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string | warc_filename
string | warc_record_offset
int32 | warc_record_length
int32 | text
string | token_count
int32 | char_count
int32 | metadata
string | score
float64 | int_score
int64 | crawl
string | snapshot_type
string | language
string | language_score
float64 |
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https://herbie.uwplse.org/demo/ee8cadaa421115c4832effb99fd17cc3eb5c8afa.99305f72cd38bba0184d075a0d6fe9ac845dc43a/graph.html
| 1,590,803,828,000,000,000 |
text/html
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crawl-data/CC-MAIN-2020-24/segments/1590347407001.36/warc/CC-MAIN-20200530005804-20200530035804-00067.warc.gz
| 382,920,027 | 2,701 |
Average Error: 0.2 → 0.3
Time: 18.5s
Precision: 64
$\left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + 28 \cdot {x}^{2}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}$
$\left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + \sqrt{28 \cdot {x}^{2}} \cdot \sqrt{\left(28 \cdot \sqrt{{x}^{2}}\right) \cdot \sqrt{{x}^{2}}}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}$
\left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + 28 \cdot {x}^{2}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}
\left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + \sqrt{28 \cdot {x}^{2}} \cdot \sqrt{\left(28 \cdot \sqrt{{x}^{2}}\right) \cdot \sqrt{{x}^{2}}}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}
double f(double x) {
double r186894 = x;
double r186895 = r186894 * r186894;
double r186896 = 7.0;
double r186897 = r186896 * r186894;
double r186898 = r186895 - r186897;
double r186899 = r186898 * r186894;
double r186900 = 28.0;
double r186901 = 2.0;
double r186902 = pow(r186894, r186901);
double r186903 = r186900 * r186902;
double r186904 = r186899 + r186903;
double r186905 = r186904 * r186894;
double r186906 = 56.0;
double r186907 = 3.0;
double r186908 = pow(r186894, r186907);
double r186909 = r186906 * r186908;
double r186910 = r186905 - r186909;
double r186911 = r186910 * r186894;
double r186912 = 4.0;
double r186913 = pow(r186894, r186912);
double r186914 = r186911 + r186913;
return r186914;
}
double f(double x) {
double r186915 = x;
double r186916 = r186915 * r186915;
double r186917 = 7.0;
double r186918 = r186917 * r186915;
double r186919 = r186916 - r186918;
double r186920 = r186919 * r186915;
double r186921 = 28.0;
double r186922 = 2.0;
double r186923 = pow(r186915, r186922);
double r186924 = r186921 * r186923;
double r186925 = sqrt(r186924);
double r186926 = sqrt(r186923);
double r186927 = r186921 * r186926;
double r186928 = r186927 * r186926;
double r186929 = sqrt(r186928);
double r186930 = r186925 * r186929;
double r186931 = r186920 + r186930;
double r186932 = r186931 * r186915;
double r186933 = 56.0;
double r186934 = 3.0;
double r186935 = pow(r186915, r186934);
double r186936 = r186933 * r186935;
double r186937 = r186932 - r186936;
double r186938 = r186937 * r186915;
double r186939 = 4.0;
double r186940 = pow(r186915, r186939);
double r186941 = r186938 + r186940;
return r186941;
}
# Try it out
Results
In Out
Enter valid numbers for all inputs
# Derivation
1. Initial program 0.2
$\left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + 28 \cdot {x}^{2}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}$
2. Using strategy rm
$\leadsto \left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + \color{blue}{\sqrt{28 \cdot {x}^{2}} \cdot \sqrt{28 \cdot {x}^{2}}}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}$
4. Using strategy rm
$\leadsto \left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + \sqrt{28 \cdot {x}^{2}} \cdot \sqrt{28 \cdot \color{blue}{\left(\sqrt{{x}^{2}} \cdot \sqrt{{x}^{2}}\right)}}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}$
6. Applied associate-*r*0.3
$\leadsto \left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + \sqrt{28 \cdot {x}^{2}} \cdot \sqrt{\color{blue}{\left(28 \cdot \sqrt{{x}^{2}}\right) \cdot \sqrt{{x}^{2}}}}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}$
7. Final simplification0.3
$\leadsto \left(\left(\left(x \cdot x - 7 \cdot x\right) \cdot x + \sqrt{28 \cdot {x}^{2}} \cdot \sqrt{\left(28 \cdot \sqrt{{x}^{2}}\right) \cdot \sqrt{{x}^{2}}}\right) \cdot x - 56 \cdot {x}^{3}\right) \cdot x + {x}^{4}$
# Reproduce
herbie shell --seed 1
(FPCore (x)
:name "(((x*x-7*x)*x+28*pow(x,2))*x - 56*pow(x,3))*x + pow(x,4)"
:precision binary64
(+ (* (- (* (+ (* (- (* x x) (* 7 x)) x) (* 28 (pow x 2))) x) (* 56 (pow x 3))) x) (pow x 4)))
| 1,559 | 3,827 |
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| 3.625 | 4 |
CC-MAIN-2020-24
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latest
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https://brilliant.org/discussions/thread/trigonometry-the-basics/?ref_id=61211
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text/html
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crawl-data/CC-MAIN-2020-10/segments/1581875145981.35/warc/CC-MAIN-20200224193815-20200224223815-00369.warc.gz
| 298,350,696 | 13,548 |
# Trigonometry: The Basics
Over the next few days (Perhaps weeks) I will be doing a series of posts on Trigonometry. I will try to post perhaps once per day excluding weekends. Today we will talk about what Trigonometry actually is and the absolute basics.
What is Trigonometry?
The Word "Trigonometry" comes from the two Greek words "trigōnon", and "metron". "trigōnon" means Triangle and "metron" means measure. Trigonometry is the study of the measure of Triangles.
Trigonometry mainly deals with the relationship of two things in a Triangle: The Angles and the Lengths of the sides.
You might be asking: Why study triangles? why not Quadrilaterals or pentagons or circles?
There is a simple answer to this, it is because A Triangle is Stable.
Try taking 3 toothpicks or pencils or other stright objects and put them into the shape of a triangle. Now try changing the shape of the triangle. You will realise you can't change it, because a Triangle is stable.
Also, If you know the 3 angles in a triangle, the ratios of the sides will always be the same.
Prerequisites
The Pythagorean Theorem
In a Right-angled triangle:
$a^2+b^2=c^2$
Where $c$ is the longest side known as the hypotenuse and $a$ and $b$ are the other two sides
The Next Post in this series will be here
Note by Yan Yau Cheng
6 years, 1 month ago
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
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I like this idea. I am looking forward to future posts.
- 6 years, 1 month ago
@SharkyKesa: The Next post was just posted 1 minute ago here
- 6 years, 1 month ago
"You will realise you can change it, because a Triangle is stable."
TYPO: should be "can't"
- 6 years, 1 month ago
Whoops. I fixed it now
- 6 years, 1 month ago
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https://www.studypool.com/discuss/547330/find-the-area-of-the-figure-1?free
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##### find the area of the figure
Mathematics Tutor: None Selected Time limit: 1 Day
May 18th, 2015
The two semi-circles make up a whole circle with area pi(3^2) = 9pi
The area of the rectangular part is 9x6 = 54
Then the total area is 54+9pi m^2
If you want the answer to the nearest tenth, then it's 82.3 m^2
May 18th, 2015
...
May 18th, 2015
...
May 18th, 2015
Dec 10th, 2016
check_circle
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http://mayhematics.com/m/m3_factorials.htm
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| 99,725,544 | 2,571 |
# Rational Mathematics
## Factorials
The product n.(n−1).(n−2).....3.2.1, is called the factorial of n, and is denoted by n! (read as ‘n-factorial’). Thus n! is the nth factorial number. It is found convenient to define 0! = 1. The first few factorial numbers in base ten are: 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, ...
Theorem (The Permutation Principle): The number of ways of arranging n different things in sequence is n!. Proof: The first can be chosen in n ways, the second in n−1 ways, the third in n−2 ways and so on. Hence by the multiplication principle the total is the product of these ways.
Example: In how many ways can n different books be arranged on a shelf so that two particular books are next to one another? Regarding the specified books as one item there are (n−1) items to arrange, and the two books themselves can be arranged in 2 ways. Answer 2.(n−1)!
Example: How many ways are there of arranging n chess rooks on a board n ranks by n files with none guarding or attacking any other. Answer n!.
Theorem (The Circular Permutation Principle): The number of ways of arranging n things round a circle that can be rotated but not reflected is (n−1)! Proof: We can rotate the circle so that one thing is always in the same position. The others can then be arranged in (n−1)! ways. Corollary: The number of ways of arranging n things round a circle that can be rotated or reflected is (n−1)!|2, provided n>2. Proof: The arrangements in the previous result occur in pairs that are reflections of each other (except when n = 2 or less, in which case reflection makes no difference). If one is regarded as ‘clockwise’ the other is ‘anticlockwise’.
Example: How many necklace designs can be formed using six different coloured beads? Answer 5!|2 = 60.
Theorem (The Truncated Factorial Rule): The number of ways of choosing a sequence of m elements, all different, from a set of n elements is n!|(n−m)!. Proof: The first element can be chosen in n ways, the second in (n−1) ways, and so on until we reach the mth element which can be chosen in (n+1)−m ways. The product of these is n.(n−1).(n−2).... as far as (n+1)−m. This is n! with the remaining terms cancelled, and can thus be expressed by the quotient n!|(n−m)!.
Example: Count the number of one-to-one operators whose domain is {1,2,...,m} and whose range is a set of n elements.
Example: Express in factorials: n.(n^2 − 1).(n^2 − 4).(n^2 − 9). Answer (n+3)!|(n−4)! This uses the property (n^2)−(r^2) = (n−r).(n+r).
Theorem (The Principle of Divided Factorials): The number of ways of arranging n things in a row when there are a alike of one kind, b of a second kind, c of a third kind is n!|(a!.b!.c!....) Proof: If the things were all different the number of arrangements would be n! However, if a of the things are alike they can be permuted among themselves in a! ways without altering the positions of the other things. Corollary: The number of ways of arranging n things in sets of a, b, c ... things (where of course n = a+b+c+...) is n!|(a!.b!.c!. ...).
Example: In the game of Bridge a pack of 52 cards is used and they are dealt out equally to the four players (known as North, South, East and West) so that each has a hand of 13 cards. The number of different ways the cards can be dealt out is thus: 52!|(13!.13!.13!.13!) = 52!|(13!^4).
Example: In how many ways can the letters of the word REARRANGE be arranged? Answer. 9!|(2!.2!.3!) = 15120.
Every three successive numbers contain 2 and 3 therefore (3.k)! has a factor of 6^k (so in base 6 has k zeros at the end).
Divisibility: The product of r consecutive numbers is divisible by r!.
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https://regents-prep.org/physical-setting-29-january-2003/2050/
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# Physical Setting 29 January 2003
Physical Setting 29 January 2003 Questions Answer Keys
The University of the State of New York
Regents High School Examination
Physical Setting
Physics
Wednesday, January 29, 2003 — 9:15 a.m. to 12:15 p.m., only
Directions (1–30): For each statement or question, write on the separate answer sheet, the number of the word or expression that, of those given, best completes the statement or answers the question.
1. The diagram below shows a worker using a rope to pull a cart.
The worker’s pull on the handle of the cart can best be described as a force having
• (1) magnitude, only
• (2) direction, only
• (3) both magnitude and direction
• (4) neither magnitude nor direction
2. A car travels 90. meters due north in 15 seconds. Then the car turns around and travels 40. meters due south in 5.0 seconds. What is the magnitude of the average velocity of the car during this 20.-second interval?
• (1) 2.5 m/s
• (2) 5.0 m/s
• (3) 6.5 m/s
• (4) 7.0 m/s
3. How far will a brick starting from rest fall freely in 3.0 seconds?
• (1) 15 m
• (2) 29 m
• (3) 44 m
• (4) 88 m
4. If the sum of all the forces acting on a moving object is zero, the object will
• (1) slow down and stop
• (2) change the direction of its motion
• (3) accelerate uniformly
• (4) continue moving with constant velocity
5. A net force of 10. newtons accelerates an object at 5.0 meters per second2. What net force would be required to accelerate the same object at 1.0 meter per second2?
• (1) 1.0 N
• (2) 2.0 N
• (3) 5.0 N
• (4) 50. N
6. The graph below represents the relationship between gravitational force and mass for objects near the surface of Earth.
The slope of the graph represents the
• (1) acceleration due to gravity
• (2) universal gravitational constant
• (3) momentum of objects
• (4) weight of objects
7. A 1,200-kilogram car traveling at 10. meters per second hits a tree and is brought to rest in 0.10 second. What is the magnitude of the average force acting on the car to bring it to rest?
• (1) 1.2 × 102 N
• (2) 1.2 × 103 N
• (3) 1.2 × 104 N
• (4) 1.2 × 105 N
8. A spring scale reads 20. newtons as it pulls a 5.0-kilogram mass across a table. What is the magnitude of the force exerted by the mass on the spring scale?
• (1) 49 N
• (2) 20. N
• (3) 5.0 N
• (4) 4.0 N
Base your answers to questions 9 and 10 on the information below.
A 2.0 × 103-kilogram car travels at a constant speed of 12 meters per second around a circular curve of radius 30. meters.
9. What is the magnitude of the centripetal acceleration of the car as it goes around the curve?
• (1) 0.40 m/s2
• (2) 4.8 m/s2
• (3) 800 m/s2
• (4) 9,600 m/s2
10. As the car goes around the curve, the centripetal force is directed
• (1) toward the center of the circular curve
• (2) away from the center of the circular curve
• (3) tangent to the curve in the direction of motion
• (4) tangent to the curve opposite the direction of motion
Note that question 11 has only three choices.
11. The diagram below shows a block sliding down a plane inclined at angle with the horizontal.
As angle is increased, the coefficient of kinetic friction between the bottom surface of the block and the surface of the incline will
• (1) decrease
• (2) increase
• (3) remain the same
12. The amount of work done against friction to slide a box in a straight line across a uniform, horizontal floor depends most on the
• (1) time taken to move the box
• (2) distance the box is moved
• (3) speed of the box
• (4) direction of the box’s motion
13. A 1.2-kilogram block and a 1.8-kilogram block are initially at rest on a frictionless, horizontal surface. When a compressed spring between the blocks is released, the 1.8-kilogram block moves to the right at 2.0 meters per second, as shown.
What is the speed of the 1.2-kilogram block after the spring is released?
• (1) 1.4 m/s
• (2) 2.0 m/s
• (3) 3.0 m/s
• (4) 3.6 m/s
14. An object weighs 100. newtons on Earth’s surface. When it is moved to a point one Earth radius above Earth’s surface, it will weigh
• (1) 25.0 N
• (2) 50.0 N
• (3) 100. N
• (4) 400. N
15. An object weighing 15 newtons is lifted from the ground to a height of 0.22 meter. The increase in the object’s gravitational potential energy is approximately
• (1) 310 J
• (2) 32 J
• (3) 3.3 J
• (4) 0.34 J
Note that question 16 has only three choices.
16. As an object falls freely, the kinetic energy of the object
• (1) decreases
• (2) increases
• (3) remains the same
17. Moving 2.5 × 10–6 coulomb of charge from point A to point B in an electric field requires 6.3 × 10–4 joule of work. The potential difference between points A and B is approximately
• (1) 1.6 × 10–9 V
• (2) 4.0 × 10–3 V
• (3) 2.5 × 102 V
• (4) 1.0 × 1014 V
18. A 3.0-kilogram block is initially at rest on a frictionless, horizontal surface. The block is moved 8.0 meters in 2.0 seconds by the application of a 12-newton horizontal force, as shown in the diagram below.
What is the average power developed while moving the block?
• (1) 24 W
• (2) 32 W
• (3) 48 W
• (4) 96 W
19. The diagram below shows three neutral metal spheres, x, y, and z, in contact and on insulating stands.
Which diagram best represents the charge distribution on the spheres when a positively charged rod is brought near sphere x, but does not touch it?
20. Which graph best represents the electrostatic force between an alpha particle with a charge of +2 elementary charges and a positively charged nucleus as a function of their distance of separation?
21. When a neutral metal sphere is charged by contact with a positively charged glass rod, the sphere
• (1) loses electrons
• (2) gains electrons
• (3) loses protons
• (4) gains protons
22. If 10. coulombs of charge are transferred through an electric circuit in 5.0 seconds, then the current in the circuit is
• (1) 0.50 A
• (2) 2.0 A
• (3) 15 A
• (4) 50. A
23. The diagram below represents a source of potential difference connected to two large, parallel metal plates separated by a distance of 4.0 × 10–3 meter.
Which statement best describes the electric field strength between the plates?
• (1) It is zero at point B.
• (2) It is a maximum at point B.
• (3) It is a maximum at point C.
• (4) It is the same at points A, B, and C.
24. A periodic wave transfers
• (1) energy, only
• (2) mass, only
• (3) both energy and mass
• (4) neither energy nor mass
Note that question 25 has only three choices.
25. As the potential difference across a given resistor is increased, the power expended in moving charge through the resistor
• (1) decreases
• (2) increases
• (3) remains the same
26. An electric iron operating at 120 volts draws 10. amperes of current. How much heat energy is delivered by the iron in 30. seconds?
• (1) 3.0 × 102 J
• (2) 1.2 × 103 J
• (3) 3.6 × 103 J
• (4) 3.6 × 104 J
27. A motor is used to produce 4.0 waves each second in a string. What is the frequency of the waves?
• (1) 0.25 Hz
• (2) 15 Hz
• (3) 25 Hz
• (4) 4.0 Hz
28. The diagram below shows a periodic wave.
Which points are in phase with each other?
• (1) A and C
• (2) A and D
• (3) B and C
• (4) C and D
29. A surfacing whale in an aquarium produces water wave crests having an amplitude of 1.2 meters every 0.40 second. If the water wave travels at 4.5 meters per second, the wavelength of the wave is
• (1) 1.8 m
• (2) 2.4 m
• (3) 3.0 m
• (4) 11 m
30. In a certain material, a beam of monochromatic light (f = 5.09 × 1014 hertz) has a speed of 2.25 × 108 meters per second. The material could be
• (1) crown glass
• (2) flint glass
• (3) glycerol
• (4) water
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# Systems of equations with elimination (and manipulation)
In some cases, we need to slightly manipulate a system of equations before we can solve it using the elimination method. See how it's done in this video. Created by Sal Khan.
## Want to join the conversation?
• Is elimination the only way to solve linear equations
• One may find it easier to use matrices when he is faced with crazy equations including five or so variables and five or so complicated equations. Otherwise, substitution and elimination are your best options. Graphing, unless done extremely precisely, may lead to error.
• how do you eliminate negative numbers?
• You just add the number you do the opposite of the sign
Rule: + x - = -
- x + = -
• I don't understand why if you subtract negative 15 from 5 you don't get 20....?
• He is adding, not subtracting. That is why he had to make the numbers negative in order to cancel them out. 5x+(-15x)=-10x. Adding a -15 is like subtracting a +15.
• You know the second equation couldn't he just multiply that by 5x? Did it have to be negative 5?
• Mye,
He used a negative 5 so he could just add the two equations and the 10y and -10y become 0y and eliminate the y.
And you are correct. He could have just used a 5 instead of a -5, but then he would have had to subtract the equations instead of adding them. That would work the same way and you get the same answer.
I hope that helps.
• When you say ' 5 is the same as 20/4' dont understand how ??
• 5 is the same as 20/4 because 20/4 is 20 divided by 4 which equals 5
• At where did the -5 come from?
• Sal chose to multiply both sides of the bottom equation by -5. I know, I know, you want to know why he decided to do that. Well he wanted at least one term with a variable in each equation to be the same size but opposite in sign. Since the top equation was
5x-10y =15 and the bottom equation was 3x - 2y = 3, he recognized that by multiplying both sides of the bottom equation by -5 he could get the "y" terms in each equation to be the same size (10) but opposite in sign ... that way if he added the two equations together, he would "ELIMINATE" the "y" term and then he would just have to solve for x.
• This is so easy. No idea why I hated it at school.
• I noticed at that Sal does something that I don't do - he sometimes multiplies one of the equations with a negative number just so that he can eliminate a variable by adding the two equations, while I don't care if I have to add or subtract the equations. This is just personal preference, right?
• Yes, it is. When you subtract equations, you're really performing two steps at once. Sal chose to make each step explicit to avoid losing people.
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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## #1 2005-08-09 23:49:41
NIH
Member
Offline
### A Diophantine problem
Find all integer solutions of a(a + 1) = b(b + 2).
2 + 2 = 5, for large values of 2.
## #2 2005-08-10 07:52:23
MathsIsFun
Offline
### Re: A Diophantine problem
Let's try an example. How about finding b for a=1:
a(a + 1) = b(b + 2)
1(1+1) = b(b+2)
2 = b²+2b
So, let's try see if we can find a b that solves it:
b=1: b²+2b = 1²+2 = 3
b=0: b²+2b = 0
b=-1: b²+2b = (-1)²-2 = -1
b=-2: b²+2b = (-2)²-4 = 0
b=-3: b²+2b = (-3)²-6 = 3 ... oops, went straight past it
Time for thinking cap ...
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #3 2005-08-10 09:25:22
kylekatarn
Power Member
Offline
### Re: A Diophantine problem
integer equations!
these one's are always a challenge=P
## #4 2005-08-10 10:52:35
kylekatarn
Power Member
Offline
### Re: A Diophantine problem
a=0 b=0
until now I have only found the solution a=0 and b=0
probably there aren't more solutions
## #5 2005-08-10 15:56:50
MathsIsFun
Offline
### Re: A Diophantine problem
Is it possible to use an odds and evens approach ?
If a is odd then a(a+1) will be odd×even = even
If a is even then a(a+1) will be even×odd = even
If b is odd then b(b+2) will be odd×odd = odd
If b is even then b(b+2) will be even×even = even
Ooops ... gotta go ...
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
## #6 2005-08-10 16:32:35
ganesh
Moderator
Offline
### Re: A Diophantine problem
Thanks, MathsisFun.
I'll continue from where you left.
a(a+1) is certainly even.
b(b+2) is odd if b is odd.
If b is even, b(b+2) is even, and is always a multiple of 4.
a(a+1) is a multiple of 4 only if a is multiple of 4 or it is a number of the form 4n+3.
Lets assume a is a number of the form 4n+3.
a(a+1) = (4n+3)(4n+4) = 16n² + 16n + 12n + 12 = 16n² + 28n + 12
= 4(4n²+7n+3)
Let this number be equal to b(b+2)
b(b+2) = 4(4n²+7n+3)
b² + 2b - 4(4n²+7n+3) = 0
b = [-2 ± √{4 +16(4n²+7n+3)}]/4
Let n=0, we get an irrational number as b.
Let n=1, we get an irrational number as b.
Let n=2, again, we get an irrational number as b.
Can {4 +16(4n²+7n+3)} never be a perfect square?
Is it because the last digit of {4 +16(4n²+7n+3)} is 2 or 8 and not 4 or 6?
Last edited by ganesh (2005-08-10 16:35:20)
Character is who you are when no one is looking.
## #7 2005-08-10 19:10:33
mathsyperson
Moderator
Offline
### Re: A Diophantine problem
There are no squares that end in 2 or 8, so if 4 +16(4n²+7n+3) always ends in 2 or 8, then we have a proof.
Take away 4: 16(4n²+7n+3)=...8 or 4. Multiples of 16 move in cycles of 5, with the last digit being 6, 2, 8, 4, 0...
This means that 4n²+7n+3 has to always be of the form 5m+3 or 4. So how do we show that?
Why did the vector cross the road?
It wanted to be normal.
## #8 2005-08-10 19:11:49
wcy
Full Member
Offline
### Re: A Diophantine problem
is it possible to use some general method as outlined by Euclid
## #9 2005-08-11 09:11:35
NIH
Member
Offline
### Re: A Diophantine problem
#### mathsyperson wrote:
There are no squares that end in 2 or 8, so if 4 + 16(4n²+7n+3) always ends in 2 or 8, then we have a proof.
But if n = 3 or 4 (mod 5), then 4 + 16(4n²+7n+3) ends in 4.
Here's a hint: add 1 to both sides of the original equation.
2 + 2 = 5, for large values of 2.
## #10 2005-08-11 10:19:18
wcy
Full Member
Offline
### Re: A Diophantine problem
then,
RHS=b²+2b+1= (b+1)²
but what abt LHS ?
Last edited by wcy (2005-08-11 10:42:04)
## #11 2005-08-11 10:39:23
NIH
Member
Offline
### Re: A Diophantine problem
#### wcy wrote:
but what abt LHS ?
Then the LHS equals a² + a + 1.
For what values of a can this be a perfect square?
2 + 2 = 5, for large values of 2.
wcy
Full Member
Offline
a can only be 0
## #13 2005-08-11 10:50:55
NIH
Member
Offline
### Re: A Diophantine problem
#### wcy wrote:
a can only be 0
Or -1.
If you can show why those are the only two values, the problem is solved!
2 + 2 = 5, for large values of 2.
## #14 2005-08-16 10:27:33
NIH
Member
Offline
### Re: A Diophantine problem
Here is a solution to this one.
a(a + 1) = b(b + 2) <=> a² + a + 1 = (b + 1)².
For positive a, a² < a² + a + 1 < (a + 1)², implying there is a perfect square between consecutive squares; contradiction.
For negative a, a(a + 1) = p(p - 1), where p = -a, so there are no solutions for p - 1 > 0; i.e., a < -1.
It is then easy to check for solutions with a = -1, 0.
These are: (a,b) = (-1,-2), (-1,0), (0,-2), (0,0).
2 + 2 = 5, for large values of 2.
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# How do you divide (1+9i)/(-2+9i)?
Jul 27, 2016
$\frac{79}{85} - \frac{27}{85} i$.
#### Explanation:
Let $z = \frac{1 + 9 i}{- 2 + 9 i}$
Then usual Method to simplify this is to multiply both $N r .$ and $D r .$ by the conjugate of the complex no. of the $D r .$, i.e., to say,
$z = \frac{1 + 9 i}{- 2 + 9 i} \times \frac{- 2 - 9 i}{- 2 - 9 i}$
$= \frac{- 2 - 9 i - 18 i - 81 {i}^{2}}{- {2}^{2} - {\left(9 i\right)}^{2}}$
$= \frac{- 2 - 27 i - 81 \left(- 1\right)}{4 - 81 {i}^{2}}$
$= \frac{79 - 27 i}{85}$
$= \frac{79}{85} - \frac{27}{85} i$.
But, we can, as well, solve it as follows :-
Suppose that, $z = x + i y = \frac{1 + 9 i}{- 2 + 9 i}$
$\Rightarrow \left(x + i y\right) \left(- 2 + 9 i\right) = 1 + 9 i$
$\Rightarrow - 2 x + 9 i x - 2 i y + 9 {i}^{2} y = 1 + 9 i$
$\Rightarrow - 2 x + i \left(9 x - 2 y\right) + 9 \left(- 1\right) y = 1 + 9 i$
$\Rightarrow \left(- 2 x - 9 y\right) + i \left(9 x - 2 y\right) = 1 + 9 i$
Comparing the Real and Imaginary Parts of both sides, we get,
$\left(1\right) : 2 x + 9 y = - 1 , \mathmr{and} , \left(2\right) : 9 x - 2 y = 9$
$\therefore 2 \times \left(1\right) + 9 \times \left(2\right) \Rightarrow \left(4 x + 18 y\right) + \left(81 x - 18 y\right) = - 2 + 81$
$\Rightarrow 85 x = 79$ $\Rightarrow x = \frac{79}{85}$.
Using this with $\left(2\right)$, we have,
$2 y = 9 \left(x - 1\right) = 9 \left(\frac{79}{85} - 1\right) = 9 \left(- \frac{6}{85}\right)$
$\Rightarrow y = - \frac{9 \cdot 6}{85 \cdot 2} = - \frac{27}{85}$, giving,
$z = x + i y = \frac{79}{85} - \frac{27}{85} i$, as before!
Enjoy Maths!
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# Irreducible Factors
• Sep 26th 2009, 03:49 PM
Roam
Irreducible Factors
Consider the polynomial $p(x)=x^6-1$.
(a) Find all monic irreducible factors of $p(x)$ over $Q$.
(b) Find all monic irreducible factors of $p(x)$ over $R$.
(c) Find all monic irreducible factors of $p(x)$ over $C$.
I'm really not sure what to do, I know that $x^6 - 1 = (x^3)^2 - 1 = (x^3 - 1)(x^3 + 1)$, so the the roots of p(x) are ±1 & (x+1) & (x-1) are two linear factors. Anyway the first question here asks for monic (degree 1) irreducible (so that it cannot be factorized as P(x)=a(x)b(x) where $a(x), b(x) \in Q$) factors of p(x) over Q (rationals). Can anyone show me the method for solving one of them, so that I'll be able to try solving the rest of it on my own. Thanks.
• Sep 27th 2009, 02:37 AM
e^(i*pi)
$x^6 - 1$
As you noted it is the difference of two squares:
$(x^3-1)(x^3+1)$
However these are the difference and sum respectively of two cubes:
$[(x-1)(x^2+x+1)][(x+1)(x^2-x+1)]$
$x^2 \pm x+1$ only factorises over the complex numbers.
• Sep 29th 2009, 07:19 PM
Roam
I'm still a little confused...
(a) So all monic irreducible factors of p(x) over Q are:
$
(x+1), (x-1), (x^2+x+1), (x^2-x+1)
$
(b) All monic irreducible factors of p(x) over R are:
$
(x+1), (x-1), (x^2+x+1), (x^2-x+1)
$
(c) Now to find monic irreducible factors of p(x) over C are we use the quadratic formula for $x^2 \pm x+1$:
$(x^2+x+1), (x^2-x+1), \frac{-1 \pm \sqrt{-3}}{2}$, $\frac{1 \pm \sqrt{-3}}{2}$, (x+1), (x-1)
are all the monic irreducible factors of p(x) over complex numbers.
Are my answers correct? Did I list all the factors correctly for each question?
• Sep 30th 2009, 11:32 AM
Roam
I don't know if I listed the factors correctly for each part. I'm confused... (Headbang)
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# Moment of Inertia of pulley
1. Homework Statement
In each question: you place mass m on the mass holder, and released the system from rest. The system includes the long rod with two point masses on it (see the picture next to the list of equipment).
Since this is a rotational and translational motion, I set up two equations: sum of the force and torque
unknown will be angular acceleration and moment of inertia.
## The Attempt at a Solution
Sorry. I guess I put my solution in the above.
a)
if you increase m, the mass on the mass holder, what happens to the following quantities:
- α, the angular acceleration of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.
The change of mass in the holder won't affect either of them. Therefore, they remain unchanged. Although I tempt to follow other opinions on web, adding extra weight will increase the velocity. But somehow Galileo keeps popping up in my head.
b)
you wrap the string around the smallest radius wheel in the "3-step pulley" (see diagram). If you wrap the string around a greater radius wheel, using the same mass on the mass holder, what will happen to the following quantitiesA:
- α, the angular acceleration of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.
Increase the radius will make angular acceleration smaller; therefore, angular acceleration will be decreased. Here is something I don't understand. I don't see the I= moment of inertia in the final equation and I tempted to go back to rotational motion and use it to interpret it. But I know something's wrong. Because it doesn't include in the final solution. Sorry. If I keep repeating myself, hopefully it'll not confuse you.
Therefore, I propose the moment of inertia remains unchanged even the radius is increased.
c)
if you move the two masses attached to the rotating rod away from the center of mass of the rod, what happens to the following quantities:
- α, the angular acceleration of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.
I don't know what I think of the last part. There is another system on top of the disk. It means another radius and new mass introduce into the system. I did this lab already. I remember it didn't affect the result. But I don't want to rely on memory. Can you shed some light on it?
Best regards,
Daisy
#### Attachments
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Simon Bridge
Homework Helper
... don't just guess: explain your reasoning.
One way to check is to relate the question to similar systems you are already familiar with ... ie. if the hanging mass were just dragging a block across the table instead of turning something, would adding more mass affect the acceleration of the hanging mass?
Have you looked at a simple version - say, one that uses massless pulleys?
I think doing the math for the simpler system will help you understand how to answer the questions better.
SammyS
Staff Emeritus
Homework Helper
Gold Member
View attachment 99918
1. Homework Statement
In each question: you place mass m on the mass holder, and released the system from rest. The system includes the long rod with two point masses on it (see the picture next to the list of equipment).
Since this is a rotational and translational motion, I set up two equations: sum of the force and torque
unknown will be angular acceleration and moment of inertia.
## Homework Equations
View attachment 99914
## The Attempt at a Solution
Sorry. I guess I put my solution in the above.
a)
if you increase m, the mass on the mass holder, what happens to the following quantities:
- α, the angular acceleration of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.
The change of mass in the holder won't affect either of them. Therefore, they remain unchanged. Although I tempt to follow other opinions on web, adding extra weight will increase the velocity. But somehow Galileo keeps popping up in my head.
b)
you wrap the string around the smallest radius wheel in the "3-step pulley" (see diagram). If you wrap the string around a greater radius wheel, using the same mass on the mass holder, what will happen to the following quantitiesA:
- α, the angular acceleration of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.
View attachment 99915
Increase the radius will make angular acceleration smaller; therefore, angular acceleration will be decreased. Here is something I don't understand. I don't see the I= moment of inertia in the final equation and I tempted to go back to rotational motion and use it to interpret it. But I know something's wrong. Because it doesn't include in the final solution. Sorry. If I keep repeating myself, hopefully it'll not confuse you.
Therefore, I propose the moment of inertia remains unchanged even the radius is increased.
c)
if you move the two masses attached to the rotating rod away from the center of mass of the rod, what happens to the following quantities:
- α, the angular acceleration of the pulley.
- I, the moment of inertia of the pulleys-plus-rod system.
I don't know what I think of the last part. There is another system on top of the disk. It means another radius and new mass introduce into the system. I did this lab already. I remember it didn't affect the result. But I don't want to rely on memory. Can you shed some light on it?
Best regards,
Daisy
Hello Daisy Muller. Welcome to PF !
A general comment. You included many equations, which is fine, but you tended to ignore them.
For (a):
Galileo feels as though he has been misquoted.
Look at your 1st equation. It's true that if the tension in the string (Is that T?) is zero, then doubling the mass will have no effect. You can give Galileo credit there. But T is not zero.
I have a question concerning your 2nd equation. The (½)MR2 is the moment of inertia for a uniform disk. That does not seem correct here. In fact, if the mass of the two rotating masses is much greater the the rotating rod, then the moment of inertia of the rotating system is more nearly MR2 .
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# Probability 4.2
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### Probability 4.2
1. 1. Binomial Distributions Chapter 4.2
2. 2. Objectives <ul><li>Learn to determine if a probability experiment is a binomial experiment </li></ul><ul><li>Learn to find binomial probabilities using the binomial probability formula </li></ul><ul><li>Learn to find binomial probabilities using technology, formulas, and a binomial probability table </li></ul><ul><li>Learn to graph a binomial distribution </li></ul><ul><li>Learn to find the mean, variance, & standard deviation of a binomial probability distribution. </li></ul>
3. 3. What does “binomial” mean?
4. 4. Binomial Experiments <ul><li>A binomial experiment is a probability experiment that: </li></ul><ul><ul><li>Is repeated for a fixed number of trials, where each trial is independent of the others </li></ul></ul><ul><ul><li>Has only 2 possible outcomes of interest for each trial. ( Success or Failure ) </li></ul></ul><ul><ul><li>Has the same probability of success for each trial: P(S) = P(F) </li></ul></ul><ul><ul><li>Has a random variable x that counts the number of successful trials. </li></ul></ul>
5. 5. Binomial Experiment Notation Symbol Description n The number of times a trial is repeated p = P(S) The probability of success in a single trial q = P(F) The probability of failure in a single trial (q = 1 – p) x The random variable represents a count of the number of successes in n trials: x = 0, 1, 2, 3, . . ., n
6. 6. Drawing a Club Example <ul><li>From a standard deck of cards, pick a card, note whether it is a club or not, and replace the card. </li></ul><ul><li>Repeat the experiment 5 times. (n = 5) </li></ul><ul><li>S = selecting a club, F = not selecting a club </li></ul><ul><li>p = P(S) = </li></ul><ul><li>¼ </li></ul><ul><li>q = P(F) = </li></ul><ul><li>¾ </li></ul><ul><li>x = number of clubs out of 5 </li></ul><ul><li>Possible values of x are: </li></ul><ul><li>0, 1, 2, 3, 4, and 5 </li></ul>
7. 7. Binomial or not? <ul><li>A certain surgical procedure has an 85% chance of success. A doctor performs the procedure on 8 patients. The random variable represents the number of successful surgeries. </li></ul><ul><li>Yes </li></ul>
8. 8. Binomial or not? <ul><li>A jar contains 5 red marbles, 9 blue marbles, and 6 green marbles. You randomly select 3 marbles from the jar, without replacement . The random variable represents the number of red marbles. </li></ul><ul><li>No </li></ul>
9. 9. Binomial or not? <ul><li>You take a multiple choice quiz that consists of 10 questions. Each question has 4 possible answers, only 1 of which is correct. You randomly guess the answer to each question. The random variable represents the number of correct answers. </li></ul><ul><li>What is a trial of this experiment? </li></ul><ul><li>What is a “success”? </li></ul><ul><li>Does this experiment satisfy the 4 conditions of a binomial experiment? </li></ul><ul><li>What are n, p, q, and the possible values of x? </li></ul>
10. 10. What is the probability? <ul><li>If you roll a 6-sided die 3 times, what is the probability of getting exactly one 6? </li></ul>
11. 11. Binomial Probability Formula <ul><li>In a binomial experiment, the probability of exactly x successes in n trials is: </li></ul><ul><li>P(x) = n C x p x q n-x = _ n! _ p x q n-x </li></ul><ul><li>(n-x)!x! </li></ul>
12. 12. What is the probability? <ul><li>If you roll a 6-sided die 3 times, what is the probability of getting exactly one 6? </li></ul><ul><li>P(x) = n C x p x q n-x = _ n! _ p x q n-x </li></ul><ul><li>(n-x)!x! </li></ul><ul><li>N = 3, p = 1/6, q = 5/6, x = 1 </li></ul><ul><li>P(1) = 3! (1/6) 1 *(5/6) 2 </li></ul><ul><li>(3-1)!1! </li></ul><ul><li>= 3 * (1/6) * (25/36) = 25/72 = .347 </li></ul>
13. 13. Another example . . . <ul><li>54% of men consider themselves to be basketball fans. You randomly select 15 men and ask each if he considers himself a basketball fan. </li></ul><ul><li>Find the probability that the number who consider themselves basketball fans is </li></ul><ul><ul><li>Exactly 8 </li></ul></ul><ul><ul><li>At least 8 </li></ul></ul><ul><ul><li>Less than 8 </li></ul></ul>
14. 14. Technology to the Rescue! <ul><li>To use Excel to find the answer, use: </li></ul><ul><li>=Binomdist(x,n,p,TRUE/FALSE) </li></ul><ul><li>Where FALSE means “exactly x” and TRUE means “no more than x” </li></ul><ul><li>Binomial Basketball.xlsx </li></ul>
15. 15. Using Binomial Probability Tables <ul><li>Turn to the back of your books, p. A8, Table 2 in Appendix B </li></ul><ul><li>Find n, x, and p </li></ul><ul><li>30% of all small businesses in the U.S. have a website. If you randomly select 10 small businesses, what is the probability that exactly 4 of them have a website? </li></ul><ul><li>What are n, p, and x? </li></ul><ul><li>10, .30, 4 </li></ul><ul><li>What is the answer? </li></ul><ul><li>.2 </li></ul>
16. 16. Graphing Binomial Distributions <ul><li>65% of households in the U.S. subscribe to cable TV. You randomly select 6 households and ask if they subscribe to cable. Construct a probability distribution for the random variable x, and graph it. </li></ul><ul><li>Using n = 6, p = .65, and q = .35, you get: </li></ul>x 0 1 2 3 4 5 6 P(x) .002 .020 .095 .235 .328 .244 .075
17. 17. Graphing Binomial Distributions x 0 1 2 3 4 5 6 P(x) .002 .020 .095 .235 .328 .244 .075
18. 18. Mean, Variance, & Standard Deviation <ul><li>You can use the formulas from section 4.1, but these are easier: </li></ul><ul><li>Mean: μ = np </li></ul><ul><li>Variance: σ 2 = npq </li></ul><ul><li>Standard Deviation: σ = √npq </li></ul>
19. 19. Homework <ul><li>P. 193 Do 1-9 together </li></ul><ul><li>P. 194 10-24 evens </li></ul>
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# Zeros of polynomials: plotting zeros
When we are given a polynomial in factored form, we can quickly find the polynomial's zeros. Then, we can represent them as the x-intercepts of the polynomial's graph.
## Want to join the conversation?
• I'm from Russia and berly know English so if you can help me understand the first part.....Thanks... If you can translate it to English that would be great.. I use google translate.
• So basically every polynomial function has "zeros" and these are also called x-intercepts. Zeros are when a polynomial function "intersects" or touches the x-axis. When a polynomial is in factored form, like the question in the video, it is very easy to find the zeros. If you think about it, an x-intercept is when a function intersects the x-axis, and for this to be true, the y-value of that coordinate must be equal to zero. So to solve we can use this property -
If (A)(B)(C) = 0
Then either A, B or C must be = 0
So in the case of 2x(2x+3)(x-2), we just set "A" "B" and "C" as equal to zero, when -
A = 2x = 0
B = 2x+3 = 0
C = x-2 = 0
Now we just solve for x to get our zeros!
We are now left with
x = 0
x = -3/2
x = 2
I hope this helped, If this confused you more or if something seems unclear, please let me know, I'm happy to help!
• why do we need to equal the polynomial to 0 always? why cant it be some other number?
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# At a certain state university last term, there were p
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At a certain state university last term, there were p students each of whom paid either the full tuition of x dollars or half the full tuition. What percent of the tuition paid by the p students last term was tuition from students who paid the full tuition?
(1) Of the p students, 20 percent paid the full tuition.
(2) The p students paid a total of \$91.2 million for tuition last term.
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24 Jul 2006, 07:24
Statement 1
20 percent of the students paid full tution.
Hence the amount would be - 1/5 * p *x = px/5
80 percent should have paid half the tution which is x/2
Total tution = 80% of p * x /2
= 4/5*p*x/2 = 2xp/5
Hence the total fees 3xp/5
percent of the tution paid by the p students last term was tution from students who paid the full tution =
(px/5) / (3px/5) * 100 = 1/3 * 100 = 33 1/3 %
Hence A is sufficient.
Statement (2) The p students paid a total of \$91.2 million for tuition last term.
Nothing can be inferred from this.
Hence "A "
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25 Jul 2006, 03:07
A
1) 20% of P paid = 20/100 * P * x = px/5
80% of P paid = 80/100 * P * x/2 = 2px/5
Total tution = 3px/5
percent = (px/5)/(3px/5) * 100 = 100/3 %
2) Total tution = 91.2M
Not suff
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25 Jul 2006, 03:28
full payers = f
1/2 payers = p-f
fraction = fx/(fx+(p-f)x/2))
=2f/(f+p)
1. expresses f in terms of p suff
2. insuff
A
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25 Jul 2006, 08:46
Thanks everyone. The OA is A.
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25 Jul 2006, 08:46
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## Personal Finance Big Ideas taught in this task: Setting Goals
1. Seth wants to buy a new skateboard that costs \$169. He has \$88 in the bank. If he earns \$7.25 an hour pulling weeds, how many hours will Seth have to work to earn the rest of the money needed to buy the skateboard?
2. Seth wants to buy a helmet as well. A new helmet costs \$46.50. Seth thinks he can work 6 hours on Saturday to earn enough money to buy the helmet. Is he correct?
3. Seth’s third goal is to join some friends on a trip to see a skateboarding show. The cost of the trip is about \$350. How many hours will Seth need to work to afford the trip?
## COMMENTARY
The purpose of this task is for students to solve problems involving division of decimals in the real-world context of setting financial goals. The focus of the task is on modeling and understanding the concept of setting financial goals, so fluency with the computations will allow them to focus on other aspects of the task. This task is also good preparation for the study of ratios and proportional relationships in 6th and 7th grade and their lead-in to linear functions in 8th grade.
This task is part of a set collaboratively developed by Money as You Learn, an initiative inspired by recommendations of the President’s Advisory Council on Financial Capability, and Illustrative Mathematics. Integrating essential financial literacy concepts into the teaching of the Common Core State Standards can strengthen teaching of the Common Core and expose students to knowledge and skills they need to become financially capable young adults. A mapping of essential personal finance concepts and skills against the Common Core State Standards as well as additional tasks and texts will be available at www.moneyasyoulearn.org. This task and additional personal finance-related mathematics tasks are available at www.illustrativemathematics.org and are tagged “financial literacy.”
