this works for me.
``````>>> int(21 / 5)
4
>>> int(21 / 5) + (21 % 5 > 0)
5
``````
The first part becomes 4 and the second part evaluates to "True" if there is a remainder, which in addition True = 1; False = 0. So if there is no remainder, then it stays the same integer, but if there is a remainder it adds 1.
• Nice. You can also use `//` for integer division, so this becomes `21 // 5 + (21 % 5 > 0)`. – naught101 Aug 19 '15 at 13:04
• This is the best solution if only integers are involved. No unnecessary `float`s. Nice. – Nico Schlömer Jul 16 '17 at 21:39
If working with integers, one way of rounding up is to take advantage of the fact that `//` rounds down: Just do the division on the negative number, then negate the answer. No import, floating point, or conditional needed.
``````rounded_up = -(-numerator // denominator)
``````
For example:
``````>>> print(-(-101 // 5))
21
``````
• What about when you don't need to perform any math operation? I.e. you just have one number. – Klik Sep 8 '16 at 6:33
• @Klik: then you can just divide by 1 ==> -( -num // 1) and you are getting your answer :-) Have a nice day! David Bau: very nice proposal! – Marco smdm Apr 4 '17 at 8:57
• I timed all the answers in here and this was five times faster than the next best (math.ceil). @Andreas had the same time – mini totent Jul 20 '17 at 18:23
• @minitotent That is not surprising since it's simple integer division and a couple of single-cycle operations. This is the sort of answer that gets you a job: Understanding not only the language, but all the layers of abstractions beneath it. – Nearoo Jan 6 '19 at 18:44
• Nice! I've always used `(num + den - 1) // den`, which is fine for `int` inputs with positive denominators, but fails if even a single non-integral `float` is involved (either numerator or denominator); this is more magical looking, but works for both `int`s and `float`s. For small numerators, it's also faster (on CPython 3.7.2), though oddly, when only the numerator is large enough that array based math is needed, your approach is slower; not clear why this is, since the division work should be similar and two unary negations should be cheaper than addition + subtraction. – ShadowRanger Aug 7 '19 at 16:17
You might also like numpy:
``````>>> import numpy as np
>>> np.ceil(2.3)
3.0
``````
I'm not saying it's better than math, but if you were already using numpy for other purposes, you can keep your code consistent.
Anyway, just a detail I came across. I use numpy a lot and was surprised it didn't get mentioned, but of course the accepted answer works perfectly fine.
• Using numpy is nice too. The easiest would be with math since it is already part of python built in libraries. It makes more sense. Instead as you mentioned if you use a lot numpy for other issues, then it makes sense and consistent to use numpy.ceil :-) Good hint! – Marco smdm Apr 4 '17 at 8:52
Use `math.ceil` to round up:
``````>>> import math
>>> math.ceil(5.4)
6.0
``````
NOTE: The input should be float.
If you need an integer, call `int` to convert it:
``````>>> int(math.ceil(5.4))
6
``````
BTW, use `math.floor` to round down and `round` to round to nearest integer.
``````>>> math.floor(4.4), math.floor(4.5), math.floor(5.4), math.floor(5.5)
(4.0, 4.0, 5.0, 5.0)
>>> round(4.4), round(4.5), round(5.4), round(5.5)
(4.0, 5.0, 5.0, 6.0)
>>> math.ceil(4.4), math.ceil(4.5), math.ceil(5.4), math.ceil(5.5)
(5.0, 5.0, 6.0, 6.0)
``````
• The input does not necessarily need to be a float if using python 3: `ceil()` will take care of it internally – guival Feb 24 '17 at 10:33
The syntax may not be as pythonic as one might like, but it is a powerful library.
https://docs.python.org/2/library/decimal.html
``````from decimal import *
print(int(Decimal(2.3).quantize(Decimal('1.'), rounding=ROUND_UP)))
``````
I am surprised nobody suggested
``````(numerator + denominator - 1) // denominator
``````
for integer division with rounding up. Used to be the common way for C/C++/CUDA (cf. `divup`)
• Relevant only for statically typed languages. If the denominator is a float you're dead. – Bharel Jul 6 '17 at 12:31
• This also only works consistently if the denominator is positive; if the denominator is negative, you need to add `1` instead of subtracting it, or flip the signs of both numerator and denominator before performing the math. – ShadowRanger Aug 7 '19 at 16:28
Be shure rounded value should be float
``````a = 8
b = 21
print math.ceil(a / b)
>>> 0
``````
but
``````print math.ceil(float(a) / b)
>>> 1.0
``````
Try this:
``````a = 211.0
print(int(a) + ((int(a) - a) != 0))
``````
• Clever. The `((int(a) - a) != 0)` expression returns `1` whenever `a` needs to be rounded up. You may want to expand your answer and explain how this work. – Tom Aranda Dec 4 '17 at 23:01
• @TomAranda Can anyone explain how a boolean expression evaluates to a value please? – Bowen Liu Dec 18 '18 at 20:51
``````>>> def roundup(number):
... return round(number+.5)
>>> roundup(2.3)
3
>>> roundup(19.00000000001)
20
``````
This function requires no modules.