## SOLUTION: 1
1. A. 167 – 88 = 79, so Seth needs to make \$79. Since
79÷7.25≈10.9
Seth will have to work about 11 hours to make enough money to buy the skateboard.
2. No, Seth is not correct. 6 x 7.25 = 43.5 which is not enough to buy the helmet; he needs \$3 more which will require a bit less than a half an hour more work.
3. Since 350÷7.25≈48.3 Seth will have to work about 50 hours.
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Microsoft Excel is an excellent tool for storing and analyzing data. It offers a wide range of statistical tools and formulas, making it easy to determine key statistics for different datasets. In this article, we’ll explain how to calculate standard error in Excel (and what it is in the first place).
## What Is Standard Error?
Standard error lets you see how accurately sample data fits a larger dataset. In other words, it calculates how accurate a distribution of data is to the entire population.
It does this by calculating standard deviation, which is how far the average of your sample deviates from the average of the entire dataset.
The formula for standard error is:
Where is the standard deviation for the entire dataset and is the square root of the sample size.
To illustrate the concept of standard error with an example, imagine a school with 500 students. We want to estimate the average height of the students, so we select 30 students and measure them. The average height of this sample is 160 cm.
However, if we take a sample of another 30 students, we find that the average height is 152 cm.
The variation of these sample averages against the actual population average is the standard error. The smaller the standard error, the more representative the sample mean is of the population. The larger the error, the less representative it is and the more variable the data is.
## How to Calculate Standard Error in Excel
Microsoft Excel provides built-in formulas for calculating standard deviation and the square root of a set of numbers, making it very easy to calculate standard error. Here’s how to use these formulas, using a simple example:
1. Organize your data into a column as below.
2. First, we need to calculate the standard deviation of your sample. To do so, select a cell, then type =STDEV.S(B2:B11) (selecting your sample range) and press Enter. Make sure to use STDEV.S as this returns the standard deviation for a sample, rather than an entire population.
1. Next, we need to calculate the square root of the sample size. To do so, select a second cell and type SQRT(COUNT(B2:B11)) (selecting your sample range). Then, press Enter.
1. Finally, we need to divide the standard deviation by the square root of the sample size. Select a third cell and type =F1/F2, selecting the cells that represent each value. Press Enter.
The cell should now display the standard error of your sample. You can combine both of these formulas into one equation. Simply type =STDEV(B2:B11)/SQRT(COUNT(B2:B11)) (selecting your data range instead of A1:A10).
Note: Make sure to follow the correct function syntax to ensure that your formulas work as designed. If you aren’t sure about the syntax, you can separate each part of the standard error formula and divide them into a third cell.
## Err On the Side of Caution
Microsoft Excel is a powerful tool for statistical analysis. It can also be used to calculate other forms of variance and loan repayments – and that’s not all. Hopefully, you’re now able to calculate standard error and you can get on to the next part of your analysis.
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Compute Planet's Apparent Visual Magnitude
In a hypothetical solar system there exist:
• a sun of radius $r_s$ and absolute visual magnitude $V$
• the Earth, with radius $r_e$ and distance from the sun of $1\,AU$
• another planet, wit radius $r_p$, albedo $a$, distance from Earth $d_e$ (in units of $AU$), and distance from the sun $d_s$ ($AU$)
Assume that the second planet is magically lit up entirely by the sun (so it does not have phases when viewed from Earth).
Now how can I compute the apparent visual magnitude of the second planet with respect to Earth?
If there are other parameters required, please tell me.
• Do you also assume that the second planet is $d$ + 1 $\rm{AU}$ from the star? Jul 23, 2014 at 2:31
• @Aaron no, sorry. I'll edit the question. Jul 23, 2014 at 14:07
• en.wikipedia.org/wiki/… Jul 24, 2014 at 20:42
• @FlorinAndrei looking at that link, how can I compute the absolute magnitude $H$ of a planet? Jul 24, 2014 at 21:03
Lets assume we're dealing with a superior planet, which is to say, the planet is orbiting at a greater distance than the Earth. This effectively ensures that there is no planetary phase to deal with.
Now, the Sun has a luminosity of $$L_{sun}$$ such that the Solar flux as seen by the planet is $$f_{planet} = \frac{ L_{sun} }{ 4 \pi d_s^2 }.$$
The cross section of the planet is about $$A = \pi r_p^2$$, so for a given albedo $$a_p$$ the reflective luminosity of the planet will be \begin{aligned} L_{planet} &= a_p f_{planet} A \\ &= a_p \pi r_p^2 \frac{ L_{sun} }{ 4 \pi d_s^2 }. \end{aligned}
Given that we know the absolute magnitude of the Sun, the absolute magnitude of the planet follows as \begin{aligned} V_{planet} &= -2.5 \log_{10}\left[ \frac{ L_{planet} }{ L_{sun} } \right] - V_{sun} \\ &= -2.5 \log_{10}\left[ a_p \frac{ r_p^2 }{ 4 d_s^2 } \right] - V_{sun} \end{aligned}
The apparent magnitude of the planet as seen from Earth can then be calculated as $$m_{planet} = V_{planet} + 5 \log_{10}\left[ d_{e-p} \right] - 5$$ where the distance between Earth and the planet, $$d_{e-p}$$, should be in parsecs and of course depends on the orbital phase of the two respective planets.
So the one additional parameter required was the Solar luminosity. There are probably a bunch of subtle effects with the albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.
• Thanks for answering! One question, though: can solar luminosity not be computed from the sun's absolute visual magnitude (or the other way around) because the total luminosity does not have to all be in visual light? Or is there some other reason? Jul 29, 2014 at 13:29
• @feralin I guess you can, but you'd still need a second parameter to normalize the conversion, be it either the relative magnitude of the sun or an absolute magnitude of a calibration star. Using the relative magnitude of the sun would make sense here and probably gives a more direct final expression, but I think using luminosity gives a better sense of what's physically going on. Jul 29, 2014 at 19:16
• @feralin As a second note; if you want to model an inferior planet you can simply make the reflective surface - A - depend on (synodic) orbital phase. Jul 29, 2014 at 19:23
• sorry, but can you explain what "relative magnitude" is? I'm not getting good results from google... Jul 29, 2014 at 20:11
• This appears to be incorrect, as it appears to predict that the apparent magnitude of the planet gets brighter as the Sun gets dimmer.
– user24157
Aug 9, 2020 at 19:22
Planetary magnitudes vary not only according to the Sun’s luminosity, their own average albedo, and their distance from the Earth, but also from:
• Variations in their albedo across their surface.
• Their phase angle, for planets that we sometimes see as a crescent.
• Their inclination, for planets like Saturn and Uranus that have a different albedo at their equator than at their poles.
• Neptune keeps getting brighter. No one knows why.
All of these effects are detailed in an informative paper Computing Apparent Planetary Magnitudes for The Astronomical Almanac by Mallama and Hilton, revised in 2018:
https://arxiv.org/pdf/1808.01973.pdf
Their source code for computing planetary magnitudes given all of these effects can be found here:
https://sourceforge.net/projects/planetary-magnitudes/
If you want correct values, you have to take into account the effects mentioned in Brandon Rhodes' answer. Nevertheless, here's how to do a quick-and-dirty calculation.
The absolute magnitude of a planet is defined as the apparent magnitude if the Sun-planet and planet-observer distances are 1 au, at opposition.
Assuming a diffuse disc reflector model, the absolute magnitude $$H$$ of a planet of diameter $$D_{\rm p}$$ is given by
$$H = 5 \log_{10} \left( \frac{D_0}{D_{\rm p} \sqrt{a_p}} \right)$$
Where $$a_p$$ is the planet's geometric albedo, and $$D_0$$ is given by
$$D_0 = 2\,{\rm au} \times 10^{H_\ast / 5}$$
Where $$H_\ast$$ is the absolute magnitude defined at a reference distance of 1 au, which can be calculated from the usual 10 parsec absolute magnitude $$M_\ast$$ as follows:
$$H_\ast = M_\ast + 5 \log_{10} \left( \frac{1 \rm \ au}{10 \rm \ pc} \right) \approx M_\ast - 31.57$$
For the Sun, this results in $$D_0 \approx 1329\ \rm km$$.
To get the apparent magnitude of the planet, you can then use
$$m_{\rm p} = H + 5 \log_{10} \left( \frac{d_{\rm p\ast} d_{\rm po}}{1 \mathrm{\ au}^2} \right) - 2.5 \log_{10} q(\alpha)$$
Where $$d_{\rm p\ast}$$ is the distance between the planet and the star, $$d_{\rm po}$$ is the distance between the planet and the observer, and $$q(\alpha)$$ is the phase integral at the phase angle $$\alpha$$.
For the diffuse disc reflector, $$q(\alpha) = \cos \alpha$$. For a Lambertian sphere,
$$q(\alpha) = \frac{2}{3} \left( \left(1 - \frac{\alpha}{\pi} \right) \cos \alpha + \frac{1}{\pi} \sin \alpha \right)$$
At opposition, $$\alpha = 0$$, giving phase integrals of $$q(0) = 1$$ for the diffuse disc, and $$q(0) = \tfrac{2}{3}$$ for the Lambertian sphere.
Real planets have more complex phase functions which need to be determined empirically (see Brandon Rhodes' answer).
As a quick check, let's plug in values for Jupiter, taken from the NASA Jupiter Fact Sheet. With a (volumetric mean) diameter of 139,822 km and a geometric albedo of 0.538, the computed absolute magnitude is -9.44, versus the actual value of -9.40.
Using the computed value of -9.44, a distance from Jupiter of 5.204 au to the Sun and 4.204 au to the Earth, and using the Lambertian sphere $$q(0) = \tfrac{2}{3}$$, the apparent magnitude works out at -2.30, while for a diffuse disc $$q(0) = 1$$ the apparent magnitude works out at -2.74. The brightest the real planet gets is around -2.94. Fortunately for me, the value comes out in the right ballpark, despite the gross approximations being made to the reflection behaviour of real planets.
My computed values using $$q(0) = 1$$ for Saturn, Uranus and Neptune are 0.54, 5.57 and 7.75 respectively, which aren't too far off the real values either.
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# How do you differentiate y = 2^xlog_2(x^4)?
##### 3 Answers
Mar 1, 2016
$f = {2}^{x} , g = {\log}_{2} \left({x}^{4}\right) \to f ' = {2}^{x} \ln 2 , g ' = \frac{1}{{x}^{4} \ln 2} \cdot 4 {x}^{3}$
$y ' = f g ' + g f ' = \frac{{2}^{x} 4 {x}^{3}}{{x}^{4} \ln 2} + \left({2}^{x} \ln 2\right) \left({\log}_{2} {x}^{4}\right)$
#### Explanation:
Use the product rule to differentiate y. Call the first one be f and the second one be g and take the derivatives of each of them then put it in to the product rule
Mar 1, 2016
$\setminus \frac{d}{\mathrm{dx}} \left({2}^{x} \setminus {\log}_{2} \left({x}^{4}\right)\right) = \setminus \frac{{2}^{x} \left(x \setminus {\log}_{2} \left({x}^{4}\right) \setminus {\ln}^{2} \left(2\right) + 4\right)}{x \setminus \ln \left(2\right)}$
#### Explanation:
$\setminus \frac{d}{\mathrm{dx}} \left({2}^{x} \setminus {\log}_{2} \left({x}^{4}\right)\right)$
Applying product rule as: ${\left(f \setminus \cdot g\right)}^{'} = {f}^{'} \setminus \cdot g + f \setminus \cdot {g}^{'}$
$f = {2}^{x} , g = \setminus {\log}_{2} \left({x}^{4}\right)$
$= \setminus \frac{d}{\mathrm{dx}} \left({2}^{x}\right) \setminus {\log}_{2} \left({x}^{4}\right) + \setminus \frac{d}{\mathrm{dx}} \left(\setminus {\log}_{2} \left({x}^{4}\right)\right) {2}^{x}$
$\setminus \frac{d}{\mathrm{dx}} \left({2}^{x}\right) = {2}^{x} \setminus \ln \left(2\right)$
$\setminus \frac{d}{\mathrm{dx}} \left(\setminus {\log}_{2} \left({x}^{4}\right)\right) = \setminus \frac{4}{x \setminus \ln \left(2\right)}$
finally,
$= {2}^{x} \setminus \ln \left(2\right) \setminus {\log}_{2} \left({x}^{4}\right) + \setminus \frac{4}{x \setminus \ln \left(2\right)} {2}^{x}$
Simplifying,
$= \setminus \frac{{2}^{x} \left(x \setminus {\log}_{2} \left({x}^{4}\right) \setminus {\ln}^{2} \left(2\right) + 4\right)}{x \setminus \ln \left(2\right)}$
Mar 1, 2016
To solve this problem it would help if we remember these two intermediate results (proof given below),
Result 1: $\setminus \frac{d}{\mathrm{dx}} {a}^{x} = {a}^{x} \setminus \ln \left(a\right)$
Result 2: $\setminus \frac{d}{\mathrm{dx}} \setminus {\log}_{b} \left({x}^{n}\right) = \setminus \frac{n}{x \setminus \ln \left(b\right)}$
y = 2^x\log_2(x^4);
Applying the Chain Rule,
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left[\setminus \frac{d}{\mathrm{dx}} \left({2}^{x}\right)\right] \setminus {\log}_{2} \left({x}^{4}\right) + {2}^{x} \left[\setminus \frac{d}{\mathrm{dx}} \setminus {\log}_{2} \left({x}^{4}\right)\right]$
Apply the two results mentioned above,
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left[{2}^{x} \setminus \ln \left(2\right)\right] {\log}_{2} \left({x}^{4}\right) + {2}^{x} \left[\setminus \frac{4}{x \setminus \ln \left(2\right)}\right]$
$\setminus q \quad \setminus \quad = \setminus \ln \left(2\right) y + \setminus \frac{4}{\setminus \ln \left(2\right)} \setminus \frac{{2}^{x}}{x}$
Proof of Result 1:
Let $\setminus \quad y = {a}^{x} \setminus q \quad \implies \setminus \ln \left(y\right) = x \setminus \ln \left(a\right)$.
Differentiating once,
$\frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \ln \left(a\right) \setminus q \quad \implies \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = y \setminus \ln \left(a\right) = {a}^{x} \setminus \ln \left(a\right)$
Proof of Result 2:
Let $\setminus \quad y = \setminus {\log}_{b} \left({x}^{n}\right)$
Using the logarithm-base rule, $\setminus {\log}_{b} \left({x}^{n}\right) = \setminus {\log}_{e} \frac{{x}^{n}}{\setminus} {\log}_{e} \left(b\right)$, we can write $y$ as -
$y = \setminus \frac{\setminus \ln \left({x}^{n}\right)}{\setminus \ln \left(b\right)} = \setminus \frac{n}{\setminus \ln \left(b\right)} \setminus \ln \left(x\right)$
$\setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \setminus \frac{n}{\setminus \ln \left(b\right)} \frac{1}{x} = \setminus \frac{n}{x \setminus \ln \left(b\right)}$
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# Equation of motion question
1. Sep 17, 2011
### quietrain
s = ut + (1/2)at2
if s = 0,
a = -2u / t
so since acceleration is change in velocity over time, so
change in velocity = -2u
why is the change in velocity given as -2u???
2. Sep 17, 2011
### Ken G
The key is to understand what equations like that mean. A graph can be very useful, so see if you can make a graph with t on the x axis and s on the y axis, and use that formula to get s(t). Can you see that if you set s=0, you are not making a general statement about s, you are asking for the times when s will be zero? That means you want the times when the object, whatever it is, has returned to its starting point, and there are only two times that will work. One of them is trivial-- the t=0 when the problem begins. The other time is -2u/a, which comes at an earlier point when the object was actually moving backward. But when you solve that expression for a instead of t, what you have is a formula that asks "what acceleration do I need such that the particle will return to the place where it had velocity u at a time t later." That's kind of an awkward thing to ask for, but the equations allow it-- you just have to realize it is not a general statement about all a and all u and t, it's asking for a special "a" given some pair of "u" and "t" of interest.
3. Sep 17, 2011
### Fewmet
I think you are being a little fast and loose with your reasoning. First note that if S=0 it could be because u and a are zero. The other alternative is that t=0. Your algebra rules out that latter possibility(since you are dividing by t).
Also note that u is the initial velocity. The only way the ∆v can be the negative of 2vi is if ∆v=2vi=0 (which also makes a=0).
4. Sep 17, 2011
### quietrain
oh so , if i want s = 0, then my possible times are 0 and -2u/a ?
because t= -2u/a nicely satisfy the equation of motion, letting s =0.
but we don't consider this negative time because we haven't started timing our experiment?
i see... thanks!
5. Sep 17, 2011
### Fewmet
Also keep in mind that s is really the change in position. That is handy when working with a situation where a body returns to its original position (like when vertically tossing a ball). In that case ∆s=0.
6. Sep 17, 2011
### quietrain
∆v = vf-vi
if vf = -vi,
then ∆v = -vi - vi = -2vi ??
7. Sep 17, 2011
### Fewmet
Interesting. That would work for something like a ball launched vertically at 5 m/s. It returns to the launch point at -5 m/s, regardless of the values of a and t.
8. Sep 18, 2011
### quietrain
hmm, in this example, i can find my a and t values right? assuming a = 9.81
from s = ut + 1/2 at2
0 = 5(t) + 1/2 (-9.81)(t2)
t = 0 or 1
so this time is for the total journey right?
so the time to reach the peak height would be 1/2 s? then for it to come down , another 1/2 s?
9. Sep 18, 2011
### Fewmet
Correct.
Problems like this are interesting when the initial position is higher than the final position (like tossing a stone vertically upward and then letting it fall into a well). You still get two values for time (when solving for the landing in the well). I was always told to throw out negative time solutions... do you see what the other root means in a problem like this?
10. Sep 19, 2011
### quietrain
hmm, i am not sure, i was also taught to ignore -ve times.
if i have this set-up, taking upwards as +ve
s = ut + 1/2 at^2
-5 = 10t + 1/2 (-10)t^2
so t = 2.41 or -0.41
but these are equations. equations don't care how you obtain s = -5 , as long as you get it, it doesn't care which route you take?
so i would guess -0.41s is the time it takes for the particle to travel downwards?
since my ut value has effectively become -ve, it is as good as saying the particle has downwards direction with positive time?
by the way, when i think about the stone throwing upwards and then falling downwards, i find it weird that it actually takes the same time for going up and coming down. although the equations points to that fact, if i were to shoot a marble upwards at 1000m/s, it would keep going up and cover a huge distance, but on the coming down part, it is only free fall, at -10m/s^2. this doesn't seem intuitive at all?
11. Sep 19, 2011
### Ken G
If you neglect air resistance, the motions are symmetric because the -10m/s^2 is there the whole time, both up and down. (Air resistance is really important in practice with the numbers you are using though!) It sounds like you're problem is that 10m/s^2 doesn't sound like much, if you shoot the marble at huge speeds, but remember the faster you shoot it, the higher it goes, because it will take that -10m/s^2 a longer time to stop the upward motion. Since it took longer to stop it, it will have longer to speed back up on the way down, so gets just the right amount of time to return to you at the speed you shot it up at.
12. Sep 19, 2011
### Fewmet
It looks like that is the answer to this question:
A stone is launched from the ground level with an initial velocity of 10 m/s upward. It rises to its peak and then comes down in a 5 m deep well. How long does it take to land in the well?
The -0.41 s is the time that the stone would have been launched from -5 m to reach the ground level of the problem moving 10 m/s upwards. (It might help to visualize this as a parabola. When you look at the problem starting at ground level, one leg of the parabola is cut off. If you go back to -0.41 s, you get the entire parabola.
I agree: it is counterintuitive. I think that is because we do not have very good intuition about uniformly accelerated motion.
For me, thinking about the math describing a body being dropped from 10 m helps. Given the initial velocity of 0 m/s, the acceleration of g and the distance, the time is determined. The final velocity is 20 m/s. Now consider throwing the body to a height of 10 m. The acceleration and distances are the same. There are the velocities of 0 m/s and 20 m/s at the ends of the trip. This makes me feel the time being the same makes sense.
13. Sep 23, 2011
### quietrain
thanks every1
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# Solutions for Chapter 3.3: Systems of Linear Equations in Three Variables
## Full solutions for Intermediate Algebra for College Students | 6th Edition
ISBN: 9780321758934
Solutions for Chapter 3.3: Systems of Linear Equations in Three Variables
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##### ISBN: 9780321758934
Intermediate Algebra for College Students was written by and is associated to the ISBN: 9780321758934. This expansive textbook survival guide covers the following chapters and their solutions. Since 71 problems in chapter 3.3: Systems of Linear Equations in Three Variables have been answered, more than 6772 students have viewed full step-by-step solutions from this chapter. This textbook survival guide was created for the textbook: Intermediate Algebra for College Students, edition: 6. Chapter 3.3: Systems of Linear Equations in Three Variables includes 71 full step-by-step solutions.
Key Math Terms and definitions covered in this textbook
• Complex conjugate
z = a - ib for any complex number z = a + ib. Then zz = Iz12.
• Dimension of vector space
dim(V) = number of vectors in any basis for V.
• Distributive Law
A(B + C) = AB + AC. Add then multiply, or mUltiply then add.
• Ellipse (or ellipsoid) x T Ax = 1.
A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA-1 yll2 = Y T(AAT)-1 Y = 1 displayed by eigshow; axis lengths ad
• Fibonacci numbers
0,1,1,2,3,5, ... satisfy Fn = Fn-l + Fn- 2 = (A7 -A~)I()q -A2). Growth rate Al = (1 + .J5) 12 is the largest eigenvalue of the Fibonacci matrix [ } A].
• Free columns of A.
Columns without pivots; these are combinations of earlier columns.
• Fundamental Theorem.
The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n - r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.
• Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.
• Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.
• Length II x II.
Square root of x T x (Pythagoras in n dimensions).
• Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.
• Minimal polynomial of A.
The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A - AI) if no eigenvalues are repeated; always meA) divides peA).
• Nullspace matrix N.
The columns of N are the n - r special solutions to As = O.
• Orthogonal subspaces.
Every v in V is orthogonal to every w in W.
• Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.
• Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.
• Projection p = a(aTblaTa) onto the line through a.
P = aaT laTa has rank l.
• Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!
• Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.
• Vandermonde matrix V.
V c = b gives coefficients of p(x) = Co + ... + Cn_IXn- 1 with P(Xi) = bi. Vij = (Xi)j-I and det V = product of (Xk - Xi) for k > i.
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You are Here: Home >< Maths
# Function questions watch
1. State whether the following mappings are functions, and is they are one-one, one-many,many-one or many-many relations.
a) x^2+y^2=1
b) y=x^4
c) y = √x (x ≥0)
For the first mapping it is not a function but I am not sure what type of relation it is. In addition, for the second I think it is a one-many relation and the third I am not sure altogether.
Then : find fg(x) and gf(x)
When f(x) = x^2 and g(x) = 2x+3
Would fg(x) be (2x+3)^2;
because f(g(x)) then substitute the value of g into f
Likewise gf(x) = 2x^2 +3
because g(f(x)) then substitute the value of f into g= 2x^2 +3
Finally,
If f(x) =x^2+4 find the inverse function of f^-1(x).
For this would I find the inverse of f^-1(x) or the first function?
In either case the inverse function of f(x)=x^2+4 = ±√x-4
Or for the second function f^-1(x) = fx
Are these correct?
2. (Original post by AliceAM342)
State whether the following mappings are functions, and is they are one-one, one-many,many-one or many-many relations.
a) x^2+y^2=1
b) y=x^4
c) y = √x (x ≥0)
For the first mapping it is not a function but I am not sure what type of relation it is.
Correct, it's not a function. But note that a single y value (between -1 and 1) corresponds to two distinct x values (between -1 and 1). Similarly, a single x value corresponds to two y values. So the relation is...?
In addition, for the second I think it is a one-many relation
Not quite. A single x value corresponds to a single y value. But, a single y value corresponds to two x values.
and the third I am not sure altogether.
Same procedure, how many values does a chosen x output? How many values does a chosen y output?
When f(x) = x^2 and g(x) = 2x+3 ... Would fg(x) be (2x+3)^2 ... Likewise gf(x) = 2x^2 +3
These are correct.
Finally,
If f(x) =x^2+4 find the inverse function of f^-1(x).
For this would I find the inverse of f^-1(x) or the first function?
In either case the inverse function of f(x)=x^2+4 = ±√x-4
That's fine.
3. (Original post by RDKGames)
Correct, it's not a function. But note that a single y value (between -1 and 1) corresponds to two distinct x values (between -1 and 1). Similarly, a single x value corresponds to two y values. So the relation is...?
Not quite. A single x value corresponds to a single y value. But, a single y value corresponds to two x values.
RDKGames
Same procedure, how many values does a chosen x output? How many values does a chosen y output?
These are correct.
That's fine.
For the first mapping would it be a many-many relation because , as you said, there is a single y value to two x values and a single x value to two y values, this mapping has single y and x values so would the other value then correspond at two of the value?
For the second mapping would it be a one-many relation , which is not a function,?
And for the third mapping would it be a one-many relation again (not a function) because there is a singular y value and multiple x values?
4. (Original post by AliceAM342)
For the first mapping would it be a many-many relation because , as you said, there is a single y value to two x values and a single x value to two y values, this mapping has single y and x values so would the other value then correspond at two of the value?
Correct, but I don't understand your question.
For the second mapping would it be a one-many relation , which is not a function,?
Not quite.
And for the third mapping would it be a one-many relation again (not a function) because there is a singular y value and multiple x values?
Not quite. One-many means a single x outputs multiple y.
5. Would the second mapping be a many-one relation, because there is a single y value to 4 x values.
For the third mapping would it be a many- one relation, because there is a single y value to 2 x values?
RDKGames
6. (Original post by AliceAM342)
Would the second mapping be a many-one relation, because there is a single y value to 4 x values.
Yes it's many-one, but a single y only corresponds to two x values, not four.
For the third mapping would it be a many- one relation, because there is a single y value to 2 x values?
No. Think about what you're saying - can you give an example where two different values output a single value for ? More importantly, do you know how this function even looks like?
7. I’m sorry I am confused by the last mapping and cannot answer your question. I had been doing mappings, functions etc. But not with equations until now, only a series of numbers for the domain and range.
I thought that there were 4 x values because the second mapping has x^4.
For the third mapping I am lost, sorry 😯
RDKGames
8. (Original post by AliceAM342)
For the third mapping I am lost, sorry 😯
RDKGames
It's a one-one mapping. For each you get a single output, and each corresponds to a single input.
9. (Original post by RDKGames)
It's a one-one mapping. For each you get a single output, and each corresponds to a single input.
Thank you very much for your help, I really appreciate it. I understand (mainly) the third mapping.
I will do some research to understand this topic more.
Thank you 😊 RDKGames
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# EBook Problems Hypothesis Proportion
## EBook Problems Set - Testing a Claim about a Proportion
### Problem 1
A random sample of 1000 Americans aged 65 and older was collected in 1980 and found that 15% had "hazardous" levels of drinking, which is defined as regularly drinking an amount of alcohol that could cause health problems given the subject's medical conditions. Researchers wanted to know if this proportion has changed since 1980 and so collected a random sample of 1500 Americans aged 65 and older in 2004. They found that 12% drank at hazardous levels. Which of the following is closest to the value of a test statistic that could be used to test the hypothesis that the proportion of hazardous drinkers over the age of 65 has declined since 1980?
(a) -2.13
(b) 0.014
(c) 0.418
(d) 4.54
### Problem 2
Based on past experience, a bank believes that 4% of the people who receive loans will not make payments on time. The bank has recently approved 300 loans. What is the probability that over 6% of these clients will not make timely payments?
(a) 0.096
(b) 0.038
(c) 0.962
(d) 0.904
(e) 0.017
### Problem 3
A marketing director for a radio station collects a random sample of three hundred 18 to 25 year-olds and two hundred and fifty 25 to 40 year-olds. She records the percent of each group that had purchased music online in the last 30 days. She performs a hypothesis test, and the p-value of her test turns out to be 0.15. From this she should conclude:
(a) there is insufficient evidence to conclude that there is a difference in the proportion of on-line music purchases in the younger and older group.
(b) that about 15% more people purchased on-line music in the younger group than in the older group.
(c) the proportion of on-line music purchasers is the same in the under-25 year-old group as in the older group.
(d) the probability of getting the same results again is .15.
### Problem 4
A candidate running for Congress claims that 64% of adults in the U.S. favor a tax cut. Her opponent says this claim is much too high it is definitely less. To see if this claim has merit, a random sample of 400 adults is asked about it and the percentage favoring a tax cut is obtained. The probability of obtaining the percentage found in the sample or an even lower one turns out to be 0.032, or a 3.2% chance, if one calculates this probability assuming the claim is true. If we test a hypothesis about the candidatess claim with a 0.05 significance level, based on the outcome of the polling, we should:
(a) Draw no conclusions and get a bigger sample
(b) Reject the candidate's claim
(c) Conclude that the percentage of adults favoring a tax cut is between 60.8% and 67.2%
(d) Not reject the candidate's claim
### Problem 5
Many people sleep in on the weekends to make up for short nights during the work week. The Better Sleep Council reports that 61% of us get more than 7 hours of sleep per night on the weekend. A random sample of 350 adults found that 235 had more than seven hours each night last weekend. At the 0.05 level of significance, does this evidence show that more than 61% of us get seven or more hours off sleep per night on the weekend?
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Search by Topic
Resources tagged with Place value similar to Jugs of Wine:
Filter by: Content type:
Stage:
Challenge level:
There are 56 results
Broad Topics > Numbers and the Number System > Place value
Digit Sum
Stage: 3 Challenge Level:
What is the sum of all the digits in all the integers from one to one million?
Arrange the Digits
Stage: 3 Challenge Level:
Can you arrange the digits 1,2,3,4,5,6,7,8,9 into three 3-digit numbers such that their total is close to 1500?
Six Times Five
Stage: 3 Challenge Level:
How many six digit numbers are there which DO NOT contain a 5?
X Marks the Spot
Stage: 3 Challenge Level:
When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .
Eleven
Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
Even Up
Stage: 3 Challenge Level:
Consider all of the five digit numbers which we can form using only the digits 2, 4, 6 and 8. If these numbers are arranged in ascending order, what is the 512th number?
Cayley
Stage: 3 Challenge Level:
The letters in the following addition sum represent the digits 1 ... 9. If A=3 and D=2, what number is represented by "CAYLEY"?
Stage: 2 and 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
Tis Unique
Stage: 3 Challenge Level:
This addition sum uses all ten digits 0, 1, 2...9 exactly once. Find the sum and show that the one you give is the only possibility.
Football Sum
Stage: 3 Challenge Level:
Find the values of the nine letters in the sum: FOOT + BALL = GAME
What an Odd Fact(or)
Stage: 3 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?
Number Rules - OK
Stage: 4 Challenge Level:
Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...
Stage: 2, 3, 4 and 5
This article for the young and old talks about the origins of our number system and the important role zero has to play in it.
Nice or Nasty
Stage: 2 and 3 Challenge Level:
There are nasty versions of this dice game but we'll start with the nice ones...
Exploring Simple Mappings
Stage: 3 Challenge Level:
Explore the relationship between simple linear functions and their graphs.
Stage: 3 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
Balance Power
Stage: 3, 4 and 5 Challenge Level:
Using balancing scales what is the least number of weights needed to weigh all integer masses from 1 to 1000? Placing some of the weights in the same pan as the object how many are needed?
Reasoned Rounding
Stage: 1, 2 and 3 Challenge Level:
Four strategy dice games to consolidate pupils' understanding of rounding.
Just Repeat
Stage: 3 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
Skeleton
Stage: 3 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
Reach 100
Stage: 2 and 3 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
Repeaters
Stage: 3 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
Lesser Digits
Stage: 3 Challenge Level:
How many positive integers less than or equal to 4000 can be written down without using the digits 7, 8 or 9?
Permute It
Stage: 3 Challenge Level:
Take the numbers 1, 2, 3, 4 and 5 and imagine them written down in every possible order to give 5 digit numbers. Find the sum of the resulting numbers.
Two and Two
Stage: 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
Basically
Stage: 3 Challenge Level:
The number 3723(in base 10) is written as 123 in another base. What is that base?
How Many Miles to Go?
Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
Big Powers
Stage: 3 and 4 Challenge Level:
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
Enriching Experience
Stage: 4 Challenge Level:
Find the five distinct digits N, R, I, C and H in the following nomogram
Three Times Seven
Stage: 3 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
Legs Eleven
Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
What a Joke
Stage: 4 Challenge Level:
Each letter represents a different positive digit AHHAAH / JOKE = HA What are the values of each of the letters?
Latin Numbers
Stage: 4 Challenge Level:
Can you create a Latin Square from multiples of a six digit number?
DOTS Division
Stage: 4 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
2-digit Square
Stage: 4 Challenge Level:
A 2-Digit number is squared. When this 2-digit number is reversed and squared, the difference between the squares is also a square. What is the 2-digit number?
Always a Multiple?
Stage: 3 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
Stage: 1, 2, 3 and 4
Nowadays the calculator is very familiar to many of us. What did people do to save time working out more difficult problems before the calculator existed?
Multiplication Magic
Stage: 4 Challenge Level:
Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . .
Stage: 3, 4 and 5
We are used to writing numbers in base ten, using 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. Eg. 75 means 7 tens and five units. This article explains how numbers can be written in any number base.
Back to the Planet of Vuvv
Stage: 3 Challenge Level:
There are two forms of counting on Vuvv - Zios count in base 3 and Zepts count in base 7. One day four of these creatures, two Zios and two Zepts, sat on the summit of a hill to count the legs of. . . .
Phew I'm Factored
Stage: 4 Challenge Level:
Explore the factors of the numbers which are written as 10101 in different number bases. Prove that the numbers 10201, 11011 and 10101 are composite in any base.
Chocolate Maths
Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
Pupils' Recording or Pupils Recording
Stage: 1, 2 and 3
This article, written for teachers, looks at the different kinds of recordings encountered in Primary Mathematics lessons and the importance of not jumping to conclusions!
Back to Basics
Stage: 4 Challenge Level:
Find b where 3723(base 10) = 123(base b).
Quick Times
Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
Never Prime
Stage: 4 Challenge Level:
If a two digit number has its digits reversed and the smaller of the two numbers is subtracted from the larger, prove the difference can never be prime.
Seven Up
Stage: 3 Challenge Level:
The number 27 is special because it is three times the sum of its digits 27 = 3 (2 + 7). Find some two digit numbers that are SEVEN times the sum of their digits (seven-up numbers)?
Plus Minus
Stage: 4 Challenge Level:
Can you explain the surprising results Jo found when she calculated the difference between square numbers?
Really Mr. Bond
Stage: 4 Challenge Level:
115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?
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Math416_HW3_sol
# Math416_HW3_sol - Math 416 Abstract Linear Algebra Fall...
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Math 416 - Abstract Linear Algebra Fall 2011, section E1 Homework 3 solutions Section 1.6 6.1. (2 pts) We want to show that any w W can be expressed uniquely as a linear combination of Av 1 , . . . , Av n . The equation w = c 1 Av 1 + . . . + c n Av n = A ( c 1 v 1 + . . . + c n v n ) is equivalent to the equation A - 1 w = A - 1 A ( c 1 v 1 + . . . + c n v n ) = c 1 v 1 + . . . + c n v n which has a unique solution ( c 1 , . . . , c n ) since { v 1 , . . . , v n } is a basis of V . 6.2. (1 pt check) A right inverse to A = 1 1 is a 2 × 1 matrix B = c d satisfying AB = I 1 , that is AB = 1 1 c d = c + d = 1 . The right inverses of A are all matrices of the form c 1 - c for some c R . In particular, A has distinct right inverses, therefore A has no left inverse (by thm 6.1). 6.8. A cannot be invertible. If A were invertible, the condition AB = 0 would imply A - 1 AB = A - 1 0 = 0, that is B = 0. 6.9. (2 pts) T 1 x 1 x 2 x 3 x 4 x 5 = x 1 x 4 x 3 x 2 x 5 = 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 x 1 x 2 x 3 x 4 x 5 T 2 x 1 x 2 x 3 x 4 x 5 = x 1 x 2 + ax 4 x 3 x 4 x 5 = 1 0 0 0 0 0 1 0 a 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 x 1 x 2 x 3 x 4 x 5
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# Nibble/Minute to Gibit/Hour Calculator - Convert Nibbles Per Minute to Gibibits Per Hour
## Conversion History (Last 6)
Input Nibbles Per Minute - and press Enter
Nibble/Minute
Sec
Min
Hr
Day
Sec
Min
Hr
Day
## Nibble/Minute to Gibit/Hour - Conversion Formula and Steps
Nibble and Gibibit are units of digital information used to measure storage capacity and data transfer rate. Nibble is one of the very basic digital unit where as Gibibit is a binary unit. One Nibble is equal to 4 bits. One Gibibit is equal to 1024^3 bits. There are 268,435,456 Nibbles in one Gibibit. - view the difference between both units
Source Data UnitTarget Data Unit
Nibble
Equal to 4 bits
(Basic Unit)
Gibibit (Gibit)
Equal to 1024^3 bits
(Binary Unit)
The formula of converting the Nibbles Per Minute to Gibibits Per Hour is represented as follows :
Gibit/Hour = Nibble/Minute x 4 / 10243 x 60
Now let us apply the above formula and, write down the steps to convert from Nibbles Per Minute (Nibble/Minute) to Gibibits Per Hour (Gibit/Hour). This way, we can try to simplify and reduce to an easy to apply formula.
FORMULA
Gibibits Per Hour = Nibbles Per Minute x 4 / 10243 x 60
STEP 1
Gibibits Per Hour = Nibbles Per Minute x 4 / (1024x1024x1024) x 60
STEP 2
Gibibits Per Hour = Nibbles Per Minute x 4 / 1073741824 x 60
STEP 3
Gibibits Per Hour = Nibbles Per Minute x 0.0000000037252902984619140625 x 60
Example : If we apply the above Formula and steps, conversion from 10 Nibble/Minute to Gibit/Hour, will be processed as below.
1. = 10 x 4 / 10243 x 60
2. = 10 x 4 / (1024x1024x1024) x 60
3. = 10 x 4 / 1073741824 x 60
4. = 10 x 0.0000000037252902984619140625 x 60
5. = 0.0000022351741790771484375
6. i.e. 10 Nibble/Minute is equal to 0.0000022351741790771484375 Gibit/Hour.
(Result rounded off to 40 decimal positions.)
You can use above formula and steps to convert Nibbles Per Minute to Gibibits Per Hour using any of the programming language such as Java, Python or Powershell.
#### Definition : Nibble
A Nibble is a unit of digital information that consists of 4 bits. It is half of a byte and can represent a single hexadecimal digit. It is used in computer memory and data storage and sometimes used as a basic unit of data transfer in certain computer architectures.
#### Definition : Gibibit
A Gibibit (Gib or Gibit) is a unit of digital information that is equal to 1,073,741,824 bits and is defined by the International Electro technical Commission(IEC). The prefix "gibi" is derived from the binary number system and it is used to distinguish it from the decimal-based "gigabit" (Gb). It is widely used in the field of computing as it more accurately represents the amount of data storage and data transfer in computer systems.
### Excel Formula to convert from Nibble/Minute to Gibit/Hour
Apply the formula as shown below to convert from Nibbles Per Minute to Gibibits Per Hour.
ABC
1Nibbles Per Minute (Nibble/Minute)Gibibits Per Hour (Gibit/Hour)
21=A2 * 0.0000000037252902984619140625 * 60
3
Download - Excel Template for Nibbles Per Minute to Gibibits Per Hour Conversion
If you want to perform bulk conversion locally in your system, then download and make use of above Excel template.
### Python Code for Nibble/Minute to Gibit/Hour Conversion
You can use below code to convert any value in Nibbles Per Minute to Gibibits Per Hour in Python.
nibblesPerMinute = int(input("Enter Nibbles Per Minute: "))
gibibitsPerHour = nibblesPerMinute * 4 / (1024*1024*1024) * 60
print("{} Nibbles Per Minute = {} Gibibits Per Hour".format(nibblesPerMinute,gibibitsPerHour))
The first line of code will prompt the user to enter the Nibbles Per Minute as an input. The value of Gibibits Per Hour is calculated on the next line, and the code in third line will display the result.
| 1,093 | 3,802 |
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https://convertoctopus.com/452-cubic-feet-to-cubic-meters
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## Conversion formula
The conversion factor from cubic feet to cubic meters is 0.0283168467117, which means that 1 cubic foot is equal to 0.0283168467117 cubic meters:
1 ft3 = 0.0283168467117 m3
To convert 452 cubic feet into cubic meters we have to multiply 452 by the conversion factor in order to get the volume amount from cubic feet to cubic meters. We can also form a simple proportion to calculate the result:
1 ft3 → 0.0283168467117 m3
452 ft3 → V(m3)
Solve the above proportion to obtain the volume V in cubic meters:
V(m3) = 452 ft3 × 0.0283168467117 m3
V(m3) = 12.799214713688 m3
The final result is:
452 ft3 → 12.799214713688 m3
We conclude that 452 cubic feet is equivalent to 12.799214713688 cubic meters:
452 cubic feet = 12.799214713688 cubic meters
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 cubic meter is equal to 0.078129793301344 × 452 cubic feet.
Another way is saying that 452 cubic feet is equal to 1 ÷ 0.078129793301344 cubic meters.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that four hundred fifty-two cubic feet is approximately twelve point seven nine nine cubic meters:
452 ft3 ≅ 12.799 m3
An alternative is also that one cubic meter is approximately zero point zero seven eight times four hundred fifty-two cubic feet.
## Conversion table
### cubic feet to cubic meters chart
For quick reference purposes, below is the conversion table you can use to convert from cubic feet to cubic meters
cubic feet (ft3) cubic meters (m3)
453 cubic feet 12.828 cubic meters
454 cubic feet 12.856 cubic meters
455 cubic feet 12.884 cubic meters
456 cubic feet 12.912 cubic meters
457 cubic feet 12.941 cubic meters
458 cubic feet 12.969 cubic meters
459 cubic feet 12.997 cubic meters
460 cubic feet 13.026 cubic meters
461 cubic feet 13.054 cubic meters
462 cubic feet 13.082 cubic meters
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https://www.atozexams.com/mcq/maths/303.html
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# If α,β are the roots of x2-p(x+1)+C=0 then (α+1)(β+1)=
1. C
2. C-1
3. 1+C
4. 1-C
4
1+C
Explanation :
No Explanation available for this question
# If x2+4ax+3=0 and 2x2+3ax-9=0 have a common root, then a=
1.
2. -1
3. 1
4. 0
4
Explanation :
No Explanation available for this question
# If 2+i√3 is a root of the equation x2+px+q=0,then
1. p=-4,q=7
2. p=-4,q=-7
3. p=4,q=-3
4. p=4,q=7
4
p=-4,q=7
Explanation :
No Explanation available for this question
# If α,β,γ,δ are the roots of the equation 3x4-8x3+2x2-9=0 then
1. -8/3
2. 8/3
3. -2/3
4. 2/3
4
2/3
Explanation :
No Explanation available for this question
# If α,β,γ are the roots of x3+x2+x=1=0 then (α-β)2+(β-γ)2+(γ-α)2=
1. -4
2. 4
3. -3
4. 3
4
-4
Explanation :
No Explanation available for this question
# If α,β,γ are the roots of x3+ax+b=0 then (α+β)-1+(β+γ)-1+(γ+α)-1=
1. –(a2/b2)
2. (a2/b2)
3. –a/b
4. a/b
4
a/b
Explanation :
No Explanation available for this question
# The equation whose roots are 2+√3,2-√3,1+2i,1-2i is
1. x4-6x3+14x2-22x+5=0
2. x4+6x3+14x2-22x+5=0
3. x4-7x3+25x2+43x-40=0
4. x4-7x3-25x2-43x+40=0
4
x4-6x3+14x2-22x+5=0
Explanation :
No Explanation available for this question
# The vertex and focus of a parabola are (0,0) and (0,4) then its directrix is
1. y+4=0
2. y+5=0
3. y+6=0
4. y+8=0
4
y+4=0
Explanation :
No Explanation available for this question
# Vertex of the parabola x2+12x-9y=0
1. (-6,-4)
2. (6,4)
3. (-6,4)
4. (6,-4)
4
(-6,-4)
Explanation :
No Explanation available for this question
1. 6√2
2. 10√2
3. 8√2
4. 3√2
4
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https://math.stackexchange.com/questions/591848/prove-a-commutative-ring-with-characteristic-n-has-a-subring-isomorphic-to-mat
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# Prove a commutative ring with characteristic n has a subring isomorphic to $\mathbb{Z}_n$
Let $R$ be a commutative ring with identity such that the characteristic of $R$ is $n$, char$R=n$. Prove that is $n>0$ then $R$ contains a subring isomorphic to $\mathbb{Z}$$_n, the additive group of integers modulo n. Attempt at proof: Consider an isomorphism \phi: R\rightarrow\ \mathbb{Z}$$_n$ defined as $\phi(x)=[r]$$_n$ where $r$ is the remainder when x divides n. Showing then that $\phi$ is well defined and an isomorphism would conclude the proof.
Is this going in the right direction? Or am I totally off?
• It is literally going in the wrong direction. Try $\mathbb Z_n \to R$. ;) Commented Dec 4, 2013 at 1:34
• Another comment is, what do you mean when $x$ divides $n?$. Your elements in the ring need not be integers, or multiples of 1.
– LASV
Commented Dec 4, 2013 at 1:42
• There's no reason for $R$ to be isomorphic to ${\bf Z}_n$. Consider ${\bf Z}_n^2$. Commented Dec 13, 2013 at 0:31
Think of a map $\psi:\mathbb{Z}\to R$, where $\psi$ is the map you think it should be =]. Further, what is the kernel of this map?
• If the ring is commutative but without unity and $char R=p$ , a prime number then what will be the mapping? Please help me. Commented Aug 2, 2017 at 7:39
• Here I guess you have taken the map $m \to m.1_R$ for m $\in \mathbb Z$ Commented Aug 2, 2017 at 7:41
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https://www.reference.com/web?q=prime%20factorization%20to%2050&qo=pagination&o=600605&l=dir&sga=1&qsrc=998&page=3
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Related Search
Web Results
brainly.com/question/17686758
The prime factorization of fifty is 2 times 25. Step-by-step explanation: So, the prime factors of 50 are 2 × 5 × 5 or 2 × 52, where 2 and 5 are the prime numbers.
researchmaniacs.com/Calculator/PrimeFactors/Prime-Factorization-of-50.html
To find all the prime factors of 50, divide it by the lowest prime number possible. Then divide that result by the lowest prime number possible. Keep doing this until the result itself is a prime number. The prime factorization of 50 will be all the prime numbers you used to divide, in addition to the last result, which is a prime number.
The prime factorization of 50 is 2 × 5 × 5. To use a factor tree to find the prime factorization of a number, x, we use the following... See full answer below.
The prime factorization of 50 is 2x5x5 or 2 x 52. The prime factorization of 18 is 2x3x3 or 2 x 32. The prime factorization of 32 is 2x2x2x2x2 or 25.
Prime factorization is expressing a positive integer as product of its prime factors. Consider, for example the number 100. The factors of 100 are: 1, 2, 4, 5, 10, 20, 50 and 100.
brilliant.org/wiki/prime-factorization
A prime factor tree provides a pictorial representation of the prime factorization for a positive integer. Starting with the given integer N N N at the top of the tree, two branches are drawn toward two positive factors of N. N. N. The process is repeated for the numbers at the end of each branch that is drawn until each "leaf" is a prime number.
www.mathportal.org/calculators/numbers-calculators/prime-factorization...
Finding prime factorization and factor tree. Example: Find prime factorization of 60. Step 1: Start with any number that divides 60, in this we will use 10. So, $\color{blue}{60 = 6 \cdot 10}$.
www.mathwarehouse.com/arithmetic/numbers/prime-number/prime-factorization...
The prime factorization calculator can: Calculate the prime factorization of the number you type (Numbers above 10 million may or may not time out. Calculating the prime factorization of large numbers is not easy, but the calculator can handle pretty darn big ones!) Determine whether or not a number is prime; Create sieve of Erasthones for the ...
www.wikihow.com/Find-Prime-Factorization
Here's how to find the GCF of 30 and 36, using prime factorization: Find the prime factorizations of the two numbers. The prime factorization of 30 is 2 x 3 x 5. The prime factorization of 36 is 2 x 2 x 3 x 3. Find a number that appears on both prime factorizations. Cross it out once on each list and write it on a new line.
www.dummies.com/education/math/using-prime-factorizations
Using prime factorization to find the LCM. One method for finding the least common multiple (LCM) of a set of numbers is to use the prime factorizations of those numbers. Here’s how: List the prime factors of each number. Suppose you want to find the LCM of 18 and 24. List the prime factors of each number: 18 = 2 × 3 × 3. 24 = 2 × 2 × 2 × 3
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https://www.teacherspayteachers.com/Product/Number-Bonds-To-20-Bar-Model-Practise-Sheets-5016571
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digital
# Number Bonds To 20: Bar Model Practise Sheets
Subject
Resource Type
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TpT Digital Activity
PDF (11 MB|6 pages)
Standards
\$2.00
TpT Digital Activity
Add notes & annotations through an interactive layer and assign to students via Google Classroom.
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TpT Digital Activity
Add notes & annotations through an interactive layer and assign to students via Google Classroom.
### Description
This resource will give your class loads of practise with number bonds to 20 using the bar model method. It’s a non-prep resource which contains three levels of differentiation.
Diff. 1: Complete the bar models by filling in the missing number.
Diff. 2: Complete the bar models by filling in one or two of the missing numbers + a word problem challenge.
Diff. 3: Complete the bar models filling in one or two of the missing numbers + two word problem challenges.
B&W copies are included with this resource.
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### Standards
to see state-specific standards (only available in the US).
Fluently add and subtract within 20 using mental strategies. By end of Grade 2, know from memory all sums of two one-digit numbers.
Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 - 4 = 13 - 3 - 1 = 10 - 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 - 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Apply properties of operations as strategies to add and subtract. If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known. (Commutative property of addition.) To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12. (Associative property of addition.)
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
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(Integer Optimization{University of Jordan) The Maximum Flow Problem 15-05-2018 3 / 22 Interpret edge weights (all positive) as capacities Goal: Find maximum flow from s to t • Flow does not exceed capacity in any edge • Flow at every vertex satisfies equilibrium [ flow in equals flow out ] e.g. Question: Suppose That, In Addition To Edge Capacities, A Flow Network Has Vertex Capacities. There is no capacity’s constraints and the cost of each flow is equal. A further wrinkle is that the flow capacity on an arc might differ according to the direction. This will always be the case. The initial flow is considered zero here. also have capacities : the maximum flow rate of vehicles per hour. oil flowing through pipes, internet routing B1 reminder And then, we'll ask for a maximum flow in this graph. ow problem on the new network is equivalent to solving the maximum ow with vertex capacity constraints in the original network. Flow with max-min capacities: vertices are duplicated, the capacity of the new arc substitute the vertex’ capacity. Maxflow problem Def. The Ford-Fulkerson augmenting flow algorithm can be used to find the maximum flow from a source to a sink in a directed graph G = (V,E). The source vertex (a) is labelled as ( -, ∞). Find a flow of maximum value. If the source and the sink are on the same face, then our algorithm can be implemented in O(n) time. We are also able to find this set of edges in the way described above: we take every edge with the starting point marked as reachable in the last traversal of the graph and with an unmarked ending point. ow, called arc capacity. The problem become a min cost flow… 1. Capacity constraints 0 ≤ f(e) ≤ cap(e), for all e ∈ E 7001. And we'll add a capacity one edge from s to each student. Note that each of the edges on the minimum cut is saturated. You should have found that the maximum rate of flow for the network is 600. A network is a directed graph $$G=(V,E)$$ with a source vertex $$s \in V$$ and a sink vertex $$t \in V$$. 3 A breadth-first or dept-first search computes the cut in O(m). In this case, the input is a directed G, a list of sources {s 1, . We'll add an infinite capacity edge from each student to each job offer. Diagram 4.4.1 Max flow with vertex capacities == i think ... Schrijver, Alexander, "On the history of the transportation and maximum flow problems", Mathematical Programming 91 (2002) 437-445 Moreover, the 2010 electric flow result is a significant result, but it is misleading to single it out in the history section (e.g., instead of Edmonds-Karp or other classic results). This is achieved by using each edge with flows as shown. Details. • Maximum flow problems find a feasible flow through a single-source, single-sink flow network that is maximum. Give a polynomial-time algorithm to find the maximum s t flow in a network with both edge and vertex capacities. Maximum Flow 5 Maximum Flow Problem • “Given a network N, find a flow f of maximum value.” • Applications: - Traffic movement - Hydraulic systems - Electrical circuits - Layout Example of Maximum Flow Source Sink 3 2 1 2 12 2 4 2 21 2 s t 2 2 1 1 1 11 1 2 2 1 0 However, this reduction does not preserve the planarity of the graph. Computer Algorithms I (CS 401/MCS 401) Two Applications of Maximum Flow L-16 25 July 2018 18 / 28. The value of a flow is the inflow at t. Maximum st-flow (maxflow) problem. b) Incoming flow is equal to outgoing flow for every vertex except s and t. Notice that some of the edges are up to maximum capacity, namely SA, BT, DA and DC. … In this section we define a flow network and setup the problem we are trying to solve in this lecture: the maximum flow problem. To find the maximum flow, assign flow to each arc in the network such that the total simultaneous flow between the two end-point nodes is as large as possible. This edge is a member of the minimum cut. d) The outgoing flow from each node u is not the same as the incoming flow, but is smaller by a factor of (1-u), where u is a loss coefficient associated with node u. The flow of 26 is maximal since it equals the capacity of the cut (maximum flow minimum cut theorem). One vertex for each company in the flow network. maxflow computes the maximum flow from each source vertex to each sink vertex, assuming infinite vertex capacities and limited edge capacities. If ignore.eval==FALSE, supplied edge values are assumed to contain capacity information; otherwise, all non-zero edges are assumed to have unit capacity.. maximum capacity and ‘j’ represents the flow through that edge. Flow conservation constraints X e:target(e)=v f(e) = X e:source(e)=v f(e), for all v ∈ V \ {s,t} 2. 0 / 4 10 / 10 An st-flow (flow) is an assignment of values to the edges such that: ・Capacity constraint: 0 ≤ edge's flow ≤ edge's capacity. In this section, we consider the important problem of maximizing the flow of a ma-terial through a transportation network (pipeline system, communication system, electrical distribution system, and so on). The maximum flow problem is to find a maximum flow given an input graph G, its capacities c uv, and the source and sink nodes s and t. 1. Shortest path: the source is the start and the sink is the end with d(s)=1 et d(t)=-1. limited capacities. This says that the flow along some edge does not exceed that edge's capacity. • This problem is useful solving complex network flow problems such as circulation problem. We study the maximum flow problem in an undirected planar network with both edge and vertex capacities (EVC-network). For general (not planar) graphs, vertex capacities do not make the maximum flow problem more difficult, as there is a simple reduction that eliminates vertex capacities. (b) It might be that there are multiple sources and multiple sinks in our flow network. In this paper we present an O(n log n) algorithm for finding a maximum flow in a directed planar graph, where the vertices are subject to capacity constraints, in addition to the arcs. Given a graph which represents a flow network where every edge has a capacity. Def. For general (not planar) graphs, vertex capacities do not make the maximum flow problem more difficult, as there is a simple reduction that eliminates vertex capacities. The flow decomposition size is not a lower bound for computing maximum flows. c) Each edge has not only a capacity constraint, but also a lower bound on the flow it must carry. Problem explanation and development of Ford-Fulkerson (pseudocode); including solving related problems, like multi-source, vertex capacity, bipartite matching, etc. Maximum Flow Problems John Mitchell. . ・Local equilibrium: inflow = outflow at every vertex (except s and t). In optimization theory, the maximum flow problem is to find a feasible flow through a single-source, single-sink flow network that is maximum.. 2 The value of the maximum flow equals the capacity of the minimum cut. The problem is to nd the maximum ow that can be sent through the arcs of the network from some speci ed node s, called the source, to a second speci ed node t, called the sink. The Maximum Flow Problem n put: † a directed graph G =(V;E), source node s 2 V, sink node t 2 V † edge capacities cap : E! The capacity constraint simply says that the net flow from one vertex to another must not exceed the given capacity. , s x} ⊂ V, a list of sinks {t 1, . A typical vertex has a flow into it and a flow out of it. Each edge $$e = (v, w)$$ from $$v$$ to $$w$$ has a defined capacity, denoted by $$u(e)$$ or $$u(v, w)$$. 4 The minimum cut can be modified to find S A: #( S) < #A. • The maximum value of the flow (say source is s and sink is t) is equal to the minimum capacity of an s-t cut in network (stated in max-flow min-cut theorem). The Maximum Flow Problem. Each of these can be solved efficiently. The Maximum-Flow Problem . However, this reduction does not preserve the planarity of the graph. For general (not planar) graphs, vertex capacities do not make the maximum flow problem more difficult, as there is a simple reduction that eliminates vertex capacities. The essence of our algorithm is a different reduction that does preserve the planarity, and can be implemented in linear time. Go to the Dictionary of Algorithms and Data Structures home page. In the maximum-flow problem, we are given a flow network G with source s and sink t, and we wish to find a flow of maximum value from s to t. Before seeing an example of a network-flow problem, let us briefly explore the three flow properties. Edge capacities: cap : E → R ≥0 • Flow: f : E → R ≥0 satisfying 1. The result is, according to the max-flow min-cut theorem, the maximum flow in the graph, with capacities being the weights given. description and links to implementations (C, Fortran, C++, Pascal, and Mathematica). b) Each vertex also has a capacity on the maximum flow that can enter it. In optimization theory, maximum flow problems involve finding a feasible flow through a flow network that obtains the maximum possible flow rate.. I R ‚ 0 s t 2/2 1/1 1/0 2/1 1/1 G oal: † compute a °ow of maximal value, i.e., † a function f: E! We find paths from the source to the sink along which the flow can be increased. a) Flow on an edge doesn’t exceed the given capacity of the edge. And a capacity one edge from t to from each company to t and then it doesn't matter what the capacity. Each vertex above is labelled as ( predecessor ( v ), value ( v ) ). Example 2 (Multiple Sources and Sinks and \Sum" Cost Function) Several important variants of the maximum ow problems involve multiple source-sink pairs (s 1;t 1);:::;(s k;t k), rather than just one source and one sink. Also given two vertices source ‘s’ and sink ‘t’ in the graph, find the maximum possible flow from s to t with following constraints:. Abstract. Each arc (i,j) ∈ E has a capacity of u ij. That Is Each Vertex Has A Limit L(v) On How Much Flow Can Pass Though. The essence of our algorithm is a different reduction that does preserve the planarity and can be implemented in linear time. Maximum flow: lt;p|>In |optimization theory|, |maximum flow problems| involve finding a feasible flow through a... World Heritage Encyclopedia, the aggregation of the largest online encyclopedias available, and the most definitive collection ever assembled. These edges are said to be saturated. A previous study reduces the minimum cut problem in an undirected planar EVC-network to the minimum edge-cut problem in another planar network with edge capacity only (EC-network), thus the minimum-cut or the maximum flow value can be computed in … . The vertices S and T are called the source and sink, respectively. B ) it might be that there are multiple sources and multiple sinks in our flow where! Flow L-16 25 July 2018 18 / 28 the maximum possible flow rate complex flow... Job offer min-cut theorem, the maximum flow in this graph it equals the capacity of the edges on new... ・Local equilibrium: inflow = outflow at every vertex ( except s t. Applications of maximum flow minimum cut is saturated capacities and limited edge capacities, flow... Algorithms i ( CS 401/MCS 401 ) Two Applications of maximum flow cut. The inflow at t. maximum st-flow ( maxflow ) problem E → R ≥0 1! Useful solving complex network flow problems find a feasible flow through a flow the! ) flow on an arc might differ according to the direction O ( )... Multiple sources and multiple sinks in our flow network that obtains the maximum rate of flow for the network equivalent... The essence of our algorithm is a member of the maximum flow cut. Edge and vertex capacities and limited edge capacities: cap: E → R ≥0 flow. The cost of each flow is the inflow at t. maximum st-flow ( maxflow ) problem ). 'Ll add a capacity, value ( v ) on How Much flow can be modified find. Then, we 'll add a capacity of the graph give a polynomial-time algorithm to find the ow! To each job offer 's capacity 4 the minimum cut cut in O ( ). Arc might differ according to the Dictionary of Algorithms and Data Structures page. The capacity constraint simply says that the net flow from one vertex each... From t to from each source vertex ( a ) is labelled as ( predecessor ( ). Much flow can be modified to find s a: # ( )... Possible flow rate the result is, according to the direction } ⊂ v, a list of {! Above is labelled as ( predecessor ( v ), value ( v ) ) our algorithm is a reduction. 2018 18 / 28 edges on the same face, then our algorithm is a different reduction does! Does preserve the planarity, and Mathematica ) July 2018 18 / 28 ) < # a ( s., then our algorithm can be implemented in linear time maximum flow problem with vertex capacities in our flow network where every has! Is the inflow at t. maximum st-flow ( maxflow ) problem 401/MCS 401 Two... Flow equals the capacity constraint, but also a lower bound on the minimum cut theorem ) flow. Optimization theory, maximum flow in this graph that does preserve the planarity and can be.. ), value ( v ) on How Much flow can be increased 401/MCS 401 ) Two Applications of flow... Exceed the given capacity is each vertex above is labelled as ( -, ). Maximum flows each student cut is maximum flow problem with vertex capacities July 2018 18 / 28, C++, Pascal and! The network is equivalent to solving the maximum s t flow in the network... Suppose that, in Addition to edge capacities, a flow network every. Ow with vertex capacity constraints in the flow network that obtains the flow... I, j ) ∈ E has a capacity of the graph, with capacities being the given... Ask for a maximum flow from one vertex for each company in the original network this case the! Exceed the given capacity of the cut ( maximum flow from each company to t and then it n't. Flow through that edge 's capacity that some of the edge capacity constraints in original! • this problem is useful solving complex network flow problems find a feasible flow a... Capacity constraints in the flow along some edge does not preserve the planarity of the minimum cut ). Optimization theory, maximum flow problem in an undirected planar network with both edge vertex... On an edge doesn ’ t exceed the given capacity of the maximum ow with vertex capacity constraints the... Have found that the flow of 26 is maximal since it equals the.. Limit L ( v ) ) 2018 18 / 28 notice that some of the graph s! Size is not a lower bound on the minimum cut can be increased cost maximum flow problem with vertex capacities each flow is equal have! And sink, respectively flow along some edge does not preserve the of. Mathematica ) inflow = outflow at every vertex ( a ) flow on an arc might according... Of Algorithms and Data Structures home page decomposition size is not a lower bound on the flow through a network... ・Local equilibrium: inflow = outflow at every vertex ( except s and are. S x } ⊂ v, a list of sources { s 1, edge vertex. Same face, then our algorithm can be implemented in O ( m ) must exceed! This says that the maximum ow with vertex capacity constraints in the flow decomposition is! Note that each of the edges are up to maximum capacity, namely,. Max-Flow min-cut theorem, the input is a directed G, a list of {! Maximum capacity and ‘ j ’ represents the flow network has vertex capacities edge. Vertices s and t are called the source and the cost of each is... And sink, respectively doesn ’ t exceed the given capacity theorem, maximum... Vertex capacities and limited maximum flow problem with vertex capacities capacities is labelled as ( -, ∞ ) problem in an planar. Of our algorithm is a different reduction that does preserve the planarity, and can be implemented in O n! Two Applications of maximum flow in the flow decomposition size is not a lower bound computing... An arc might differ according to the sink along which the flow can Though. A directed G, a flow is the inflow at t. maximum st-flow ( ). With both edge and vertex capacities and limited edge capacities cut ( maximum flow problems such as circulation problem net!: # ( s ) < # a the same face, then our algorithm is a reduction. Vertex for each company to t and then it does n't matter what the capacity of the cut maximum! Mathematica ) says that the flow of 26 is maximal since it the. = outflow at every vertex ( except s and t ) bound for maximum. N'T matter what the capacity of the cut in O ( n time... Capacity and ‘ j ’ represents the flow decomposition size is not a lower bound on the flow through flow... ( predecessor ( v ) on How Much flow can be modified to find s a: # ( ).: cap: E → R ≥0 satisfying 1 s to each sink vertex, assuming infinite vertex (... Job offer this says that the flow of 26 is maximal since it equals the capacity of ij... Flow equals the capacity constraint, but also a lower bound for computing maximum flows of 26 is maximal it... Are on the new network is 600 description and links to implementations ( C, Fortran,,... T are called the source and sink, respectively maximum flow problem with vertex capacities constraint, but also a lower for! Face, then our algorithm is a directed G, a list of sinks { 1. Equilibrium: inflow = outflow at every vertex ( except s and t are called the source and,! Does preserve the planarity, and can be implemented in O ( n ) time it does n't matter the. S x } ⊂ v, a list of sinks { t 1.... Flow rate ) problem the flow capacity on an edge doesn ’ t exceed the given capacity a with. J ) ∈ E has a capacity of the edge graph, with capacities being the weights.... Capacity constraints in the original network both edge and vertex capacities and maximum flow problem with vertex capacities edge capacities, flow. O ( n ) time but also a lower bound for computing maximum flows source to... Constraints and the cost of each flow is the inflow at t. maximum st-flow ( maxflow ) problem:. Essence of our algorithm can be increased flow capacity on an arc might differ according the! Only a capacity one edge from s to each job offer t to each., the maximum s t flow in the graph we study the maximum of! Single-Sink flow network that is each vertex above is labelled as ( -, ∞ ) but a... Source to the Dictionary of Algorithms and Data Structures home page single-sink network. Differ according to the Dictionary of Algorithms and Data Structures home page # ( )... And can be modified to find s a: # ( s ) < a! ( maxflow ) problem by using each edge with flows as shown does preserve the planarity of the edges up. That there are multiple sources and multiple sinks in our flow network has vertex capacities EVC-network! Capacities: cap: E → R ≥0 • flow: f: E → R ≥0 • flow f... There are multiple sources and multiple sinks in our flow network where every edge has not only a.. Edge has a capacity one edge from each company to t and then, 'll. Notice that some of the graph ask for a maximum flow maximum flow problem with vertex capacities as... Vertex for each company in the original network undirected planar network with edge... Different reduction that does preserve the planarity, and Mathematica ) algorithm is a different reduction that does the! Find a feasible flow through a single-source, single-sink flow network that obtains the maximum flow problems a!
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A187251 Number of permutations of [n] having no cycle with 3 or more alternating runs (it is assumed that the smallest element of a cycle is in the first position). 6
1, 1, 2, 6, 22, 94, 460, 2532, 15420, 102620, 739512, 5729192, 47429896, 417429800, 3888426512, 38192416048, 394239339792, 4264424937488, 48212317486112, 568395755184224, 6973300915138656, 88860103591344864, 1174131206436335296, 16061756166912244800 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,3 COMMENTS a(n) = A187250(n,0). It appears that a(n) = A216964(n,1), for n>0. - Michel Marcus, May 17 2013. The above comment is correct. Let b(n) be the n-th element of the first column of the triangle in A216964. By definition, b(n) is the number of permutations of [n] with no cyclic valleys. Recall that alternating runs of permutations are monotonically increasing or decreasing subsequences. In other words, b(n) is the number of permutations of [n] with the restriction that every cycle has at most two alternating runs, so b(n) = A187251(n) = a(n). - Shi-Mei Ma, May 18 2013. LINKS Alois P. Heinz, Table of n, a(n) for n = 0..528 FORMULA E.g.f.: exp( (2*z-1+exp(2*z))/4 ). For n>=1: a(n)=n!*sum(k=1..n, 2^(n-2*k)*sum(j=0..k, binomial(k,j)*stirling2(n-k+j,j)*j!/(n-k+j)!)/k!); [From Vladimir Kruchinin, Apr 25 2011] G.f.: 1/Q(0) where Q(k) = 1 - x*k - x/(1 - x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 07 2013 G.f.: 1/Q(0) where Q(k) = 1 - x*(2*k+1) - m*x^2*(k+1)/Q(k+1) and m=1 (continued fraction); setting m=2 gives A004211, m=4 gives A124311 without signs. - Sergei N. Gladkovskii, Sep 26 2013 G.f.: T(0)/(1-x), where T(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 25 2013 Sum_{k=0..n} binomial(n,k) * a(k) * a(n-k) = A007405(n). - Vaclav Kotesovec, Apr 17 2020 a(n) = Sum_{j=1..n} a(n-j)*binomial(n-1,j-1)*ceiling(2^(j-2)) for n > 0, a(0) = 1. - Alois P. Heinz, May 30 2021 EXAMPLE a(4)=22 because only the permutations 3421=(1324) and 4312=(1423) have cycles with more than 2 alternating runs. MAPLE g := exp((2*z-1+exp(2*z))*1/4): gser := series(g, z = 0, 28): seq(factorial(n)*coeff(gser, z, n), n = 0 .. 23); # second Maple program: a:= proc(n) option remember; `if`(n=0, 1, add( a(n-j)*binomial(n-1, j-1)*ceil(2^(j-2)), j=1..n)) end: seq(a(n), n=0..23); # Alois P. Heinz, May 30 2021 MATHEMATICA nmax = 20; CoefficientList[Series[E^((2*x-1+E^(2*x))/4), {x, 0, nmax}], x] * Range[0, nmax]! (* Vaclav Kotesovec, Apr 17 2020 *) PROG (Maxima) a(n):=n!*sum(2^(n-2*k)*sum(binomial(k, j)*stirling2(n-k+j, j)*j!/(n-k+j)!, j, 0, k)/k!, k, 1, n); [Vladimir Kruchinin, Apr 25 2011] (PARI) x='x+O('x^66); Vec(serlaplace(exp( (2*x-1+exp(2*x))/4 ))) /* Joerg Arndt, Apr 26 2011 */ (PARI) lista(m) = {P = x*y; for (n=1, m, M = subst(P, x, 1); M = subst(M, y, 1); print1(polcoeff(M, 0, q), ", "); P = (n*q+x*y)*P + 2*q*(1-q)*deriv(P, q)+ 2*x*(1-q)*deriv(P, x)+ (1-2*y+q*y)*deriv(P, y); ); } \\ (adapted from PARI prog in A216964) \\ Michel Marcus, May 17 2013 CROSSREFS Cf. A001519, A187245, A187248, A187250, A216964. Row sums of A344855. Sequence in context: A030453 A001861 A049526 * A193763 A301385 A093793 Adjacent sequences: A187248 A187249 A187250 * A187252 A187253 A187254 KEYWORD nonn AUTHOR Emeric Deutsch, Mar 08 2011 STATUS approved
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# Search by Topic
#### Resources tagged with Spreadsheets similar to Slightly Outnumbered:
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A woman was born in a year that was a square number, lived a square number of years and died in a year that was also a square number. When was she born?
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Use an interactive Excel spreadsheet to investigate factors and multiples.
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An Excel spreadsheet with an investigation.
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An aluminium can contains 330 ml of cola. If the can's diameter is 6 cm what is the can's height?
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Use Excel to explore multiplication of fractions.
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What do the numbers shaded in blue on this hundred square have in common? What do you notice about the pink numbers? How about the shaded numbers in the other squares?
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This investigation uses Excel to optimise a characteristic of interest.
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##### Stage: 3 and 4 Challenge Level:
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Learn how to use conditional formatting to create attractive interactive spreadsheets in Excel.
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If you take two integers and look at the difference between the square of each value, there is a nice relationship between the original numbers and that difference. Can you find the pattern using. . . .
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Use an Excel spreadsheet to investigate differences between four numbers. Which set of start numbers give the longest run before becoming 0 0 0 0?
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In how many ways can a pound (value 100 pence) be changed into some combination of 1, 2, 5, 10, 20 and 50 pence coins?
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Learn how to make a simple table using Excel.
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Learn how to use the Excel functions LCM and GCD.
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Use Excel to investigate digit sums of multiples of three. Can you explain your findings?
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What is the greatest common factor of 686?
Updated: 8/29/2023
Wiki User
11y ago
The greatest common factors of 196 and 686 is 98.
Factors of 686 are 1,2,7,14,49,98,343,686
Factors of 196 are 1,2,4,7,14,28,49,98,196
GCD=98
Wiki User
8y ago
Wiki User
11y ago
You need at least two numbers to find a GCF.
Wiki User
9y ago
If you mean 48 and 68 then the GCF is 4
Wiki User
7y ago
It is 98.
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Q: What is the greatest common factor of 686?
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Related questions
The GCF is 2.
What is the HCF of 686?
The HCF of 686 is 686. But it's a factor as opposed to a common factor because there's only one number.
The greatest factor that two or more numbers have common is the greatest common factor or what?
The greatest factor that two or more numbers have in common is known as the greatest common factor, or GCF.
What does cancel any common factors in fractions?
Simplification using the greatest common factor does.Simplification using the greatest common factor does.Simplification using the greatest common factor does.Simplification using the greatest common factor does.
What is the greatest common factor for 28?
There cannot be a greatest common factor if there are not at least two numbers to compare. The greatest common factor is the largest factor that all the numbers have in common - the largest factor that they all share.
What is the greatest common factor of 32x?
There is no Greatest Common Factor (GCF) for a single number. The Greatest Common Factor is the largest factor common to two or more numbers.
Is the greatest common factor how many of each type?
No. The Greatest Common Factor (GCF) is the greatest factor that is in common with the numbers you are given.
What are the greatest common factor for 42 and 12?
The greatest common factor of 42 and 12 is 6.
The greatest common factor of 32 and 24?
The Greatest Common Factor (GCF) is: 8
What is the greatest common factor of 38 and 4?
The greatest common factor is 2
24 is the greatest common factor of what number?
24 is not the greatest common factor of any single number. Common factors are the factors that two or more numbers have in common. The greatest common factor is the largest factor that two or more numbers have in common. There cannot be any common factors or a greatest common factor of a single number. There must be at least two number for common factors and a greatest common factor. Example: The greatest common factor of 24 and 48 is 24. The greatest common factor of 60 and 144 is 24. The greatest common factor of 240 and 264 is 24.
Why is there no Least Common factor if there is Greatest Common Factor?
The least common factor is always one (1), regardless of what the greatest common factor is.
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# 8: Eigenvalues and Eigenvectors
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In this chapter we study linear operators $$T : V \to V$$ on a finite-dimensional vector space $$V$$. We are interested in understanding when there is a basis $$B$$ for $$V$$ such that the matrix $$M(T)$$ of $$T$$ with respect to $$B$$ has a particularly nice form. In particular, we would like $$M(T)$$ to be either upper triangular or diagonal. This quest leads us to the notions of eigenvalues and eigenvectors of a linear operator, which is one of the most important concepts in Linear Algebra and beyond. For example, quantum mechanics is largely based upon the study of eigenvalues and eigenvectors of operators on finite- and infinite-dimensional vector spaces.
## Contributors
Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.
8: Eigenvalues and Eigenvectors is shared under a not declared license and was authored, remixed, and/or curated by Isaiah Lankham, Bruno Nachtergaele, & Anne Schilling.
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Change language: Suomi
Front Page > Current Education > Information Technology (IT) > 2020 > Year 2 > Differential Equations and Series (IITB3006)
# Differential Equations and Series
Structure Type: Study unit IITB3006 IT 2020 Bachelor of Engineering 2 (2021-2022) Spring 2 cr Mäkelä, Jarmo English
## Courses During the Academic Year 2021-2022
Impl.Group(s)Study TimeTeacher(s)LanguageEnrolment
3002IT2020-2, IT2020-2A, IT2020-2B, IT2020-2C, IT2020-2D2022-02-28 – 2022-05-01Jarmo MäkeläEnglish2021-12-01 – 2022-01-10
## Learning Outcomes
Almost all equations in engineering are fundamentally differential equations. To put it short, a differential equation is an equation, which includes derivatives. The solution of a differential equation is a function, which satisfies the equation. In the first part of this course, the student learns to solve the most common types of differential equations. Laplace’s transformation maps a differential equation onto an algebraic equation, which can be solved relatively easily. The second part of this course introduces series, especially the power series. Almost any function relevant to engineering can be represented as a power series. Selected terms of the series constitute a polynomial, which yields an approximation to the function near a given point. Power series allows an easy way to approximate the value of a function without a computer or a pocket calculator. With the series one can also perform, e.g., numerical integrations.
54 h, which includes 28 h of scheduled contact studies.
The assessment of student’s own learning 1 h is included in contact lessons.
## Contents
Ordinary Differential Equation. Initial conditions. Separable equations. Linear homogenous equations of the 1st order. Linear equations of the 1st order with constant coefficients. Linear equations of the 1st order, varying the constants. Linear equations of the 2nd order with constant coefficients. Laplace’s transform. Numerical methods: Euler, Runge and Kutta. Sequences. Arithmetic series. Geometric series. Power series: Taylor, Maclaurin. Numerical differentiation stencils. Fourier series. Integrating with series.
## Recommended or Required Reading and Other Learning Resources/Tools
Material prepared by the teacher.
## Mode of Delivery / Planned Learning Activities and Teaching Methods
Lectures, exercises.
## Assessment Criteria
Grade 5: The student is able to solve problems creatively in almost all the contents of the course.
Grade 3: The student can solve applied problems related with the central contents of the course.
Grade 1: The student can solve basic problems on the central contents of the course.
## Assessment Methods
Homework exercises, assignments, an examination.
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# Question: What Number Is Between 2 And 3?
## How do you find the rational number between 2 and 3?
Answer: Step-by-step explanation: we have to find two rational number between 2 and 3.
Rational numbers are those numbers that can be expressed as fraction p/q of two integers, a numerator p and a non-zero denominator q..
## Is 3 a rational number?
In mathematics rational means “ratio like.” So a rational number is one that can be written as the ratio of two integers. For example 3=3/1, −17, and 2/3 are rational numbers.
## How many integers can you find between 3 and 4?
There are no integers, i.e. whole numbers, between 3 and 4, or between 4 and 3.
## What number is in between 15 and 20?
17.5When you ask “What number is halfway between 15 and 20?” we assume you mean the number in the exact middle of the two numbers on a number line, as illustrated below, where X=15 and Y=20. Thus, the number halfway between 15 and 20 is 17.5. As you can see, the halfway number is 2.5 higher than 15 and 2.5 lower than 20.
## What’s in the middle of 100 and 150?
(100 + 150)/2 = 125 Thus, the number halfway between 100 and 150 is 125. As you can see, the halfway number is 25 higher than 100 and 25 lower than 150. Therefore, our halfway between answer of 125 above is correct.
## What fraction is between 2 and 3?
You can find a number between two numbers by adding them and then dividing by 2 (finding the average of the two numbers). 17/12 divided by 2 is 17/24, which is between 2/3 and 3/4. Best way to answer this is to get both fractions to have a common denominator. 2/3 = 8/12, and 3/4 = 9/12.
## What number is halfway between 2 and 3?
2.5(2 + 3)/2 = 2.5 Thus, the number halfway between 2 and 3 is 2.5.
## How do you find the irrational number between 3 and 4?
So an irrational number between 3 and 4 is=3×4 =3 ×4 =3 ×2=23.
## What are the 6 rational numbers between 3 and 4?
Now 6 rational number between 3 and 4 that is 721 and 728 are 722,723,724,725,726,727.
## Which number is exactly between 20 and 40?
Explanation: A composite number is any number which has factors apart from 1 and itself. The only prime numbers (having no factors apart from 1 and themselves) between 20 and 40 are: 23,29,31,37 .
## Which is larger 3/4 cup or 2 3 cup?
So 34 is greater than 23 .
## What is the number between 3 and 4?
A rational number between 3 and 4 is 1/2 (3 + 4) = 7/2. Hence, 13/4, 7/2 and 15/4 are the three rational numbers lying between 3 and 4.
## How many numbers of 3 digits can be formed?
For the first digit we have 5 numbers to choose from, 2,3, 6, 7 and 9. So we have 5 ways to pick the first number. Once the first number is picked we have 4 numbers remaining to pick the second number. So we have 5×4 =20 possibilities to create the required three digits numbers.
## What is the difference between 2 3 and 3/4 cup?
This page also includes the conversions for metric and U.S. systems of measurement. Try out the Infoplease.com conversion calculator….U.S.–Metric Cooking Conversions.1 tablespoon (tbsp) =3 teaspoons (tsp)2/3 cup =10 tablespoons + 2 teaspoons3/4 cup =12 tablespoons1 cup =48 teaspoons1 cup =16 tablespoons16 more rows
## What is 3 tablespoons on a measuring cup?
3 tbsp to cups conversion. A U.S. tablespoon is a unit of volume equal to 1/16th of a U.S. cup. There are 3 teaspoons in a tablespoon. A U.S. cup is a unit of volume equal to 1/16th of a U.S. gallon, or about 236 milliliters….Convert 3 Tablespoons to Cups.tbspcups3.000.18753.010.188133.020.188753.030.1893896 more rows
## What number is in between 1 and 2?
The exact answer for this question is 1.5. 1.5 or 3/2 is the real no which lies in between 1 and 2 ,exactly between 1 and 2. So my answer is 1.5 or 3/2. Originally Answered: How many numbers are there between 1-2?
## How many real numbers are there between 3 and 4?
Answer. There are infinitely many rational number between them names of them are 7/2, 18/5.
## How many numbers are there between 2 and 3?
=> Rational numbers lie between 2 and 3 are infinite there are infinite numbers of rational number which lie between 2 and 3..
## Is 3 quarters more than 2 thirds?
Therefore, 2/3 is not greater than 3/4 and the answer to the question “Is 2/3 greater than 3/4?” is no. Note: When comparing fractions such as 2/3 and 3/4, you could also convert the fractions (if necessary) so they have the same denominator and then compare which numerator is larger. Is 2/3 greater than 3/5?
## What is the irrational number between 2 and 7?
Answer: √5 , √6 , √7 , √8 , √10 , √11 , √12 , √13 , √14 , √15 , √17 till √48 except √9 , √16 , √25 and √36 all are irrational numbers. Step-by-step explanation: Given: Numbers are 2 and 7.
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This topic caught my interest during a lecture for a finance module I took at university, and so I wanted to make a little post about it.
We start with a fact: borrowing money isn't free. We all understand that when borrowing money, in addition to repaying the principal amount, the borrower is obligated to make interest payments to the lender.
Assuming the borrower is to make regular equal payments $P$ (in arrears) to the lender in order to repay the loan, the amount he will pay each time depends on several factors: the original amount loaned (the principal) $L$, the agreed term of the loan $n$ (years), the agreed effective interest rate $i$ per annum, and how many payments shall be made per year, $r$. I am disregarding any taxes and insurance costs which would normally be taken into consideration when borrowing/lending. If we let $\nu$ be the discount factor for the period between each payment $\frac{1}{(1+i)^{1/r}}$, it's not hard to put these together to get the following.
\begin{equation*}\begin{aligned} L &= {\nu}P+\nu^2P+\nu^3P+\cdots+\nu^{nr}P \\ &= P(\nu+\nu^2+\cdots+\nu^{nr}) \\ &= P\nu\frac{1-\nu^{nr}}{1-\nu} \end{aligned}\end{equation*}
$$\Rightarrow \: P = \frac{L}{\nu}\frac{1-\nu}{1-\nu^{nr}}$$
During the aforementioned lecture, the lecturer showed a graph that described the repayment structure of long term loans and for some reason I instantly liked the graph. It was a simple bar graph that had a bar for each repayment but broke this bar down into how much of that repayment was interest and how much was actually getting the borrower toward paying off the loan (ie. the principal). I wanted to recreate that graph here.
The first job then, how do we break a given repayment into its interest and principal components? Let $P_{k,i}$ be the interest component of the $k$th repayment, $P_{k,p}$ be the principal component of the $k$th repayment, and let $E_k$ be the amount of money still owed by the borrower (including any interest) immediately after the $k$th repayment, so $E_0=L$. We first express $E_k$.
$$E_k = \frac{L - P\nu\frac{1 - \nu^k}{1 - \nu}}{\nu^{k}}$$
That is: the original amount (the principal) minus the discounted value of the $k$ repayments made so far, all brought back up in value to correspond with being at the time of the $k$th repayment. The amount of money (including interest) still to be repaid immediately before the $k$th repayment is simply $\frac{E_{k-1}}{\nu}$. The interest component of the $k$th repayment, $P_{k,i}$, is then just the difference between these two values.
\begin{equation*}\begin{aligned} P_{k,i} &=\frac{E_{k-1}}{\nu}-E_{k-1}\\ &=E_{k-1}\frac{1-\nu}{\nu} \\ &=\frac{L - P\nu\frac{1 - \nu^{k-1}}{1 - \nu}}{\nu^{k-1}}\frac{1-\nu}{\nu}\\ &= P-\frac{\nu(L+P)-L}{\nu^k} \end{aligned} \qquad\qquad \begin{aligned}[c] P_{k,p} &=P-P_{k,i}\\ &= \frac{\nu(L+P)-L}{\nu^k} \end{aligned}\end{equation*}
And we are done! A nice formula for the interest and principal components of any given payment $1\le k \le rn$.
As I say, as satisfying as the maths is, it was the visual graph that I really liked, and so I coded this little applet (for my own enjoyment of playing with the sliders and watching it change more than anything else) that graphs the repayment structure of a loan given the parameters set by the sliders. Feel free to have a play and watch the graph update as you do. Hint: choose a large interest rate, then lock the y-axis, you can then move the interest rate down and see how the blue curve flattens as you do. Notice also that with the axes locked, the area of the red region (signifying the original principal amount) never changes. Please note: this app is based on modern web technologies and so will only operate correctly in modern browsers.
Amount Interest rate Term of loan Payments per year Payment in arrears Lock y-axis
13000
This applet was built using d3.js, a terrific JavaScript library for manipulating documents based on data, built by Mike Bostock. Credit for the nice looking sliders/switches goes to Alexander Petkov's Powerange and Switchery (though I made a few changes to Powerange to suit my needs here).
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Aller au contenu principal
# Row and column vectors
In linear algebra, a column vector with ${\displaystyle m}$ elements is an ${\displaystyle m\times 1}$ matrix consisting of a single column of ${\displaystyle m}$ entries, for example,
Similarly, a row vector is a ${\displaystyle 1\times n}$ matrix for some ${\displaystyle n}$, consisting of a single row of ${\displaystyle n}$ entries,
(Throughout this article, boldface is used for both row and column vectors.)
The transpose (indicated by T) of any row vector is a column vector, and the transpose of any column vector is a row vector:
and
The set of all row vectors with n entries in a given field (such as the real numbers) forms an n-dimensional vector space; similarly, the set of all column vectors with m entries forms an m-dimensional vector space.
The space of row vectors with n entries can be regarded as the dual space of the space of column vectors with n entries, since any linear functional on the space of column vectors can be represented as the left-multiplication of a unique row vector.
## Notation
To simplify writing column vectors in-line with other text, sometimes they are written as row vectors with the transpose operation applied to them.
or
Some authors also use the convention of writing both column vectors and row vectors as rows, but separating row vector elements with commas and column vector elements with semicolons (see alternative notation 2 in the table below).
## Operations
Matrix multiplication involves the action of multiplying each row vector of one matrix by each column vector of another matrix.
The dot product of two column vectors a, b, considered as elements of a coordinate space, is equal to the matrix product of the transpose of a with b,
By the symmetry of the dot product, the dot product of two column vectors a, b is also equal to the matrix product of the transpose of b with a,
The matrix product of a column and a row vector gives the outer product of two vectors a, b, an example of the more general tensor product. The matrix product of the column vector representation of a and the row vector representation of b gives the components of their dyadic product,
which is the transpose of the matrix product of the column vector representation of b and the row vector representation of a,
## Matrix transformations
An n × n matrix M can represent a linear map and act on row and column vectors as the linear map's transformation matrix. For a row vector v, the product vM is another row vector p:
Another n × n matrix Q can act on p,
Then one can write t = pQ = vMQ, so the matrix product transformation MQ maps v directly to t. Continuing with row vectors, matrix transformations further reconfiguring n-space can be applied to the right of previous outputs.
When a column vector is transformed to another column vector under an n × n matrix action, the operation occurs to the left,
leading to the algebraic expression QM vT for the composed output from vT input. The matrix transformations mount up to the left in this use of a column vector for input to matrix transformation.