• What if your number is `3`, then it would round up to `4` which may or may not be what someone wants – Bolboa Oct 20 '18 at 23:01
• This only works in 99% of all cases. You didn't think this through properly. Such solutions should be avoided at all costs. – Nearoo Jan 6 '19 at 18:47
The above answers are correct, however, importing the `math` module just for this one function usually feels like a bit of an overkill for me. Luckily, there is another way to do it:
``````g = 7/5
g = int(g) + (not g.is_integer())
``````
`True` and `False` are interpreted as `1` and `0` in a statement involving numbers in python. `g.is_interger()` basically translates to `g.has_no_decimal()` or `g == int(g)`. So the last statement in English reads `round g down and add one if g has decimal`.
• And if you feel fancy, you can use `int(g) + (g % 1 > 0)` instead ;-) – Nearoo May 5 '17 at 3:23
• `from math import ceil` seems to fix importing the entire math module :) – SH7890 Jan 25 '19 at 21:50
• @SH7890 I'm afraid that line isn't much different to `import math` in terms of what happens behind the scenes. It just drops all symbols except `ceil`. – Nearoo Jan 27 '19 at 16:52
Without importing math // using basic envionment:
a) method / class method
``````def ceil(fl):
return int(fl) + (1 if fl-int(fl) else 0)
def ceil(self, fl):
return int(fl) + (1 if fl-int(fl) else 0)
``````
b) lambda:
``````ceil = lambda fl:int(fl)+(1 if fl-int(fl) else 0)
``````
For those who want to round up `a / b` and get integer:
Another variant using integer division is
``````def int_ceil(a, b):
return (a - 1) // b + 1
>>> int_ceil(19, 5)
4
>>> int_ceil(20, 5)
4
>>> int_ceil(21, 5)
5
``````
I'm surprised I haven't seen this answer yet `round(x + 0.4999)`, so I'm going to put it down. Note that this works with any Python version. Changes made to the Python rounding scheme has made things difficult. See this post.
Without importing, I use:
``````def roundUp(num):
return round(num + 0.49)
testCases = list(x*0.1 for x in range(0, 50))
print(testCases)
for test in testCases:
print("{:5.2f} -> {:5.2f}".format(test, roundUp(test)))
``````
Why this works
From the docs
For the built-in types supporting round(), values are rounded to the closest multiple of 10 to the power minus n; if two multiples are equally close, rounding is done toward the even choice
Therefore 2.5 gets rounded to 2 and 3.5 gets rounded to 4. If this was not the case then rounding up could be done by adding 0.5, but we want to avoid getting to the halfway point. So, if you add 0.4999 you will get close, but with enough margin to be rounded to what you would normally expect. Of course, this will fail if the `x + 0.4999` is equal to `[n].5000`, but that is unlikely.