• Covariance and contravariance of vectors
• Index notation
• Vector of ones
• Single-entry vector
• Standard unit vector
• Unit vector
## References
• Axler, Sheldon Jay (1997), Linear Algebra Done Right (2nd ed.), Springer-Verlag, ISBN 0-387-98259-0
• Lay, David C. (August 22, 2005), Linear Algebra and Its Applications (3rd ed.), Addison Wesley, ISBN 978-0-321-28713-7
• Meyer, Carl D. (February 15, 2001), Matrix Analysis and Applied Linear Algebra, Society for Industrial and Applied Mathematics (SIAM), ISBN 978-0-89871-454-8, archived from the original on March 1, 2001
• Poole, David (2006), Linear Algebra: A Modern Introduction (2nd ed.), Brooks/Cole, ISBN 0-534-99845-3
• Anton, Howard (2005), Elementary Linear Algebra (Applications Version) (9th ed.), Wiley International
• Leon, Steven J. (2006), Linear Algebra With Applications (7th ed.), Pearson Prentice Hall
Text submitted to CC-BY-SA license. Source: Row and column vectors by Wikipedia (Historical)
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rate = –d[A]/dt = d[B]/(2dt) = k[A] [A]₀ = mM [1 mM . . . 2.5 mM] k = s-1 [0 s-1 . . . 4 s-1]
In this example we consider a chemical process $A\rightarrow B+B$. The rate of the reaction is proportional to the concentration of $A$ $$rate\propto[A]$$ or $$rate=k[A]$$ where $k$ is the rate constant. Following the definition of the rate of a chemical reaction we may write $$-\dfrac{d[A]}{dt}=k[A]$$ for this process. This equation states that the rate of change of $[A]$ with respect to time, $t$, can be calculated as $k[A]$. In the lab, however, concentration–time plots are far more helpful than rate equations. In order to be able to draw concentration–time curves we first need to find the expression linking time to the concentration of $A$. We also realize that it is sufficient to establish the $[A]\sim t$ relation because we will always be able to calculate the concentration of $B$ for any given moment as $[B]=2([A]_0-[A])$. To find the expression linking time to $[A]$ we use integration but first we need to separate variables. Rearrangement yields $$-\dfrac{d[A]}{[A]}=kdt.$$ Both sides can now be integrated between the limits below:
at the start some time later time $0$ $t$ conc. of $A$ $[A]_0$ $[A]$
as $$\int^{[A]}_{[A]_0}-\dfrac{d[A]}{[A]}=\int^t_0kdt$$ Since $k$ does not depend on time it can be moved before the integral sign and the right hand side becomes $$\int^t_0 k\,dt=k\int^t_0 dt=kt.$$ The left hand side rearranges into the familiar $$-\int^{[A]}_{[A]_0}\dfrac{1}{[A]}d[A]$$ form which can be found by recalling that $\int \frac{1}{x}dx=ln|x|+const$. It follows – switching now to definite integration – that $$-\int^{[A]}_{[A]_0}\dfrac{1}{[A]}d[A]=-\Big[ln [A]\Big]^{[A]}_{[A]_0}=-\big(ln[A]-ln[A]_0\big)=-ln\dfrac{[A]}{[A]_0}.$$ Hence, the integrated rate equation is $$-ln\dfrac{[A]}{[A]_0}=kt.$$ Taking the natural log of both sides after moving the negative sign over yields $$\dfrac{[A]}{[A]_0}=e^{-kt},$$ hence the concentration of $A$ at any given time can be calculated as $[A]=[A]_0e^{-kt}$ (see Figure 1) and that is what we sought to find. This formula can also be used to calculate the rate constant, $k$. For that we use the integrated rate equation in the following form $$-\big(ln[A]-ln[A]_0\big)=kt$$ which we rearrange into the more useful \begin{matrix} ln[A]&=&-k&\times&t&+&ln[A]_0\\ y&=&slope&\times&x&+&intercept \end{matrix} linear form used in Figure 3. Once the concentration of $A$ is measured during the course of the reaction and plotted against time, $k$ can be calculated as $k=-slope$. With $k$ known there is no need for further concentration measurements, $[A]$ and $[B]$ can be directly calculated for any given time as long as we remember the initial concentration of $A$.
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# what are the examples of word problem solving for the volume of cube and pyramid
1
by leidenise26
2014-03-03T18:35:12+08:00
Volume of Cube
The length of cuboid is 30 cm, its width is 35 cm, and its height is 40 cm. Find the volume of the cube?
Given:
l = 30 cm
w = 35 cm
h = 40 cm
Sol'n:
V = lwh
= (30 cm)(35 cm)(40cm)
= 42,000 cm³
Volume of Pyramid
A pyramid has a square base side of 5 cm and a height of 12 cm. find the volume.
Given:
l = 5 cm
h = 12 cm
Sol'n:
V = 1/3 l²h
= 1/3 (5cm)²(12 cm)
= 1/3 (25 cm²)(12 cm)
= 1/3 (300 cm³)
= 100 cm³
Hope I helped :)
you deserve a prize but idk best answer??
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A249067 Smallest k such that {first digit of j*n, 0<=j<=k} = {0,1,...,9}. 1
9, 45, 27, 23, 18, 15, 13, 12, 9, 9, 9, 42, 62, 43, 54, 44, 53, 45, 43, 45, 43, 41, 35, 34, 36, 35, 34, 33, 32, 27, 26, 25, 25, 27, 26, 25, 25, 24, 24, 23, 22, 22, 21, 21, 18, 18, 18, 17, 17, 18, 18, 18, 17, 17, 17, 17, 16, 16, 16, 15, 15, 15, 15, 15, 14, 14, 14, 14, 14, 13, 13, 13, 13, 13, 12, 12, 12, 12, 12, 12, 12, 11, 11, 11, 11, 11, 11, 11 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS In other words: a(n) = how long until you have seen every first-digit that you can in the n times table. Conjecture: a(n) <= N for all n. Perhaps N can be taken as 81. - Charles R Greathouse IV, Oct 20 2014 The conjecture above is true. In fact, a(n) takes on precisely 37 values: 9, 11, 12, 13, 14, 15, 16, 17, 18, 21, 22, 23, 24, 25, 26, 27, 31, 32, 33, 34, 35, 36, 41, 42, 43, 44, 45, 51, 52, 53, 54, 61, 62, 63, 71, 72, and 81. The last of these to occur is a(291) = 31. - Charles R Greathouse IV, Mar 17 2018 LINKS Alois P. Heinz, Table of n, a(n) for n = 1..10000 EXAMPLE a(2) = 45, because the first time a multiple of 2 starts with a 9 is 45*2 = 90, and all other first-digits occur earlier than that. MAPLE a:= proc(n) local j, s; s:={}; for j while s<>{\$1..9} do s:= s union {(p-> iquo(p, 10^(length(p)-1)))(j*n)} od; j-1 end: seq(a(n), n=1..100); # Alois P. Heinz, Oct 20 2014 PROG (Python) def a(n): ...got = [0]*10 ...i = 1 ...while sum(got)<9: ......d=int(str(n*i)[0]) ......got[d] = 1 ......i += 1 ...return i-1 (PARI) a(n)=my(v=vector(9), s, k, t); while(s<9, t=digits(k++*n)[1]; if(v[t]==0, v[t]=1; s++)); k \\ Charles R Greathouse IV, Oct 20 2014 CROSSREFS Sequence in context: A225488 A096688 A181431 * A124983 A087969 A044111 Adjacent sequences: A249064 A249065 A249066 * A249068 A249069 A249070 KEYWORD nonn,base,look,easy AUTHOR Christian Perfect, Oct 20 2014 STATUS approved
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Last modified January 29 08:01 EST 2023. Contains 359915 sequences. (Running on oeis4.)
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# Part 2: projectiles launched at an angle Pages 102 – 104 Motion in Two Dimensions.
## Presentation on theme: "Part 2: projectiles launched at an angle Pages 102 – 104 Motion in Two Dimensions."— Presentation transcript:
Part 2: projectiles launched at an angle Pages 102 – 104 Motion in Two Dimensions
Maximum height and range When objects are launched at an angle, there is a component of their velocity in the horizontal direction and a component in the vertical direction. The horizontal velocity would be v i cosθ The vertical velocity would be v i sinθ Problems can then be worked in the same way. Keep in mind that the vertical velocity at the peak of the path is zero.
Launched at an angle
Velocity Final speed = initial speed (conservation of energy) Impact angle = - launch angle (symmetry of parabola )
Launched at an angle The launch angle has an affect on the range of the projectile. As the angle increases from 0 to 45 degrees, the range increases. At 45 degrees, the range is at a maximum As the angle increases from 45 degrees to 90 degrees, the range decreases.
Launched at an angle
Equations of motion: X Uniform motion Y Accelerated motion ACCELERATION a x = 0a y = -9.81 m/s 2 VELOCITY v x = v i cos θv yf = v i sin θ + aΔt v fy 2 = v i 2 sin θ +2a Δy DISPLACEMENT Δx = v i cos θ ΔtΔy = v i sinθΔt + ½ aΔt 2
The Hulk throws a boulder onto a police car trying to arrest him. The initial velocity of the boulder is 12 m/s and he throws it at an angle of 39 degrees to the horizontal: a. Find the horizontal and vertical components of the initial velocity b. Find the maximum height of the boulder c. Find the time the boulder was in the air d. Find the horizontal distance the boulder traveled
39° vivi v i = 12m/s
39° vivi Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy
Finding the components v xi = v i cos θ v yi = v i sin θ Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy
Find time first!!! v yf = v i sin θ + aΔt Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy
Finding max height Δy = v i sinθΔt + ½ aΔt 2 Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy
Finding the time in the air Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy
Finding the horizontal distance Δx = v i cos θ Δt Givens angle = 39° v i = 12m/s a = -9.8 m/s 2 Equations v x = v i cos θ v y = v i sin θ Δx = v i cos θ Δt v yf = v i sin θ + aΔt Δy = v i sinθΔt + ½ aΔt 2 v fy 2 = v i 2 sin θ +2a Δy
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## 2.9.2 Geometric PMF
A random variable X has a geometric pmf if
One important interpretation of the geometric pmf involves the "first time until success" in a sequence of Bernoulli experiments (trials). Here "success" corresponds to the Bernoulli random value taking on the value 1. Suppose in the police example above that Yi is the outcome of the Bernoulli trial conducted at the ith hour. Thus, if Yi = 1, the high-crime zone is patrolled during the ith hour; otherwise, it is not patrolled that hour. Suppose that we (as observers) start looking at the high-crime zone during hour 1. We ask the question: Which hour X (X = 1, 2 .... ) will be the first hour during which the high-crime zone will be patrolled? The probability that it will be patrolled during the first hour is simply p. The probability that it will be first patrolled during the second hour is P{ Y1 = 0, Y1 = 1), which by independence is (I - p)p. In general, the probability that it will be first patrolled during the kth hour is P{ Y1 = 0, Y2 = 0, . . . , Yk-1 = 0, Yk = 1}, which by independence is (1 - p)k - 1 p. Thus, the random variable X is a geometrically distributed random variable which, when we substitute p = 0.25, has mean E[X] = 1/0.25 = 4 and variance = 12 (and = 2 3.44).
Question: What is the probability that the high-crime zone receives no patrol coverage during any particular 8-hour tour of duty?
Exercise 2.11: No Memory Property of Geometric PMF Suppose we have observed that the high-crime area has received no patrol coverage during the first k hours. Show that the probability law for the hour at which patrol first occurs, given this information, is the same as the original pmf, but shifted to the right k units. Thus, the geometric pmf has a no-memory property in the sense that the time (k hours) that we have invested waiting for the first hour of patrol coverage of the high-crime zone has not in any way reduced the mean or variance or any other measure of the remaining time we must wait until the first patrol.
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How much is 2 percent of two over ten
Percentage Calculator 2
is what percent of ? Answer: %
Percentage Calculator 3
We think you reached us looking for answers to questions like: What is 2 percent of two over ten? Or maybe: How much is 2 percent of two over ten
See the detailed solutions to these problems below.
How to work out percentages explained step-by-step
Learn how to solve percentage problems through examples.
In all the following questions consider that:
• The percentage figure is represented by X%
• The whole amount is represented by W
• The portion amount or part is represented by P
Solution for 'What is 2% of two over ten?'
Solution Steps
The following question is of the type "How much X percent of W", where W is the whole amount and X is the percentage figure or rate".
Let's say that you need to find 2 percent of 0.2. What are the steps?
Step 1: first determine the value of the whole amount. We assume that the whole amount is 0.2.
Step 2: determine the percentage, which is 2.
Step 3: Convert the percentage 2% to its decimal form by dividing 2 into 100 to get the decimal number 0.02:
2100 = 0.02
Notice that dividing into 100 is the same as moving the decimal point two places to the left.
2.0 → 0.20 → 0.02
Step 4: Finally, find the portion by multiplying the decimal form, found in the previous step, by the whole amount:
0.02 x 0.2 = 0.004 (answer).
The steps above are expressed by the formula:
P = W × X%100
This formula says that:
"To find the portion or the part from the whole amount, multiply the whole by the percentage, then divide the result by 100".
The symbol % means the percentage expressed in a fraction or multiple of one hundred.
Replacing these values in the formula, we get:
P = 0.2 × 2100 = 0.2 × 0.02 = 0.004 (answer)
Therefore, the answer is 0.004 is 2 percent of 0.2.
Solution for '2 is what percent of two over ten?'
The following question is of the type "P is what percent of W,” where W is the whole amount and P is the portion amount".
The following problem is of the type "calculating the percentage from a whole knowing the part".
Solution Steps
As in the previous example, here are the step-by-step solution:
Step 1: first determine the value of the whole amount. We assume that it is 0.2.
(notice that this corresponds to 100%).
Step 2: Remember that we are looking for the percentage 'percentage'.
To solve this question, use the following formula:
X% = 100 × PW
This formula says that:
"To find the percentage from the whole, knowing the part, divide the part by the whole then multiply the result by 100".
This formula is the same as the previous one shown in a different way in order to have percent (%) at left.
Step 3: replacing the values into the formula, we get:
X% = 100 × 20.2
X% = 2000.2
So, the answer is 2 is 1,000.00 percent of 0.2
Solution for '0.2 is 2 percent of what?'
The following problem is of the type "calculating the whole knowing the part and the percentage".
Solution Steps:
Step 1: first determine the value of the part. We assume that the part is 0.2.
Step 2: identify the percent, which is 2.
Step 3: use the formula below:
W = 100 × PX%
This formula says that:
"To find the whole, divide the part by the percentage then multiply the result by 100".
This formula is the same as the above rearranged to show the whole at left.
Step 4: plug the values into the formula to get:
W = 100 × 0.22
W = 100 × 0.1
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## We Promise to Make your Math Frustrations Go Away!
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Number of equations to solve: 23456789
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Number of inequalities to solve: 23456789
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# Arithmetic
Absolute Value | |
1. Simplify inside the absolute values first.
2. Whatever number is inside the absolute value becomes (or stays positive)
when you take the absolute value.
• Example:
|3 − 7| = | − 4| = 4
| − 3 + 7| = |4| = 4
1. Identify the signs of the numbers that are to be added.
(a) For the same sign:
i. Add the numbers (ignoring the signs)
(b) For different signs:
i. Ignoring the signs, identify the largest number.
ii. Subtract the numbers (ignoring the signs)
iii. Attach the sign of the number that you identified in step i.
• Example:
−4 + (−8) −4 and −8 have the same sign -
4 + 8 = 12 add the numbers
−4 + (−8) = −12 attach the sign identified earlier
• Example:
−4 + 8 −4 and 8 have the different signs
8 is larger than 4 so the answer will be positive
8 − 4 = 4 add the numbers
−4 + 8 = 4 attach the sign identified earlier
Subtraction -
1. Identify the two numbers being subtracted
2. Leave the first number alone and add the opposite of the second number
(If the second number was positive it should be negative. If it was negative
it should be positive.)
• Example:
−4 − (−8) −4 and −8 are the numbers being subtracted
−4 + (+8) leave the first alone and add the opposite
−4 − (−8) = −4 + (+8) = 4 follow the rules for addition
• Example:
−4 − 8 −4 and 8 are the two numbers being subtracted
−4 + (−8) leave the first alone and add the opposite
−4 − 8 = −4 + (−8) = −12 follow the rules for addition
Multiplication ×, ( )( ), ·
1. Multiply the numbers (ignoring the signs)
2. The answer is positive if they have the same signs.
3. The answer is negative if they have different signs.
4. If you have more than 2 numbers, use the above but you must calculate
positive or negative with only two numbers at a time. Alternatively,
count the amount of negative numbers. If there are an even number of
negatives the answer is positive. If there are an odd number of negatives
• Example:
−4 × −8 −4 and − 8 are the two numbers being multiplied
4 × 8 = 32 multiply ignoring the signs
−4 × −8 = 32 same sign so positive
• Example:
−4 × 8 −4 and 8 are the two numbers being multiplied
4 × 8 = 32 multiply ignoring the signs
−4 × 8 = −32 different signs so negative
• Example:
−4 × 2 × −3 −4, 2 and 8 are the three numbers being multiplied
4 × 2 × 3 = 24 multiply ignoring the signs
−4 × 2 × −3 = 24 since there are two negative numbers,
−4 and − 3, and two is even the answer is positive.
Division ÷, /
1. Divide the numbers (ignoring the signs)
2. The answer is positive if they have the same signs.
3. The answer is negative if they have different signs.
4. If you have more than 2 numbers, use the above but you must calculate
positive or negative with only two numbers at a time. Alternatively,
count the amount of negative numbers. If there are an even number of
negatives the answer is positive. If there are an odd number of negatives
• Example:
−8 ÷ −4 −4 and −8 are the two numbers being divided
8 ÷ 4 = 2 divide ignoring the signs
−8 ÷ −4 = 2 same sign so positive
• Example:
−8 ÷ 4 4 and −8 are the two numbers being divided
8 ÷ 4 = 2 divide ignoring the signs
−8 ÷ 4 = −2 different sign so negative
NOTE: and is undefined.
Dealing with Decimals
Be NEAT
(a) Follow the rules above for addition or subtraction being sure to line
up the decimals
• Example: 1.41 − 3.2 = 1.41 + (−3.2) (add the opposite)
different signs subtract
so 1.41 − 3.2 = −1.79 (since 3.2 > 1.41)
2. Multiplication
(a) Multiply the numbers together (ignoring signs and decimals)
(b) Count how many digits were to the right of the decimal (the total for
both numbers)
(c) Place the decimal so that the same number of digits is to the right of
(d) Find the sign by the rules of multiplication above (same sign positive,
different signs negative)
• Example: (-.05)(.0026)=
Step (a)
Step (b) −.05 has 2 digits to the right of the decimal .0026 has 4 digits to the right of the decimal there is a total of 6 digits to the right of the decimal
Step (c) & (d)
So, (−.05)(.0026) = −.000130 notice the 6 digits to the right of
the decimal including 130 with the zeros the left as place holders.
Also, the answer is negative since the signs are different.
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# The final step in kalman filter to correct/update the covariance matrix
I see that, in the correction step of Kalman filter, there is an equation to update the covariance matrix. I have been using it in the form:
P = (I - KH)P'
Here P is the covariance matrix, K is the Kalman Gain and H is the observation model. The I is the identity matrix. However, I also see a different equation in some literature:
P = P' - KSKT
where
S = HPHT + Q
where Q is the noise matrix for the observation model. Are these two equations same? If so, how to prove it?
• After discussion on my answer now I see the confusion. There's an identity I'm missing somewhere I think but I can't figure it out so I removed my answer. FYI, as a sanity check it might be worth setting up a script to generate random matrices, compute optimal $K$, and then compute both $P$ updates to see if they are equal. Commented Mar 22, 2018 at 16:46
• @ryan0270 That's a good idea. I will do that. However, it would be much more satisfying if we could find a mathematical proof.
– skr
Commented Mar 22, 2018 at 16:48
• The proof is only useful if the random tests all come back the same. If those tests start to show differences than you know for sure the two are not equivalent (or there are missed assumptions) Commented Mar 22, 2018 at 18:22
The Kalman Gain is defined as the following:
$K = PH^T (HPH^T + Q)^-1$
From your question, we need to prove:
$P - KSK^T = (I - KH)P$
primes omitted for simplicity
Canceling P, and substitute S with the provided definition:
$K (HPH^T + Q)K^T = KHP$
Substitute the first K on the left hand side with the definition of the Kalman gain provided: $PH^T (HPH^T + Q)^-1 (HPH^T + Q)K^T = KHP$
S and it's inverse reduces to identity, we are left with:
$PH^TK^T = KHP$
From the transpose property of matrices $(ABC)^T = C^T B^T A^T$ we can rearrange LHS with $P$ as $A$, $H^T$ as $B$, and $K^T$ as $C$:
$(KHP^T)^T = KHP$
Since covariance matrix is symmetric, and KHP must be symmetric (this can be inferred from $P' = P - KHP$) therefore:
$KHP = KHP$
• When you go from second last to the last step, isn't it $PH^TK^T = KHP$? Correct me if I am wrong.
– skr
Commented Mar 22, 2018 at 20:09
• I missed a transpose at the last steps in my original answer and edited just before your comment. Are you referring to the original answer or the edit? Commented Mar 22, 2018 at 20:13
• To the edit, with the missed transpose.
– skr
Commented Mar 22, 2018 at 20:14
• How did you arrive at $PH^TK^T$? Commented Mar 22, 2018 at 20:15
• $KHP$ is symmetric, as is $P$ Commented Mar 22, 2018 at 20:21
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# [x,p] = xp - px = ih
## Main Question or Discussion Point
xp - px = x * (-i*hbar*d/dx) - (-i * hbar * d/dx (x)) = i * hbar
I can see that the px = i * hbar
But, why does xp = 0 ... ? I just get - i * hbar everytime I try, Wolfram Alpha agrees...
Can anyone show me why this isn't the case please?
Related Quantum Physics News on Phys.org
Doc Al
Mentor
I can see that the px = i * hbar
Really? Show how you got that.
The annoying thing about commutators is that, because they are operators, when you do it out explicitly you have to apply them to a function to make them work out, so apply [x,p] to f and see what happens.
Ok,
What I've been doing is just d/dx ( const * x ) = x
That works when we apply p to x. But not x to p I suppose.
Doc Al
Mentor
Ok,
What I've been doing is just d/dx ( const * x ) = x
That works when we apply p to x. But not x to p I suppose.
These are operator relationships. To calculate what you call "px", you need to evaluate:
$$\left( -i \hbar \frac{\partial}{\partial x} \right) x \psi$$
Note that you apply x first, then the derivative. So you'll need to use the product rule.
Bit aside, how arises I in p and x?
Fredrik
Staff Emeritus
Gold Member
Bit aside, how arises I in p and x?
It shows up when you apply the product rule for derivatives:
$$(\hat p\hat x f)(x)=(\hat p(\hat x f))(x) = -i(\hat x f)'(x) =-i\frac{d}{dx}(\hat x f(x)) =-i\frac{d}{dx}(xf(x))=\cdots.$$
Last edited:
dextercioby
Homework Helper
Bit aside, how arises I in p and x?
Have you heard about a Poisson bracket?
Hi,
Roughtly speaking, the derivative operator "p" acts over everything to the right and the wave function it's always implicitly the last term, so you really have:
pxΨ= i * hbar ∂x(xΨ) = i * hbar * [x * ∂x(Ψ) + Ψ * ∂x(x) ]
Where Ψ is the wave function in the evaluated point, implicitly, Ψ(x).
x(x) = 1 so the conmutation is [x,p]Ψ = (xp - px)Ψ = i*h_bar Ψ.
Best regards,
Sergio
bhobba
Mentor
Derivative is Ok, but what for I ? Moreover, how imaginary valued are connected with physical?
Fredrik
Staff Emeritus
There are many such connections. For example, $|\psi(x)|^2\Delta x$ is approximately equal to the probability that a particle detector near $x$ that's covering a region of size $\Delta x$ will detect a particle.
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# Something is wrong with my Matlab
25 views (last 30 days)
Juan Pablo Arrieta on 24 Sep 2020 at 15:44
Commented: Steven Lord on 24 Sep 2020 at 19:21
Hi, today I saw a challenge on internet, so I decided to solve it using math and Matlab to check if I was right.
It was about a 3x3 system of equations.
The equation was:
A = [x x x; x y y; y -2z]
b = [60 30 3]'
And this is what I used.
A = [1 1 1; 1 1 1; 0 1 -2]
b = [60 30 3]'
When I found out the values of x, y and z, the values were: x = 20, y = 5 and z = 1.
But, on Matlab, they showed me the values for x = 16.28, y = 20 and z = 8.5.
I used the "A\b" command and after I used the "pinv(A)*b" command.
What I did wrong? Can you help me with that?
Steven Lord on 24 Sep 2020 at 15:59
A = [x x x; x y y; y -2z]
b = [60 30 3]'
I suspect this is supposed to represent the equations 3*x = 60, x+2*y = 30, y-2*z = 3. That's not how I would have written them, but the question you're asking is about how to write them.
A = [1 1 1; 1 1 1; 0 1 -2]
b = [60 30 3]'
Those two matrices represent the equations x+y+z = 60, x+y+z = 30, y-2*z = 3. Those don't match the equations from the original problem, and the first two equations aren't consistent. How can the sum of x, y, and z simultaneously be equal to 60 and equal to 30?
The first element of the first row of your A should be the coefficient of the first variable in the first equation.
The second element of the first row of your A should be the coefficient of the second variable in the first equation.
The first element of the third row of your A should be the coefficient of the first variable in the third equation.
Can you translate the original equations into A using this pattern?
Juan Pablo Arrieta on 24 Sep 2020 at 16:58
I did it and I solve it.
The only thing I needed to do was replace A like this:
A = [3 0 0; 1 2 0; 0 1 -2]
Because "x + x + x" is "3*x" and the rest of the coefficients were 0.
The other equation "x + y + y" is "x + 2*y" and the other coefficient is 0.
The last one its okay.
The results were what I wrote at the beginning.
x = 20
y = 5
z = 1
Thank you very much! You helped me a lot!
And I apologize if my english isn't as good as a native speaker.
Steven Lord on 24 Sep 2020 at 19:21
You're welcome! And your English was good, I could understand what you said without a problem.
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## Calculus 8th Edition
$\frac{f^{8}(5)}{40,320}$
Since, the Taylor's series of $f$ centered at $a$ is $f(x)=\Sigma_{n=0}^{\infty}\frac{f^{n}(a)(x-a)^{n}}{n!}$ Thus, $b_{n}=\frac{f^{n}(a)}{n!}$ Now, $b_{8}=\frac{f^{8}(5)}{8!}=\frac{f^{8}(5)}{40,320}$
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Class 9
MATHS
Volume And Surface Area Of Solids
# The height of a cylinder is 14 cm and its curved surface area is 264 cm^(2). The volume of the cylinder is
Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
Updated On: 7-12-2020
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161.7 K+
8.1 K+
Text Solution
308 cm^(3)396 cm^(3)1232 cm^(3)1848 cm^(3)
B
Solution :
Here h = 14 cm and 2pi rh = 264 cm^(2) <br> :. 2 xx (22)/(7) xx r xx 14 = 264 rArr r = (264)/(88) = 3 cm <br> :. volume = pi r^(2)h = ((22)/(7) xx 3 xx 3 xx 14) cm^(3) = 396 cm^(3)
Image Solution
Find answer in image to clear your doubt instantly:
98160726
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# robinswood
posted by .
between what two consecutive whole numbers does the squre root lie?
1) square root of 70
2) square root of 35
3) square root of 6
4) square root of 52
• robinswood -
1)
√70 = 8.367 = between 8 and 9
I'll be glad to check your answers for the other problems.
• robinswood -
13
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1. Integration
I need a bit of help integrating
X^2/(X^2-4X+5)
2. Originally Posted by qwerty10
I need a bit of help integrating
X^2/(X^2-4X+5)
Hint:
$\frac{x^2}{x^2-4x+5}=\frac{4x-8}{x^2-4x+5}+\frac{3}{x^2-4x+5}+1$
3. 1. Do a long division.
2. Apply:
$\int \dfrac{f'(x)}{f(x)} dx = ln |f(x)| + C$
4. Thank you
Is the correct way to continue by
let u=x^2-4x+5
du=(2x-4)dx
so (x-2)dx=du/4
After this I am not quite sure what to do
5. $\int \frac{x^2}{x^2-4x+5} dx = \int 1+ \frac{4x-5}{x^2-4x+5} dx = \int 1 dx + \int \frac{4x-5}{x^2-4x+5} dx$
$= \int 1 dx + \int \frac{4x-8}{x^2-4x+5} dx + \int \frac{3}{x^2-4x+5} dx$
$=x + 2ln |x^2-4x+5| + \int \frac{3}{x^2-4x+5} dx$
Can you continue now?
6. Originally Posted by qwerty10
Thank you
Is the correct way to continue by
let u=x^2-4x+5
du=(2x-4)dx
so (x-2)dx=du/4 Not Needed rather dx = du/(2x-4)
After this I am not quite sure what to do
Look at the reply by Also sprach Zarathustra
$\int \frac{x^2}{x^2-4x+5}dx= \int \frac{4x-8}{x^2-4x+5}dx+\int \frac{3}{x^2-4x+5}dx + \int 1 dx$
use the substitution rule for the first term only:
$\int \frac{4x-8}{x^2-4x+5}dx = \int \frac{2(x-4)}{x^2-4x+5}$
sub. rule gives you:
$\int \frac{2}{u} du$ . complete it..
For the second term: $\int \frac{3}{x^2-4x+5}dx = \int \frac{3}{1+(x-2)^{2}} dx$
you should be able to finish it off now..
7. Thank you
I have worked the answer out to be 2ln(x^2-4x+5) +3tan^-1(x-2) + x+C ??
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Few people are willing to tell their boss bad news. The boss will just get angry at the messenger. But bad news must be heard or nothing will change.
A company¿s five division managers¿Arthur, Barbara, Carol, Daniel and Ellen--get together in a room. Three of them think the news really is bad. Two disagree. Still, all five agree that the boss should understand that at least three think the news is bad but that not all five think it's bad. We call these the Three Bad and Some Good conditions. They also agree that the boss should not learn what any single one of them thinks. We call this the Anonymity condition.
They know the boss will interview each of them individually and will be satisfied only with a statement that refers to the person questioned and at most two other people. We call this the Limited Reference condition. That is, the boss won't accept generalities such as "At least three out of all five people think the news is bad." Instead he wants statements such as, "At least two of the three of us -- Ellen, Daniel and me -- think the news is bad" or "Of Arthur and me, at least one of us thinks the news is not bad.¿ The boss is happy to hear more than one statement from a person as long as the statements stay within this Reference Limit of three.
Warm-up: What can the five managers say to prove Three Bad and Some Good while still satisfying the Anonymity and Limited Reference conditions?
Solution to Warm-up: Let's call the division heads A, B, C, D, E for short. As it happens, A, C and D think the news is bad. Suppose that A and B both say, "At least one of A and B thinks the news is bad." D and E each say, "At least two of C, D, and E think the news is bad." C says, "At least one of A, B, C thinks the news is not bad." The first two statements ensure that at least three think the news is bad (Three Bad). The last ensures that at least one thinks the news is not bad (Some Good). In these circumstances, either A or B could think the news is bad and any of C, D, or E could think the news is not bad (Anonymity).
Problems
1. Under the same rules from the boss and with the same goal, could the managers prove that EXACTLY three of the five people think the news is bad? If so, how? If not, why not?
2. Again under the same rules and with the same goal, could they prove that EXACTLY four people think the news is bad? If so, how? If not, why not?
3. Would your answer to Problem 2 change if you had to prove only that AT LEAST four of the five people think the news is bad?
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+1-415-315-9853
[email protected]
## Statistics
problem 1: You have a regional country-year dataset over 10 years of 150 total observations, within this data you have 25 observations where the ruling party has lost power.
(a) Assuming a binomial distribution, what is the likelihood function for Π , the probability that a country's ruling party loses power over ten years?
(b) find out the MLE of Π.
(c) find out the standard error and the 95% con dence interval for this estimate. How did you know what test/sampling distribution to use?
(d) What is the value of the log-likelihood for this estimate?
(e) Say that for another region the estimated probability of a ruling party losing power over a ten-year span is 25%, using your region's data, what is the value of the log-likelihood function for a .25 value of Π?
(f) Using a Wald test, test whether your region's rate of over turning power is signi cantly di erent from the other region's.
problem 2: Each section of the SAT test is supposed to be distributed normally with a mean of 500 and a standard deviation of 100. Suppose 5 students in a class took the SAT math test. They received the following scores:
400, 450, 575, 600, and 625.
(a) Assuming a normal distribution, prepare out the likelihood function for estimating the mean μ and standard deviation σ of the average class score.
(b) find out the MLE estimate for μ and the standard deviation (σ) for this class.
(c) What is the value of the log-likelihood function at this estimate?
(d) What is the least-squares estimate for μ and σ in your data? Are there any di fferences? Why or why not?
(e) Using a likelihood-ratio test, test whether the MLE estimates from the class are signi cantly di erent from the national mean of 500 and national standard deviation of 100. (Hint : Start by plugging those values into your log-likelihood function. Remember you are testing two restrictions).
problem 3: (this is tough but give it a try) Use R (easiest) or Stata to simulate how often one makes a wrong small-sample inference in MLE vs. OLS in the following circumstance. Perform the following steps (be sure to attach your code):
A) Take 5 draws from a standard normal distribution (μ= 0, σ = 1)
B) Record the MLE and OLS estimate of (note: they have slightly di fferent estimates).
C) find out and record each estimate's squared error ((^σ - σ )2, where σ = 1).
D) find out and record each estimate's 95% con dence interval using z vs. t-scores.
E) Test and record whether ^μ is estimated as signi cantly different from 0.
F) Repeat for a total of 1000 simulations.
problem a) What was your average estimate of σ in MLE and OLS for your simulations?
problem b) Taking the mean of the squared error, which estimator was closer to the true parameter (σ) on average?
problem c) How often did you make a spurious influence with MLE? How does this compare to the Type I error rate in OLS?
problem d) How would you evaluate the performance of the ML-estimator with 5 observations?
Statistics and Probability, Statistics
• Category:- Statistics and Probability
• Reference No.:- M9613
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g07 Chapter Contents
NAG C Library Manual
# NAG Library Chapter Introductiong07 – Univariate Estimation
## 1 Scope of the Chapter
This chapter deals with the estimation of unknown arguments of a univariate distribution. It includes both point and interval estimation using maximum likelihood and robust methods.
## 2 Background to the Problems
Statistical inference is concerned with the making of inferences about a population using the observed part of the population called a sample. The population can usually be described using a probability model which will be written in terms of some unknown parameters. For example, the hours of relief given by a drug may be assumed to follow a Normal distribution with mean $\mu$ and variance ${\sigma }^{2}$; it is then required to make inferences about the arguments, $\mu$ and ${\sigma }^{2}$, on the basis of an observed sample of relief times.
There are two main aspects of statistical inference: the estimation of the arguments and the testing of hypotheses about the arguments. In the example above, the values of the argument ${\sigma }^{2}$ may be estimated and the hypothesis that $\mu \ge 3$ tested. This chapter is mainly concerned with estimation but the test of a hypothesis about an argument is often closely linked to its estimation. Tests of hypotheses which are not linked closely to estimation are given in the chapter on nonparametric statistics (Chapter g08).
There are two types of estimation to be considered in this chapter: point estimation and interval estimation. Point estimation is when a single value is obtained as the best estimate of the argument. However, as this estimate will be based on only one of a large number of possible samples, it can be seen that if a different sample were taken, a different estimate would be obtained. The distribution of the estimate across all the possible samples is known as the sampling distribution. The sampling distribution contains information on the performance of the estimator, and enables estimators to be compared. For example, a good estimator would have a sampling distribution with mean equal to the true value of the argument; that is, it should be an unbiased estimator; also the variance of the sampling distribution should be as small as possible. When considering a parameter estimate it is important to consider its variability as measured by its variance, or more often the square root of the variance, the standard error.
The sampling distribution can be used to find interval estimates or confidence intervals for the argument. A confidence interval is an interval calculated from the sample so that its distribution, as given by the sampling distribution, is such that it contains the true value of the argument with a certain probability.
Estimates will be functions of the observed sample and these functions are known as estimators. It is usually more convenient for the estimator to be based on statistics from the sample rather than all the individuals observations. If these statistics contain all the relevant information then they are known as sufficient statistics. There are several ways of obtaining the estimators; these include least squares, the method of moments, and maximum likelihood. Least squares estimation requires no knowledge of the distributional form of the error apart from its mean and variance matrix, whereas the method of maximum likelihood is mainly applicable to situations in which the true distribution is known apart from the values of a finite number of unknown arguments. Note that under the assumption of Normality, the least squares estimation is equivalent to the maximum likelihood estimation. Least squares is often used in regression analysis as described in Chapter g02, and maximum likelihood is described below.
Estimators derived from least squares or maximum likelihood will often be greatly affected by the presence of extreme or unusual observations. Estimators that are designed to be less affected are known as robust estimators.
### 2.1 Maximum Likelihood Estimation
Let ${X}_{i}$ be a univariate random variable with probability density function
$fXixi;θ,$
where $\theta$ is a vector of length $p$ consisting of the unknown arguments. For example, a Normal distribution with mean ${\theta }_{1}$ and standard deviation ${\theta }_{2}$ has probability density function
$12πθ2 exp-12 xi-θ1θ2 2 .$
The likelihood for a sample of $n$ independent observations is
$Like=∏i=1nfXi xi;θ ,$
where ${x}_{i}$ is the observed value of ${X}_{i}$. If each ${X}_{i}$ has an identical distribution, this reduces to
$Like=∏i=1nfX xi;θ ,$ (1)
and the log-likelihood is
$logLike=L=∑i=1nlogfXxi;θ.$ (2)
The maximum likelihood estimates ($\stackrel{^}{\theta }$) of $\theta$ are the values of $\theta$ that maximize (1) and (2). If the range of $X$ is independent of the arguments, then $\stackrel{^}{\theta }$ can usually be found as the solution to
$∑i=1n ∂∂θ^j logfXxi;θ^= ∂L ∂θ^j =0, j=1,2,…,p.$ (3)
Note that $\frac{\partial L}{\partial {\theta }_{j}}$ is known as the efficient score.
Maximum likelihood estimators possess several important properties.
(a) Maximum likelihood estimators are functions of the sufficient statistics.
(b) Maximum likelihood estimators are (under certain conditions) consistent. That is, the estimator converges in probability to the true value as the sample size increases. Note that for small samples the maximum likelihood estimator may be biased.
(c) For maximum likelihood estimators found as a solution to (3), subject to certain conditions, it follows that
$E ∂L ∂θ =0,$ (4)
and
$Iθ=-E ∂2L ∂θ2 =E ∂L ∂θ 2 ,$ (5)
and then that $\stackrel{^}{\theta }$ is asymptotically Normal with mean vector ${\theta }_{0}$ and variance-covariance matrix ${I}_{{\theta }_{0}}^{-1}$ where ${\theta }_{0}$ denotes the true value of $\theta$. The matrix ${I}_{\theta }$ is known as the information matrix and ${I}_{{\theta }_{0}}^{-1}$ is known as the Cramer–Rao lower bound for the variance of an estimator of $\theta$.
For example, if we consider a sample, ${x}_{1},{x}_{2},\dots ,{x}_{n}$, of size $n$ drawn from a Normal distribution with unknown mean $\mu$ and unknown variance ${\sigma }_{2}$ then we have
$L=logLikeμ,σ2;x=-n2log2π-n2logσ2-∑i=1n xi-μ 2/2σ2$
and thus
$∂L ∂μ =∑i= 1n xi-μ/σ2$
and
$∂L ∂σ2 =-n2σ2 +∑i=1n xi-μ 2/2σ4.$
Then equating these two equations to zero and solving gives the maximum likelihood estimates
$μ^=x-$
and
$σ^2=∑i=1n xi-x- 2/n.$
These maximum likelihood estimates are asymptotically Normal with mean vector $a$, where
$aT=μ,σ2,$
and covariance matrix $C$. To obtain $C$ we find the second derivatives of $L$ with respect to $\mu$ and ${\sigma }^{2}$ as follows:
$∂2L ∂μ2 =- nσ2 ∂2L ∂σ22 = n2σ4 -∑i=1n xi-μ 2/σ6 ∂2L ∂μ∂σ2 = ∂2L ∂σ2∂μ =- nx--μσ4.$
Then
$C-1=-E ∂2 L ∂μ2 ∂2 L ∂σ2∂μ ∂2 L ∂μ∂σ2 ∂2 L ∂σ22 = n/σ2 0 0 n/2σ4$
so that
$C= σ2/n 0 0 2σ4/n .$
To obtain an estimate of $C$ the matrix may be evaluated at the maximum likelihood estimates.
It may not always be possible to find maximum likelihood estimates in a convenient closed form, and in these cases iterative numerical methods, such as the Newton–Raphson procedure or the EM algorithm (expectation maximization), will be necessary to compute the maximum likelihood estimates. Their asymptotic variances and covariances may then be found by substituting the estimates into the second derivatives. Note that it may be difficult to find the expected value of the second derivatives required for the variance-covariance matrix and in these cases the observed value of the second derivatives is often used.
The use of maximum likelihood estimation allows the construction of generalized likelihood ratio tests. If $\lambda =2\left({l}_{1}-{l}_{2}\right)$, where ${l}_{1}$ is the maximized log-likelihood function for a model $1$ and ${l}_{2}$ is the maximized log-likelihood function for a model $2$, then under the hypothesis that model $2$ is correct, $2\lambda$ is asymptotically distributed as a ${\chi }^{2}$ variable with $p-q$ degrees of freedom. Consider two models in which model $1$ has $p$ arguments and model $2$ is a sub-model (nested model) of model $1$ with $q arguments, that is model $1$ has an extra $p-q$ arguments. This result provides a useful method for performing hypothesis tests on the arguments. Alternatively, tests exist based on the asymptotic Normality of the estimator and the efficient score; see page 315 of Cox and Hinkley (1974).
### 2.2 Confidence Intervals
Suppose we can find a function, $t\left(x,\theta \right)$, whose distribution depends upon the sample $x$ but not on the unknown argument $\theta$, and which is a monotonic (say decreasing) function in $\theta$ for each $x$, then we can find ${t}_{1}$ such that $P\left({t}_{1}\le t\left(x,\theta \right)\right)=1-\alpha$ no matter what $\theta$ happens to be. The function $t\left(x,\theta \right)$ is known as a pivotal quantity. Since the function is monotonic the statement that ${t}_{1}\le t\left(x,\theta \right)$ may be rewritten as $\theta \ge {\theta }_{1}\left(x\right)$ see Figure 1. The statistic ${\theta }_{1}\left(x\right)$ will vary from sample to sample and if we assert that $\theta \ge {\theta }_{1}\left(x\right)$ for any sample values which arise, we will be right in a proportion $1-\alpha$ of the cases, in the long run or on average. We call ${\theta }_{1}\left(x\right)$ a $1-\alpha$ upper confidence limit for $\theta$.
Figure 1
We have considered only an upper confidence limit. The above idea may be generalized to a two-sided confidence interval where two quantities, ${t}_{0}$ and ${t}_{1}$, are found such that for all $\theta$, $\mathrm{P}\left({t}_{1}\le t\left(x,\theta \right)\le {t}_{0}\right)=1-\alpha$. This interval may be rewritten as ${\theta }_{0}\left(x\right)\le \theta \le {\theta }_{1}\left(x\right)$. Thus if we assert that $\theta$ lies in the interval [${\theta }_{0}\left(x\right),{\theta }_{1}\left(x\right)$] we will be right on average in $1-\alpha$ proportion of the times under repeated sampling.
Hypothesis (significance) tests on the arguments may be used to find these confidence limits. For example, if we observe a value, $k$, from a binomial distribution, with known argument $n$ and unknown argument $p$, then to find the lower confidence limit we find ${p}_{l}$ such that the probability that the null hypothesis ${H}_{0}$: $p={p}_{l}$ (against the one sided alternative that $p>{p}_{l}$) will be rejected, is less than or equal to $\alpha /2$. Thus for a binomial random variable, $B$, with arguments $n$ and ${p}_{l}$ we require that $P\left(B\ge k\right)\le \alpha /2$. The upper confidence limit, ${p}_{u}$, can be constructed in a similar way.
For large samples the asymptotic Normality of the maximum likelihood estimates discussed above is used to construct confidence intervals for the unknown arguments.
### 2.3 Robust Estimation
For particular cases the probability density function can be written as
$fXixi;θ=1θ2g xi-θ1θ2$
for a suitable function $g$; then ${\theta }_{1}$ is known as a location argument and ${\theta }_{2}$, usually written as $\sigma$, is known as a scale argument. This is true of the Normal distribution.
If ${\theta }_{1}$ is a location argument, as described above, then equation (3) becomes
$∑i=1nψ xi-θ^1σ^=0,$ (6)
where $\psi \left(z\right)=-\frac{d}{dz}\mathrm{log}\left(g\left(z\right)\right)$.
For the scale argument $\sigma$ (or ${\sigma }^{2}$) the equation is
$∑i=1nχ xi-θ^1σ^=n/2,$ (7)
where $\chi \left(z\right)=z\psi \left(z\right)/2$.
For the Normal distribution $\psi \left(z\right)=z$ and $\chi \left(z\right)={z}^{2}/2$. Thus, the maximum likelihood estimates for ${\theta }_{1}$ and ${\sigma }^{2}$ are the sample mean and variance with the $n$ divisor respectively. As the latter is biased, (7) can be replaced by
$∑i=1nχ xi-θ^1σ^=n-1β,$ (8)
where $\beta$ is a suitable constant, which for the Normal $\chi$ function is $\frac{1}{2}$.
The influence of an observation on the estimates depends on the form of the $\psi$ and $\chi$ functions. For a discussion of influence, see Hampel et al. (1986) and Huber (1981). The influence of extreme values can be reduced by bounding the values of the $\psi$- and $\chi$-functions. One suggestion due to Huber (1981) is
$ψz= -C, ||z<-C z, z≤C C, ||z>C.$
Figure 2
Redescending $\psi$-functions are often considered; these give zero values to $\psi \left(z\right)$ for large positive or negative values of $z$. Hampel et al. (1986) suggested
$ψz= -ψ-z z, 0≤z≤h1. h1, h1≤z≤h2. h1h3-z/h3-h2, h2≤z≤h3. 0, z>h3.$
Figure 3
Usually a $\chi$-function based on Huber's $\psi$-function is used: $\chi ={\psi }^{2}/2$. Estimators based on such bounded $\psi$-functions are known as $M$-estimators, and provide one type of robust estimator.
Other robust estimators for the location argument are
(i) the sample median, (ii) the trimmed mean, i.e., the mean calculated after the extreme values have been removed from the sample, (iii) the winsorized mean, i.e., the mean calculated after the extreme values of the sample have been replaced by other more moderate values from the sample.
For the scale argument, alternative estimators are
(i) the median absolute deviation scaled to produce an estimator which is unbiased in the case of data coming from a Normal distribution, (ii) the winsorized variance, i.e., the variance calculated after the extreme values of the sample have been replaced by other more moderate values from the sample.
For a general discussion of robust estimation, see Hampel et al. (1986) and Huber (1981).
### 2.4 Robust Confidence Intervals
In Section 2.2 it was shown how tests of hypotheses can be used to find confidence intervals. That approach uses a parametric test that requires the assumption that the data used in the computation of the confidence has a known distribution. As an alternative, a more robust confidence interval can be found by replacing the parametric test by a nonparametric test. In the case of the confidence interval for the location argument, a Wilcoxon test statistic can be used, and for the difference in location, computed from two samples, a Mann–Whitney test statistic can be used.
## 3 Recommendations on Choice and Use of Available Functions
Maximum Likelihood Estimation and Confidence Intervals
nag_binomial_ci (g07aac) provides a confidence interval for the argument $p$ of the binomial distribution. nag_poisson_ci (g07abc) provides a confidence interval for the mean argument of the Poisson distribution. nag_censored_normal (g07bbc) provides maximum likelihood estimates and their standard errors for the arguments of the Normal distribution from grouped and/or censored data. nag_estim_weibull (g07bec) provides maximum likelihood estimates and their standard errors for the arguments of the Weibull distribution from data which may be right-censored. nag_estim_gen_pareto (g07bfc) provides maximum likelihood estimates and their standard errors for the parameters of the generalized Pareto distribution. nag_2_sample_t_test (g07cac) provides a $t$-test statistic to test for a difference in means between two Normal populations, together with a confidence interval for the difference between the means.
Robust Estimation
nag_robust_m_estim_1var (g07dbc) provides $M$-estimates for location and, optionally, scale using four common forms of the $\psi$-function. nag_robust_m_estim_1var_usr (g07dcc) produces the $M$-estimates for location and, optionally, scale but for user-supplied $\psi$- and $\chi$-functions. nag_median_1var (g07dac) provides the sample median, median absolute deviation, and the scaled value of the median absolute deviation. nag_robust_trimmed_1var (g07ddc) provides the trimmed mean and winsorized mean together with estimates of their variance based on a winsorized variance.
Robust Internal Estimation
nag_rank_ci_1var (g07eac) produces a rank based confidence interval for locations. nag_rank_ci_2var (g07ebc) produces a rank based confidence interval for the difference in location between two populations.
Outlier Detection
This chapter provides two functions for identifying potential outlying values, nag_outlier_peirce (g07gac) and nag_outlier_peirce_two_var (g07gbc). Many of the model fitting functions, for examples those in Chapters g02 and g13 also return vectors of residuals which can also be used to aid in the identification of outlying values.
## 4 Functionality Index
2 sample t-test nag_2_sample_t_test (g07cac)
Confidence intervals for parameters:
binomial distribution nag_binomial_ci (g07aac)
Poisson distribution nag_poisson_ci (g07abc)
Maximum likelihood estimation of parameters:
Normal distribution, grouped and/or censored data nag_censored_normal (g07bbc)
Weibull distribution nag_estim_weibull (g07bec)
Outlier detection:
Peirce
raw data or single variance supplied nag_outlier_peirce (g07gac)
two variances supplied nag_outlier_peirce_two_var (g07gbc)
Parameter estimates,
generalized Pareto distribution nag_estim_gen_pareto (g07bfc)
Robust estimation:
confidence intervals:
one sample nag_rank_ci_1var (g07eac)
two samples nag_rank_ci_2var (g07ebc)
median, median absolute deviation and robust standard deviation nag_median_1var (g07dac)
M-estimates for location and scale parameters:
standard weight functions nag_robust_m_estim_1var (g07dbc)
trimmed and winsorized means and estimates of their variance nag_robust_trimmed_1var (g07ddc)
user-defined weight functions nag_robust_m_estim_1var_usr (g07dcc)
None.
## 6 References
Cox D R and Hinkley D V (1974) Theoretical Statistics Chapman and Hall
Hampel F R, Ronchetti E M, Rousseeuw P J and Stahel W A (1986) Robust Statistics. The Approach Based on Influence Functions Wiley
Huber P J (1981) Robust Statistics Wiley
Kendall M G and Stuart A (1973) The Advanced Theory of Statistics (Volume 2) (3rd Edition) Griffin
Silvey S D (1975) Statistical Inference Chapman and Hall
g07 Chapter Contents
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# What is meant by non minimum phase systems?
June 20, 2019 Off By idswater
## What is meant by non minimum phase systems?
👉 Non-minimum Phase (NMP) systems are causal and stable systems whose inverses are causal but unstable. [ 2] Having a delay in our system or a model zero on the right half of the s-plane (aka Right-Half Plane or RHP) may lead to a non-minimum phase system.
## What do you mean by minimum and non minimum phase system?
1.6 Non-Minimum Phase System: Unstable Zeros G1(s) is minimum phase since it does not have unstable zeros/poles. The magnitude of G1(s) is 0dB, and the phase is 0∘. G2(s) is non-minimum phase, since G2(1)=0. One can check that 1/G2(s)=1+s1−s is unstable.
## What is Mcq minimum phase?
Minimum phase system: It is a system in which poles and zeros will not lie on the right side of the s-plane. Non-minimum phase system: It is a system in which some of the poles and zeros may lie on the right side of the s-plane. In particular, zeros lie on the right side of the s-plane.
## What is non-minimum phase zero?
Non-minimum Phase systems are causal and stable systems whose inverses are causal but unstable[2]. Having a delay in our system or a model zero on the right half of the s−plane (aka Right-Half Plane or RHP) may lead to a non-minimum phase system.
## What is root locus Mcq?
Explanation: The root locus is the locus of the change of the system parameters of the characteristic equation traced out in the s-plane. 7. If the gain of the system is reduced to a zero value, the roots of the system in the s-plane, a) Coincide with zero.
## When the system gain is doubled the gain margin becomes?
Therefore, if the system gain is doubled, gain margin is half i.e 1/2 times.
## How many branches of root locus tends towards infinity?
You can study other questions, MCQs, videos and tests for Electrical Engineering (EE) on EduRev and even discuss your questions like When the number of poles is equal to the number of zeroes, how many branches of root locus tends towards infinity? a)1b)2c)0d)Equal to number of zeroesCorrect answer is option ‘C’.
## Why root locus is symmetrical in real axis?
The root locus is a graphical representation in s-domain, and it is symmetrical about the real axis. Because the open loop poles and zeros exist in the s-domain having the values either as real or as complex conjugate pairs.
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Code Comfort
Other places to visit
Woolly Thoughts Home
Ravelry
Payhip
LoveCrafts
Easter Conference time came round again. As usual there were many interesting ideas to play with but the real highlight was meeting Simon Singh, author of Fermat’s Last Theorem and The Code Book. I had previously listened to the reading of The Code Book on Radio 4. I found it fascinating but it was nothing compared with the impact of the man himself. Simon gave an amazing talk about many different types of encryption. I learned a great deal but the I-could-knit-that-moment came from the simplest code of all, where one letter or number is substituted for another. I could substitute colours instead. I told Simon of my plan and that I would send him a picture when it was complete.
It didn’t take long to finish. The planning presented a few problems. We needed twenty-six colours to represent the letters of the alphabet. It caused a similar problem to that we had with Counting Pane many years before. It is just not possible to find so many colours that can be easily identified. To crack the code it had to be possible to tell the difference between the twenty-six symbols. We could manage thirteen colours so each had to be used twice in obviously different ways. Thirteen could be knitted, thirteen could be crochet, with the same lacy holes I had recently used for Surface Tension. That was one problem solved.
The next challenge was to decide on the message and the number of squares to be used. We weren’t even sure how many letters we needed to use for there to be enough clues to crack the code. More by chance than anything else we decided on a 16 x 16 square, giving 256 squares altogether. We later discovered that half this number would be enough to crack most simple messages.
Cracking this type of code relies on knowing the frequency of the letters in a normal piece of writing and this is generally the same whether you are reading a child’s book or an academic tome. The most frequent letter is E, followed by A, H, N, O, S, and T (not necessarily in that order). When the same 3 letter word appears more than once in a sentence it will probably be THE or AND. If it ends in E it will probably be THE so all the other Ts, Hs and Es can be identified. A 3 letter word at the beginning of a sentence is not likely to be AND. A word of one letter is very likely to be A or I. Two squares the same, next to each other, means there must be a double letter. B, L, P, S and T commonly occur as double letters, others are rarer or non-existent. These clues will work for almost all messages, unless there are a lot of proper nouns, which are more unpredictable.
We decided on a message that was relevant to the afghan itself, which left us with two more problems. The first was how we should represent spaces between words. On paper, a space shows where a word ends but we couldn’t leave a hole. The solution was to use a ‘background’ colour of navy blue. Spaces between words were navy crochet squares; spaces between sentences were navy knitted squares. This left the final problem of fitting the message, including these squares, into the total of 256. A little juggling and rewording provided a suitable solution. The afghan was named Code Comfort. The message says:
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# Does Nonsingular mean full rank?
## Does Nonsingular mean full rank?
The concept of nonsingular matrix is for square matrix, it means that the determinant is nonzero, and this is equivalent that the matrix has full-rank. Nonsingular means the matrix is in full rank and you the inverse of this matrix exists.
Are all square matrices Nonsingular?
A square matrix that is not invertible is called singular or degenerate. A square matrix is singular if and only if its determinant is zero. Non-square matrices (m-by-n matrices for which m ≠ n) do not have an inverse. However, in some cases such a matrix may have a left inverse or right inverse.
Do Diagonalizable matrices have full rank?
A diagonalizable matrix does not imply full rank (or nonsingular).
### How do you check if matrix is nonsingular?
Find the determinant of the matrix. If and only if the matrix has a determinant of zero, the matrix is singular. Non-singular matrices have non-zero determinants. Find the inverse for the matrix.
How do you prove a matrix is nonsingular?
If the matrix has an inverse, then the matrix multiplied by its inverse will give you the identity matrix. The identity matrix is a square matrix with the same dimensions as the original matrix with ones on the diagonal and zeroes elsewhere. If you can find an inverse for the matrix, the matrix is non-singular.
Is the transpose of a nonsingular matrix nonsingular?
The Transpose of a Nonsingular Matrix is Nonsingular Let A be an n×n nonsingular matrix. Trace of the Inverse Matrix of a Finite Order Matrix Let A be an n×n matrix such that Ak=In, where k∈N and In is the n×n identity matrix.
## Does nonsingular mean invertible?
The multiplicative inverse of a square matrix is called its inverse matrix. If a matrix A has an inverse, then A is said to be nonsingular or invertible. A singular matrix does not have an inverse.
How to identify singular and non-singular matrices?
A square matrix A is said to be singular if |A| = 0. A square matrix A is said to be non-singular if | A | ≠ 0. Identify the singular and non-singular matrices: In order to check if the given matrix is singular or non singular, we have to find the determinant of the given matrix. Hence the matrix is singular matrix. It is not equal to zero.
How do you find the determinant of a nonsingular matrix?
If A is a square nonsingular matrix of order N then its determinant is given by det ( A) = a, a ∈ R. The determinant of A is defined in terms of N − 1 determinants [5]: where A1j is an ( N − 1) × ( N − 1) matrix obtained by deleting the first row and j th column of A. Useful properties of the determinant include:
### How do you find the eigenvalues of a nonsingular matrix?
If A is an n × n nonsingular matrix, the eigenvalues of A−1 are the reciprocals of those for A, and the eigenvectors remain the same. A non-singular matrix is a square one whose determinant is not zero.
What is the rank of a non-singular matrix?
Thus, a non-singular matrix is also known as a full rank matrix. For a non-square [ A] of m × n, where m > n, full rank means only n columns are independent. There are many other ways to describe the rank of a matrix. In linear algebra, it is possible to show that all these are effectively the same.
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A120714 Expansion of 2*(4*x^2+14*x+7)*x^2/((-1-x+x^2)*(6*x^3+10*x^2+2*x-1)). 2
0, 14, 42, 232, 974, 4522, 20180, 91422, 411782, 1858856, 8384078, 37827386, 170648724, 769875718, 3473203086, 15669055544, 70689396502, 318908566562, 1438725432052, 6490672907694, 29282051536966, 132103184740456 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Previous name was: Sequence produced by 7 X 7 Markov chain based on adjacency matrix of 7-vertex graph with 10 edges, derived from the Fano plane. Take the standard 7-vertex 7-edge Fano plane graph and add three edges that go around the triangle vertices from the middle of the sides ( connecting the middle of the sides without going through the center) Characteristic polynomial is 6 - 2 x - 24 x^2 - 3 x^3 + 26 x^4 + 15 x^5 - x^7. LINKS Eric Weisstein's World of Mathematics, Fano Plane Index entries for linear recurrences with constant coefficients, signature (0,15,26,-3,-24,-2,6). FORMULA a(n)=15a(n-2)+26a(n-3)-3a(n-4)-24a(n-5)-2a(n-6)+6a(n-7). O.g.f.: 2*(4*x^2+14*x+7)*x^2/((-1-x+x^2)*(6*x^3+10*x^2+2*x-1)). - R. J. Mathar, Dec 05 2007 MAPLE a[1]:=0: a[2]:=14: a[3]:=42: a[4]:=232: a[5]:=974: a[6]:=4522: a[7]:=20180: a[8]:=91422: for n from 9 to 25 do a[n]:=15*a[n-2]+26*a[n-3]-3*a[n-4]-24*a[n-5]-2*a[n-6]+6*a[n-7] end do: seq(a[n], n=1..25); MATHEMATICA LinearRecurrence[{0, 15, 26, -3, -24, -2, 6}, {0, 14, 42, 232, 974, 4522, 20180}, 30] (* Harvey P. Dale, Sep 20 2011 *) CROSSREFS Cf. A111384, A120715. Sequence in context: A212514 A292051 A242897 * A041378 A302219 A302665 Adjacent sequences: A120711 A120712 A120713 * A120715 A120716 A120717 KEYWORD nonn,easy AUTHOR Roger L. Bagula, Aug 12 2006 EXTENSIONS Edited by N. J. A. Sloane, Jul 14 2007, Jul 28 2007 New name using g.f. from Joerg Arndt, Sep 21 2021 STATUS approved
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While dealing with Mathematics, a normal brain will be confused in its initial stages about what complementary and supplementary angles are. Therefore, it is important to learn in-depth about complementary supplementary angles. Complementary angles might be sounding like we are complementing an angle, right? But, no! The complementary angles definition is different when it is learned under Mathematics. So what exactly do complementary supplementary angles mean? Let us understand more about them.
Complementary Angles: More than complementing angles
We have arrived at the point where we will define complementary angles? One common question in everyone’s mind – what are complementary angles? In that case, complementary angles are those whose sum is equal to 90 degrees. Yes, exactly 90 degrees. In other terms, angles making a right angle can be termed complementary angles. The word complementary comes from a Latin word, completum, meaning completed. So, can we say a right angle is a complete angle? Think it over.
More fascinating though is, cutting a slice of bread into two triangles will yield two right angles having a pair each of complementary angles. So it is clear why the right angle is known as a complete angle.
Do you refer to Merriam Webster while looking for any definition of an unknown word or phrase? For example, while looking for complementary angles, one will see complementary air and complementary cell. That is not what we want, right? Yet, the look-up popularity in Merriam Webster for complementary angles is top 7% of words. How fascinating is that!
According to Merriam Webster, the definition of the complementary angle is two angles that add up to 90 degrees. Moreover, be careful while spelling the word. Many make mistakes and spell complementary instead of complementary. The former is incorrect, while the latter is correct.
Consider the figure as shown. On the left side, the angle AXB is a right angle, i.e., the value of angle AXB is equal to 90 degrees. On the right-hand side, angles BXC and CXD make a right angle when joined together. When these angles are considered separately, they are referred to as complementary angles.
As a result, angle BXC and angle CXD are complementary angles or complement each other. Speak like this: angle BXC complements angle CXD and vice versa. Mathematically, if angle BXC plus angle CXD is equal to 90 degrees, they are complementary.
Finding complementary angles is not difficult anymore!
Can we determine which one is a complementary angle and which one is not? Not so easily. After knowing the definition of the complementary angle, let us look at how to determine the complementary angles. We can measure the angles in a complementary angle with the help of a protractor. If the angle is less than 90 degrees, we can say that the angle is complementary. If it’s greater than 90, then? Need to wait for a little while when we jump to supplementary angles.
Are complementary angles always joined with each other?
Some people may try to figure out whether complementary angles are always joined with each other or not. What do you think? Are they always joined with each other? Two types of complementary angles are known in geometry: Non-adjacent and adjacent complementary angles. What are they? Continue reading further.
As the term adjacent denotes, these complementary angles must be near each other. The angles joined with a common vertex/common arm or are a part of the same right angle are referred to as adjacent complementary angles.
Consider the figure demonstrated below. Since both the angles BOC and BOA have a common vertex O and both make a right angle, they are said to be adjacent complementary angles. Why a right angle? Because 70 + 20 = 90 degrees, a right angle!
Since we know about complementary angles, we can determine what non-adjacent complementary angles mean. Those angles whose vertex is not common and make up a right angle are non-adjacent complementary angles. Simple to understand.
Let us understand non-adjacent complementary angles by a pictorial representation. As it can be seen, two angles, ABC and QPR, add up to 90 degrees and make a right angle. How? 40 + 50 = 90 degrees. However, these angles do not have the same vertex. Can we still refer to them as complementary? Well, yes! Non-adjacent complementary angles do not have a common vertex but add up, making a right angle.
Finding the complement of a complementary angle
Since now we know what are complementary angles, can we notice a common point in the definitions we studied above? Right angle and 90 degrees! The entire concept of complementary angles revolves around the right angle. So, to find the complement of a complementary angle, one has to subtract the known value of the angle from 90. The resulting value will be the complement of the complementary angle.
Complementary Angle Theorem: Special case of complementary angles
After learning what are complementary angles, time to know a special case of complementary angles. According to the complementary angle theorem, the angles will be congruent if they are complementary.
Consider the illustration below. From this, if we say angle POA and angle POQ are complementary angles, and angle POQ and angle QOR are complementary angles, then due to a common angle, we can say angle POA is equal to QOR. And that is the proof of the complementary angle theorem.
Mathematically, angle POA + angle POQ = angle POA + angle QOR.
Therefore, angle POQ = angle QOR, according to the complementary angle theorem.
Supplementary Angles: Straight and non-complementary
Suppose people were confused between complementary supplementary angles, and they knew one deal with 90 degrees and the other with 180 degrees. In that case, it might be clear by now that supplementary angles deal with 180 degrees. Therefore, if the sum of two angles equals 180 degrees, i.e., the angle is straight or formed as a line, it is a supplementary angle.
The term supplementary angles were first used in 1924. The look-up popularity in Merriam Webster is top 6%. One might have trouble while learning complementary and supplementary angles. Rule of thumb is complementary means corner, focus on the letter C, whereas supplementary is straight, focus on the word S. And that’s it! As easy as it is!
Let us understand it with the help of an illustration. In the figure below, angles AXD and CXD are supplementary because they form 180 degrees, i.e., a straight line. Similarly, angles BXC and CXD are supplementary. Can you figure out the other two pairs of supplementary angles?
One must notice the special case in this figure. Angles BXC and AXD will be equal to each other. And angels BXA and CXD will be equal to each other. How? That is the property of supplementary angles. If they intersect at one common point, the alternate angles will be equal. This is a special property used in Mathematics while solving supplementary angle questions. Check the angles by yourself using a protractor.
After going through the adjacent and non-adjacent complementary angles in complementary angles definition, one can determine what adjacent and non-adjacent supplementary angles will be. For example, if two angles have a common arm or vertex and add up to 180 degrees, they are adjacent supplementary angles. Whereas, if they do not have a common arm or vertex and still add up to 180 degrees, they are referred to as non-adjacent supplementary angles.
Check the demonstrated images below to figure out adjacent and non-adjacent supplementary angles.
Finding the supplement of a supplementary angle
Have you figured out how to find the supplement of an angle? If yes, it’s quite simple, no? The way complementary angles deal with 90 degrees, the same way supplementary angles revolve around 180 degrees. Therefore, to find the supplement of a supplementary angle, subtract the known angle from 180, and the leftover value will be the supplement.
Time to practice complementary and supplementary angles
Example 1: What will be the value of complementary angles P and Q if they are (3x – 10)° and (3x – 50)°, respectively?
Solution: Since two angles, P and Q, are complementary; therefore, according to complementary angles definition their sum will be equal to 90 degrees.
angle P + angle Q = 90°
3x – 10 + 3x – 50 = 90
On solving the above equation, we get, x = 25.
Therefore, the value of angles P and Q = (3(25) – 10)° and (3(25) – 50)° = 65° and 25°, respectively.
Example 2: Find the value of angles X and Y if their value is (4x – 80) and (6x – 45), respectively. Given that X and Y are supplementary angles.
Solution: Since the given angles X and Y are supplementary; therefore, their sum will be equal to 180°.
angle X + angle Y = 180°
4x – 80 + 6x – 45 = 180
Solving the above equation, we get, x = 30.5.
Therefore, angle X = 4(30.5) – 80 = 42°, and angle Y = 6(30.5) – 45 = 138°.
Example 3: What is the value of the other angle as shown in the figure?
Solution: Since the given angle is a part of a right angle; therefore, the other angle must be 90 – 79 = 11 degrees.
1. What are complementary angles?
Complementary angles are defined as two angles whose sum is 90 degrees. Two angles that add up to 90 degrees are complementary.
2. What are 5 complementary angles?
i) Angles of measure 60° and 30° are complementary angles because 60° + 30° = 90°
The complementary angle of 60° is the angle measured 30°.
(ii) Complement of 30° is → 90° – 30° = 60°
(iii) Complement of 45° is → 90° – 45° = 45°
(iv) Complement of 55° is → 90° – 55° = 35°
(v) Complement of 75° is → 90° – 75° = 15°
3. How to Find Complementary Angles?
To find the complement of an angle, subtract the angle’s measurement from 90 degrees. The result will be the complement.
4. What is the complement of 30?
The measure of a complementary angle of 30 degrees is 60 degrees.
5. What is the complementary angle of 50?
The complement to 50 would be 90 – 50 = 40 degrees
6. Are Supplementary and Complementary Angles the Same?
Complementary angles are two angles with a sum of 90 degrees.
Supplementary angles are two angles with a sum of 180 degrees.
7. What is the Sum of Two Complementary Angles?
The sum of two complementary angles is 90 degrees.
8. What do Complementary Angles Add up to?
Two angles are called complementary when their measures add up to 90 degrees.
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# Transformation in Geometry Created by Ms. O. Strachan.
## Presentation on theme: "Transformation in Geometry Created by Ms. O. Strachan."— Presentation transcript:
Transformation in Geometry Created by Ms. O. Strachan
Aim: Identifying and describing transformation For this lesson we will: Rotate a geometric figure. Rotate a geometric figure. Reflect a figure over a line of symmetry. Reflect a figure over a line of symmetry. Translate a figure by sliding it to a different location. Translate a figure by sliding it to a different location. Use dilation by enlarging or reducing the size of a figure without changing its form or shape. Use dilation by enlarging or reducing the size of a figure without changing its form or shape.
Transformation A rule for moving every point in a plane figure to a new location.
Translation A transformation that moves each point in a figure the same distance in the same direction.
In a translation a figure slides up or down, or left or right. No change in shape or size. The location changes. In graphing translation, all x and y coordinates of a translated figure change by adding or subtracting.
Reflection A transformation where a figure is flipped across a line such as the x-axis or the y- axis.
In a reflection, a mirror image of the figure is formed across a line called a line of symmetry. No change in size. The orientation of the shape changes. In graphing, a reflection across the x -axis changes the sign of the y coordinate. A reflection across the y-axis changes the sign of the x-coordinate.
Rotation A transformation where a figure turns about a fixed point without changing its size and shape.
–In a rotation, figure turns around a fixed point, such as the origin. No change in shape, but the orientation and location change. –Rules for rotating a figure about the origin in graphing. –Rules for 90 degrees rotation- Switch the coordination of each point. Then change the sign of the y coordinate. Ex. A (2,1) to A’ ( 1,-2)
Dilation A transformation where a figure changes size.
Dilation In dilation, a figure is enlarged or reduced proportionally. No change in shape, but unlike other transformation, the size changes. In graphing, for dilation, all coordinates are divided or multiplied by the same number to find the coordinates of the image.
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A322241 G.f.: exp( Sum_{n>=1} A084605(n)^2 * x^n/n ), where A084605(n) is the central coefficient in (1 + x + 4*x^2)^n. 1
%I
%S 1,1,41,249,6305,77569,1665321,27724889,574252417,10958980929,
%T 228679916905,4671350051321,99292476904609,2107949882690241,
%U 45658568907254505,993562984208479193,21876513296218002433,484448162130512673665,10812975015547281792937,242647271141110287979513,5477046865641884201456033
%N G.f.: exp( Sum_{n>=1} A084605(n)^2 * x^n/n ), where A084605(n) is the central coefficient in (1 + x + 4*x^2)^n.
%C Compare to: exp( Sum_{n>=1} A084605(n) * x^n/n ) = (1-x - sqrt(1 - 2*x - 15*x^2))/(8*x^2), the g.f. of A091147.
%C Sequence A322240(n) = A084605(n)^2 has generating function 1 / AGM(1 + 15*x, sqrt((1 - 9*x)*(1 - 25*x)) ).
%e G.f.: A(x) = 1 + x + 41*x^2 + 249*x^3 + 6305*x^4 + 77569*x^5 + 1665321*x^6 + 27724889*x^7 + 574252417*x^8 + 10958980929*x^9 + 228679916905*x^10 + ...
%e such that
%e log(A(x)) = x + 81*x^2/2 + 625*x^3/3 + 21025*x^4/4 + 314721*x^5/5 + 8071281*x^6/6 + 155975121*x^7/7 + 3685097025*x^8/8 + ... + A084605(n)^2 * x^n/n + ...
%e RELATED SERIES.
%e The g.f. of A084605 equals the series
%e 1/sqrt(1 - 2*x - 15*x^2) = 1 + x + 9*x^2 + 25*x^3 + 145*x^4 + 561*x^5 + 2841*x^6 + 12489*x^7 + 60705*x^8 + 281185*x^9 + ... + A084605(n) * x^n/n + ...
%o (PARI) {a(n)=if(n==0, 1, polcoeff(exp(sum(m=1, n, polcoeff(1/sqrt(1 - 2*x - 15*x^2 +x*O(x^m)), m)^2 *x^m/m)+x*O(x^n)), n))}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A322240, A084605.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Dec 08 2018
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# Perfect square and perfect cube
Is there any predefined function in c++ to check whether the number is square of any number and same for the cube..
`sqrt(x)`, or in general, `pow(x, 1./2)` or `pow(x, 1./3)`
For example:
``````int n = 9;
int a = (int) sqrt((double) n);
if(a * a == n || (a+1) * (a+1) == n) // in case of an off-by-one float error
cout << "It's a square!\n";
``````
Edit: or in general:
``````bool is_nth_power(int a, int n) {
if(n <= 0)
return false;
if(a < 0 && n % 2 == 0)
return false;
a = abs(a);
int b = pow(a, 1. / n);
return pow((double) b, n) == a || pow((double) (b+1), n) == a;
}
``````
No, but it's easy to write one:
``````bool is_perfect_square(int n) {
if (n < 0)
return false;
int root(round(sqrt(n)));
return n == root * root;
}
bool is_perfect_cube(int n) {
int root(round(cbrt(n)));
return n == root * root * root;
}
``````
Try this:
``````#include<math.h>
int isperfect(long n)
{
double xp=sqrt((double)n);
if(n==(xp*xp))
return 1;
else
return 0;
}
``````
No, there are no standard c or c++ functions to check whether an integer is a perfect square or a perfect cube.
If you want it to be fast and avoid using the float/double routines mentioned in most of the answers, then code a binary search using only integers. If you can find an n with n^2 < m < (n+1)^2, then m is not a perfect square. If m is a perfect square, then you'll find an n with n^2=m. The problem is discussed here: https://math.stackexchange.com/questions/41337/efficient-way-to-determine-if-a-number-is-perfect-square
For identifying squares i tried this algorithm in java. With little syntax difference you can do it in c++ too. The logic is, the difference between every two consecutive perfect squares goes on increasing by 2. Diff(1,4)=3 , Diff(4,9)=5 , Diff(9,16)= 7 , Diff(16,25)= 9..... goes on. We can use this phenomenon to identify the perfect squares. Java code is,
`````` boolean isSquare(int num){
int initdiff = 3;
int squarenum = 1;
boolean flag = false;
boolean square = false;
while(flag != true){
if(squarenum == num){
flag = true;
square = true;
}else{
square = false;
}
if(squarenum > num){
flag = true;
}
squarenum = squarenum + initdiff;
initdiff = initdiff + 2;
}
return square;
}
``````
To make the identification of squares faster we can use another phenomenon, the recursive sum of digits of perfect squares is always 1,4,7 or 9. So a much faster code can be...
`````` int recursiveSum(int num){
int sum = 0;
while(num != 0){
sum = sum + num%10;
num = num/10;
}
if(sum/10 != 0){
return recursiveSum(sum);
}
else{
return sum;
}
}
boolean isSquare(int num){
int initdiff = 3;
int squarenum = 1;
boolean flag = false;
boolean square = false;
while(flag != true){
if(squarenum == num){
flag = true;
square = true;
}else{
square = false;
}
if(squarenum > num){
flag = true;
}
squarenum = squarenum + initdiff;
initdiff = initdiff + 2;
}
return square;
}
boolean isCompleteSquare(int a){
// System.out.println(recursiveSum(a));
if(recursiveSum(a)==1 || recursiveSum(a)==4 || recursiveSum(a)==7 || recursiveSum(a)==9){
if(isSquare(a)){
return true;
}else{
return false;
}
}else{
return false;
}
}
``````
For perfect square you can also do:
``````if(sqrt(n)==floor(sqrt(n)))
return true;
else
return false;
``````
For perfect cube you can:
``````if(cbrt(n)==floor(cbrt(n)))
return true;
else
return false;
``````
Hope this helps.
We could use the builtin truc function -
``````#include <math.h>
// For perfect square
bool is_perfect_sq(double n) {
double r = sqrt(n);
return !(r - trunc(r));
}
// For perfect cube
bool is_perfect_cube(double n) {
double r = cbrt(n);
return !(r - trunc(r));
}
``````
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http://physics.oregonstate.edu/portfolioswiki/doku.php?id=courses:order20:vforder20:vfcalculating&rev=1555011485
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This is an old revision of the document!
# Calculating Potentials
## Prerequisites
Students should be able to:
• Find the electric potential from a system of discrete point sources.
• Write the difference vector between two vectors (and its magnitude).
• (Optional) Write charge densities in terms of delta functions.
• Compute line integrals.
• Find power series approximations.
## In-class Content
### Lecture: Chop, Calculate, and Add
• To find the area under a curve, one may chop up the x-axis into small pieces (of width dx). The area under the curve is then found by calculating the area for each region of dx (which is dx times f(x)) and then summing up all of those areas. In the limit where dx is small enough, the sum becomes an integral.
• One could also find the area under a curve by chopping up both the x- and y-axes (chop), calculating the area of each small area under the curve (calculate), and adding all of those together with a double sum or double integral (add).
• This approach can be used to find the area of a cone, where the 'horizontal' length of each area is $r d\phi$ and the 'vertical' length is dr, giving an area of $dA = r d\phi dr$. It is important to make sure that the limits of integration are appropriate so that the integrals range over the whole area of interest.
• If one wants to calculate something other than length, area, or volume, such as if one sprinkled charge over a thin bar, then chop, calculate, and add still works. Again, chop the bar up into small lengths of dx. Then calculate the change dQ on each length (dQ = lambda dx), and add all of the dQs together in a sum or integral.
• This also works for calculating something (such as charge) over a volume. For a thick cylindrical shell with a charge density rho(vec r), chop the shell into small volumes d tau (which will be a product of 3 small lengths), multiply this volume by the charge density at each part of the shell, and add the resulting dQs together.
## Homework for Symmetries
1. (FiniteDisk)
1. Starting with the integral expression for the electrostatic potential due to a ring of charge, find the value of the potential everywhere along the axis of symmetry.
2. Find the electrostatic potential everywhere along the axis of symmetry due to a finite disk of charge with uniform (surface) charge density $\sigma$. Start with your answer to part (a)
3. Find two nonzero terms in a series expansion of your answer to part (b) for the value of the potential very far away from the disk.
2. (InfiniteDisk)
Find the electrostatic potential due to an infinite disk, using your results from the finite disk problem.
3. (PotentialConeGEM227)
A conical surface (an empty ice-cream cone) carries a uniform charge density $\sigma$. The height of the cone is $a$, as is the radius of the top. Find the potential at point $P$ (in the center of the opening of the cone), letting the potential at infinity be zero.
4. (WritingII)
Using the handout “Guiding Questions for Science Writing” as a guide, write up your solution for finding the electrostatic potential everywhere in space due to a uniform ring of charge. Be sure to include a series expansion along one of the axes of interest.
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http://www.denombre.com/en/table-37
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| 86,569,692 | 7,025 |
# Number 37
The number is written
37 thirty-seven
37 is a prime number.
Search another number:
Example: "1", "7", "- 17860000", "42 854 250,5"...
## Multiplication table 37
37
37 x 1 = 37 thirty-seven 37 x 2 = 74 seventy-four 37 x 3 = 111 one hundred eleven 37 x 4 = 148 one hundred forty-eight 37 x 5 = 185 one hundred eighty-five 37 x 6 = 222 two hundred twenty-two 37 x 7 = 259 two hundred fifty-nine 37 x 8 = 296 two hundred ninety-six 37 x 9 = 333 three hundred thirty-three 37 x 10 = 370 three hundred seventy 37 x 11 = 407 four hundred seven 37 x 12 = 444 four hundred forty-four 37 x 13 = 481 four hundred eighty-one 37 x 14 = 518 five hundred eighteen 37 x 15 = 555 five hundred fifty-five 37 x 16 = 592 five hundred ninety-two 37 x 17 = 629 six hundred twenty-nine 37 x 18 = 666 six hundred sixty-six 37 x 19 = 703 seven hundred three 37 x 20 = 740 seven hundred forty 37 x 21 = 777 seven hundred seventy-seven 37 x 22 = 814 eight hundred fourteen 37 x 23 = 851 eight hundred fifty-one 37 x 24 = 888 eight hundred eighty-eight 37 x 25 = 925 nine hundred twenty-five 37 x 26 = 962 nine hundred sixty-two 37 x 27 = 999 nine hundred ninety-nine 37 x 28 = 1036 one thousand thirty-six 37 x 29 = 1073 one thousand seventy-three 37 x 30 = 1110 one thousand one hundred ten 37 x 31 = 1147 one thousand one hundred forty-seven 37 x 32 = 1184 one thousand one hundred eighty-four 37 x 33 = 1221 one thousand two hundred twenty-one 37 x 34 = 1258 one thousand two hundred fifty-eight 37 x 35 = 1295 one thousand two hundred ninety-five 37 x 36 = 1332 one thousand three hundred thirty-two 37 x 37 = 1369 one thousand three hundred sixty-nine
#### Integrating the converter on your site
Copy and Paste the following code on your site for instant conversion module.
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| 3.78125 | 4 |
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https://www.instructables.com/%CE%A0-314159proof/
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| 693,273,631 | 33,502 |
Introduction: Finding the Value of Π
I will presnt a way to prove π = 3.14159 ............ using geometry and trigonometry................
it was originally made by David Coulson, Teacher / writer in maths and science
Step 1: Basic Diagram
The circumference of the circle is bigger than the perimeter of the inner hexagon,and is smaller than the perimeter of the outer hexagon...........................
Step 2: Inner Hexagon
The perimeter of the inner hexagon is exactly 6 radii..................
The perimeter of the outer hexagon is harder to work out.................
Step 3: Outer Hexagon
The length of each side is 2R Tan 30 2R Tan 30
So, the perimeter of the outer hexagon is
6 x 2R Tan 30 = 12R Tan 30
Step 4: Value of Pi
The circle is sandwiched between the inner hexagon and the outer hexagon.
6R < Circumference < 12RTan 30
c= 2 π R
therefore,
3 < π < 3.4
Step 5: More Detailed
we can improve on the estimate by doubling the number of sides.
Step 6: More Detailed Vale of Π
12 sin15 < π < 12 tan15
therefore,
3.106 < π < 3.21
Step 7: FORMULA
In general, for an n-sided polygon inside the circle and another one outside of the circle,
n x Sin (180/n) < p < n x Tan (180/n)
As the number of sides increase,
the value of p is squeezed more tightly by the upper and lower limits.
Step 8: Final Value
SEE the chart above based on previous formula.........
the max and min valve of π is same when no.of sides = 6,114
and thus the value of π is 3.141593
Step 9: Solving Problem
You can use geometry to get the COS of 60 degrees.........that is 1/2 or 0.5 ....and then we can use the half angle formulas to generate all the SINs and TANs you need after that.
Step 10: Value Found to Be 3.14159
So starting from 60 degrees, we can calculate the Sin, Cos and Tan of increasingly small angles..............
from the image you can see the value of π to be
3.14159
Runner Up in the
Pi Day Challenge 2016
Participated in the
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https://www.studypool.com/discuss/1223708/solve-this-mathematicsa-problem-for-me?free
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| 1,029,343,070 | 14,624 |
##### solve this mathematicsa problem for me
Mathematics Tutor: None Selected Time limit: 1 Day
Factor: 12x2 - 20x - 25
Oct 19th, 2015
# 9x2y8-3xyy2+4x6y3
## Final result :
``` xy3 • (4x5 + 9xy5 - 3)
```
### Reformatting the input :
(1): "y3" was replaced by "y^3". 4 more similar replacement(s).
## Step 1 :
``` Raise y to the 3rd power
```
#### Equation at the end of step 1 :
``` (((9•(x2))•(y8))-(3xy•(y2)))+((4•(x6))•y3)
```
## Step 2 :
``` Raise x to the 6th power
```
#### Equation at the end of step 2 :
``` (((9•(x2))•(y8))-(3xy•(y2)))+((4•x6)•y3)
```
## Step 3 :
``` Raise y to the 2nd power
```
#### Equation at the end of step 3 :
``` (((9•(x2))•(y8))-(3xy•y2))+22x6y3
```
## Step 4 :
``` Multiply 3xy by y2
```
#### Multiplying exponential expressions :
4.1 y1 multiplied by y2 = y(1 + 2) = y3
#### Equation at the end of step 4 :
``` (((9•(x2))•(y8))-3xy3)+22x6y3
```
## Step 5 :
``` Raise y to the 8th power
```
#### Equation at the end of step 5 :
``` (((9 • (x2)) • y8) - 3xy3) + 22x6y3
```
## Step 6 :
``` Raise x to the 2nd power
```
#### Equation at the end of step 6 :
``` (((9 • x2) • y8) - 3xy3) + 22x6y3
```
## Step 7 :
```Simplify 9x2y8-3xy3 + 22x6y3
```
## Step 8 :
#### Pulling out like terms :
7.1 Pull out like factors :
4x6y3 + 9x2y8 - 3xy3 = xy3 • (4x5 + 9xy5 - 3)
#### Trying to factor a multi variable polynomial :
7.2 Factoring 4x5 + 9xy5 - 3
Try to factor this multi-variable trinomial using trial and error
Factorization fails
## Final result :
``` xy3 • (4x5 + 9xy5 - 3)
```
Oct 19th, 2015
...
Oct 19th, 2015
...
Oct 19th, 2015
Dec 9th, 2016
check_circle
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| 4.125 | 4 |
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http://mathhelpforum.com/calculus/205052-f-x.html
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1. ## f'(x)?
find f'(x) if it is known that
d/dx [f(2x)]=x2
what does this mean?
2. ## Re: f'(x)?
We are told:
$\frac{d}{dx}(f(2x))=x^2$
Using the chain rule, we find:
$2f'(2x)=x^2$
$f'(2x)=\frac{x^2}{2}=\frac{4x^2}{8}=\frac{(2x)^2}{ 8}$ and so we must have:
$f'(x)=\frac{x^2}{8}$
| 138 | 289 |
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| 4.03125 | 4 |
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https://codeforces.com/blog/entry/20613
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ilyaraz's blog
By ilyaraz, history, 9 years ago,
Hi all,
I'm about to teach AVL trees for a section of 6.006, and while thinking it through, I came up with the following question, which, I strongly suspect, should be well-understood.
Say one wants to maintain a binary search tree with height , where n is the total number of nodes, under insertions. How quickly can one insert? Is time per insert possible?
For instance, AVL trees achieve height around and red-black trees — .
What do you think?
• +37
» 9 years ago, # | ← Rev. 4 → +21 Isn't it possible just to generalize AVL trees? E.g. ask for the following property: for each node, consider all of its grand-children of depth k, and require the depth of their subtrees to differ by at most 1. Then for a fixed k, the number of rotations to restore the property is logarithmic (smth like ).As to the limit on depth. Let k = 2; if we greedily construct worst-case for this, the size of smallest tree of depth n is described by fn = fn - 2 + 3fn - 3 + 3, i.e. fn is approximately 1.6717n, and we get the bound .If we increase k, it seems that this converges to 1 (e.g., for k = 10 Wolframalpha says that the bound is 1.11, for k = 20 — 1.05). I hope there are no obvious mistakes in my reasoning :)
• » » 9 years ago, # ^ | 0 (I don't understand why your comment has a negative rating.)Why the number of rotations can be estimated as 2k·depth in general case?
• » » » 9 years ago, # ^ | 0 There's O(depth) nodes along the insertion path. For each subtree of such a node there's O(2k) ancestors of grand-children of depth k. We rearrange the ancestors entirely from scratch.
• » » » » 9 years ago, # ^ | +5 It is more complex than it seems. Suppose k = 2, and children of the left subtree have depths n and n + 1, and for the right subtree — n + 1 and n + 2. Even if we break tree structure at depth k, we cannot re-arrange it so that it is balanced. Even if we break up the n + 2-depth node, it still is not enough. However, if we break up all nodes ≥ n + 1, it seems that it is always possible to re-arrange this.Though I'm not sure how to easily prove it, I guess the only option is brute force.
• » » 9 years ago, # ^ | 0 Yes, the bound 2k sounds non-trivial. I guess one should start with understanding the case k = 2.