• Using 0.4999, it will fail to give a correct result for any input in between ???.0000 and ???.0001 (open interval), not just exactly ???.0001. For instance, if you try it with 3.00005, you will get a result of 3 instead of the expected 4. Of course you can decrease the likelihood of this happening by adding more and more digits up to the maximum precision of floats, but what's the point to that if there are more robust and intuitive solutions at hand, like using `math.ceil()`? – blubberdiblub Nov 14 '16 at 9:08
• @blubberdiblub In my answer I state `Without importing I use:`. I've also mentioned that it will fail if the `x + 0.4999` is equal to `[n].5000`. – Klik Nov 14 '16 at 22:23
• Yes, you state in your answer that your solution is without importing, but I don't see the value of it. The `math` module and `math.ceil()` is in the standard library, so available everywhere for all practical purposes without installing extra stuff. And regarding your mention of when it fails, this is incomplete in your answer, as it fails for a whole interval, not just for a single point. Technically, you could argue you are correct, as you say if and not iff, but it will make the impression on the casual reader that it is less likely than it really is. – blubberdiblub Nov 17 '16 at 5:59
To do it without any import:
``````>>> round_up = lambda num: int(num + 1) if int(num) != num else int(num)
>>> round_up(2.0)
2
>>> round_up(2.1)
3
``````
I know this is from quite a while back, but I found a quite interesting answer, so here goes:
``````-round(-x-0.5)
``````
This fixes the edges cases and works for both positive and negative numbers, and doesn't require any function import
Cheers
• This will still rounds down `-round(-x-0.3) = x` – Diblo Dk Jun 19 '15 at 20:44
when you operate 4500/1000 in python, result will be 4, because for default python asume as integer the result, logically: 4500/1000 = 4.5 --> int(4.5) = 4 and ceil of 4 obviouslly is 4
using 4500/1000.0 the result will be 4.5 and ceil of 4.5 --> 5
Using javascript you will recieve 4.5 as result of 4500/1000, because javascript asume only the result as "numeric type" and return a result directly as float
Good Luck!!
• That's only true in Python 2.x. In Python 3, division with a single `/` always results in a float, so `4500/1000` is always 4.5. – Nearoo Jun 19 '18 at 12:41
If you don't want to import anything, you can always write your own simple function as:
```def RoundUP(num): if num== int(num): return num return int(num + 1) ```
• This does not work if num is 2.05. You have to have at least as many digits with a 9 as your input, leaving you with a 0.999... which is 1. But then your corner case 2 is rounded up again. -- Well, I guess there is a reason why math.ceil is there. – Johannes Maria Frank Mar 28 '17 at 0:01
In case anyone is looking to round up to a specific decimal place:
``````import math
def round_up(n, decimals=0):
multiplier = 10 ** decimals
return math.ceil(n * multiplier) / multiplier
``````
You can use floor devision and add 1 to it. 2.3 // 2 + 1
• or use `ceil()` instead of weirdly doing the opposite and then compensating – guival Feb 24 '17 at 10:34
• This won't work. For example: `from math import ceil; assert 4 // 2 + 1 == ceil(4 / 2)` – Carl Thomé Jul 26 '17 at 15:36
I think you are confusing the working mechanisms between `int()` and `round()`.
`int()` always truncates the decimal numbers if a floating number is given; whereas `round()`, in case of `2.5` where `2` and `3` are both within equal distance from `2.5`, Python returns whichever that is more away from the 0 point.
``````round(2.5) = 3
int(2.5) = 2
``````
• "rounding up" means that e.g. `2.3` gets turned into `3`, which happens in neither of your examples. – Nearoo Jun 19 '18 at 12:43
My share
I have tested `print(-(-101 // 5)) = 21` given example above.
Now for rounding up:
``````101 * 19% = 19.19
``````
I can not use `**` so I spread the multiply to division:
``````(-(-101 //(1/0.19))) = 20
``````
I'm basically a beginner at Python, but if you're just trying to round up instead of down why not do:
``````round(integer) + 1
``````
• This will not work for any integer i where 2.5 < integer < 3. The desired value after rounding up is 3 but your expression will turn it into 4. – Pranav Shukla Jun 1 '16 at 2:52
• I think you mean `round(integer + 0.5)` This is what I often do – Klik Sep 8 '16 at 5:33 | 4,439 | 14,870 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2020-10 | latest | en | 0.826622 |
http://www.java-forums.org/new-java/64124-comparing-elements-two-arrays-print.html | 1,433,259,395,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1433195035877.3/warc/CC-MAIN-20150601214355-00039-ip-10-180-206-219.ec2.internal.warc.gz | 467,320,014 | 3,265 | # Comparing elements in two arrays
• 10-21-2012, 06:50 PM
Comparing elements in two arrays
Good afternoon everyone.
I am trying to compare the int elements in two arrays for a homework assignment. I just need to confirm something - I do not need any actual code answer posted, ok?
I am trying to see which array has the higher number overall when comparing each element. Does ArrayA have more "wins" than ArrayB, essentially.