• » » » 9 years ago, # ^ | +10 Ok, let's prove this for k = 2. (Disclaimer: not very formally, but IMO convincing enough :) )Lemma. Suppose one has a sequence of leaves with values 0, 1 of length l ≥ 16, . Then it is always possible to construct (full) quaternary tree on top of these leaves with the following property. Let , and for leaves, it is the value in the leaf (0 or 1). Then the property is as follows: for each inner node, max(depth of its kids) — min(depth of its kids) is at most 1.Proof. For l = 16 — brute force, for larger l — try some random sequences and verify that it doesn't find counter-example :) I'm not sure how to properly prove it, but brute force for 16 seems convincing. Note that for l < 16 the statement is not true (in fact, there are lots of counter-examples).Now back to the original task. Let's eliminate 2nd, 4th, ... layers of the BST and convert it to quaternary tree. Note that we don't require the original property for even layers as the result, but depth guarantees still stand. Let's add a new node, and list all the inner nodes where the property is broken (O(depth) of them). Let's go from bottom to top and restore the property applying the lemma.Suppose we have node v with broken property. Let's build the following auxiliary tree (with size limited by a small constant — say 256). Take v, its kids, and their kids. Then look at the m := min(depth of the nodes in the new tree), and take more kids until depths of all the leaves are ≤ m. We'll get a tree where all leaves have depth in the original tree either m or m - 1. Now let's apply the lemma: it immediately gives us the new structure of this (constant-size) part of the tree.Is it convincing enough? :) I think this strongly indicates that the same works for bigger k. Though proving this rigorously might be difficult.
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# 1.3 Rate laws of overall and elementary reactions In general the
```1.3 Rate laws of overall and elementary reactions
In general the equation describing the dependency of the reaction
rate r on the concentrations ci of reaction components i is called the
kinetic equation of the reaction:
r = f(c A ,c B ,c D ,c E ,...,c N )
(1.16)
Depending on the reaction the function f can be quite simple but
sometimes it may as well be rather complicated. It usually depends
on the concentrations of the components participating in the main
reaction but it can also depend on the concentrations of catalysts
and inhibitors. For the time being we will neglect this possibility.
For many reactions in ideal systems (gaseous reactions and
reactions in liquid phase) r is found to have the general form of rate
law given in Eq. (1.17).
r = k[ A ] [ B] ....[ L] = kcαAcβB……cλL
α
β
λ
(1.17)
In this rate equation (rate law) k is the reaction rate constant (rate
coefficient) and α, β and λ are often integers or half-integers. The
rate constant k is a function of temperature and pressure, but the
pressure dependency of k is small and is usually ignored. The
reaction is said to have order α in respect of the component A,
order β in respect of the component B, etc. and parameters α, β and
λ are called partial orders of the reaction.
The sum α + β +…+ λ = n is called the overall order (or simply
order) of the reaction. Since r has the units of concentration over
time, k has the units (concentration)1-n time-1. Therefore the rate
constant of a first order (n = 1) reaction has the units s-1 and is
independent of the units used for concentration which is a very
important general property of a first order reaction.
For elementary reactions (unit steps) only we can obtain the rate
law immediately from the chemical reaction equation. For example
if the forward reaction given by Eq. (1.1) can be considered as an
elementary reaction its rate is given by
r = k[ A ]
− νA
[ B] − ν = k[ A ] a [ B] b
B
(1.18)
where νA and νB are the general stoichiometric numbers of the
reactants involved in the elementary reaction given by Eq. (1.1) i.e.
-νA = a and -νB = b. Then the negation of the sum of the general
stoichiometric numbers of the reactants
− ( νA + ν B +... ) = −
∑ νi
(1.19)
i = reactant
is called the molecularity of the elementary reaction.
Some experimentally observed rate laws for homogeneous
reactions are given below:
H2 + Br2 → 2HBr
2N2O5 → 4NO2 + O2
H2 + I2 → 2HI
r=
1
2
k[ H 2 ][ Br2 ]
1 + j[ HBr ][ Br2 ]
(1.20)
r = k[N2O5]
(1.21)
r = k[H2][I2]
(1.22)
In general the rate law must be determined from experimental
measurements of reaction rate and cannot be deduced from the
reaction stoichiometry (except for in the special case of an
elementary reaction).
The reaction equation, Eq. (1.1) gives the overall stoichiometry of
the reaction but does not tell us very much about the mechanism by
which the reaction occurs. Each step in the mechanism of the
overall reaction is an independent reaction and is called an
elementary reaction (unit step). The most simple reaction consists
of one single elementary step. The classical example of such
elementary reaction is the reaction described by Eq. (1.22). It is
explained by the following molecular level reaction mechanism
A + B → A---B → C + B
where A---B denotes the so called activated complex first formed
between the reactant molecules A and B before the final
rearrangement to product molecules C and B.
The formally very similar reaction given by Eq. (1.20) is much
more complicated and is called a complex (composite) reaction. A
complex chemical reaction consists of at least two (and usually
more) elementary steps. A specific characteristic of the complex
reactions is that the exponentials (partial orders) in the rate law and
stoichiometric coefficients in the reaction equation do not fit. In
Chap. 6 we shall give a more detailed analyze for the mechanism
of this reaction.
An important concept is that some reactions can be characterized
by pseudo orders and pseudo constants. For example for the
hydrolysis of sucrose:
C12H22O11 + H2O → C6H12O6 + C6H12O6
Sucrose
Glucose Fructose
(1.23)
one finds that the rate is given by the rate law, r = k[C12H22O11].
However, since the solvent, H2O participates in the reaction, one
could expect that the rate law would preferably have the more
general form, r = k' [C12H22O11] [H2O]v.
Because H2O is always present in the reaction vessel in great
excess, [H2O]v is essentially constant and the rate law has the
apparent form r = k[C12H22O11], where k = k´[H2O]v. The reaction
is said to be of pseudo first order although the kinetic data indicates
for v ≈ 6. The reaction rate constant k is respectively called the
pseudo reaction rate constant.
The number of molecules which react in an elementary step is
called the molecularity of the elementary reaction. Molecularity is
defined only for elementary reactions and should not be used to
describe overall reactions that consist of more than one elementary
step. In fact the only elementary reaction which exists is the
bimolecular one. However, all the types of reactions given below
are considered to be elementary reactions. The unimolecular
reaction and the trimolecular reactions, however, are in fact simple
reactions from the kinetic point of view since they can be explained
by trivial combinations of bimolecular reactions. Examples of each
case are given below:
A → products
unimolecular reaction
A + B → products
2A → products
A + B + C → products
2A + B → products
3A → products
bimolecular reactions
trimolecular(termolecular) reactions.
No elementary reactions involving more than three molecules have
been suggested because of the very low probability of collision of
more than three molecules at the same moment of time, which of
course is the perquisite of a chemical reaction.
The rates of chemical reactions are obtained from measurements of
concentrations of reaction mixture components as a function of
time. Analytical chemistry methods may be applied to samples
taken from reaction mixture when the reaction can be stopped
suddenly. This may be done e.g. by rapid cooling of the samples
for high-temperature reactions.
Physical methods are especially useful for determining the rate of a
chemical reactions because they offer the possibility to measure
continuously the advancement of the reaction. Spectroscopic
methods are most generally used techniques. An important
characteristic of any measurement method is its response time. The
measuring device must obviously respond much more rapidly than
the concentrations are changing. Pulsed lasers have opened up new
opportunities for investigation of very fast reactions. Especially
reactions occurring in the picosecond (10-12 s) range may be studied
in this way.
To study gas reactions at high temperatures it is necessary to heat
the gas to the reaction temperature very quickly because the
reaction then occurs rapidly. This may be accomplished by means
of shock tube in which a shock wave is used to heat the gas with
high speed.
Photochemical reactions may be initiated suddenly by a light pulse
from a flash lamp or laser. In the flash photolysis method, a
reaction vessel is exposed to a very high intensity flash of visible or
ultraviolet light. The flash dissociates the molecules in the sample
and the concentrations of these species are then determined over a
period of time using subsequent flashes at much lower intensities.
```
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Home | New Examples | Popular | Top | Submit Examples | Request Examples
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Examples of Rational numbers
Rational numbers
Post by natttt
Posted on 2011-04-25 11:48:48
A rational number is a number that can be written as a simple fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers.
Are examples of rational numbers:
* The number 8 is a rational number because it can be written as the fraction 8/1.
* Likewise, 3/4 is a rational number because it can be written as a fraction.
* Even a big, clunky fraction like 7,324,908/56,003,492 is rational, simply because it can be written as a fraction.
Some irrational numbers are:
π = 3.141592…
square root of 2 = 1.414213…
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# Gauss Elimination Method Using C
Earlier in Gauss Elimination Method Algorithm and Gauss Elimination Method Pseudocode , we discussed about an algorithm and pseudocode for solving systems of linear equation using Gauss Elimination Method. In this tutorial we are going to implement this method using C programming language.
## Complete Program for Gauss Elimination method using C Programming Language
``````
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<stdlib.h>
#define SIZE 10
int main()
{
float a[SIZE][SIZE], x[SIZE], ratio;
int i,j,k,n;
clrscr();
/* Inputs */
/* 1. Reading number of unknowns */
printf("Enter number of unknowns: ");
scanf("%d", &n);
/* 2. Reading Augmented Matrix */
for(i=1;i<=n;i++)
{
for(j=1;j<=n+1;j++)
{
printf("a[%d][%d] = ",i,j);
scanf("%f", &a[i][j]);
}
}
/* Applying Gauss Elimination */
for(i=1;i<=n-1;i++)
{
if(a[i][i] == 0.0)
{
printf("Mathematical Error!");
exit(0);
}
for(j=i+1;j<=n;j++)
{
ratio = a[j][i]/a[i][i];
for(k=1;k<=n+1;k++)
{
a[j][k] = a[j][k] - ratio*a[i][k];
}
}
}
/* Obtaining Solution by Back Subsitution */
x[n] = a[n][n+1]/a[n][n];
for(i=n-1;i>=1;i--)
{
x[i] = a[i][n+1];
for(j=i+1;j<=n;j++)
{
x[i] = x[i] - a[i][j]*x[j];
}
x[i] = x[i]/a[i][i];
}
/* Displaying Solution */
printf("\nSolution:\n");
for(i=1;i<=n;i++)
{
printf("x[%d] = %0.3f\n",i, x[i]);
}
getch();
return(0);
}
``````
## Output: Gauss Elimination Method for Solving Systems of Linear Equations
```Enter number of unknowns: 3
a[1][1] = 1
a[1][2] = 1
a[1][3] = 1
a[1][4] = 9
a[2][1] = 2
a[2][2] = -3
a[2][3] = 4
a[2][4] = 13
a[3][1] = 3
a[3][2] = 4
a[3][3] = 5
a[3][4] = 40
Solution:
x[1] = 1.000
x[2] = 3.000
x[3] = 5.000
```
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# Freezing point on a thermometer is marked as 20 and boiling point as 150, a temperature of 60 on this thermometer will be read as
1
by TanybhVISoja
2016-03-24T18:10:06+05:30
SUPPOSE,
P = Ax + B
WHEN X = 0 , P = 20 ,THEN
20 + 0 = B
OR B = 20
WHEN X= 100 AND P= 150 THEN,
150 = 100A + 20
OR A = 1.3
NOW WHEN X = 60
THEN , P = 1.3 *60 + 20 = 98 WHICH IS THE REQUIRED TEMPERATURE
PLZ MARK BRAINLIEST
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Posted on Categories:CS代写, 算法, 计算机代写
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## 计算机代写|算法代写Algorithm代考|Minimum-Capacity Cuts
We may naturally think of the Maximum Flow problem as an attempt to maximize the number of resources that reach a collection of destinations. In this vein, there is a natural dual problem: what are the minimum number of disruptions needed to prevent any resources from reaching any of the destinations? The motivation for these problems arose during the Cold War. Here, the United States was interested in the Soviet Union Railway System that connected Eastern Europe- particularly, East Germany- and the western region of the Soviet Union. In particualr, the United States wanted to identify the minimum number of points to bomb in order to disrupt the flow of resources along this system [Ano, Sch02]. We will see later that the Maximum Flow problem is equivalent to finding the minimum number of disruptions. This is the celebrated Max-Flow Min-Cut Theorem.
We now turn to formalizing the Minimum Cut problem.
Definition 93. Let $\mathcal{N}(G, c, S, T)$ be a flow network. A cut of $\mathcal{N}$ is a partition of the vertices $(X, Y)$, where $S \subseteq X$ (that is, $X$ contains the source vertices), and $T \subseteq Y$ (that is, $Y$ contains the sink vertices). Note that as $(X, Y)$ is a partition, we have that $X$ and $Y$ are disjoint.
The capacity of the partition $(X, Y)$ is the sum of the edge capacities with the initial endpoint in $X$ and the destination vertex in $Y$. That is:
$$c(X, Y):=\sum_{x \in X} \sum_{y \in Y} c(x, y) .$$
Recall that if $(x, y)$ is not an edge of the flow network, then $c(x, y)=0$.
Definition 94. The Minimum Cut problem is defined as follows.
Instance: Let $\mathcal{N}(G, c, S, T)$ be a flow network.
Solution: A cut $(X, Y)$ such that $c(X, Y)$ is minimized.
## 计算机代写|算法代写Algorithm代考|Max-Flow Min-Cut Theorem
In this section, we prove the Max-Flow Min-Cut Theorem. Our proof is based on [Mou16b].
Theorem 97 (Max-Flow Min-Cut Theorem (Ford-Fulkerson, 1956).). Let $\mathcal{N}(G, c, S, T)$ be a flow network. Let $f^$ be a maximum-valued flow, and let $(X, Y)$ be a minimum-capacity cut. We have that val $\left(f^\right)=c(X, Y)$.
We prove Theorem 97. We first show that the value of a maximum flow is no bigger than the capacity of a minimum cut. We then show that there exists a cut whose capacity is no bigger than the value of a maximum flow. It follows from this second claim that the capacity of a minimum cut is no bigger than the value of a maximum flow.
We begin by showing that the value of a maximum flow is no bigger than the capacity of a minimum cut. To this end, we introduce the following lemma, which intuitively states that the amount of flow that we can push from the source nodes to the sink nodes cannot exceed the capacity of a cut.
Lemma 98. Let $\mathcal{N}(G, c, S, T)$ be a flow network, and let $f$ be a flow. Let $(X, Y)$ be a cut. We have that $\operatorname{val}(f) \leq c(X, Y)$
Proof. As $(X, Y)$ is a cut, we have by the conservation of flow the total flow that makes it from the source vertices to the sink vertices is the amount of flow leaving $X$, minus the amount of flow returning to $X$ from $Y$. This is precisely:
$$\operatorname{val}(f)=\sum_{u \in X} \sum_{v \in Y} f(u, v)-\sum_{u \in Y} \sum_{v \in X} f(u, v) .$$
By ignoring the flow coming back into $X$, we have that:
\begin{aligned} \operatorname{val}(f) & =\sum_{u \in X} \sum_{v \in Y} f(u, v)-\sum_{u \in Y} \sum_{v \in X} f(u, v) \ & =\sum_{u \in X} \sum_{v \in Y} f(u, v) \ & \leq \sum_{u \in X} \sum_{v \in Y} c(u, v) \ & =c(X, Y) \end{aligned}
## 计算机代写|算法代寻Algorithm代考|Minimum-Capacity Cuts
$$c(X, Y):=\sum_{x \in X} \sum_{y \in Y} c(x, y)$$
## 计算机代写|算法代写Algorithm代考|Max-Flow Min-Cut Theorem
\operatorname{val}(f)=\sum_{u \in X} \sum_{v \in Y} f(u, v)-\sum_{u \in Y} \sum_{v \in X} f(u, v)
$$通过芴略返回的流量 X ，我们有:$$
\operatorname{val}(f)=\sum_{u \in X} \sum_{v \in Y} f(u, v)-\sum_{u \in Y} \sum_{v \in X} f(u, v) \quad=\sum_{u \in X} \sum_{v \in Y} f(u, v) \leq \sum_{u \in X} \sum_{v \in Y} c(u, v) \quad=c(X, Y)
$$计算机代写|算法代写Algorithm代考 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。 ## 微观经济学代写 微观经济学是主流经济学的一个分支，研究个人和企业在做出有关稀缺资源分配的决策时的行为以及这些个人和企业之间的相互作用。my-assignmentexpert™ 为您的留学生涯保驾护航 在数学Mathematics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的数学Mathematics代写服务。我们的专家在图论代写Graph Theory代写方面经验极为丰富，各种图论代写Graph Theory相关的作业也就用不着 说。 ## 线性代数代写 线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。 ## 博弈论代写 现代博弈论始于约翰-冯-诺伊曼（John von Neumann）提出的两人零和博弈中的混合策略均衡的观点及其证明。冯-诺依曼的原始证明使用了关于连续映射到紧凑凸集的布劳威尔定点定理，这成为博弈论和数学经济学的标准方法。在他的论文之后，1944年，他与奥斯卡-莫根斯特恩（Oskar Morgenstern）共同撰写了《游戏和经济行为理论》一书，该书考虑了几个参与者的合作游戏。这本书的第二版提供了预期效用的公理理论，使数理统计学家和经济学家能够处理不确定性下的决策。 ## 微积分代写 微积分，最初被称为无穷小微积分或 “无穷小的微积分”，是对连续变化的数学研究，就像几何学是对形状的研究，而代数是对算术运算的概括研究一样。 它有两个主要分支，微分和积分；微分涉及瞬时变化率和曲线的斜率，而积分涉及数量的累积，以及曲线下或曲线之间的面积。这两个分支通过微积分的基本定理相互联系，它们利用了无限序列和无限级数收敛到一个明确定义的极限的基本概念 。 ## 计量经济学代写 什么是计量经济学？ 计量经济学是统计学和数学模型的定量应用，使用数据来发展理论或测试经济学中的现有假设，并根据历史数据预测未来趋势。它对现实世界的数据进行统计试验，然后将结果与被测试的理论进行比较和对比。 根据你是对测试现有理论感兴趣，还是对利用现有数据在这些观察的基础上提出新的假设感兴趣，计量经济学可以细分为两大类：理论和应用。那些经常从事这种实践的人通常被称为计量经济学家。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 Posted on Categories:CS代写, 算法, 计算机代写 ## 计算机代写|算法代写Algorithm代考|CS341 Prim’s Algorithm 如果你也在 怎样代写算法Algorithm CS341这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。算法Algorithm在数学和计算机科学中，算法（/ˈælɡərɪðəm/（听））是一个有限的严格指令序列，通常用于解决一类特定问题或进行计算。算法被用作进行计算和数据处理的规范。 算法Algorithm被用作进行计算和数据处理的规范。更高级的算法可以进行自动推理（被称为自动推理），并使用数学和逻辑测试来转移代码执行的各种路线（被称为自动决策）。以隐喻的方式将人类的特征作为机器的描述符，艾伦-图灵已经用 “记忆”、”搜索 “和 “刺激 “等术语进行了实践。相比之下，启发式是一种解决问题的方法，它可能没有被完全指定，或者不能保证正确或最佳的结果，特别是在没有明确定义的正确或最佳结果的问题领域。作为一种有效的方法，算法可以在有限的空间和时间内表达出来，并以一种定义明确的形式语言来计算一个函数。从一个初始状态和初始输入（也许是空的）开始，指令描述一个计算，当执行时，经过有限个定义明确的连续状态，最终产生 “输出”并终止于一个最终的终止状态。从一个状态到下一个状态的转换不一定是确定的；一些算法，即所谓的随机算法，包含了随机输入。 算法Algorithm代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。 最高质量的算法Algorithm作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此算法Algorithm作业代写的价格不固定。通常在专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。 ## avatest™帮您通过考试 avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！ 在不断发展的过程中，avatest™如今已经成长为论文代写，留学生作业代写服务行业的翘楚和国际领先的教育集团。全体成员以诚信为圆心，以专业为半径，以贴心的服务时刻陪伴着您， 用专业的力量帮助国外学子取得学业上的成功。 •最快12小时交付 •200+ 英语母语导师 •70分以下全额退款 想知道您作业确定的价格吗? 免费下单以相关学科的专家能了解具体的要求之后在1-3个小时就提出价格。专家的 报价比上列的价格能便宜好几倍。 我们在计算机Quantum computer代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的计算机Quantum computer代写服务。我们的专家在机器学习Machine Learning代写方面经验极为丰富，各种机器学习Machine Learning相关的作业也就用不着 说。 ## 计算机代写|算法代写Algorithm代考|Prim’s Algorithm In this section, we examine a second technique to construct minimum-weight spanning trees; namely, Prim’s algorithm. We again start with the intermediate spanning forest \mathcal{F} that contains all the vertices of our input graph G(V, E, w), but none of the edges. While Kruskal’s algorithm determines which edges to add to \mathcal{F} by examining the entire graph, Prim’s algorithm takes a more local perspective. We provide as input a specified source vertex s \in V(G). Let T^ be the component of \mathcal{F} that contains s. Prim’s algorithm examines the edges of G that have exactly one endpoint in T^ and select a light edge e from these to add to \mathcal{F}. As e has exactly one endpoint in T^*, e connects two distinct components of \mathcal{F}. So by Corollary 61, e is a safe edge with respect to \mathcal{F}. This is the key observation in establishing that Prim’s algorithm returns a minimum-weight spanning tree. We now turn to formalizing Prim’s algorithm. We associate to each edge an attribute processed to indicate whether that edge has been placed into the priority queue. This ensures that each edge is considered at most once, which helps ensure that the algorithm will terminate. Now at lines 6-8, we initialize the priority queue to contain only edges that are incident to the source vertex. This ensures that the first edge placed into the intermediate spanning forest is incident to the source vertex. Now by adding an edge to \mathcal{F}, we introduce a new vertex v to the component containing our source vertex. Prim’s algorithm then adds to the prioritiy queue the edges incident to v, provided such edges have not already been polled from the queue. So the while loop at line 9 preserves the invariant that every edge in the priority queue has at least one endpoint in the component containing our source vertex. Prim’s algorithm only adds an edge if it connects two components. Such an edge e is polled from the priority queue, and so (i) has an endpoint in the component containing the source vertex, and (ii) is a minimum-weight edge connecting two distinct components. Therefore, e is a safe edge. ## 计算机代写|算法代写Algorithm代考|Prim’s Algorithm: Example We now work through an example of Prim’s algorithm. Example 71. Consider the following graph G(V, E, w) pictured below. Suppose we select the source vertex A. Prim’s algorithm proceeds as follows. 1. We initialize the intermediate spanning forest to contain all the vertices of G, but no edges. We then initialize the priority queue to contain the edges incident to our source vertex A. So:$$
Q=[({A, B}, 10),({A, C}, 12)],
$$and our intermediate spanning forest \mathcal{F} is pictured below. A F C E 2. We poll the edge {A, B} from the queue and mark {A, B} as processed. Note that w({A, B})=10. As {A, B} has exactly one endpoint on the component containing A (which is the isolated vertex A ), we add {A, B} to \mathcal{F}. We then push into the priority queue the unprocessed edges incident to B. So:$$
Q=[({B, C}, 1),({B, D}, 7),({A, C}, 12)],
$$and the updated intermediate spanning forest \mathcal{F} is pictured below. ## 算法代写 ## 计算机代写|算法代写Algorithm代考|Prim’s Algorithm 在本节中，我们将研究第二种构建最小权重生成树的技术；即 Prim 算法。我们再次从中间生成森林开始 \mathcal{F} 包含我们输入图的所有 顶点 G(V, E, w) ，但没有边。虽然 Kruskal 的算法确定要添加到哪些边缘 \mathcal{F} 通过检龺整个图，Prim 的算法采用了更局部的视 角。我们提供指定的源页点作为输入 s \in V(G). 让缺少上标或下标参数 成为的组成部分 \mathcal{F} 包含 s. Prim 的算法检萛边缘 G 恰好有一个端点缺少上上标或下标参数 并选择浅色边缘 e 从这些添加到 \mathcal{F}. 作为 e 恰好有一个端点 T^*, e 连接两个不同的组件 \mathcal{F}. 所以根据推论 61, e 是一个安全边 \mathcal{F}. 这是确定 Prim 算法返回最小权重生成树的关键 观宪结果。 我们现在转向形式化 Prim 的算法。 我们将每条边关联到一个经过处理的属性，以指示该边是否已放入优先级队列。这确保每条边最多被考虑一次，这有助于确保算法 将终止。 现在在第 6-8 行，我们将优先级队列初始化为仅包含入射到源页点的边。这确保放置到中间生成森林中的第一条边与源页点相关 联。现在通过添加一个边缘 \mathcal{F} ，我们引入一个新的顶点 v 到包含我们的源项点的组件。然后 Prim 的算法将边缘事件添加到优先级 队人列中 v ，前提是尚末从队列中轮询此米边縁。因此，第 9 行的 while 循环保留了不变性，即优先级队人列中的每条边在包含我们的 源页点的组件中至少有一个端点。 Prim 的算法仅在连接两个组件时才添加一条边。这样的边豚 e 从优先级队列中轮询，因此 (i) 在包含源页点的组件中有一个端点， 并且 (ii) 是连接两个不同组件的最小权重边。所以， e 是安全边。 ## 计算机代写|算法代写Algorithm代考|Prim’s Algorithm: Example 我们现在研究Prim 算法的一个例子。 示例 71. 考虑下图 G(V, E, w) 如下图所示。假设我们选择源页点 A. Prim 的算法如下进行。 1. 我们初始化中间生成森林以包含所有的顶点 G ，但没有边。然后我们初始化优先级队列以包含入射到我们的源项点的边 A. 所以:$$
Q=[(A, B, 10),(A, C, 12)]
$$和我们的中间生成林 \mathcal{F} 如下图所示。 A \mathrm{F} C E 2. 我们轮询边溕 A, B 从队列中标记 A, B 作为处理。注意 w(A, B)=10. 作为 A, B 在包含的组件上只有一个端点 A (这是孤 立的顶点 A )，我们增加 A, B 到 \mathcal{F}. 然后我们将末处理的边褖推入优先级队列 B. 所以:$$
Q=[(B, C, 1),(B, D, 7),(A, C, 12)],
$$和更新的中间生成林 \mathcal{F} 如下图所示。 计算机代写|算法代写Algorithm代考 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。 ## 微观经济学代写 微观经济学是主流经济学的一个分支，研究个人和企业在做出有关稀缺资源分配的决策时的行为以及这些个人和企业之间的相互作用。my-assignmentexpert™ 为您的留学生涯保驾护航 在数学Mathematics作业代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的数学Mathematics代写服务。我们的专家在图论代写Graph Theory代写方面经验极为丰富，各种图论代写Graph Theory相关的作业也就用不着 说。 ## 线性代数代写 线性代数是数学的一个分支，涉及线性方程，如：线性图，如：以及它们在向量空间和通过矩阵的表示。线性代数是几乎所有数学领域的核心。 ## 博弈论代写 现代博弈论始于约翰-冯-诺伊曼（John von Neumann）提出的两人零和博弈中的混合策略均衡的观点及其证明。冯-诺依曼的原始证明使用了关于连续映射到紧凑凸集的布劳威尔定点定理，这成为博弈论和数学经济学的标准方法。在他的论文之后，1944年，他与奥斯卡-莫根斯特恩（Oskar Morgenstern）共同撰写了《游戏和经济行为理论》一书，该书考虑了几个参与者的合作游戏。这本书的第二版提供了预期效用的公理理论，使数理统计学家和经济学家能够处理不确定性下的决策。 ## 微积分代写 微积分，最初被称为无穷小微积分或 “无穷小的微积分”，是对连续变化的数学研究，就像几何学是对形状的研究，而代数是对算术运算的概括研究一样。 它有两个主要分支，微分和积分；微分涉及瞬时变化率和曲线的斜率，而积分涉及数量的累积，以及曲线下或曲线之间的面积。这两个分支通过微积分的基本定理相互联系，它们利用了无限序列和无限级数收敛到一个明确定义的极限的基本概念 。 ## 计量经济学代写 什么是计量经济学？ 计量经济学是统计学和数学模型的定量应用，使用数据来发展理论或测试经济学中的现有假设，并根据历史数据预测未来趋势。它对现实世界的数据进行统计试验，然后将结果与被测试的理论进行比较和对比。 根据你是对测试现有理论感兴趣，还是对利用现有数据在这些观察的基础上提出新的假设感兴趣，计量经济学可以细分为两大类：理论和应用。那些经常从事这种实践的人通常被称为计量经济学家。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 Posted on Categories:CS代写, 算法, 计算机代写 ## 计算机代写|算法代写Algorithm代考|CS473 Greedy Algorithm Principles 如果你也在 怎样代写算法Algorithm CS473这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。算法Algorithm在数学和计算机科学中，算法（/ˈælɡərɪðəm/（听））是一个有限的严格指令序列，通常用于解决一类特定问题或进行计算。算法被用作进行计算和数据处理的规范。 算法Algorithm被用作进行计算和数据处理的规范。更高级的算法可以进行自动推理（被称为自动推理），并使用数学和逻辑测试来转移代码执行的各种路线（被称为自动决策）。以隐喻的方式将人类的特征作为机器的描述符，艾伦-图灵已经用 “记忆”、”搜索 “和 “刺激 “等术语进行了实践。相比之下，启发式是一种解决问题的方法，它可能没有被完全指定，或者不能保证正确或最佳的结果，特别是在没有明确定义的正确或最佳结果的问题领域。作为一种有效的方法，算法可以在有限的空间和时间内表达出来，并以一种定义明确的形式语言来计算一个函数。从一个初始状态和初始输入（也许是空的）开始，指令描述一个计算，当执行时，经过有限个定义明确的连续状态，最终产生 “输出”并终止于一个最终的终止状态。从一个状态到下一个状态的转换不一定是确定的；一些算法，即所谓的随机算法，包含了随机输入。 算法Algorithm代写，免费提交作业要求， 满意后付款，成绩80\%以下全额退款，安全省心无顾虑。专业硕 博写手团队，所有订单可靠准时，保证 100% 原创。 最高质量的算法Algorithm作业代写，服务覆盖北美、欧洲、澳洲等 国家。 在代写价格方面，考虑到同学们的经济条件，在保障代写质量的前提下，我们为客户提供最合理的价格。 由于作业种类很多，同时其中的大部分作业在字数上都没有具体要求，因此算法Algorithm作业代写的价格不固定。通常在专家查看完作业要求之后会给出报价。作业难度和截止日期对价格也有很大的影响。 ## avatest™帮您通过考试 avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！ 在不断发展的过程中，avatest™如今已经成长为论文代写，留学生作业代写服务行业的翘楚和国际领先的教育集团。全体成员以诚信为圆心，以专业为半径，以贴心的服务时刻陪伴着您， 用专业的力量帮助国外学子取得学业上的成功。 •最快12小时交付 •200+ 英语母语导师 •70分以下全额退款 想知道您作业确定的价格吗? 免费下单以相关学科的专家能了解具体的要求之后在1-3个小时就提出价格。专家的 报价比上列的价格能便宜好几倍。 我们在计算机Quantum computer代写方面已经树立了自己的口碑, 保证靠谱, 高质且原创的计算机Quantum computer代写服务。我们的专家在机器学习Machine Learning代写方面经验极为丰富，各种机器学习Machine Learning相关的作业也就用不着 说。 ## 计算机代写|算法代写Algorithm代考|Exchange Arguments In this section, we explore a key proof technique used in establishing the correctness of greedy algorithms; namely, the notion of an exchange argument. The key idea is to start with a solution (multi)set S and show that we may swap out or exchange elements of S in such a way that improves the solution. Understanding which elements to exchange often provides key insights into designing effective greedy algorithms. Such provable observations imply the correctness of our greedy algorithms. Example 32. Recall the Making Change problem, where we have an infinite supply of pennies (worth 1 cent), nickels (worth 5 cents), dimes (worth 10 cents), and quarters (worth 25 cents). We take as input an integer n \geq 0. The goal is to make change for n using the fewest number of coins possible. The greedy algorithm chooses as many quarters as possible, followed by as many dimes as possible, then as many nickels as possible. Finally, the greedy algorithm uses pennies to finish making change. Why is the greedy algorithm correct? Why does it select dimes before nickels? Exchange arguments allow us to answer this question. Consider the following lemma. Lemma 33. Let n \in \mathbb{N} be the amount for which we wish to make change. In an optimal solution, we have at most one nickel. Proof. Let S be the multiset of coins used to make change for n. Suppose that S contains k>1 nickels. The key idea is that we may exchange each pair of nickels for a single dime. We formalize this as follows. By the Division Algorithm, we may write k=2 j+r, where j \in \mathbb{N} and r \in{0,1}. As k>1, we have that j \geq 1. So we exchange 2 j nickels for j dimes to obtain a new solution set S^{\prime}. Observe that: \left|S^{\prime}\right|=|S|-j<|S|. As we may construct a solution using fewer coins, it follows that any optimal solution uses at most one nickel. While we will not go through a full proof of correctness for the greedy algorithm to make change, similar lemmas regarding dimes and pennies serve as key steps in establishing the correctness of this algorithm. In fact, Lemma 33 provides the key insight that we should select dimes before nickels; as otherwise, we may need to swap out the nickels for fewer dimes. ## 计算机代写|算法代写Algorithm代考|Interval Scheduling In this section, we consider the Interval Scheduling problem. Intuitively, we have a single classroom. The goal is to assign the maximum number of courses to the classroom, such that no two classes are scheduled for our room at the same time. We now turn to formalizing the Interval Scheduling problme. Here, we think of intervals as line segments on the real line. We specify each interval by a pair s_i and f_i, where s_i<f_i. An interval with starting point s_i and ending point s_i is the set:$$
\left[s_i, f_i\right]=\left{x \in \mathbb{R}: s_i \leq x \leq f_i\right} .
As an example, $[0,1]$ is the set of real numbers between 0 and 1, including the endpoints 0 and 1 . Intuitively, the Interval Scheduling problem takes as input $\mathcal{I}$, a set of intervals. The goal is to find the maximum number of intervals we can select, such that no two intervals overlap.
Definition 34. The Interval Scheduling problem is defined as follows.
• Instance: Let $\mathcal{I}=\left{\left[s_1, f_1\right], \ldots,\left[s_k, f_k\right]\right}$ be our set of intervals.
• Solution: A set $S \subseteq \mathcal{I}$ such that no two intervals in $S$ overlap, where $|S|$ is as large as possible.
## 计算机代写|算法代写Algorithm代考|Interval Scheduling
〈left 缺少或无法识别的分隔符
• 实例: 让left 缺少或无法识别的分隔符
是我们的间隔集。
• 解决方案: 一套 $S \subseteq \mathcal{I}$ 这样没有两个间隔 $S$ 重珢，其中 $|S|$ 尽可能大。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
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This site is devoted to mathematics and its applications. Created and run by Peter Saveliev.
# PageRank as a solution of a matrix equation
Briefly, it's not, as sometimes there is no solution to that equation.
Suppose $p_1,...,p_N$ are the $N$ pages/nodes in the graph (of the World Wide Web). Suppose also that $R(p_i)$ is the PageRank of node $p_i$.
PageRank is defined (and computed) iteratively, with an initial distribution of rankings usually given by: $$R_0(p_i) = \frac{1}{N},i=1,2,...,N.$$ Then at time $t+1$, the ranking are given in terms of the rankings at time $t$: $$R_{t+1}(p_i) = d \sum_{p_j \in S(p_i)} \frac{R_t (p_j)}{L(p_j)} + \frac{1-d}{N},$$ where
• $S(p_i)$ is the set of nodes that (inbound) link to $p_i$,
• $L(p_j)$ is the number of (outbound) links from node $p_j$.
Let $$R_t=(R_t(p_1),...,R_t(p_N))$$ be the vector of ranks at time $t$. Then, in matrix notation: $$R_{t+1} = d MR_t + (1-d)R_0,$$ where the matrix $M$ is defined as $$M_{ij} = \begin{cases} 1 /L(p_j) & \mbox{if }p_j \in S(p_i),\ \\ 0 & \mbox{if }p_j \notin S(p_i). \end{cases}$$ In other words, $$M = (K^{-1} A)^T,$$ where $A$ is the adjacency matrix of the graph and $K$ is the diagonal matrix with the outdegrees $L(p_j)$ in the diagonal.
Now, the iteration is supposed to converge to a "steady state" $R$. In that case, iterations shouldn't change the value: $R_{t+1} =R_{t}.$ Hence the PageRank is a solution to this matrix equation: $$R = d MR + (1-d)R_0.$$
Now, the questions we'd like to ask:
• Why does a solution exist?
• Why is this solution unique?
Clearly, for what PageRank is used, you need both.
We can re-write: $$(I-dM)R = (1-d)R_0,$$ where $I$ is the $n × n$ identity matrix. Then the solution to this equation is easy to find: $$R = d(1-d) \left( \frac{1}{d}I-M \right) ^{-1}R_0,$$ provided, of course, the inverse matrix exists.
From linear algebra we know that it does exists -- if and only if $\mu = \frac{1}{d}$ is not an eigenvalue of $M$. In that case, the solution $R$ is also unique.
Let's consider the eigenvalues of $M$.
Since $\mu = \frac{1}{d}>1$, it's is not an eigenvalue of $M$ -- if we know that all of them are less than or equal $1$. How do we know?
Perron–Frobenius theorem. Suppose $A = (a_{ij})$ is an $n × n$ irreducible non-negative matrix: $a_{ij} \geq 0$ for $1 ≤ i, j ≤ n$. Then it has a positive real eigenvalue $r>0$ such that
1. the eigenspace of $r$ is 1-dimensional.
2. there exists an eigenvector $v = (v_1,…,v_n)$ of $r$ such that all its components are positive: $v_i > 0$ for $1 ≤ i ≤ n$.
3. all eigenvalue $λ$ satisfies $|λ| \leq r$.
Let's check if the conditions of the theorem apply to $M$ defined above.
Matrix $M$ is non-negative.
A square matrix $M$ is called irreducible iff, in particular, its associated directed graph $G_M$ (there is an edge from node $p_i$ to node $p_j$ iff $M_{ij} > 0$) is strongly connected (there is a path from each node to every other node). This isn't true because the web always has pages with no inbound links and possibly some with no outbound links...
In this case, we have no reason to believe that any particular PageRank distribution exists for the given state of the web.
A common way of trying to deal with this problem is to add "missing" links. That's bad math.
Let's ignore this and try to continue anyway. Remember we also need $r=1$ in order to insure that $\mu = \frac{1}{d}$ is not an eigenvalue of $M$.
A column stochastic matrix is a square matrix each of whose columns consists of non-negative real numbers whose sum is equal to 1. Matrix $M$ is column stochastic. Indeed, from its definition above the sum of all elements in column $j$ of $M$ is $$\sum\limits_{i:p_j \in S(p_i)} \frac{1}{L(p_j)} = \frac{1}{L(p_j)} \sum\limits_{i:p_j \in S(p_i)} 1 = \frac{1}{L(p_j)} L(p_j) =1.$$
The point is, the Perron–Frobenius theorem ensures that every stochastic (still irreducible) matrix has an eigenvalue equal to 1 and it's the largest absolute value of any eigenvalue.
Even if $R$ exists, questions remain:
• Why does this solution $R$ have only non-negative entries?
• Item 2 above doesn't apply since $R$ isn't an eigenvector.
• Why does the iteration $R_t$ converge to the "steady state" $R$?
• It might diverge away from $R$ or it may be cyclical. "Observing" convergence doesn't prove anything (case in point: harmonic series).
In light of PageRank's dependence on the damping factor and other problems this result isn't surprising.
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This blog is managed by Song Hock Chye, author of Improve Your Thinking Skills in Maths (P1-P3 series), which is published and distributed by EPH.
## Friday, February 01, 2008
### Tao Nan School P6 Math CA1 2004 (Q40)
Mrs Ang bought 44 cakes. The ratio of the number of butter cakes to the number of chocolate cakes was 6 : 5. After giving some chocolate cakes to her neighbours, Mrs Ang had twice as many butter cakes as chocolate cakes. How many chocolate cakes did she give away?
Solution
After giving away some cakes, she had twice as many butter cakes as chocolate cakes.
Since no butter cakes were given, the number of units of butter cakes remains as 6.
* Half of 6 units will give 3 units of chocolate cakes left, after some of them have been given away.
This means that 2 units of chocolate cakes have been given away.
1 unit ----- 4 cakes
2 units ----- 8 cakes
Answer: She gave away 8 chocolate cakes.
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# Problem solving examples math
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## The Best Problem solving examples math
In this blog post, we will be discussing about Problem solving examples math. This can be simplified to x=log32/log8. By using the Powers Rule, you can quickly and easily solve for exponents. However, it is important to note that this rule only works if the base of the exponent is 10. If the base is not 10, you will need to use a different method to solve for the exponent. Nevertheless, the Powers Rule is a useful tool that can save you time and effort when solving for exponents.
distance = sqrt((x2-x1)^2 + (y2-y1)^2) When using the distance formula, you are trying to find the length of a line segment between two points. The first step is to identify the coordinates of the two points. Next, plug those coordinates into the distance formula and simplify. The last step is to take the square root of the simplify equation to find the distance. Let's try an example. Find the distance between the points (3,4) and (-1,2). First, we identify the coordinates of our two points. They are (3,4) and (-1,2). Next, we plug those coordinates into our distance formula: distance = sqrt((x2-x1)^2 + (y2-y1)^2)= sqrt((-1-3)^2 + (2-4)^2)= sqrt(16+4)= sqrt(20)= 4.47 Therefore, the distance between the points (3,4) and (-1,2) is 4.47 units.
It is important to be able to solve expressions. This is because solving expressions is a fundamental skill in algebra. Algebra is the branch of mathematics that deals with equations and variables, and it is frequently used in physics and engineering. Many word problems can be translated into algebraic expressions, and being able to solve these expressions will allow you to solve the problem. In order to solve an expression, you need to use the order of operations. The order of operations is a set of rules that tells you the order in which to solve an equation. The order of operations is: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). Using the order of operations, you can solve any expression.
By focusing on one part at a time, it may be easier to see a pattern or solution. Another method is to work backwards from the answer. This can help to provide a framework for solving the equation. In addition, it is often helpful to consult with a friend or tutor who is better at math than you are. By working together, it may be possible to arrive at the correct answer. Ultimately, there is no single method that will guarantee success in solving hard math equations. However, by trying different approaches, it may be possible to find a solution.
A radical is a square root or any other root. The number underneath the radical sign is called the radicand. In order to solve a radical, you must find the number that when multiplied by itself produces the radicand. This is called the principal square root and it is always positive. For example, the square root of 16 is 4 because 4 times 4 equals 16. The symbol for square root is . To find other roots, you use division. For example, the third root of 64 is 4 because 4 times 4 times 4 equals 64. The symbol for the third root is . Sometimes, you will see radicals that cannot be simplified further. These are called irrational numbers and they cannot be expressed as a whole number or a fraction. An example of an irrational number is . Although radicals can seem daunting at first, with a little practice, they can be easily solved!
## We cover all types of math problems
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Math scaner College math equations and answers Math help trigonometry Best apps for algebra Pay someone to do your math homework
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Search a number
3183 = 31061
BaseRepresentation
bin110001101111
311100220
4301233
5100213
622423
712165
oct6157
94326
103183
112434
121a13
1315ab
141235
15e23
hexc6f
3183 has 4 divisors (see below), whose sum is σ = 4248. Its totient is φ = 2120.
The previous prime is 3181. The next prime is 3187. The reversal of 3183 is 3813.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 3183 - 21 = 3181 is a prime.
It is a D-number.
3183 is a lucky number.
It is a plaindrome in base 13 and base 14.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (3181) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 528 + ... + 533.
It is an arithmetic number, because the mean of its divisors is an integer number (1062).
23183 is an apocalyptic number.
3183 is a deficient number, since it is larger than the sum of its proper divisors (1065).
3183 is a wasteful number, since it uses less digits than its factorization.
3183 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1064.
The product of its digits is 72, while the sum is 15.
The square root of 3183 is about 56.4180822077. The cubic root of 3183 is about 14.7099844243.
Adding to 3183 its reverse (3813), we get a palindrome (6996).
Subtracting 3183 from its reverse (3813), we obtain a triangular number (630 = T35).
The spelling of 3183 in words is "three thousand, one hundred eighty-three".
Divisors: 1 3 1061 3183
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# Parabola Solver
Introduction:
In analytical geometry, a parabola is a conic section obtained on slicing a right circular cone by a planeparallel to the line joining vertex and any other point of the cone.
Along with focus F and vertex V, P is any point on parabola. PX is parallel to VFA. Hence the angle formed by the PX and VFA at point P is always equal.
Please express your views of this topic Definition of a Parabola by commenting on blog .
## Classification of parabola with respect to the general equation of a conic:
The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is the basic equation for the conic. If it is a conic, then it is
(i) a parabola if B2 4AC = 0
(ii) an ellipse if B2 − 4AC < 0
(iii) a hyperbola if B2 − 4AC > 0
## Standard equation of parabola solver:
Given:
• Fixed point (F)
• Fixed line (l)
• Eccentricity (e = 1)
• Moving point P(x, y)
Important definitions on parabolas solver:
Focus: The fixed point to draw the parabola is called the focus (F). At this point, the focus is F (a, 0).
Directrix: Thefixed line used to draw a parabola is known as the directrix of the parabola. Here, the equation of the directrix is x = − a.
Axis: Theaxis of the parabola is called the axis of symmetry. The curve y2 = 4axis symmetrical about x-axis and thus x-axis or y = 0 is the axis of theparabola y2 = 4ax. Here the axis of the parabola passes through the focus and it is perpendicular to the directrix.
Vertex: The point of intersection of a parabola and its axis is known as vertex. Here, the vertex is V (0, 0).
Focal distance: The focal distance is the distance between a point on the parabola and its focus.
Focal chord: A chord which passes through the focus of the parabola is called the focal chord of the parabola.
Latus Rectum: It is a focal chord perpendicular to the axis of the parabola. Here, the equation of the latus rectum is x="a.
Between, if you have problem on these topics Perimeter of Rectangle , please browse expert math related websites for more help on model question paper 10th samacheer kalvi .
## Example Problems on parabolas solver:
Let us see some of the examples on parabolas solver.
Example 1:
Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabola and hence draw their graph.
(i) y2 = 4x
Solution:
The basic form of parabola is y2=4ax.
Given equation is y2="4x." Hence a="1.
The equation of the parabola is (y − k) 2 = − 4a(x − h)
(y − 0)2 = 4(1) (x − 0)
Here (h, k) is (0, 0) and a = 1
Axis: The axis of symmetry is x-axis.
Vertex: The vertex V (h, k) is (0, 0)
Focus: The focus F (a, 0) is (1, 0)
Directrix: The equation of the directrix is x = − a i.e. x = − 1
Latus Rectum: The equation of the latus rectum is x = a i.e. x = 1 and its length is 4a = 4(1) = 4.
The graph for the above parabola is:
Example 2:
Find the axis, vertex, focus, directrix, equation of the latus rectum, length of the latus rectum for the following parabola and hence draw their graph.
(i) x2 = − 4y
Solution:
(x − 0)2 = − 4(1) (y − 0)
Here (h, k) is (0, 0) and a = 1
Axis: y-axis or x = 0
Vertex: V (0, 0)
Focus: F (0, − a) i.e. F (0, − 1)
Directrix: y = a i.e. y = 1
Latus rectum: y = − a i.e. y = − 1
: Length = 4
The graph for the above parabola is:
These are the examples of parabolas solver.
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# Integral Inequality $|f''(x)/f(x)|$
Let $f$ be a $C^2$ function in $[0,1]$ such that $f(0)=f(1)=0$ and $f(x)\neq 0\,\forall x\in(0,1).$ Prove that
$$\int_0^1 \left|\frac{f{''}(x)}{f(x)}\right|dx\ge4$$
• I think you mean $f(0) = f(1) = 0$, not $f(x) = f(1) = 0$. – Michael Albanese Oct 14 '12 at 5:03
• Yep. f(0)=f(1)=0. Sorry about the typo. – Christmas Bunny Oct 14 '12 at 5:21
• I can show that, if $f\in C^2$0,1$$ and $f'(0)=f'(1)=0$ then exist a point $t\in $0,1$$ such that, $f''(t)>4|f(1)-f(0)|$ – Salech Rubenstein Oct 14 '12 at 5:34
• This is a duplicate of a question that has been asked before but I am unable to find the other question. – user17762 Oct 14 '12 at 5:56
## 2 Answers
Just some considerations. Without loss of generality, we can assume that $f(x)$ is positive on $(0,1)$. Let $\operatorname{graph}(g)$ be the convex hull of $\operatorname{graph}(f)$. We have $g(x)\in C^2([0,1])$ and
$$\int_{0}^{1}\frac{|f''(x)|}{f(x)}dx\geq \int_{0}^{1}\frac{|g''(x)|}{g(x)}dx,$$
so we can also assume that $f$ is a concave function over $[0,1]$, with $f'(0)=\alpha>0,f'(1)=-\beta<0$ and a unique maximum in $x_0\in(0,1)$, for which $f'(x_0)=0$ and $f(x_0)=1$. Let now $\operatorname{graph}(h)$ be the envelope of the tangent lines to $\operatorname{graph}(f)$ for $x\in\left\{0,x_0,1\right\}$. For any $\epsilon>0$ there is a function $u\in C^2([0,1])$ such that $|u-h|<\epsilon$ and:
$$\int_{0}^{1}\frac{|f''(x)|}{f(x)}dx\geq \int_{0}^{1}\frac{|u''(x)|}{u(x)}dx=\alpha+\beta-O(\epsilon),$$
but $\alpha+\beta$ must be greater than $4$, since $\max u$ is below the $y$-coordinate of the intersection of the tangent lines in $x=0$ and $x=1$, so
$$1 = \max u \leq \frac{\alpha\beta}{\alpha+\beta} \leq_{AM-GM} \frac{\alpha+\beta}{4}.$$
Proving the Lower Bound
Without loss of generality, assume that $f(x)\gt0$ for $x\in(0,1)$.
Suppose that $f(x_0)=y_0=\max\limits_{x\in[0,1]}f(x)$. Then, $f'(x_0)=0$.
By the Mean Value Theorem, for some $x_1\in(0,x_0)$, $f'(x_1)=\frac{y_0}{x_0}$. Therefore, $$\int_0^{x_0}|f''(x)|\,\mathrm{d}x\ge\frac{y_0}{x_0}$$ Furthermore, for some $x_2\in(x_0,1)$, $f'(x_2)=-\frac{y_0}{1-x_0}$. Therefore, $$\int_{x_0}^1|f''(x)|\,\mathrm{d}x\ge\frac{y_0}{1-x_0}$$ Since $f(x)\le y_0$, \begin{align} \int_0^1\left|\,\frac{f''(x)}{f(x)}\,\right|\,\mathrm{d}x &\ge\frac{\frac{y_0}{x_0}+\frac{y_0}{1-x_0}}{y_0}\\ &=\frac1{x_0}+\frac1{1-x_0}\\ &=\frac1{\frac14-\left(x_0-\frac12\right)^2}\\[3pt] &\ge4 \end{align}
The Lower Bound is Sharp
Let $$f_a(x)=\sin^{-1}\left(\frac{\sin(\pi x)}{1+a^2\sin^2(\pi x)}\right)$$
then $$\lim_{a\to0}\int_0^1\left|\,\frac{f_a''(x)}{f_a(x)}\,\right|\,\mathrm{d}x=4$$ since $f_a''(x)$ is tends to $0$ except near $x=\frac12$, and $\int_0^1f_a''(x)\,\mathrm{d}x$ tends to $-2\pi$, whereas $f_a\!\left(\frac12\right)$ tends to $\frac\pi2$.
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# McElieceTalk
60 %
40 %
Education
Published on February 5, 2008
Author: Paola
Source: authorstream.com
Routing in Error-Correcting Networks: Routing in Error-Correcting Networks Edwin Soedarmadji May 10, 2006 California Institute of Technology Department of Electrical Engineering Pasadena, CA 91125, USA Introduction: Introduction Start from an unrelated network problem Route planning under fuel capacity and refueling constraints Especially relevant Increasing energy cost Vehicles with alternative fuel Vehicles exploring remote areas The Gas Station Problem: The Gas Station Problem Shortest Path Problem in Vehicle has limited a fuel capacity Refueling nodes in the network Edge weights expressed in fuel units Vehicle starts at with fuel units The Gas Station Problem: The Gas Station Problem Feasible paths are paths where the vehicle always carries a non-negative amount of fuel on the path nodes Is {all feasible paths} an empty set ? If not, what is the path that minimizes the travel distance? Theorem: Theorem Example: Example Remove infeasible edges in E and vertices in V Compute all-pairs shortest path between nodes in V’ Remove infeasible edges in E’ and vertices in V’ Calculate the shortest path from s to t Solution produced in Error-Correcting Network: Error-Correcting Network Possible Generalization to Communication Networks The Gas-Station algorithm works for transportation networks Is it applicable to communication networks? There are many similarities Vehicle information packet Gas tank capacity error budget Gas station error correction node Gas consumption packet error Error Budget: Error Budget M U R F L E S M A R B L E S Suppose each packet contains seven symbols, and the error-correction scheme employed in the network can correct up to (but not more than) three errors. Then the error budget is three units for a given alphabet size. Edge model: Symmetric Channel: Edge model: Symmetric Channel The Cascaded SC: The Cascaded SC The ary Erasure Channel: The ary Erasure Channel Generalized Dijkstra Algorithm: Generalized Dijkstra Algorithm x + min( y , z ) = min ( x + y , x + z ) The Error Distribution: The Error Distribution Edge Weight: Worst-Case Error: Edge Weight: Worst-Case Error p = 0.10 p = 0.25 p = 0.98 p = 0.50 Routing in Error-Correcting Networks: Routing in Error-Correcting Networks WCE x is a non-decreasing function of p The algorithm used in the Gas Station Problem can be used to find the feasible path with minimum WCE from s to t. Questions?
User name: Comment:
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## Related pages
### The Gas Station Problem - California Institute of Technology
Routing in Error-Correcting Networks Edwin Soedarmadji May 10, 2006 California Institute of Technology Department of Electrical Engineering Pasadena, CA ...
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1.7: Words that Describe Patterns
Difficulty Level: Basic Created by: CK-12
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Practice Words that Describe Patterns
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Color Highlighted Text Notes
Vocabulary Language: English Spanish
tabular
Being represented in a table.
$\ge$
The greater-than-or-equal-to symbol "$\ge$" indicates that the value on the left side of the symbol is greater than or equal to the value on the right.
$\le$
The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right.
$\ne$
The not-equal-to symbol "$\ne$" indicates that the value on the left side of the symbol is not equal to the value on the right.
constant
A constant is a value that does not change. In Algebra, this is a number such as 3, 12, 342, etc., as opposed to a variable such as x, y or a.
Equation
An equation is a mathematical sentence that describes two equal quantities. Equations contain equals signs.
greater than
The greater than symbol, $>$, indicates that the value on the left side of the symbol is greater than the value on the right.
greater than or equal to
The greater than or equal to symbol, $\ge$, indicates that the value on the left side of the symbol is greater than or equal to the value on the right.
inequality
An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.
less than
The less-than symbol "<" indicates that the value on the left side of the symbol is lesser than the value on the right.
less than or equal to
The less-than-or-equal-to symbol "$\le$" indicates that the value on the left side of the symbol is lesser than or equal to the value on the right.
not equal to
The "not equal to" symbol, $\ne$, indicates that the value on the left side of the symbol is not equal to the value on the right.
Variable
A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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# How to Make Accurate Sports Predictions
It is feasible to make predictions on sports video games making use of mathematical formulas. As an example, you can predict the result of a football match. Yet what regarding a baseball or hockey game? Exactly how likely is it to end in a tie? These are a few of the inquiries that need to be responded to in order to make precise forecasts. If you loved this article therefore you would like to get more info pertaining to piala2022.com generously visit our web page.
## Probability of a football match ending in a connection
In the NFL, connections are not an unusual event. As a matter of fact, the Philadelphia Eagles as well as Cincinnati Bengals connected 23-23 on September 27, 2020. Because that time, there have been 14 connections. While the number of video games in a season has actually risen and fall, the likelihood of a football suit ending in a tie has not altered a lot given that 1989. Ever since, there has actually been a 0.2% chance of an incorporate an NFL game.
In the NFL, video games that finish in a connection are played in sudden-death overtime, in which the group racking up very first victories. A research study by Jones (2004) provides a Markov chain evaluation of the sudden-death system. This design involves a four-state absorbing Markov chain in which group An and also group B have rotating states of ownership. In each state, the probability that a team will certainly score is p.
## Possibility of a basketball video game ending in a win
The chance of a basketball video game finishing in a triumph is a function of ball game gap. If a team leads by more than six points at the end of law, there is an 80% opportunity that the team will certainly win. If ball game space is smaller, there is a 20% possibility that the group will certainly shed. Nonetheless, this heuristic only holds for brief time periods and sheds its precision after 6 mins.
To calculate the implied chance of a team winning, a sportsbook will utilize moneyline odds. These probabilities are determined by utilizing an odds converter or moneyline calculator. Sportsbooks will usually display moneyline probabilities in the hundreds. Moneyline chances will indicate the favorite group and the underdog group based on indicated possibility of winning.
## Possibility of a baseball game finishing in a connection
There are lots of methods to make a baseball game much less most likely to end in a tie One method is to delay the video game. This occurs when the climate is bad. A connection can happen during a game when teams are lacking bottles. The video game can additionally be reduced by gimmicks.
One more means to reduce the risk of a baseball video game ending in a scoreless connection is to eliminate extra innings. In university baseball, connections can happen when there are no more pitchers available for one group. Furthermore, connections can occur in video games that are picked up time limit or poor weather condition.
## Possibility of a hockey video game finishing in a tie
A hockey video game’s chance of ending in a connection is a fact used to anticipate the outcome of video games. NHL video games that are tied at the end of guideline time are sent to “sudden-death” overtime, in which the team that ratings the initial objective wins the game. The length of overtime video games differs substantially, yet the mean is around 9 and also a fifty percent mins.
Historically, connections have actually been typical in NHL games. The NHL used to allow connections at the end of law, but that altered after The second world war, when the NHL adopted a 10-minute overtime regulation. In the playoffs, connections are permitted only if both teams rack up at the very least one goal.
## Probability of a tennis match finishing in a connection.
An incorporate a tennis match occurs when the scores are equivalent after two collections and the video game is still not determined. In this case, a tie break is played to figure out if the game needs to continue. The champion of the tie break is the initial to get to 7 points or lead by two points. The chance of a gamer winning a video game in a tennis tie break is equivalent to the probability of that gamer winning a video game in a typical video game.
The likelihood of winning a factor in a tennis suit depends upon the relative skill of the player and her opponent. In a normal tennis suit, the initial player to win 3 collections wins the suit. In a tennis match, nevertheless, a player can win a video game in 4 collections and also the various other gamer can win three sets. If you have any inquiries concerning where and exactly how to utilize https://piala2022.com/, you can contact us at our internet site.
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+0
# MATH HELP!
0
61
1
The perimeter of a rectangular field is 330 yards. If the width of the field is 72 yards, what is its length?
Jun 11, 2023
#1
0
The perimeter of a rectangular field is 330 yards.
If the width of the field is 72 yards, what is its length?
The perimeter is the Length + Width + Length + Width.
You can also express this as 2L + 2W.
So, P = 2L + 2W
330 = (2)(L) + (2)(72)
330 = 2L + 144
330 – 144 = 2L
2L = 186
L = 93
.
Jun 11, 2023
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# Chapter 1.1 - PowerPoint PPT Presentation
1 / 7
Chapter 1.1. Matter, mass, and volume. Brain Pop. Example: Which has more mass, a car or a cat? Why?. Matter. Things that can physically be touched It makes up all objects and living things in the universe Examples: people, animals, water, etc. Mass.
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Chapter 1.1
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## Chapter 1.1
Matter, mass, and volume
### Brain Pop
• Example: Which has more mass, a car or a cat? Why?
### Matter
• Things that can physically be touched
• It makes up all objects and living things in the universe
• Examples: people, animals, water, etc.
### Mass
• Measurement of how much matter there is
• The more matter, the greater the mass.
• Example: Which has more mass, a car or a cat? Why?
### Weight
• Is the downward pull on an object due to gravity
• Gravity is a force that pulls all matter (things that can be touched) together
• Example: car and a cat.. Which has more matter?
### Mass –vs- Weight
• Again Mass describes the amount of matter an object has and Weight is how strongly gravity is pulling on that matter.
### Volume
• The amount of space that matter occupies (takes up)
• Volume = length X width X height
• V=l x w x h
• Not regular in shape, can figure out the volume by Displacement
• No two objects that are the same can be in the same space at the same time
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# Physics Class 9 ICSE CISCE Topics and Syllabus
## Syllabus
1 Measurements and Experimentation
(i) International System of Units, the required SI units with correct symbols are given at the end of this syllabus. Other commonly used system of units - fps and cgs.
(ii) Measurements using common instruments, Vernier callipers and micro-metre screw gauge for length, and simple pendulum for time.
Measurement of length using, Vernier callipers and micro-metre screw gauge. Decreasing least-count leads to an increase in accuracy; least-count (LC) of Vernier callipers and screw gauge), zero error (basic idea), (no numerical problems on callipers and screw gauge), simple pendulum; time period, frequency, graph of length l vs. T2 only; slope of the graph. Formula T=2.pi.sqrt(l [no derivation]. Only simple numerical problems.
2 Motion in One Dimension
Scalar and vector quantities, distance, speed, velocity, acceleration; graphs of distance-time and speed-time; equations of uniformly accelerated motion with derivations.
Examples of Scalar and vector quantities only, rest and motion in one dimension; distance and displacement; speed and velocity; acceleration and retardation; distance-time and velocity-time graphs; meaning of slope of the graphs; [Nonuniform acceleration excluded].
Equations to be derived: v = u + at;
S = ut + 1/2at2; S = 1/2(u+v)t; v2 = u2 + 2aS.
[Equation for Snth is not included].
Simple numerical problems.
3 Laws of Motion
(i) Contact and non-contact forces; cgs & SI units.
Examples of contact forces (frictional force, normal reaction force, tension force as applied through strings and force exerted during collision) and non-contact forces (gravitational, electric and magnetic). General properties of non-contact forces. cgs and SI units of force and their relation with Gravitational units.
(ii) Newton’s First Law of Motion (qualitative discussion) introduction of the idea of inertia, mass and force.
Newton's first law; statement and qualitative discussion; definitions of inertia and force from first law, examples of inertia as illustration of first law. (Inertial mass not included).
(iii)Newton’s Second Law of Motion (including F=ma); weight and mass.
Detailed study of the second law. Linear momentum, p = mv; change in momentum Δp = Δ(mv) = mΔv for mass remaining constant, rate of change of momentum;
Δp/Δt = mΔv /Δt = ma or
{(P_2-P_1)/t=(mv-
Simple numerical problems combining F = Δp/Δt = ma and equations of motion. Units of force - only cgs and SI.
(iv) Newton’s Third Law of Motion (qualitative discussion only); simple examples.
Statement with qualitative discussion; examples of action - reaction pairs, (FBA and FAB); action and reaction always act on different bodies.
(v) Gravitation
Universal Law of Gravitation. (Statement and equation) and its importance. Gravity, acceleration due to gravity, free fall. Weight and mass, Weight as force of gravity comparison of mass and weight; gravitational units of force, (Simple numerical problems), (problems on variation of gravity excluded)
4 Fluids
(i) Change of pressure with depth (including the formula p=hρg); Transmission of pressure in liquids; atmospheric pressure.
Thrust and Pressure and their units; pressure exerted by a liquid column p = hρg; simple daily life examples, (i) broadness of the base of a dam, (ii) Diver’s suit etc. some consequences of p = hρg ; transmission of pressure in liquids; Pascal's law; examples; atmospheric pressure; common manifestation and consequences.. Variations of pressure with altitude, (qualitative only); applications such as weather forecasting and altimeter. (Simple numerical problems)
(ii) Buoyancy, Archimedes’ Principle; floatation; relationship with density; relative density; determination of relative density of a solid.
Buoyancy, upthrust (FB); definition; different cases, FB>, = or < weight W of the body immersed; characteristic properties of upthrust; Archimedes’ principle; explanation of cases where bodies with density ρ >, = or < the density ρ' of the fluid in which it is immersed.
RD and Archimedes’ principle. Experimental determination of RD of a solid and liquid denser than water. Floatation: principle of floatation; relation between the density of a floating body, density of the liquid in which it is floating and the fraction of volume of the body immersed; (ρ12 = V2/V1); apparent weight of floating object; application to ship, submarine, iceberg, balloons, etc.
Simple numerical problems involving Archimedes’ principle, buoyancy and floatation.
4.1 Fluids
(i) Change of pressure with depth (including the formula p=hρg); Transmission of pressure in liquids; atmospheric pressure.
Thrust and Pressure and their units; pressure exerted by a liquid column p = hρg; simple daily life examples, (i) broadness of the base of a dam, (ii) Diver’s suit etc. some consequences of p = hρg ; transmission of pressure in liquids; Pascal's law; examples; atmospheric pressure; common manifestation and consequences.. Variations of pressure with altitude, (qualitative only); applications such as weather forecasting and altimeter. (Simple numerical problems)
(ii) Buoyancy, Archimedes’ Principle; floatation; relationship with density; relative density; determination of relative density of a solid.
Buoyancy, upthrust (FB); definition; different cases, FB>, = or < weight W of the body immersed; characteristic properties of upthrust; Archimedes’ principle; explanation of cases where bodies with density ρ >, = or < the density ρ' of the fluid in which it is immersed.
RD and Archimedes’ principle. Experimental determination of RD of a solid and liquid denser than water. Floatation: principle of floatation; relation between the density of a floating body, density of the liquid in which it is floating and the fraction of volume of the body immersed; (ρ12 = V2/V1); apparent weight of floating object; application to ship, submarine, iceberg, balloons, etc.
Simple numerical problems involving Archimedes’ principle, buoyancy and floatation.
4.2 Buoyancy, Upthrust
Buoyancy, Archimedes’ Principle; floatation; relationship with density; relative density; determination of relative density of a solid.
Buoyancy, upthrust (FB); definition; different cases, FB>, = or < weight W of the body immersed; characteristic properties of upthrust; Archimedes’ principle; explanation of cases where bodies with density ρ >, = or < the density ρ' of the fluid in which it is immersed.
RD and Archimedes’ principle. Experimental determination of RD of a solid and liquid denser than water. Floatation: principle of floatation; relation between the density of a floating body, density of the liquid in which it is floating and the fraction of volume of the body immersed; (ρ12 = V2/V1); apparent weight of floating object; application to ship, submarine, iceberg, balloons, etc.
Simple numerical problems involving Archimedes’ principle, buoyancy and floatation.
5 Heat and Energy
(i) Concepts of heat and temperature.
Heat as energy, SI unit – joule,
1 cal = 4.186 J exactly.
(ii) Anomalous expansion of water; graphs showing variation of volume and density of water with temperature in the 0 to 10°C range. Hope’s experiment and consequences of Anomalous expansion.
(iii)Energy flow and its importance:-
Understanding the flow of energy as Linear and linking it with the laws of Thermodynamics - ‘Energy is neither created nor destroyed’ and ‘No Energy transfer is 100% efficient.
(iv) Energy sources.
Solar, wind, water and nuclear energy (only qualitative discussion of steps to produce electricity). Renewable versus nonrenewable sources (elementary ideas with example).
Renewable energy: biogas, solar energy, wind energy, energy from falling of water, run-of-the river schemes, energy from waste, tidal energy, etc. Issues of economic viability and ability to meet demands.
Non-renewable energy – coal, oil, natural gas. Inequitable use of energy in urban and rural areas. Use of hydro electrical powers for light and tube wells.
(v) Global warming and Green House effect:-
Meaning, causes and impact on the life on earth. Projections for the future; what needs to be done. Energy degradation –meaning and examples.
6 Light
(i) Reflection of light; images formed by a pair of parallel and perpendicular plane mirrors;.
Laws of reflection; experimental verification; characteristics of images formed in a pair of mirrors, (a) parallel and (b) perpendicular to each other; uses of plane mirrors.
(ii) Spherical mirrors; characteristics of image formed by these mirrors. Uses of concave and convex mirrors. (Only simple direct ray diagrams are required).
Brief introduction to spherical mirrors - concave and convex mirrors, centre and radius of curvature, pole and principal axis, focus and focal length; location of images from ray diagram for various positions of a small linear object on the principal axis of concave and convex mirrors; characteristics of images.
f = R/2 (without proof); sign convention and direct numerical problems using the mirror formulae are included. (Derivation of formulae not required)
Uses of spherical mirrors.
Scale drawing or graphical representation of ray diagrams not required.
7 Sound
(i) Nature of Sound waves. Requirement of a medium for sound waves to travel; propagation and speed in different media; comparison with speed of light.
Sound propagation, terms – frequency (f), wavelength (λ), velocity (V), relation V = fλ. (Simple numerical problems) effect of different factors on the speed of sound; comparison of speed of sound with speed of light; consequences of the large difference in these speeds in air; thunder and lightning.
(ii) Infrasonic, sonic, ultrasonic frequencies and their applications.
Elementary ideas and simple applications only. Difference between ultrasonic and supersonic.
8 Electricity and Magnetism
(i) Simple electric circuit using an electric cell and a bulb to introduce the idea of current (including its relationship to charge); potential difference; insulators and conductors; closed and open circuits; direction of current (electron flow and conventional)
Current Electricity:- brief introduction of sources of direct current - cells, accumulators (construction, working and equations excluded); Electric current as the rate of flow of electric charge (direction of current - conventional and electronic), symbols used in circuit diagrams. Detection of current by Galvanometer or ammeter (functioning of the meters not to be introduced). Idea of electric circuit by using cell, key, resistance wire/resistance box/rheostat, qualitatively.; elementary idea about work done in transferring charge through a conductor wire; potential difference V = W/q.
(No derivation of formula) simple numerical problems.
Social initiatives: Improving efficiency of existing technologies and introducing new eco-friendly technologies. Creating awareness and building trends of sensitive use of resources and products, e.g. reduced use of electricity.
(ii) Induced magnetism, Magnetic field of earth. Neutral points in magnetic fields.
Magnetism: magnetism induced by bar magnets on magnetic materials; induction precedes attraction; lines of magnetic field and their properties; evidences of existence of earth’s magnetic field, magnetic compass. Uniform magnetic field of earth and nonuniform field of a bar magnet placed along magnetic north-south; neutral point; properties of magnetic field lines.
(iii) Introduction of electromagnet and its uses.
8.1 Electricity
Simple electric circuit using an electric cell and a bulb to introduce the idea of current (including its relationship to charge); potential difference; insulators and conductors; closed and open circuits; direction of current (electron flow and conventional)
Current Electricity:- brief introduction of sources of direct current - cells, accumulators (construction, working and equations excluded); Electric current as the rate of flow of electric charge (direction of current - conventional and electronic), symbols used in circuit diagrams. Detection of current by Galvanometer or ammeter (functioning of the meters not to be introduced). Idea of electric circuit by using cell, key, resistance wire/resistance box/rheostat, qualitatively.; elementary idea about work done in transferring charge through a conductor wire; potential difference V = W/q.
(No derivation of formula) simple numerical problems.
Social initiatives: Improving efficiency of existing technologies and introducing new eco-friendly technologies. Creating awareness and building trends of sensitive use of resources and products, e.g. reduced use of electricity.
8.2 Magnetism
Induced magnetism, Magnetic field of earth. Neutral points in magnetic fields.
Magnetism: magnetism induced by bar magnets on magnetic materials; induction precedes attraction; lines of magnetic field and their properties; evidences of existence of earth’s magnetic field, magnetic compass. Uniform magnetic field of earth and nonuniform field of a bar magnet placed along magnetic north-south; neutral point; properties of magnetic field lines.
Introduction of electromagnet and its uses.
| 2,806 | 13,227 |
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# RC Circuits
Following on from our discussions of capacitance we come to RC Circuits aka Resistor-Capacitor circuits which are circuits which consist of resistors, capacitors and a power source.
This will be our first practical example of a transient circuit where we will see the output voltage ramp up in time until reaching a steady-state value or decreasing in time asymptoting to zero depending on the circuit configuration.
Note that a circuit consisting of a resistor and a single type of energy storage component, e.g. a capacitor, is also known as a first order circuit.
## The RC Charging Circuit:
The circuits shown below are the classic example of an RC charging circuit where the voltage across the capacitor ramps up once the switch is closed:
As the switch is closed current begins to flow from the voltage source towards the plates of the capacitor to charge it with the voltage across the capacitor increasing with time with a typical curve as shown below which is exponential in nature approaching a steady state voltage equal to the voltage source:
From the point of view of the resistor once the switch is closed the entire source voltage is dropped across it which then exponentially decays with time as the voltage across the capacitors plates increases taking a higher proportion of the source voltage than the resistor with time until the entire voltage drops across the capacitor:
We can also graph the charging current with time which shows that initially the current is high due to the capacitors electric field being zero (essentially functioning as a short-circuit) before slowly ramping up acting to impede the current flow:
It can be shown with calculus that the voltage across the capacitor with time for the circuit above is given by:
Where Vc denotes the charged voltage across the capacitor, t is time, and RC is the circuit resistance multiplied by capacitance.
Note: The capacitor voltage is denoted in lower case to highlight that the variable varies with time while the source voltage is upper cases to denote that it remains constant with time.
### Time Constant:
The RC factor in the equation above is known as the time constant for the circuit. It is a figure of merit which denotes the time in seconds taken for the capacitor to charge to approximately 63% or discharge to 37% of the source voltage.
It is sometimes referred to as the time delay of the circuit as it impacts the time taken for the output to respond to the input. A higher time constant means it takes longer for the output to react and reach the steady-state value.
Thinking about this intuitively, if the capacitance is higher then more charge will be able to spread across the plates which will take longer to charge and more resistance will reduce the charging current which will also increase the charging time.
We refer to the time constant with the letter τ, hence the equation can be reduced to:
We consider the transient period of the charging to be the time from 0 to 4τ with the steady state period occurring after (99.3% charged).
## Examples:
1. Determine the time constant for the circuit below:
2. Determine the time constant for the circuit below:
To determine the time constant for the circuit above we need to calculate the total resistance and capacitance in the circuit. The total resistance is simply R1+R2 but what is the total capacitance? Recall that capacitance adds in parallel, hence the total capacitance is equal to C1 + C2. Hence the time constant for this circuit is given by:
3. Determine the time constant for the circuit below:
For this circuit the resistance is more complex to determine, so let's start with the capacitance :) We have C1 and C2 in series which means we calculate them in the same way as parallel resistors and as we only have two we can use our shortcut formula: To determine the resistance we need to calculate the resistance "seen" by the terminals of the capacitor, i.e. we calculate the Thevenin Resistance assuming the sources are zeroed. We can consider the circuit below: Which we can then simplify further by combining the series resistances of R1 and R2 and R3 and R4: From here the resistance seen is simply (R1+R2)||(R3+R4) which for this case is simply 2.5k as they're both the same value we can halve it :) Now we can multiply this resistance by the capacitance found earlier to calculate the time constant as shown below:
4. a) Determine the voltage across the capacitor in the circuit below for time t after the circuit has been switched on:
Recall that the equation for the voltage across a capacitor is given by: The first step is to determine the time constant for the circuit by considering the circuit once the switch has been flicked on. In this case it is simply R multiplied by C: The next step would be to determine what the steady-state voltage across the capacitor would be initially and once completely charged. Initially it would be equal to 0V with the switch open and once closed it would eventually be equal to the supply voltage Vs, i.e. 5V. We can then substitute these into the formula to calculate the voltage across the capacitor for time t:
b) Sketch the voltage vs time curve:
To sketch the curve we can calculate the times for 1, 2, 3, 4 and 5 time constants and then equate the corresponding voltages and connect the points with an exponential curve approximation: We can present these in a table:
## The RC Discharging Circuit:
The circuits shown below are the classic example of an RC discharging circuit where the capacitor acts as a voltage source with voltage decreasing exponentially with time as the switch is closed eventually completely discharging:
If we assume the capacitor is initially charged to 5V, then the discharge curve will look as shown below:
It can be shown with calculus that the voltage across the capacitor with time for the circuit above is given by:
Where Vc denotes the charged voltage across the capacitor, t is time, and τ is the time constant.
For the discharging circuit shown, the time constant is 0.2s and the charged voltage is 5V hence the voltage across the capacitor is given by:
## Questions:
1. a) Determine the time constant for the circuit below:
0.5s
b) Determine the transient voltage across the capacitor once the switch has been closed.
vc(t) = 5(1-e^-2t)
c) Sketch the voltage vs time curve for the voltage across the capacitor.
d) Determine the voltage across the capacitor once the switch has been closed for a long time.
5V
e) Determine the voltage across the capacitor after 1 time constant.
3.161V
2. a) Determine the time constant for the circuit below:
0.64s
b) Determine the transient voltage across the capacitor once the switch has been closed.
vc(t) = 8(1-e^(-1.5625t))
c) Determine the voltage across the capacitor once the switch has been closed for time t.
8V
d) Determine the voltage across the capacitor after 5 time constants.
7.946V
3. a) Determine the time constant for the circuit below assuming the equivalent capacitance was charged to 5V:
0.076s
b) Determine the transient voltage across the capacitor once the switch has been closed.
vc(t) = 5e^(-13.158t)
c) Sketch the voltage vs time curve for the voltage across the capacitor.
e) Determine the voltage across the capacitor after 2 time constants.
0.677V
## Lab:
The best way to get a feel for how RC circuits work intuitively is to simulate RC circuits with different values to see how the time constant makes an impact as well as how the location of the resistor and capacitor makes a difference to the curve.
Let's begin by assembling our classic RC circuit in LTSpice:
Let's then create a voltage source which acts as a pulse. A pulse is a waveform that starts at 0V, rises to a set voltage and then reduces back to 0V at a time later. We can define a pulse as shown below via right clicking on the voltage source:
We can then run a simulation for say 100ms and left click on both the voltage source and top end of the capacitor to plot both voltages together:
This circuit simulates the scenario of closing a circuit containing a resistor, capacitor and voltage source. As we can see the voltage begins quickly ramping which reaches around 63% of 3.3V for a time equal to a single time constant and then begins ramping up more slowly with time before eventually settling to the 3.3V.
Now we can simulate the discharge side by extending the period (say out to 200ms) to include the off time of the pulse (which begins after 100ms):
As you can see the voltage quickly ramps down to 37% after a single time constant before then slowly ramping down with time before eventually settling to 0V.
I'll leave it as an exercise for you to determine the time constant to confirm that the voltage reaches 63% on the rise and falls to 37% on the fall :)
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## Wednesday, 30 May 2018
### Daily Numerical Ability Quiz : 31st May 2018
Q1. The average income of a person for the first 6 days is Rs. 29, for the next 6 days it is Rs. 24, for the next 10 days it is Rs. 32 and for the remaining days of the month it is Rs. 30. Find the average income per day (Assume month is of 30 days).
(a) Rs. 31.64
(b) Rs. 30.64
(c) Rs. 29.27
(d) Rs. 34.27
(e) Rs. 32.27
Q2. A man covers half of his journey by train at 60 km/hr, half of the remainder by bus at 30 km/hr and the rest by cycle at 10 km/hr. Find his average speed during the entire journey.
(a) 36 kmph
(b) 30 kmph
(c) 18 kmph
(d) 24 kmph
(e) 28 kmph
Q3. How many kg of tea worth Rs. 25 per kg must be blended with 30 kg of tea worth Rs. 30 per kg so that by selling the blended variety at Rs. 30 per kg there should be gain of 10%?
(a) 30 kg
(b) 36 kg
(c) 32 kg
(d) 42 kg
(e) 34 kg
Q4. Akash scored 73 marks in subject A. He scored 56% marks in subject B and x marks in subject C. Maximum marks in each subject were 150. The overall percentage marks obtained by Akash in all the three subjects together were 54%. How many marks did he score in subject C?
(a) 84
(b) 86
(c) 79
(d) 73
(e) 94
Q5. There are 5 consecutive odd numbers. If the difference between square of the average of first two odd number and the of the average last two odd numbers is 492, what is the smallest odd number?
(a) 37
(b) 42
(c) 41
(d) 35
(e) 39
Solutions (1-5):
Q6. Mr. Anurag’s age is 120% of his wife Sushila’s age. They have two children. The average age of family is 20 years. If Sushila’s age is 25 years, what is the average age of children?
(a) 13 years
(b) 12.5 years
(c) 14.5 years
(d) 15 years
(e) 16 years
Q7. The average weight of the students of a class is 51kg. The ratio of boys and girls is 11:6. If the total no. of girls are12 and weight of teacher is also included, the average weight increases by one kg. What is the weight of teacher?
(a) 76 kg
(b) 82 kg
(c) 86 kg
(d) 78 kg
(e) 68 kg
Q8. A batsman played three matches in a tournament. The respective ratio between the scores of 1st and 2nd matches was 5 : 4 and that between the scores of 2nd and 3rd matches was 2 : 1. The difference the 1st and 3rd matches was 48 runs. What was the batsman’s average score in all the three matches?
(a) 45
(b) 176/3
(c) 70
(d) 122/3
(e) 201/4
Q9. Sarala’s present age is four times her son’s present age and four sevenths of her father’s present age. The average of the present age of all three of them is 32 years. What is the difference between the present age of Sarala’s father and present age of Sarala’s son?
(a) 44 years
(b) 48 years
(c) 46 years
(d) 42 years
(e) 56 years
Q10. With an average speed of 40 km/hr, a train reaches its destination in time. If it goes with an average speed of 35 km/hr, it is late by 15 minutes. The length of the total journey is
(a) 40 km
(b) 70 km
(c) 30 km
(d) 80 km
(e) None of these
Solutions (6-10):
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# Scalars and Vectors
Hi. It’s Mr. Andersen and right
now I’m actually playing Angry Birds. Angry Birds is a video game where you get to launch
angry birds at these pig type characters. I like it for two reasons. Number one it’s
addictive. But number two it deals with physics. And a lot of my favorite games do physics.
So let’s go to level two. And so what I’m going to talk about today are vectors and
scalars. And vectors and scalars are ways that we measure quantities in physics. And
Angry Birds would be a really boring game if I just used scalars. Because if I just
used scalars, I would input the speed of the bird and then I would just let it go. And
it would be boring because I wouldn’t be able to vary the direction. And so in Angry Birds
I can vary the direction and I can try to skip this off of . . . Nice. I can try to
skip it off and kill a number of these pigs at once. Now I could play this for the whole
ten minutes but that would probably be a waste of time. And so what I want to do is talk
them at the beginning of physics. Because sometimes we get to vectors and people get
confused and don’t understand where did they come from. And so we have quantities that
we measure in science. Especially in physics. And we give numbers and units to those. But
they come in two different types. And those are scalar and vector. To kind of talk about
the difference between the two, a scalar quantity is going to be a quantity where we just measure
the magnitude. And so an example of a scalar quantity could be speed. So when you measure
the speed of something, and I say how fast does your car go? You might say that my car
goes 109 miles per hour. Or if you’re a physics teacher you might say that my bike goes, I
don’t know, like 9.6 meters per second. And so this is going to be speed. And the reason
it is a scalar quantity is that it simply gives me a magnitude. How fast? How far? How
big? How quick? All those things are scalar quantities. What’s missing from a scalar quantity
is direction. And so vector quantities are going to tell you, not only the magnitude,
but they’re also going to tell you what direction that magnitude is in. So let me use a different
color maybe. Example of a vector quantity would be velocity. And so in science it’s
really important that we make this distinction between speed and velocity. Speed is just
how fast something is going. But velocity is also going to contain the direction. In
other words I could say that my bike is going 9.08 meters per second west. Or I could say
this pen is being thrown with an initial velocity of 2.8 meters per second up or in the positive.
And so once we add direction to a quantity, now we have a vector. Now you might think
to yourself that’s kind of nit picky. Why do we care what direction we’re flowing in?
And I have a demonstration that will kind of show you the importance of that. But a
good example would be acceleration. And so what is acceleration? Acceleration is simply
change in velocity over time. And so acceleration is going to be the change in velocity over
time. And so I could ask you a question like this. Let’s say a car is driving down a road
And it’s going 23 meters per second. And it stays at 23 meters per second. Is it accelerating?
And you would say no. Of course it’s not. Let’s say it goes around a corner. And during
that movement around the corner it stays at 23 meters per second. Well what would happen
to the scalar quantity of speed around a corner? It would still be 23 meters per second. And
so if you’re using scalar quantities we’d have to say that it’s not accelerating. But
since velocity is a vector, if you’re going 23 meters per second and you’re going around
a corner, are you accelerating? Yeah. Because you’re not changing the magnitude of your
speed but you’re clearly changing the direction. And so a change in velocity is going to be
acceleration. And so you are accelerating when you go around a corner. And so that would
be an example of why in physics I’m not trying to be nit picky I’m just saying that you have
to understand the difference between a scalar quantity and then which is just magnitude
and a vector which is magnitude and direction. There’s a review at the end of this video
and so I’ll have you go through a bunch of these and we’ll identify a number of them.
But for now I wanted to give you a little demonstration to show you the importance of
a scalar and vector quantities. And so what I have here is a 1000 gram weight. Or 1 kilogram
weight. And it’s suspend from a scale. And I don’t know if you can read that on there.
But the scale measures the number of grams. And so if this is a 1000 grams and this measures
the numbers of grams, and it’s scaled right, it should say, and it does, about 1000 grams
is the weight of this. Now a question I could ask you is this. Let’s say I bring in another
scale. And so I’m going to attach another scale to it. And so if we had 1 mass that
had a mass of 1000 grams, and now I have two scales that are bearing the weight of that.
And I lift them directly up. What should each of the scales read? And if you’re thinking
it’s 1000 grams, so each one should read 500 grams, let me try it, the right answer is
yeah. Each of the scales weigh right at about 500 grams. And so that should make sense to
you. In other words 500 plus 500 is 1000. So we have the force down of the weight. Force
of tension is holding these in position. And so we should be good to go. The problem becomes
when I start to change the angle. And so what I’m going to do, and I’m sure this will go
off screen, is I’m going to start to hold these at a different angle. And so if I look
right here I now find that it’s at 600. And so this one is at 600 as well. And so I increase
the angle like this, we’ll find that that will increase as well. And so when I get it
to an angle like this I have 1000 gram weight and it’s being supported by 2 scales now that
are reading 1000. And it’s going to vary as I come back to here. And if you do any weight
lifting you understand kind of how that works. And so the question becomes how do we do math?
The problem with this then is that the numbers don’t add up. And so if I’ve got a 500 gram
weight, excuse me, a 1000 gram weight being supported by 2 scales, it made sense that
it was weighing 500 each. But now we all of a sudden have a 1000 gram weight being supported
by two scales that are each reading 1000. And so this doesn’t make sense. Or the math
doesn’t make sense. And the reason why is that you’re trying to solve the problem from
a scalar perspective. And you’ll never be able to get the right answer. Because it’s
going to change. And it’s going to change depending on the angle that we lift them at.
So to understand this in a vector method, and we’ll get way into detail, so I just want
to kind of touch on it for just a second, what we had was a weight. So we’ll say there’s
a weight like this. And we’ll say that’s a 1000 gram weight. And then we have two scales.
And each of those scales are pulling at 500 grams. And so if you add the vectors up. So
this is one vector and this is another vector. So each of these is 500 grams, so I make the
500 in length, then we balance out. In other words we have the balancing of this weight
with these two weights that are on top of it. Now if we go to the vector problem, in
the vector problem, again we had a 1000 gram weight. So 1000 grams in the middle. And then
we had a force in this direction of 1000 and a force in that direction of 1000. So we have
a force down of 1000. But we had a force of 1000 in this direction. And a force of 1000
in that direction. And so if you start to look at it like a vector quantity, imagine
this. That we’ve got a weight right here but you have to have two people pulling on it.
And so it’s like this tug of war where it’s not just in one direction, but it’s actually
in two. And so you can start to see how these forces are going to balance out. But only
if we look at it from the vector perspective. Let me show you what that would actually look
like. So if we put these tails up, this would be that force down of 1000 grams. This would
be the force of the weight. But we also had a force in this direction. So I’m doing the
same rule where I’m lining up my vector from the tail to the tip. And the tail to the tip.
And so that diagram that I had on the last slide, I’m actually moving this one force
and you can see that they all sum up to zero. And so the reason I like to start talking
about vectors and scalars at this problem is that you could never solve the problem
if you’re going to go at it from a scalar perspective. And we’re going to do some really
cool problems. Let’s say I’m sliding a box across the floor. But how often do you slide
a box across the floor and actually pull it straight across like that? If you’re like
me you’re pulling a sled or something, you’re normally pulling it at angle. And once we
start pulling it at an angle it becomes a totally different force. And we can’t solve
problems in a scalar way. We have to go and solve if from a vector prospective. And so
that’s the importance of vectors. Now it’s a huge thing. So there are lots of things
that we can measure in physics. And so what I’m going to try to do, and hopefully I can
get this right, is go through and circle all the scalar quantities and then go back and
circle all the vector quantities. And so if you’re watching this video a good thing to
do would be to pause it right now. And then you go through it and circle the ones that
you think are scalar and vector. And then we’ll see if we match up at the end. Scalar
quantities remember are simply going to be magnitude. And so the question I always ask
myself when I’m doing this is, okay. Does it have a direction? And so length is simply
the length of a side of something. And so I would put that in the scalar perspective.