This is my code
Code:
``` public class Compare_Two_Arrays { public static void main(String[] args) { // the integer counters are set int A_wins = 0; int B_wins = 0; // below sets up the two integer arrays that we will compare int[] intArrayA = new int[]{1, 2, 11, 12, 15, 16}; int[] intArrayB = new int[]{3, 4, 7, 8, 17, 18}; for (int i = 0; i <= intArrayA.length-1; i++) { for(int j = 0; j <= intArrayB.length-1; j++) { if (intArrayA[i] > intArrayB[j]) { A_wins = A_wins +1; } else if (intArrayA[i] < intArrayB[j]) { B_wins = B_wins + 1; } } if( A_wins > B_wins) { System.out.println(" A has the majority of wins"); } else { System.out.println(" B has the majority of wins"); } } } }```
This is what I get :
Code:
``` B has the majority of wins B has the majority of wins B has the majority of wins B has the majority of wins B has the majority of wins B has the majority of wins```
And this is where the coding error lies (since the answer above is wrong), I think.
Code:
`for(int j = 0; j <= intArrayB.length-1; j++)`
Basically, its comparing every iteration of the 2nd array to just the first element in the first array, right?
Can you confirm my suspicions?
Thanks again to all those who look this over.
• 10-21-2012, 06:54 PM
Fubarable
Re: Comparing elements in two arrays
You've got *nested* for loops, so your first of all looping through the first array, and then inside of this loop, you're using another for loop to compare the current first array item with every item in the second array. I suspect that you want to use only one for loop and compare arrayA[i] item with arrayB[i] item.
• 10-21-2012, 07:05 PM
Re: Comparing elements in two arrays
I see....
I have not done that before, truth be told.
use one for loop for the length of the array (since they are the same length), and compare each array at the element, as it loops.
I will try that.
Fubarable - thank you.
This is gonna be pretty fun to try out.
• 10-21-2012, 07:06 PM
Fubarable
Re: Comparing elements in two arrays
Quote:
Originally Posted by Adomini
I see....
use one for loop for the length of the array (since they are the same length), and compare each array at the element, as it loops.
I will try that.
Fubarable - thank you.
This is gonna be pretty fun to try out.
And I'm betting that you will succeed. You're welcome and best of luck! | 836 | 3,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2015-22 | longest | en | 0.323569 |
https://se.mathworks.com/matlabcentral/cody/problems/45663-find-the-next-binary-palindrome-number/solutions/2410651 | 1,606,637,214,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141197278.54/warc/CC-MAIN-20201129063812-20201129093812-00165.warc.gz | 458,878,403 | 18,257 | Cody
# Problem 45663. Find the next binary palindrome number
Solution 2410651
Submitted on 29 May 2020 by Binbin Qi
• Size: 22
• This is the leading solution.
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 241; y_correct = 255; assert(isequal(NextBinaryPalindrome(x),y_correct))
x = 242 x = 243 x = 244 x = 245 x = 246 x = 247 x = 248 x = 249 x = 250 x = 251 x = 252 x = 253 x = 254 x = 255
2 Pass
x = 2020; y_correct = 2047; assert(isequal(NextBinaryPalindrome(x),y_correct))
x = 2021 x = 2022 x = 2023 x = 2024 x = 2025 x = 2026 x = 2027 x = 2028 x = 2029 x = 2030 x = 2031 x = 2032 x = 2033 x = 2034 x = 2035 x = 2036 x = 2037 x = 2038 x = 2039 x = 2040 x = 2041 x = 2042 x = 2043 x = 2044 x = 2045 x = 2046 x = 2047
3 Pass
x = 10051; y_correct =10233; assert(isequal(NextBinaryPalindrome(x),y_correct))