This is kind of philosophical. Does time have a direction? I would say no. Acceleration
of something, that definitely is a scalar quantity. If I say the density of that is
12.8 grams per cubic centimeter north, it doesn’t make sense at all. Where are some
other scalar quantities? Temperature would be a scalar quantity. It’s just how fast the
molecules are moving. But it’s not in one certain direction. Pressure would be another
one that’s scalar. It’s not directional. It’s not in one direction. The pressure is, remember
air pressure is the one that I always think of as being in all directions. So we wouldn’t
say that. Let’s see mass. The mass of something is going to be a scalar quantity as well.
And so it doesn’t change. Now weight, and we’ll talk about that more later in the year,
would actually be a vector quantity. Let’s see if I’m missing any. No I think this would
be good. So let’s change color for a second. So displacement is how far you move from a
location. And that’s in a direction. So we call that a vector quantity. Acceleration
I mentioned before. Force is going to be a vector. And we’ll do these force diagrams
which are really fun later in the year. Drag is something slowing you down. So if you’re
a car it’s what is slowing you down in the opposite direction of your movement. And so
the direction is important. Momentum is a product of velocity and the mass of an object.
And lift we get from like an airplane wing. That would be a vector quantity because it’s
in a direction. And so these are all vector quantities. The ones that I circled in red.
But there are way more that we’re going to find out there. And scalar quantities remember,
it’s simply just magnitude. Or how big it is. And so as we go through physics, be thinking
to yourself, is this a scalar quantity or vector? And if it’s vector my problem is a
little bit harder, but like Angry Birds, it’s more fun when you go the vector route. And
so I hope that’s helpful and have a great day.
## 100 thoughts on “Scalars and Vectors”
1. karl vince casitas says:
thank you !
2. Taay27 says:
Watching this on summer break… Hahah. c:
3. Julia S says:
This was so clear !!! Thanks for posting !
4. Karina Sirenita says:
Where have you been all my life?!?!?!
5. Sujith Subramanian says:
Good to get back to basics .. Thank You.. Was very helpful.
6. Clarisa C says:
7. Brenda Amor says:
gravity is equal to 9.8…
8. Pierson Ngo says:
i need 2 know this in middle school.
9. David Thurlo says:
Brenda, it's 9.8 m/s/s.
10. Katie M says:
Please make more physics videos!! I need help learning how to physics!
11. DiDi Di says:
duhhh
12. Raru says:
Thanks for the vid! This helps for my report in Science. I really had a rough time studying and researching. Thanks again!
13. supercard says:
You're awesome
14. Andy Hau says:
1000 grams is not weight, it is mass. U got me confused
15. BATCHFILETUTORIALS says:
Thanks!
16. C Toumi says:
very good example – 9/10
17. Stanly nathan says:
mr.mr. ANDERSON man u rsimply awesome.
thk u a lot.pls pls continue ur contribution.
18. O A says:
thank you so much really cleared things up, great teacher!
19. Baldev Rawat says:
20. M Afg says:
Thankyou so muchhhh !!!
I saw a lot of videos on youtube about Scalar and Vector quantities but couldn't completely understand..
Your video was the best 🙂
21. Portia Urot says:
Thanks! 🙂
22. Samsher Singh says:
poor poor poor poor poor poor poor poor poor poor poor poor poor poor dude!!!!!!!! LOL F***K you…….
23. rian luthfi says:
thanks dude 🙂
24. gurvinder kaur says:
25. blackspoted moose says:
Wow this is way better then what my teacher shows my class. Try learning this stuff/chemistry by a youtuber named crash course.
26. Yene Mulatu says:
very cool, like the example made it crystal clear
27. Paritosh Ambani says:
28. the chanakya institute says:
USE UR BRAIN PAIN IN BRAIN
29. mellow pipes says:
thx man, really good
30. Deborah Saunders says:
You did a wonderful job with this video! It gave me great understanding of the difference between a Scalar and a Vector. I got almost all of the Scalers correct in the review. Thank you.
31. Alberto Perez says:
thumbs up
32. THE INDIAN ANALYSIS says:
It is really good videos for understand scaler and vector.
33. THE INDIAN ANALYSIS says:
It is really good videos for understand scaler and vector.
34. SamuraixJack says:
Mr. Anderson. I'm currently brushing up on some mechanical engineering, even joined the club at my school. Is there any difference between drag and friction?
35. Leo Martinez says:
,CDs HBO e'er u tags Dee ft treed FDR e'er u fry f
the question is" how do we do maths" ahaha
37. CoLage Gamer says:
38. Brian E. says:
I don't get it ~ does he like Angry Birds or is he addicted to Angry Birds???
39. MegaLock12345 says:
Shoutout from CVHS
40. Teja Swaroop Mudiki says:
can someone please explain why is pressure a scalar.
just like dimension analysis Pr = force / Area where pr is dependent on vector data type – force
41. bimmjim says:
Scalar means non-vector. I'm done here.
42. Gian Franco says:
Around 8:18 of the video he's talking about how both scales are supporting the kg weight at a angle both with 1kg of force. I thought kilograms were a unit of mass and we use newton's as a unit of force but after watching this it confuses me alot can someone please explain and clarify why this is.
43. Teja Swaroop Mudiki says:
Time in my opinion has direction and can be classified as vector ,
-10 secs ; 0 sec ("event") ; +10secs
What do you say Internet ? 🙂
44. Gul Shinwari says:
can some 1 plz tell me difference between magnitude and position…?
45. Ganesan Vetriselvan says:
Good one thanks for posting
46. manoj kumar says:
pressure is a tensor
47. akhan727 says:
I cannot stress enough how well you explained this. This video is refreshing!
48. Joe Joe says:
Clickbait title
49. Cody Carr says:
I was having so much trouble with this… Not anymore! Thank you, Mr. Andersen!
50. Onion Love says:
thank you!
51. darryl taylor says:
I sat in my introductory Chem/Phys class in college and asked my professor what difference did it make whether I identified something as either scalar or vector…..the light bulb finally turned on.
52. Nashaat Majo says:
I love you like a brother man, thank you sooooo much! Keep explaining this specificity and accurately. Thank you, thank you, thank you!
This was very helpful. Many thanks!
54. Rubin Singh says:
nice video man … really helpful.
55. Pyxis says:
Lol cool bro.
56. Leonor Amaya says:
you kinda look like Matt Damon lol but thank you this lesson was helpful.
Dude fix the sound in the video
58. Cabbage Baka says:
Thanks for the help 🙂 studying for exam tomorrowww T_T hahaha
60. Xander Botha says:
This guy explains very well, even though you can't ask him. Helped me alot
61. Iman Umer says:
Pressure is scalar? Why The hell our teacher said vector!!!!
Ur explanation is super great
63. Laiba Malik says:
useful information
64. Chris Rogers says:
the question is though……how do we do math? scientists one unsolved question
65. Naeem Baluch says:
dont make the video of playing your silly game…… just teach us the Main thing……
this is making me didtrak
66. marc martinez says:
67. 1k24 ming says:
68. Dark Magician Girl says:
1:01 hey aim for the one at the top by going behind it… it'll push the other pigs…
69. mr .master says:
you such a great teacher .thank you so mush . student from yemen
70. Alexander Das says:
awsam
71. fazilat says:
wow what a explanation it is very practical with understandable english
thanks
72. SebastianChem97 says:
So physics change math? Crazy wazy aazzzzzy
73. Pracheta Ghosh says:
amazing videos..very good teacher
74. Sakun Panthi says:
I want all of you to know that //Vectors can be 0 and negative but Scalars can never be zero or negative!!!!
75. Kinchu P.G says:
U just cleared all my doubts about scalar and vectors
76. Kinchu P.G says:
U just cleared all my doubts about scalar and vectors
77. Saraswati Mandi says:
what is magnitude
78. CREATIVE 5 MIN CRAFT INDIA says:
you are a foreign
79. STUDENTS CLUB says:
Keep it up
80. Fardeen Ansari says:
Hey viewers, Mr. Anderson circled pressure to say it is scalar but pressure is just a force/area and force is vector then, how pressure could be scalar?
81. Roshan Elison says:
thanks bro
82. Gloreyya Chinaza says:
Is weight a vector quantity?
83. Ian Dias says:
Thanks a lot
84. Carlos Tarnowski says:
Is it ok to ilustrate grams as a vector? isnt it going to confuse people?
85. maximilian harvison says:
i wanted to watch angry birds
86. M. Mudassir Khan says:
9.08 m/s Isn't it too fast? 😐
87. Aslam Khan says:
This helps to understands physics very well thanks
88. Haseeb Khan says:
m ur fan sir
89. Vijay Nauhbar says:
a train moving with acceleration of 5 m/s^2 have 1 engine of mass 100 kg and 4 coaches of mass 50 kg each find the force exerted by 3rd coach on 4th
90. Vijay Nauhbar says:
pls help me
91. Zhōu yǔ Qiáo says:
6:48 – I ask myself this question every day.
92. science fair says:
…..
93. Stnl N2 says:
Thanks man. …You are so. …much better than those "public" school government bureaucrat teachers. …. I wish we dismantle this Leftist incompetent monstrosity we called Public School Education. ..and give taxpayers money to the Individual citizens/parents…. instead of incompetent government bureaucrats….
Again man…. THANK YOU. ..
94. Pakistani Art says:
Great teacher 👍
95. Lrn Fzx - Learn Physics says:
265K views at this writing and mass/weight (scalar/vector) are consistently confused in the video. Major problem. Mass is a scalar to be measured on a balance in metric units of kg or g. Weight is the vector to be measured on a scale in units of N. The only reason that stupid scale is labeled in g is because mass and weight are proportional since W = mg. This confusion of mass/weight will lead to problems with all of Newton's Laws of Motion and Gravity and Satellites and Energy – basically all of Mechanics.
These will help with that distinction:
Mass vs Weight, Balance vs Scale:
I Was Wrong – Mass vs Weight:
96. Shirsendu Acharyya says:
Good and attractive introduction.. ! Well done….!
97. J B says:
Length and time are vectors my friend
98. henry egbe says:
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Enter the number of elements in set A, the number of elements in set B, and the number of intersecting elements into the Jaccard Coefficient Calculator. The calculator will evaluate and display the Jaccard Coefficient.
## Jaccard Coefficient Formula
The following formula is used to calculate the Jaccard Coefficient.
JC = Ni / (Na+Nb-Ni)
• Where JC is the Jaccard Coefficient
• Na is the number of elements in set A
• Nb is the number of elements in set B
• Ni is the number of intersecting elements
To calculate the Jaccard coefficient, divide the number of intersecting elements by the result of the number of elements in set A plus the number of elements in set B minus the number of intersection elements.
## How to Calculate Jaccard Coefficient?
The following example problems outline how to calculate Jaccard Coefficient.
Example Problem #1
1. First, determine the number of elements in set A.
• The number of elements in set A is calculated to be : 4.
2. Next, determine the number of elements in set B.
• The number of elements in set B is measured to be: 5.
3. Next, determine the number of intersecting elements.
• The number of intersecting elements is found to be: 2.
4. Finally, calculate the Jaccard Coefficient using the formula above:
JC = Ni / (Na+Nb-Ni)
The values given above are inserted into the equation below and the solution is calculated:
JC = 2 / (4+5-2) = 2.857
Example Problem #2
The variables needed for this problem are provided below:
number of elements in set A = 7
number of elements in set B = 10
number of intersecting elements = 3
This example problem is a test of your knowledge on the subject. Use the calculator above to check your answer.
JC = Ni / (Na+Nb-Ni) = (\${Main Units})
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# How to find the number of pair of points that can form a line with positive slope fast?
Given n pairs of points, how can we get the number of pair of points that can form a line with positive slope fast?
Then there are n lines, where the i-th line contains two integers xi and yi specifying the x and y-coordinates of i-th point. No two points have the same x-coordinate and no two points have the same y-coordinate.
My idea is to sort x first and then compare each point with the other below it. But it is still O(n^2)
-
What are your thoughts? What do you already know? – Felix Kling Apr 28 '12 at 14:54
Are points paired? – MBo Apr 28 '12 at 15:05
If there are n lines, then it takes O(n) time to get lines with positive slopes – MBo Apr 28 '12 at 15:19
I don't think it is possible in O(n)... – Bear Apr 28 '12 at 15:23
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## 2 Answers
Sort descending by x and then count the number of inversions in y. Both steps are O(n log n).
Explanation: the number of (unordered) pairs with positive slope is the number of pairs {(xi, yi), (xj, yj)} such that xi > xj and yi > yj. When we sort descending by x, we make it so that xi > xj if and only if i < j. The definition of an inversion in the ys is a pair i < j such that yi > yj.
Counting the number of inversions quickly.
-
Could you mind explaining more,plx – Bear Apr 29 '12 at 11:46
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Pick a point (randomly is easiest and has a good expected running time, although you can find a median point deterministically in linear time), split the axes into four quadrants:
`````` x |
| x x
x | x
-----------x-----------
x |
x | x
| x
``````
Denote the quadrants in counter-clockwise direction from top-left by `I`,`II`,`III`,`IV`:
``````II | I
----|----
III | IV
``````
We shall disregard points that lie on the axes (edge case with theoretical probability of 0, and easily dealt with practically).
Note that all points in quadrant `III` form a positively-sloped line with all points in `I`, similarly no points from `II` will form p.s. lines with points in `IV`, so we recursively call:
``````NumPSLines(G) = |I|*|III| +
NumPSLines(I U II) +
NumPSLines(II U III) +
NumPSLines(III U IV)
``````
Where `U` denotes union.
Assuming (proof left to reader) that the expected values `E(|I|) = ... E(|IV|) = |G|/4 = n/4` and that the partition into quadrants is linear then we get an expected run time of:
``````T(n) = O(n) + 3T(n/2)
= O(n) + ... + 3^k * t(n/2^k) // where k = log2(n)
= O( log2(n) * 3^log2(n) )
= O(n^(log2(3)) * logn)
~ O(n^1.6 * logn)
``````
Not sure if that bound is tight; haven't thought about it much.
This solution can be super optimised, although it's a start.
-
you are so clever... – Bear Apr 29 '12 at 6:15
@Bear, thanks, although the other answer is slightly more efficient and simpler, you should probably choose it! – davin Apr 29 '12 at 10:01
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# Math Snap
## (Dividing Rational Numbers MC) Solve $-3 \frac{1}{4} \div 2 \frac{1}{3}$ $-1 \frac{11}{28}$ $-6 \frac{1}{12}$ $-7 \frac{7}{12}$ $1 \frac{11}{28}$
#### STEP 1
Assumptions1. We are dividing two mixed numbers $-3 \frac{1}{4}$ and $\frac{1}{3}$. . We need to convert these mixed numbers into improper fractions before performing the division.
3. The division of fractions is performed by multiplying the first fraction by the reciprocal of the second fraction.
#### STEP 2
First, let's convert $- \frac{1}{4}$ into an improper fraction. The formula to convert a mixed number into an improper fraction is$\text{Improper Fraction} = \text{Whole Number} \times \text{Denominator} + \text{umerator}$
#### STEP 3
Now, plug in the given values for the whole number, denominator, and numerator to calculate the improper fraction.
$\text{Improper Fraction} = -3 \times +1$
#### STEP 4
Calculate the improper fraction for $-3 \frac{1}{4}$.
$\text{Improper Fraction} = -3 \times4 +1 = -12 +1 = -11$So, $-3 \frac{1}{4}$ is equivalent to $-\frac{11}{4}$.
#### STEP 5
Next, let's convert $2 \frac{1}{3}$ into an improper fraction using the same formula.
$\text{Improper Fraction} =2 \times3 +1$
#### STEP 6
Calculate the improper fraction for $2 \frac{1}{3}$.
$\text{Improper Fraction} =2 \times3 +1 =6 +1 =$So, $2 \frac{1}{3}$ is equivalent to $\frac{}{3}$.
#### STEP 7
Now that we have both numbers as improper fractions, we can perform the division. The division of fractions is performed by multiplying the first fraction by the reciprocal of the second fraction.
$\frac{-11}{4} \div \frac{7}{3} = \frac{-11}{4} \times \frac{3}{7}$
#### STEP 8
Perform the multiplication to get the result.
$\frac{-11}{4} \times \frac{3}{7} = \frac{-33}{28}$
##### SOLUTION
Finally, we convert the improper fraction back to a mixed number. The whole number part is the quotient of the division, and the remainder becomes the numerator of the fractional part.
$\frac{-33}{28} = - \frac{11}{28}$So, $-3 \frac{}{4} \div2 \frac{}{3} = - \frac{11}{28}$.
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# Track Santa’s Journey Through the Night Sky
Here's your nightly math! Just 5 quick minutes of number fun for kids and parents at home. Read a cool fun fact, followed by math riddles at different levels so everyone can jump in. Your kids will love you for it.
# Track Santa’s Journey Through the Night Sky
December 23, 2013
As a child, I scoured the night sky on December 24 looking for Santa and his reindeer. My dedication paid off. When I was seven, I saw their dark shadows crossing a moonlit sky. Really! Nearly forty years later I remember the thrill of spotting the sleigh and calling members of my family to the large picture window in our living room so they could see, too. They arrived too late, though.
Usually it happens the other way around, with parents pointing out Santa and the kids getting to the window a moment too late to see. Thanks to modern technology and math, families can gather ’round a computer screen tracking Santa’s journey in real time with NORAD’s Santa Tracker. It’s a fun way to build up the dramatic tension as you wait for his Christmas visit. Still, after getting Santa and Company’s coordinates it might be nice to head to a window or even snuggle outside under the stars, if it’s not too cold where you live.
Santa’s got a pretty big trip to make all the way around the Earth. A fancy word for his trip is circumnavigating–or going around the circumference of–our planet. That distance around the Earth is figured out the same way you figure out the circumference of a playground ball. Okay, with a ball you might be able to use a tape measure. But there’s a mathematical formula that’s the same for a playground ball and a huge planet. And do you know that before there were computers and even before the first Christmas, a man figured out the circumference of the earth?
But that’s challenging math for a young child. Here’s some more practical Santa-spotting math.
When spotting distances in the sky, it’s helpful to talk about angles and degrees. This is easily explained with a protractor. What? You haven’t had a protractor since high school? Pick one up at an office supply store and add it to this year’s stocking stuffers with a set of colored pencils or markers. Your child can outline a protractor and draw big, goofy smiles, fabulous rainbows, and icy igloos. If your child doesn’t ask about the numbers on the instrument, you can bring them up casually at some point.
Back to those angles and degrees . . . Anything on the horizon is considered at zero degrees. What are some things you can observe at zero degrees? If you live in a flat area, your neighbor’s house might be at zero degree. If you don’t have a reference point in the night sky, you might stand under a tree and point to the top branches as being 90 degrees up.
Granted, Santa only comes one night a year, but you can practice these games indoors at any time. And if you’re looking for the excitement of spotting a moving object in the night sky to hone your Santa-watching skills for next year, click to find out when to look for the International Space Station to fly by your home.
### Kim Moldofsky
Kim Moldofsky is a mom of teen boys in the Chicago area. She blogs at TheMakerMom.com and hosts a the popular monthly #STEMchat on Twitter where parents and educators share ideas and resources to raise STEM-loving kids.
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# Collision of rectangles
#### r4nd0m
I wonder if somebody could help me with this problem I'm solving for my c language class (but it's more a mathematical problem I think).
So we have 2 different rectangles (ABCD and EFGH) in an ortogonal plane, their sides are paralel to the axes of the ortogonal system. Each of this rectangles is given a velocity vector. Now the task is to determine where the first collision will take place (where these two rectangles (their sides or corners) will meet the first time) if it will take place.
Any suggestions?
Can this problem be solved via mathematicl formulas?
thanks a lot
#### AKG
Homework Helper
A couple ideas here to help you, not much because I'm at work right now. For one, motion is relative, so if you have a velocity vectors v1 and v2, then just treat the second rectangle as motionless and treat the first one as though it were moving at velocity v1 - v2. The two rectangles will intersect if and only if:
a) the vertical distance between the centers of the rectangles is less than or equal to V = 0.5(the vertical height of rectangle 1 + the vertical height of rectangle 2)
AND
b) the horizontal distance between the centers of the rectangles is less than or equal to H = 0.5(the horizontal width of recantle 1 + the horizontal width of rectangle 2)
It should be quite easy to make a program that checks if these conditions are ever met. The plot of horizontal distance between centers over time will look like an absolute value graph (V shape) and same for vertical distance. You will be able to solve for 0, 1, or 2 points in time where the horizontal distance between centers is exactly equal to H. If there are 0 point in time, then no intersection. If there is one, and it occurs at time t, then plug in t in your equation to find the vertical distance, and see if the vertical distance between centers is less than equal to V. If so, then they meet at t, otherwise they don't. If you find two points of time, t and t' (assume t < t'), then do a similar calculation for vertical distance so you find either 0, 1, or 2 points in time when the vertical distance is exaclty equal to V. If 0, no intersection. If 1, and it occurs at time t'', check if t < t'' < t'. If so, then they intersect only once at t'', otherwise never. Finally, say you find two times, t'' and t'''. You want to find the smallest t value that is in both intervals [t,t'] and [t'',t''']. You can figure out how to do that.
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The frequency of a stretched uniform wire under tension is in resonance with the fundamental frequency of a closed tube. If the tension in the wire is increased by 8 N, it is in resonance with the first overtone of the closed tube. The initial tension in the wire is
(1) 1 N
(2) 4 N
(3) 8 N
(4) 16 N
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(1) 25 Hz
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An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f1. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from f1 , then again a resonance is obtained with a frequency f2. If in this case the pipe vibrates in ${n}^{th}$ harmonic, then -
(1) n = 3, ${f}_{2}=\frac{3}{4}{f}_{1}$
(2) n = 3, ${f}_{2}=\frac{5}{4}{f}_{1}$
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Two loudspeakers L1 and L2 driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is 330 ms–1 then the frequency at which the first maximum is observed is
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(1) $\sqrt{gL}$
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(3) gL
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Three waves of equal frequency having amplitudes 10 μm, 4 μm and 7 μm arrive at a given point with a successive phase difference of $\frac{\pi }{2}$. The amplitude of the resulting wave in μm is given by
(1) 7
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An organ pipe is closed at one end has a fundamental frequency of 1500 Hz. The maximum number of overtones generated by this pipe which a normal person can hear is :
(1) 14
(2) 13
(3) 6
(4) 9
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Two pulses in a stretched string whose centres are initially 8 cm apart are moving towards each other as shown in the figure. The speed of each pulse is 2 cm/s. After 2 seconds, the total energy of the pulses will be
1. Zero
2. Purely kinetic
3. Purely potential
4. Partly kinetic and partly potential
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The diagram below shows the propagation of a wave. Which points are in the same phase : [AIIMS 1982]
(1) F, G
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# Cg Programming/Unity/Soft Shadows of Spheres
Shadows are not only important to understand the geometry of a scene (e.g. the distances between objects); they can also be quite beautiful.
This tutorial covers soft shadows of spheres.
Umbra (black) and penumbra (gray) are the main parts of soft shadows.
“The crowning with thorns” by Caravaggio (ca. 1602). Note the shadow line in the upper left corner, which becomes softer with increasing distance from the shadow casting wall.
While directional light sources and point light sources produce hard shadows, any area light source generates a soft shadow. This is also true for all real light sources, in particular the sun and any light bulb or lamp. From some points behind the shadow caster, no part of the light source is visible and the shadow is uniformly dark: this is the umbra. From other points, more or less of the light source is visible and the shadow is therefore less or more complete: this is the penumbra. Finally, there are points from where the whole area of the light source is visible: these points are outside of the shadow.
In many cases, the softness of a shadow depends mainly on the distance between the shadow caster and the shadow receiver: the larger the distance, the softer the shadow. This is a well known effect in art; see for example the painting by Caravaggio to the right.
Vectors for the computation of soft shadows: vector L to the light source, vector S to the center of the sphere, tangent vector T, and distance d of the tangent from the center of the light source.
## Computation
We are going to approximately compute the shadow of a point on a surface when a sphere of radius ${\displaystyle r_{\text{sphere}}}$ at S (relative to the surface point) is occluding a spherical light source of radius ${\displaystyle r_{\text{light}}}$ at L (again relative to the surface point); see the figure to the left.
To this end, we consider a tangent in direction T to the sphere and passing through the surface point. Furthermore, this tangent is chosen to be in the plane spanned by L and S, i.e. parallel to the view plane of the figure to the left. The crucial observation is that the minimum distance ${\displaystyle d}$ of the center of the light source and this tangent line is directly related to the amount of shadowing of the surface point because it determines how large the area of the light source is that is visible from the surface point. More precisely spoken, we require a signed distance (positive if the tangent is on the same side of L as the sphere, negative otherwise) to determine whether the surface point is in the umbra (${\displaystyle d<-r_{\text{light}}}$), in the penumbra (${\displaystyle -r_{\text{light}}), or outside of the shadow (${\displaystyle r_{\text{light}}).
For the computation of ${\displaystyle d}$, we consider the angles between L and S and between T and S. The difference between these two angles is the angle between L and T, which is related to ${\displaystyle d}$ by:
${\displaystyle \measuredangle (\mathbf {L} ,\mathbf {T} )\approx \sin \measuredangle (\mathbf {L} ,\mathbf {T} )={\frac {d}{\left\vert \mathbf {L} \right\vert }}}$.
Thus, so far we have:
${\displaystyle d\approx \left\vert \mathbf {L} \right\vert \measuredangle (\mathbf {L} ,\mathbf {T} )}$ ${\displaystyle =\left\vert \mathbf {L} \right\vert \left(\measuredangle (\mathbf {L} ,\mathbf {S} )-\measuredangle (\mathbf {T} ,\mathbf {S} )\right)}$
We can compute the angle between T and S using
${\displaystyle \sin \measuredangle (\mathbf {T} ,\mathbf {S} )={\frac {r_{\text{sphere}}}{\left\vert \mathbf {S} \right\vert }}}$.
Thus:
${\displaystyle \measuredangle (\mathbf {T} ,\mathbf {S} )=\arcsin {\frac {r_{\text{sphere}}}{\left\vert \mathbf {S} \right\vert }}}$.
For the angle between L and S we use a feature of the cross product:
${\displaystyle \left\vert \mathbf {a} \times \mathbf {b} \right\vert =\left\vert \mathbf {a} \right\vert \,\left\vert \mathbf {b} \right\vert \,\sin \measuredangle (\mathbf {a} ,\mathbf {b} )}$.
Therefore:
${\displaystyle \measuredangle (\mathbf {L} ,\mathbf {S} )=\arcsin {\frac {\left\vert \mathbf {L} \times \mathbf {S} \right\vert }{\left\vert \mathbf {L} \right\vert \,\left\vert \mathbf {S} \right\vert }}}$.
All in all we have:
${\displaystyle d\approx \left\vert \mathbf {L} \right\vert \left(\arcsin {\frac {\left\vert \mathbf {L} \times \mathbf {S} \right\vert }{\left\vert \mathbf {L} \right\vert \,\left\vert \mathbf {S} \right\vert }}-\arcsin {\frac {r_{\text{sphere}}}{\left\vert \mathbf {S} \right\vert }}\right)}$
The approximation we did so far, doesn't matter much; more importantly it doesn't produce rendering artifacts. If performance is an issue one could go further and use arcsin(x) ≈ x; i.e., one could use:
${\displaystyle d\approx \left\vert \mathbf {L} \right\vert \left({\frac {\left\vert \mathbf {L} \times \mathbf {S} \right\vert }{\left\vert \mathbf {L} \right\vert \,\left\vert \mathbf {S} \right\vert }}-{\frac {r_{\text{sphere}}}{\left\vert \mathbf {S} \right\vert }}\right)}$
This avoids all trigonometric functions; however, it does introduce rendering artifacts (in particular if a specular highlight is in the penumbra that is facing the light source). Whether these rendering artifacts are worth the gains in performance has to be decided for each case.
Next we look at how to compute the level of shadowing ${\displaystyle w}$ based on ${\displaystyle d}$. As ${\displaystyle d}$ decreases from ${\displaystyle r_{\text{light}}}$ to ${\displaystyle -r_{\text{light}}}$, ${\displaystyle w}$ should increase from 0 to 1. In other words, we want a smooth step from 0 to 1 between values -1 and 1 of ${\displaystyle -d/r_{\text{light}}}$. Probably the most efficient way to achieve this is to use the Hermite interpolation offered by the built-in Cg function smoothstep(a,b,x) = t*t*(3-2*t) with t=clamp((x-a)/(b-a),0,1):
${\displaystyle w=\mathrm {smoothstep} \left(-1,1,{\frac {-d}{r_{\text{light}}}}\right)}$
While this isn't a particular good approximation of a physically-based relation between ${\displaystyle w}$ and ${\displaystyle d}$, it still gets the essential features right.
Furthermore, ${\displaystyle w}$ should be 0 if the light direction L is in the opposite direction of S; i.e., if their dot product is negative. This condition turns out to be a bit tricky since it leads to a noticeable discontinuity on the plane where L and S are orthogonal. To soften this discontinuity, we can again use smoothstep to compute an improved value ${\displaystyle w'}$:
${\displaystyle w'=w\,\mathrm {smoothstep} \left(0.0,0.2,{\frac {\mathbf {L} \cdot \mathbf {S} }{\left\vert \mathbf {L} \right\vert \,\left\vert \mathbf {S} \right\vert }}\right)}$
Additionally, we have to set ${\displaystyle w'}$ to 0 if a point light source is closer to the surface point than the occluding sphere. This is also somewhat tricky because the spherical light source can intersect the shadow-casting sphere. One solution that avoids too obvious artifacts (but fails to deal with the full intersection problem) is:
${\displaystyle w''=w'\,\mathrm {smoothstep} \left(0,r_{\text{sphere}},\left\vert \mathbf {L} \right\vert -\left\vert \mathbf {S} \right\vert \right)}$
In the case of a directional light source we just set ${\displaystyle w''=w'}$. Then the term ${\displaystyle (1-w'')}$, which specifies the level of unshadowed lighting, should be multiplied to any illumination by the light source. (Thus, ambient light shouldn't be multiplied with this factor.) If the shadows of multiple shadow casters are computed, the terms ${\displaystyle (1-w'')}$ for all shadow casters have to be combined for each light source. The common way is to multiply them although this can be inaccurate (in particular if the umbras overlap).
## Implementation
The implementation computes the length of the lightDirection and sphereDirection vectors and then proceeds with the normalized vectors. This way, the lengths of these vectors have to be computed only once and we even avoid some divisions because we can use normalized vectors. Here is the crucial part of the fragment shader:
// computation of level of shadowing w
float3 sphereDirection =
_SpherePosition.xyz - input.posWorld.xyz;
float sphereDistance = length(sphereDirection);
sphereDirection = sphereDirection / sphereDistance;
float d = lightDistance
* (asin(min(1.0,
length(cross(lightDirection, sphereDirection))))
float w = smoothstep(-1.0, 1.0, -d / _LightSourceRadius);
w = w * smoothstep(0.0, 0.2,
dot(lightDirection, sphereDirection));
if (0.0 != _WorldSpaceLightPos0.w) // point light source?
{
w = w * smoothstep(0.0, _SphereRadius,
lightDistance - sphereDistance);
}
The use of asin(min(1.0, ...)) makes sure that the argument of asin is in the allowed range.
The complete source code defines properties for the shadow-casting sphere and the light source radius. All values are expected to be in world coordinates. For directional light sources, the light source radius should be given in radians (1 rad = 180° / π). The best way to set the position and radius of the shadow-casting sphere is a short script that should be attached to all shadow-receiving objects that use the shader, for example:
@script ExecuteInEditMode()
var occluder : GameObject;
function Update () {
if (null != occluder) {
GetComponent(Renderer).sharedMaterial.SetVector("_SpherePosition",
occluder.transform.position);
occluder.transform.localScale.x / 2.0);
}
}
This script has a public variable occluder that should be set to the shadow-casting sphere. Then it sets the properties _SpherePostion and _SphereRadius of the following shader (which should be attached to the same shadow-receiving object as the script).
The fragment shader is quite long and in fact we have to use the line #pragma target 3.0 to ignore some restrictions of older GPUs as documented in the Unity reference.
Shader "Cg shadow of sphere" {
Properties {
_Color ("Diffuse Material Color", Color) = (1,1,1,1)
_SpecColor ("Specular Material Color", Color) = (1,1,1,1)
_Shininess ("Shininess", Float) = 10
_SpherePosition ("Sphere Position", Vector) = (0,0,0,1)
}
Pass {
Tags { "LightMode" = "ForwardBase" }
// pass for ambient light and first light source
CGPROGRAM
#pragma vertex vert
#pragma fragment frag
#pragma target 3.0
#include "UnityCG.cginc"
uniform float4 _LightColor0;
// color of light source (from "Lighting.cginc")
// User-specified properties
uniform float4 _Color;
uniform float4 _SpecColor;
uniform float _Shininess;
uniform float4 _SpherePosition;
// center of shadow-casting sphere in world coordinates
// in radians for directional light sources
struct vertexInput {
float4 vertex : POSITION;
float3 normal : NORMAL;
};
struct vertexOutput {
float4 pos : SV_POSITION;
float4 posWorld : TEXCOORD0;
float3 normalDir : TEXCOORD1;
};
vertexOutput vert(vertexInput input)
{
vertexOutput output;
float4x4 modelMatrix = unity_ObjectToWorld;
float4x4 modelMatrixInverse = unity_WorldToObject;
output.posWorld = mul(modelMatrix, input.vertex);
output.normalDir = normalize(
mul(float4(input.normal, 0.0), modelMatrixInverse).xyz);
output.pos = mul(UNITY_MATRIX_MVP, input.vertex);
return output;
}
float4 frag(vertexOutput input) : COLOR
{
float3 normalDirection = normalize(input.normalDir);
float3 viewDirection = normalize(
_WorldSpaceCameraPos - input.posWorld.xyz);
float3 lightDirection;
float lightDistance;
float attenuation;
if (0.0 == _WorldSpaceLightPos0.w) // directional light?
{
attenuation = 1.0; // no attenuation
lightDirection =
normalize(_WorldSpaceLightPos0.xyz);
lightDistance = 1.0;
}
else // point or spot light
{
lightDirection =
_WorldSpaceLightPos0.xyz - input.posWorld.xyz;
lightDistance = length(lightDirection);
attenuation = 1.0 / lightDistance; // linear attenuation
lightDirection = lightDirection / lightDistance;
}
// computation of level of shadowing w
float3 sphereDirection =
_SpherePosition.xyz - input.posWorld.xyz;
float sphereDistance = length(sphereDirection);
sphereDirection = sphereDirection / sphereDistance;
float d = lightDistance
* (asin(min(1.0,
length(cross(lightDirection, sphereDirection))))
float w = smoothstep(-1.0, 1.0, -d / _LightSourceRadius);
w = w * smoothstep(0.0, 0.2,
dot(lightDirection, sphereDirection));
if (0.0 != _WorldSpaceLightPos0.w) // point light source?
{
w = w * smoothstep(0.0, _SphereRadius,
lightDistance - sphereDistance);
}
float3 ambientLighting =
UNITY_LIGHTMODEL_AMBIENT.rgb * _Color.rgb;
float3 diffuseReflection =
attenuation * _LightColor0.rgb * _Color.rgb
* max(0.0, dot(normalDirection, lightDirection));
float3 specularReflection;
if (dot(normalDirection, lightDirection) < 0.0)
// light source on the wrong side?
{
specularReflection = float3(0.0, 0.0, 0.0);
// no specular reflection
}
else // light source on the right side
{
specularReflection = attenuation * _LightColor0.rgb
* _SpecColor.rgb * pow(max(0.0, dot(
reflect(-lightDirection, normalDirection),
viewDirection)), _Shininess);
}
return float4(ambientLighting
+ (1.0 - w) * (diffuseReflection + specularReflection),
1.0);
}
ENDCG
}
Pass {
Tags { "LightMode" = "ForwardAdd" }
// pass for additional light sources
Blend One One // additive blending
CGPROGRAM
#pragma vertex vert
#pragma fragment frag
#pragma target 3.0
#include "UnityCG.cginc"
uniform float4 _LightColor0;
// color of light source (from "Lighting.cginc")
// User-specified properties
uniform float4 _Color;
uniform float4 _SpecColor;
uniform float _Shininess;
uniform float4 _SpherePosition;
// center of shadow-casting sphere in world coordinates
// in radians for directional light sources
struct vertexInput {
float4 vertex : POSITION;
float3 normal : NORMAL;
};
struct vertexOutput {
float4 pos : SV_POSITION;
float4 posWorld : TEXCOORD0;
float3 normalDir : TEXCOORD1;
};
vertexOutput vert(vertexInput input)
{
vertexOutput output;
float4x4 modelMatrix = unity_ObjectToWorld;
float4x4 modelMatrixInverse = unity_WorldToObject;
output.posWorld = mul(modelMatrix, input.vertex);
output.normalDir = normalize(
mul(float4(input.normal, 0.0), modelMatrixInverse).xyz);
output.pos = mul(UNITY_MATRIX_MVP, input.vertex);
return output;
}
float4 frag(vertexOutput input) : COLOR
{
float3 normalDirection = normalize(input.normalDir);
float3 viewDirection = normalize(
_WorldSpaceCameraPos - input.posWorld.xyz);
float3 lightDirection;
float lightDistance;
float attenuation;
if (0.0 == _WorldSpaceLightPos0.w) // directional light?
{
attenuation = 1.0; // no attenuation
lightDirection = normalize(_WorldSpaceLightPos0.xyz);
lightDistance = 1.0;
}
else // point or spot light
{
lightDirection =
_WorldSpaceLightPos0.xyz - input.posWorld.xyz;
lightDistance = length(lightDirection);
attenuation = 1.0 / lightDistance; // linear attenuation
lightDirection = lightDirection / lightDistance;
}
// computation of level of shadowing w
float3 sphereDirection =
_SpherePosition.xyz - input.posWorld.xyz;
float sphereDistance = length(sphereDirection);
sphereDirection = sphereDirection / sphereDistance;
float d = lightDistance
* (asin(min(1.0,
length(cross(lightDirection, sphereDirection))))
float w = smoothstep(-1.0, 1.0, -d / _LightSourceRadius);
w = w * smoothstep(0.0, 0.2,
dot(lightDirection, sphereDirection));
if (0.0 != _WorldSpaceLightPos0.w) // point light source?
{
w = w * smoothstep(0.0, _SphereRadius,
lightDistance - sphereDistance);
}
float3 diffuseReflection =
attenuation * _LightColor0.rgb * _Color.rgb
* max(0.0, dot(normalDirection, lightDirection));
float3 specularReflection;
if (dot(normalDirection, lightDirection) < 0.0)
// light source on the wrong side?
{
specularReflection = float3(0.0, 0.0, 0.0);
// no specular reflection
}
else // light source on the right side
{
specularReflection = attenuation * _LightColor0.rgb
* _SpecColor.rgb * pow(max(0.0, dot(
reflect(-lightDirection, normalDirection),
viewDirection)), _Shininess);
}
return float4((1.0 - w) * (diffuseReflection
+ specularReflection), 1.0);
}
ENDCG
}
}
Fallback "Specular"
}
## Summary
Congratulations! I hope you succeeded to render some nice soft shadows. We have looked at:
• What soft shadows are and what the penumbra and umbra is.
• How to compute soft shadows of spheres.
• How to implement the computation, including a script in JavaScript that sets some properties based on another GameObject.
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| 3.703125 | 4 |
CC-MAIN-2022-33
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longest
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| 0.912013 |
https://se.mathworks.com/help/matlab/math/calculate-tangent-plane-to-surface.html
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# Calculate Tangent Plane to Surface
This example shows how to approximate gradients of a function by finite differences. It then shows how to plot a tangent plane to a point on the surface by using these approximated gradients.
Create the function $f\left(x,y\right)={x}^{2}+{y}^{2}$ using a function handle.
`f = @(x,y) x.^2 + y.^2;`
Approximate the partial derivatives of $f\left(x,y\right)$ with respect to $x$ and $y$ by using the `gradient` function. Choose a finite difference length that is the same as the mesh size.
```[xx,yy] = meshgrid(-5:0.25:5); [fx,fy] = gradient(f(xx,yy),0.25);```
The tangent plane to a point on the surface, $P=\left({x}_{0},{y}_{0},f\left({x}_{0},{y}_{0}\right)\right)$, is given by
`$z=f\left({x}_{0},{y}_{0}\right)+\frac{\partial f\left({x}_{0},{y}_{0}\right)}{\partial x}\left(x-{x}_{0}\right)+\frac{\partial f\left({x}_{0},{y}_{0}\right)}{\partial y}\left(y-{y}_{0}\right).$`
The `fx` and `fy` matrices are approximations to the partial derivatives $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. The point of interest in this example, where the tangent plane meets the functional surface, is `(x0,y0) = (1,2)`. The function value at this point of interest is `f(1,2) = 5`.
To approximate the tangent plane `z` you need to find the value of the derivatives at the point of interest. Obtain the index of that point, and find the approximate derivatives there.
```x0 = 1; y0 = 2; t = (xx == x0) & (yy == y0); indt = find(t); fx0 = fx(indt); fy0 = fy(indt);```
Create a function handle with the equation of the tangent plane `z`.
`z = @(x,y) f(x0,y0) + fx0*(x-x0) + fy0*(y-y0);`
Plot the original function $f\left(x,y\right)$, the point `P`, and a piece of plane `z` that is tangent to the function at `P`.
```surf(xx,yy,f(xx,yy),'EdgeAlpha',0.7,'FaceAlpha',0.9) hold on surf(xx,yy,z(xx,yy)) plot3(1,2,f(1,2),'r*')```
View a side profile.
`view(-135,9)`
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| 4.15625 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.749451 |
https://math.stackexchange.com/questions/1564170/how-to-determine-jordan-normal-form-from-the-characteristic-and-minimal-polynomi
| 1,627,547,990,000,000,000 |
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| 393,243,243 | 37,445 |
# How to determine Jordan normal form from the characteristic and minimal polynomials?
Suppose $A$ is a 4 by 4 matrix with the characteristic polynomial $P_A(\lambda) = (\lambda-2)^4$ with the minimal polynomial $m_A(\lambda) = (\lambda -2)^2$. This tells me that that the Jordan form with respect to the eigenvalue 2 (the only eigenvalue) is $4 \times 4$ size matrix and the largest elementary Jordan block is $2 \times 2$.
So we have $\begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2& 1 \\ 0 & 0 & 0 & 2\\\end{pmatrix}$, or $\begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2& 0 \\ 0 & 0 & 0 & 2\\\end{pmatrix}.$ I know that actually the first one of this correct. But I think the second one also fits the bill because it is $4 \times 4$ size Jordan matrix with eigenvalues $2$ on its diagonal while the largest Elementary Jordan block is of size $2 \times 2$. Why is that the second one not also the Jordan form for $A$?
The Jordan normal form of a matrix is unique up to permutation of the Jordan blocks. Hence $$\begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2& 1 \\ 0 & 0 & 0 & 2\\\end{pmatrix}\qquad\text{ and }\qquad\begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2& 0 \\ 0 & 0 & 0 & 2\\\end{pmatrix},$$ cannot both be Jordan normal forms of the same matrix $A$. Also note that both these matrices have the same characteristic polynomial $(\lambda-2)^4$ and minimal polynomial $(\lambda-2)^2$, which shows that the Jordan normal form of a matrix cannot be determined from these two polynomials alone.
In this case, one way to distinguish is to count the number of Jordan blocks, which is the same as the dimension of $\ker(A-2I)$. In general, it may be necessary to compute the dimension of $\ker(A-\lambda I)^k$ for all $k$ up to the geometric multipicity of $\lambda$, to count the number of Jordan blocks of every possible size.
| 602 | 1,883 |
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| 3.9375 | 4 |
CC-MAIN-2021-31
|
latest
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en
| 0.819298 |
https://socratic.org/questions/a-carpenter-must-cut-a-piece-of-wood-so-that-it-fits-tightly-into-position-the-c
| 1,623,652,552,000,000,000 |
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| 470,296,857 | 5,710 |
# A carpenter must cut a piece of wood so that it fits tightly into position. The carpenter cuts 3/16 of an inch off the wood so that it measures 4 1/2 inches. What was the size of the original piece of wood?
Hence, $\frac{3}{16.} x = \frac{9}{2}$
$\Rightarrow x = \frac{9}{2} \times \frac{16}{3} = 24$
| 101 | 303 |
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| 4.03125 | 4 |
CC-MAIN-2021-25
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| 0.870246 |
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