x = 10052 x = 10053 x = 10054 x = 10055 x = 10056 x = 10057 x = 10058 x = 10059 x = 10060 x = 10061 x = 10062 x = 10063 x = 10064 x = 10065 x = 10066 x = 10067 x = 10068 x = 10069 x = 10070 x = 10071 x = 10072 x = 10073 x = 10074 x = 10075 x = 10076 x = 10077 x = 10078 x = 10079 x = 10080 x = 10081 x = 10082 x = 10083 x = 10084 x = 10085 x = 10086 x = 10087 x = 10088 x = 10089 x = 10090 x = 10091 x = 10092 x = 10093 x = 10094 x = 10095 x = 10096 x = 10097 x = 10098 x = 10099 x = 10100 x = 10101 x = 10102 x = 10103 x = 10104 x = 10105 x = 10106 x = 10107 x = 10108 x = 10109 x = 10110 x = 10111 x = 10112 x = 10113 x = 10114 x = 10115 x = 10116 x = 10117 x = 10118 x = 10119 x = 10120 x = 10121 x = 10122 x = 10123 x = 10124 x = 10125 x = 10126 x = 10127 x = 10128 x = 10129 x = 10130 x = 10131 x = 10132 x = 10133 x = 10134 x = 10135 x = 10136 x = 10137 x = 10138 x = 10139 x = 10140 x = 10141 x = 10142 x = 10143 x = 10144 x = 10145 x = 10146 x = 10147 x = 10148 x = 10149 x = 10150 x = 10151 x = 10152 x = 10153 x = 10154 x = 10155 x = 10156 x = 10157 x = 10158 x = 10159 x = 10160 x = 10161 x = 10162 x = 10163 x = 10164 x = 10165 x = 10166 x = 10167 x = 10168 x = 10169 x = 10170 x = 10171 x = 10172 x = 10173 x = 10174 x = 10175 x = 10176 x = 10177 x = 10178 x = 10179 x = 10180 x = 10181 x = 10182 x = 10183 x = 10184 x = 10185 x = 10186 x = 10187 x = 10188 x = 10189 x = 10190 x = 10191 x = 10192 x = 10193 x = 10194 x = 10195 x = 10196 x = 10197 x = 10198 x = 10199 x = 10200 x = 10201 x = 10202 x = 10203 x = 10204 x = 10205 x = 10206 x = 10207 x = 10208 x = 10209 x = 10210 x = 10211 x = 10212 x = 10213 x = 10214 x = 10215 x = 10216 x = 10217 x = 10218 x = 10219 x = 10220 x = 10221 x = 10222 x = 10223 x = 10224 x = 10225 x = 10226 x = 10227 x = 10228 x = 10229 x = 10230 x = 10231 x = 10232 x = 10233
4 Pass
x =40504; y_correct =40569; assert(isequal(NextBinaryPalindrome(x),y_correct))
x = 40505 x = 40506 x = 40507 x = 40508 x = 40509 x = 40510 x = 40511 x = 40512 x = 40513 x = 40514 x = 40515 x = 40516 x = 40517 x = 40518 x = 40519 x = 40520 x = 40521 x = 40522 x = 40523 x = 40524 x = 40525 x = 40526 x = 40527 x = 40528 x = 40529 x = 40530 x = 40531 x = 40532 x = 40533 x = 40534 x = 40535 x = 40536 x = 40537 x = 40538 x = 40539 x = 40540 x = 40541 x = 40542 x = 40543 x = 40544 x = 40545 x = 40546 x = 40547 x = 40548 x = 40549 x = 40550 x = 40551 x = 40552 x = 40553 x = 40554 x = 40555 x = 40556 x = 40557 x = 40558 x = 40559 x = 40560 x = 40561 x = 40562 x = 40563 x = 40564 x = 40565 x = 40566 x = 40567 x = 40568 x = 40569
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Here are the answers to the questions:
If m + n = 120 and m = n, whats m and n? —Explain Your Answer—a. 100 + 20b. 60 + 60c. 85 +
60 because m is equal to n. | 100 | 253 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2023-06 | latest | en | 0.713527 |
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Project Euler Problem 007
Project Euler Problem 007
import Data.List hiding (union) --003より --http://www.haskell.org/haskellwiki/Prime_numbers primes :: [Integer] primes = 2 : ([3,5..] `minus` join [[p*p, p*p+2*p..] | p <- primes']) where primes' = 3 : ([5,7..] `minus` join [[p*p, p*p+2*p..] | p <- primes']) join ((x:xs):t) = x : union xs (join (pairs t)) pairs ((x:xs):ys:t) = (x : union xs ys) : pairs t union (x:xs) (y:ys) = case (compare x y) of LT -> x : union xs (y:ys) EQ -> x : union xs ys GT -> y : union (x:xs) ys union xs [] = xs union [] ys = ys minus (x:xs) (y:ys) = case (compare x y) of LT -> x : minus xs (y:ys) EQ -> minus xs ys GT -> minus (x:xs) ys minus xs _ = xs factor :: Integer -> [Integer] factor n = factorimpl n primes where factorimpl n pri@(p:xs) = if p*p>n then [n] else if n`rem` p == 0 then p:factorimpl (n `quot` p) pri else factorimpl n xs solve :: Int -> Integer solve n = last \$ take n primes m :: Integer m = solve 10001 -- 104743 main :: IO () main = print m -- シンプルな別解 {- primes = 2:f [3] [3,5..] where f (x:xs) ys = ps ++ f(xs++ps) [z|z<-qs,z`rem`x/=0] where (ps,qs) = span(< x*x) ys m = primes !! (10^4) --104743 -- -}
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